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Kinematics of Particles—Rectilinear Motion 1
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Kinematics of Particles—Kinematics of Particles—Kinematics of Particles—Kinematics of Particles—Kinematics of Particles—Rectilinear MotionRectilinear MotionRectilinear MotionRectilinear MotionRectilinear Motion
CONCEPTS
• The motion of a particle along a straight line is known as rectilin-ear motion.
• Kinematics is the study of motion without regard to forces orother factors which influence the motion. Hence, it deals withthe geometry and time dependent aspects of motion.
• The position of any particle at anytime ‘t’ is expressed in termsof its displacement ‘x’ from fixed origin ‘O’ on the x-axis.
• The slope at any point of the graph of displacement vs. time isequal to the instantaneous velocity at that point.
dtdx
txv =
∆∆
→= 0∆timL
• The slope at any point of the graph of velocity vs. time is equalto the instantaneous acceleration at that point.
dtdv
tva =
∆∆
→= 0∆timL
• Instantaneous velocity is normally known as velocity and instan-taneous acceleration as acceleration.
2
2and;
dtxd
dtdva
dtdxv ===
Also,dtdx
dxdv
dtdva ===
Hence dxdvva =
• Displacement of a particle is the change in position of a particleduring a time interval.
∆x x xf i= −
2 Engineering Mechanics
It is a vector quantity and has both magnitude and direction.• Distance travelled by a particle depends on the path of the par-
ticle. For example, if a particle starts from A, and returns to thesame point displacement will be zero but distance travelled willbe different.
• Average velocity = = =−−
v xt
x xt t
f i
f i
∆∆
• Average acceleration = = =−−
a vt
v vt t
f i
f i
∆∆
• average speed =distance travelled
total time• If velocity is constant, it is known as uniform motion and if accel-
eration is constant, it is known as uniformly accelerated motion.• If the velocity is decreasing, the particle is said to be in retardation
or deceleration and hence acceleration is negative.• When motion is described based on the acceleration,
(1) a dvdt
dv a dtv
v t= ∫ = ∫;
0 0
(2) av dvdx
v dv a dxv
v
x
x= ∫ = ∫;
0 0• Special cases:
(a) When acceleration ‘a’ is constant
dv a dt v v at
v dv a dx v v a x x
v
v t
v
v
x
x0
0 0
00
202
02
∫ = ∫ ⇒ = +
∫ = ∫ ⇒ = + −( )
and x x v t at= + +0 021 2( )
(b) For variable acceleration, a = f(t)
dv f t dt v v f t dt
dx v dt x x vdt
v
v t t
x
x t t0
0
00
0
00
0
∫ = ∫ ⇒ = + ⋅∫
∫ = ∫ ⇒ = + ∫
( ) ( )
(c) For variable acceleration, a = f(x)
v dv f x dx v v f x dxv
v
x
x
x
x
0 0 0
202 2∫ = ∫ ⇒ = + ∫( ) ( )
Figure S 12.1
Kinematics of Particles—Rectilinear Motion 3
(d) For variable acceleration, a = f(v)
( )vfdvdt
dtdva =⇒=
( )( )
( )∫∫
∫∫
=
==
=∴
v
v
x
x
v
v
t
vfvdvdx
dxdvvvfa
vfdvdt
00
00
• Equations of motion for uniformly accelerated motion, in termsof x0 , v0 and t.
v v at x x v t atv v a x x
= + = + +∴ = + −
0 0 02
202
0
1 22
and ( )( )
In terms of displacement, s x x= −( )0 ass v t at v v as= + = +0
2 2021 2 2( ) ;
• An object thrown upward and another thrown downward experi-ence the same acceleration as the object dropped freely. In freefall motion, the object will have an acceleration downward equalto g ( . m s )29 81 .If ‘y’ is positive upward,
v v gty y v t gt
v v g y y
= −= + −= − −
0
0 02
202
0
1 22
( )( )
• Escape velocity is the minimum velocity with which a particleshould be projected, vertically upward from the surface of earthso that it does not to return to the earth.v gRe = 2 . . . when projected from surface of earth
v gRre = 2 2
0. . . when projected from a point h0 above the sur-
face of earth: where r R h0 0= + .• If a particle starts from rest, it’s initial velocity is zero. If it comes
to rest, its final velocity is zero.• Distance travelled by a particle under uniform acceleration
(a) in n th second is
d v a nn = +
−0 22 1( )
4 Engineering Mechanics
• If two particles move along the same straight line independentlythen,
ABABABABABAB aaavvvxxx −=−=−= /// and;;where ABx / represents relative position of ‘B’ with respect to A.Similarly, ABv / and ABa / represent velocity and acceleration ofB with respect to A.
• When position of one particle depends on the position of an-other particle, the motion it experiences is called dependentmotion.
• Degree of freedom represents the number of independent coor-dinates required to describe the position coordinates of particles.
• Numerical integration.
a dvdt
dv a dt
dv a dtv
v
t
t
=
=
∫ = ∫1
2
1
2
v v2 1− = area of ( )a t−curve between t1 and t2
v dxdt
dx v dt
=
=
dx v dtx
x
t
t
1
2
1
2∫ = ∫
x x2 1− = area of ( )v t−curve between t1 and t2
Example 12.1 A particle movesalong x-axis and its positon is expressedas x t t= −35 73 2. where ‘x’ is in metres and ‘t’ is in seconds.
a) Determine the position at t = 0, 1, 3, 7 seconds.b) Determine the displacement during t = 3s to t = 7s.c) Find the average velocity during t = 3s to t = 7s and instanta-
neous velocity at t = 3s, t = 7s.d) Find the average acceleration during t = 3s to t = 7s and instanta-
neous acceleration at t = 3s, t = 7s.e) Find also the distance travelled in the first four seconds
(t = 0 to t = 4s).
Figure S 12.1
Figure S 12.1
Kinematics of Particles—Rectilinear Motion 5
Solution Given: x t t= −35 73 2.
(a) Positon at t = 0, 1, 3, 7
xxxx
03 2
13 2
33 2
73 2
35 0 7 0 035 1 7 1 3535 3 7 3 31535 7 7 7 857 5
= − == − = −= − == − =
. ( ) ( ). ( ) ( ) . m. ( ) ( ) . m. ( ) ( ) . m
(b) Displacement during 3s to 7s,x x7 3 857 5 315 826− = − =. . m
(c) Average velocity and instantaneous velocity
vx xt t
x xav
f i
f i=
−−
=−−
= =( )
( ). m s7 3
7 38264
2065
v dxdt
t t
vv
= = −
= − == − =
105 14
105 3 14 3 52 5105 7 14 7 4165
2
32
72
.
. ( ) ( ) . m s
. ( ) ( ) . m s(d) Average acceleration and instantaneous acceleration
av vt t
v vav
f i
f i=
−−
=−−
=−
=( )( )
( . . ) m s27 37 3
4165 52 54
91
a dvdt
d xdt
t
aa
= = = −
= − == − =
2
2
3
7
21 14
21 3 14 4921 7 14 133
( ) m s( ) m s
2
2
(e) Distance travelled in first four seconds x t t= −35 73 2. .
to find xdxdtmax , = ⇒0 (motion in the negative direction)
dxdt
t t t= − = =105 14 0 0 1332. ; , . s
x t= = − = −1333 235 1333 7 1333 4148. . ( . ) ( . ) .
From t = 0 to t = 1.333, ( )dx dt is negative, hence the motion istowards the left.
Figure 12.1(a)
6 Engineering Mechanics
When t > 1.333s, (dx/dt) is positive hence the motion is towards theright.
Again at t = 2, x = 0 andat t x= = − =4 35 4 7 4 1123 2, . ( ) ( )
Figure 12.1(b)
∴ Distance travelled = 4.148 + 4.148 + 112 = 120.296m.Example 12.2 Murali walks at a constant speed of 8 m/s along a straightline from ‘P’ to ‘Q’ and returns along ‘QP’ from Q to P at a constantspeed of 4 m/s. Determine
(a) the average speed over the entire trip.(b) the average velocity over the entire trip.
Solution speed = distance travelled/time. From ‘P’ to ‘Q’, DistancePQ = 8t.
In return journey, time taken to cover the same distance = 2t.The distance travelled from Q to P is QP = 4(2t)
∴ = == + =
= = =
= =
total distancetime taken
total distancetime taken
Average velocity change in positiontime taken
2 162 3
163
5333
0
PQ tt t t
tt
Average speed . m s
Example 12.3 The motion of a particle along a straight line is definedby the relation, x t t= − +3 24 5 5. where ‘x’ is in metres and ‘t’ is inseconds.
Plot motion curves from t = 0s to t = 5s with ∆t = 05. s.
Solution x t tv dx dt t ta dv dt t
= − += = −= = −
3 2
24 5 5
3 96 9
.( )( )
Kinematics of Particles—Rectilinear Motion 7
Table 12.1.
t x t t3 2= − +4.5 5 a = 6t – 9(s) (m) (m/s) (m/s2)0 5 0 –9
0.5 4 –3.75 –61.0 1.5 –6 –31.5 –1.75 –6.75 02.0 –5 –6 32.5 –7.5 –3.75 63.0 –8.5 0 93.5 –7.25 5.25 124.0 –3 12 154.5 5 20.25 185.0 17.5 30 21
The motion curves x – t, v – t and a – t are shown in Fig. 12.13 next page.We know that v = (dx/dt) In Fig. 12.13, from A to B, (dx/dt) is negative andat t = 2s, dx/dt = 0 after which (dx/dt) is positive. The motion is towardsthe left during t = 0 to t = 2s. At C, (t = 1.5s), dv/dt = a is zero, after which(dx/dt) is positive.
Figure 12.02(a)(b)
8 Engineering Mechanics
Figure 12.02(c)When t > 3, motion is towards the right.
At B, when v x= = −0 8 75, . mmaxAt C, when a v= = −0 6 75, . m smax
Similarly, at t = 0, v = 0. Hence the slope of (x – t) curve at ‘A’ must be zero.Example 12.4 The motion of a particle moving along a straight line isexpressed as, x t t t= − + −3 2138 39 92 19 2. . . where ‘x’ is in meters and‘t’ is in seconds.
(a) Plot motion curves from t = 0 to t = 5s with ∆t = 1s.(b) Find x, v, a when t = 0.(c) Find x, a when v = 0.(d) Find a, v when x = 0.(e) Find x, v when a = 0.
Solution x t t t= − + −3 2138 39 92 19 2. . .
∴
v dxdt
t t
a dvdt
t
= = − +
= = −
3 27 6 39 92
6 27 6
2 . .
.
Table 12.2
t x = t3 – 13.8t2 + 39.92t – 19.2 v = 3t2–27.6t+39.92 a = 6t –27.6(s) (m) (m/s) (m/s2)0 –19.2 39.92 –27.61 7.92 15.32 –21.62 13.44 –3.28 –15.63 3.36 –15.88 –9.64 –16.32 –22.48 –3.65 –39.6 –23.08 2.4
Kinematics of Particles—Rectilinear Motion 9
(a) motion curves are shown in Fig. 12.14.(b) when t = 0 x t t t
x
v dxdt
t t
v
a d xdt
t
a
= − + −= −
= = − +
∴ =
= = −
∴ = −
3 2
02
2
2
138 39 92 19 219 2
3 27 6 39 92
0 39 92
6 27 6
0 27 6
. . ..
. .
( ) .
.
( ) .
m s.
m s2
(c) When v = 0; v dxdt
t t
t
= = − + =
=± −
=
3 27 6 39 92 0
27 6 27 6 4 3 39 92
618 7 4
2
2
. .
. ( . ) ( )( . )
. , .Substituting t = 1.8 in equation for x
x = − + − =( . ) . ( . ) . ( . ) . . m18 138 18 39 92 18 19 2 137763 2
substituting t = 1.8 in equation for a
a = − = −6 18 27 6 168( . ) . . m s2
This is also shown in Fig. 12.14At t = 7.4s;From eqn. for x
x = − + − = −( . ) . ( . ) . ( . ) . . m7 4 138 7 4 39 92 7 4 19 2 74 2563 2
From eqn. for aa = − =6 7 4 27 6 168( . ) . . m s2
(d) when x = 0x t t t= − + − =3 2138 39 92 19 2 0. . .
From Fig. 12.14, we observe that x = 0 at t = 0.6 and 3.2. Let us deter-mine using analytical method.
From Table 12.2, we observe that x = 0 between t = 0 and t = 1. UsingNewton-Raphson method explained in Appendix-C.
f t t t tf t t t
t t f tf ti i
i
i
( ) . . .( ) . .
( )( )
= − + −′ = − +
= −′+
3 2
2
1
138 39 92 19 23 27 6 39 92
10 Engineering Mechanics
Table 12.3
ti f t i( ) f ' ti( ) −′
f tf t
i
i
( )( ) ti 1+
0 –19.2 39.92 0.481 0.4810.481 –3.08 27.339 0.113 0.5940.594 –0.147 24.584 0.006 0.6
0.6 0 – – –
Hence, t t t t t bt c3 2 2138 39 92 19 2 0 6− + − = − + +. . . ( . )( ) ‘b’ and ‘c’ aredetermined using synthetic division.
0.6 1 –13.8 39.92 –19.20.6 –7.92 19.2
1 –13.2 32 0
∴ f t t t t( ) ( . )( . )= − − + =0 6 132 32 02
Other two roots of the equation are,
t =± −
=132 132 4 32
232 10
2. ( . ) ( ). ,
∴ = − − −f t t t t( ) ( . )( . )( ).0 6 32 10
Hence x = 0 at t s t s1 20 6 32= =. ; . and t s3 10= . Where t1 and t2 areshown in Fig. 12.14.
At t = 0 6. : va
= − + == − = −
3 0 6 27 6 0 6 39 92 24 446 0 6 27 6 27 24
2
2( . ) . ( . ) . . m s( . ) . . m s
Similarly, at . ; . m s; . m sat ; . m s; . m s
t v at v a
= = − = −= = =
32 17 68 8 410 6392 32 4
2
2
The time x = 0 can also be determined using Cardon’s method for cu-bic equation explained in Appendix-C.
t bt ct dt t t
3 2
3 20
138 39 92 19 2 0+ + + =− + − =. . .
where b = –13.8; c = 39.92; d = –19.2
Therefore p c b= − = − − = −2 2
339 92 138
32356. ( . ) .
Kinematics of Particles—Rectilinear Motion 11
q b cb d= −
+ =−
−−
+ −
= − + − = −
227 3
2 13827
39 92 1383
19 2
194 672 183 632 19 2 30 24
3 3( . ) ( . )( . )( . )
. . . .
( ) [ ]cos
( . )
..
.
α
α
= −−
= −−
=
∴ =
q
p2 3
30 24
2 7 8530 687
46 61
3 3
y p
y p
y p
1
2
3
2 3 3 2 7 853 1554 54
2 33
60 2 7 853 7554 14
2 33
60 2 7 853 44 36 4 00
= − = =
= − − +
= − = −
= − − −
= − − = −
.cos ( ) . cos ( . ) .
.cos . cos ( . ) .
.cos . cos ( . ) .
αα
α
t y bt y bt y b
1 12 23 3
3 54 138 3 103 14 4 6 323 4 4 6 0 6
= − = − − == − = − + == − = − + =
. ( . ). . .
. .
Hence x = 0 at t = 0.6s, 3.2s and 10s.(e) When a = 0; 6t – 27.6 = 0; t = 4.6s
xv
= − + − = −= − + = −
( . ) . ( . ) . ( . ) . . m( . ) . ( . ) . . m s4 6 138 4 6 39 92 4 6 19 2 30 24
3 4 6 27 6 4 6 39 92 2356
3 2
2
This is shown in Fig. 12.14.
12 Engineering Mechanics
Figure 12.03
Kinematics of Particles—Rectilinear Motion 13
Example 12.5 The relationship between velocity of a particle and timeis expressed as v t t= − +3 1802 . How far the particle would travel be-tween t = 4s and t = 10s and what will be the acceleration at these twotimes. Find also the distance travelled from t = 0 to t = 70s.
Solution ( )x x v dt t t dt
t t
10 44
102
4
10
3 2
4
10
3 180
33
1802
− = ∫ = − +∫
= − +
x x10 43 3 2 210 4 90 10 4 6624− = − − + − =( ) ( ) m
Therefore, particle would travel a distance of 6624 m between t = 4sand t = 10s.
advdt
t= = − +6 180
The acceleration at t = 4s and t = 10s is given byWhen t a= = − + =4 6 4 180 1564
2: ( ) m sWhen t a= = − + =10 6 10 180 12010
2: ( ) m s
Figure 12.18 (v – t) curve
To find the distance travelled during t = 0 to t = 70s, above formulashould not be applied directly. Since v t t= − + =3 180 02 . at t = 0 &t = 60s, from t = 0 to t = 60s, v is positive hence motion is towards the rightand when t > 60, v is negative and hence motion is towards the left.
∴ Distance travelled from 0 to 70 is calculated in two steps as,Since after t = 60 sec, the direction of motion is reversed.
[ ][ ]
x x t t
x x t t
60 03 2
0
60
70 603 2
60
70
90 108000
90 10000
− = − + =
− = − + =
14 Engineering Mechanics
∴ Total distance travelled during t = 0s to t = 70s is = + =108000 10000 118000.
But, if one uses the formula vdt0
70∫ directly,
[ ]x x t t70 0 90 980003 20
70− = − + = m
which is not the correct answer.Example 12.6 The acceleration of a particle is defined by the relationa t= −α 4. Knowing that v = 4 m/s when t = 2s and v = –1m/s when t = 1s,determine the constant ‘α ’. Write the equations of motion when x = 0 att = 3s.
Solution Given: a t= −α 4
t = 2s, v = 4 m/st = 1s, v = –1 m/st = 3s, x = 0
We know that
a dvdt
t
dv t dt
= = −
∴ = −
α
α
4
4( )Integrating,
( )dv t dt
v vt
t
v vt
t
v
v t
04
2 4
24
0
0
2
0
2
∫ = −∫
− = −
∴ = + −
α
α
α
( )
at t v vv
= = = + −+ =
2 4 4 2 82 12
00
: ; ( )αα
at t v vv
= = − − = + −+ =
1 1 1 2 42 6
00
: ; ( ).
αα
∴ Solving above eqns, 3 18 6α α= ⇒ = and v0 0=
∴ a t v t t= − = −6 4 3 42;
v dxdt
t t= = −3 42
Kinematics of Particles—Rectilinear Motion 15
dx t t dt= −( )3 42
Integrating the above
dx t t dt
x x t t
x x t t
x
x t=∫ −∫
− = −
= + −
0
2
0
0
3 2
03 2
3 4
33
42
2
( )
At t = 3, x = 0
∴0 3 2 3
90
3 2
0
= + −= −
xx
( ) ( )
Thus the equations of motion are given by x t t= − −3 22 9;v t t= −23 4 ; and a = 6t – 4.Example 12.7 The acceleration due to gravity of a particle falling to-wards earth is a gR r= − 2 2 where ‘r’ is the distance from the centre ofearth to the particle; ‘R’ is the radius of the earth and ‘g’ is accelerationdue to gravity. Derive an expression for escape velocity.Solution Escape velocity is the minimum velocity with which a par-ticle should be projected vertically upwards from the surface of earth sothat it does not tend to return to the earth.
Let ‘O’ be the centre of earth. Consider a particle projected from thesurface of the earth. From Fig. 12.19(a),
But
r R x
a gRr
g RR x
a dvdt
dvdx
dxdt
v dvdx
= +
= − = −+
= = =
2
2
2
2( )
( )
Equating the above two equations
v dvdx
gRR x
v dv gR dxR xve
= −+
∫ = −+
∫
2
2
02
20
( )
( )
α
Where ve = escape velocity
− =
+
=−
= −
=
v gRR x
gRR
gR
v gR
e
e
22
0
2
21
2
α
Figure 12.5(a)
16 Engineering Mechanics
If the same particle is projected from top of a tower or hill of height ‘h0’above the surface of earth as shown in Fig. 12.19(b).
av dvdx
gRR x
v dv gR dxR x
= = −+
= −+
2
2
22
( )
( )Integrating
( )
v dv gRdx
R x
v gR R x
gRR h
gRr
r R h
v h
eh
e
02
2
2 2
2
0
2
00 0
0
0
21
∫ = −+
∫
− = +
= − + = − = +
( )
Since
α
α
Therefore v gRre = 2 2
0
Hence escape velocity ‘ve’ when the particle is projected from
surface of earth, v gRe = 2Escape velocity ‘ve’ when the particle is projected from a point
‘h0’above the surface of earth, v gR re = ( )2 20 where
r R h0 0= +
Example 12.8 The acceleration of a particle moving along a straight lineis defined by a = 8 – x. The particle starts from rest at t = 0 and origin x = 0.
Determine (a) Velocity of the particle when x = 10 m, (b) the positionof the particle when velocity becomes zero and (c) Velocity of a particlewhen acceleration becomes zero.Solution Given data: a = 8 – x
Particle starts from rest and at t = 0, x = 0We know that acceleration can be written as a = f(x) since the particle
starts from rest, at t = 0, v = 0.
∴
a dvdt
f x dvdx
dxdt
advdx
v v dv a dx
= = =
= =
( )
;
Figure 12.5(b)
Kinematics of Particles—Rectilinear Motion 17
Integrating, v dv a dx x dx=∫ =∫ −∫ ( )8
v x x c
v x x c
2 2
1
2 21
28
2
16 2
= −
+
= − +where c1 = constant of integration.At t = 0, v = 0 and x c= ⇒ =0 01 .
∴ v x x2 216= − and a x= −8(a) Velocity when x = 10 m
vv2 216 10 10 60
7 746= − ==
( ) ( ). m s.
Therefore velocity of the particle when x = 10m is v = 7.746 m/s.(b) Position when velocity ‘v ’ becomes zero
v x x2 16 0= − =( )∴ x = 0 and x = 16 m
Therefore the position of the particle when velocity is zero is at x = 0and x = 16 m.
(c) Velocity when acceleration is zero. a = 8 – x = 0; x = 8.
∴ v x x2 216= −v
v2 16 8 64 64
8= − ==
( )m s
When the acceleration is zero, the velocity of the particle is given byv = 8 m/s.Example 12.9 The acceleration of a particle is defined by the relation-ship a = –v where ‘a’ is in (m s2) and ‘v’ is (m s). At t = 0,v = 30 m/s. Determine the distance travelled by the particle as it comes torest from initial velocity.Solution Given data: a = –v
When t = 0, v = 30 m/sThe acceleration of the particle is given by,
a = f(v)
advdt
f v= = ( )
∴ = =−
= −
dtdv
f vdv
vdvv( ) ( )
18 Engineering Mechanics
Integrating, dtdvv∫ = −
∫
t v ce= − +log 1when t v c vo e o= = ⇒ = =0 30 3 4011, . log
t v v v v v v ee e o e o ot= − + = ⇒ =log log log ( ) ( )
or v v eot= − ; v e t= −30
We also observe, a dvdt
dvdx
dxdt
vdv adx
dxvdv
vdv
= =
=
=−
= −( )
Integrating, x v c= − + 2.When x = 0, v = 30; c2 30=
x v e et t= − = − = −− −30 30 30 30 1( )a v ev ex e
t
t
t
= − = −== −
−
−
−
303030 1( ).
when v e t= =−0 0,∴ x = 30 m.
Example 12.10 Latha throws an apple vertically upwards from theground to Preetha standing near the window of a room 30 m above. Theapple was caught by Preetha 2s later by her outstretched hand as shown inFig. 12.21. Determine (a) Initial velocity of apple (b) the final velocityjust before it was caught.Solution Given data: Distance travelled = 30 m; time taken (t) = 2s
a g . = = −9 81 2m sVelocity of the apple is given by
v v gt v t
s v t at v t t
vvv v t
v
= − = −
= + = −
∴ = −∴ =
= −= −=
0 0
02
02
02
00
9 8112
4 905
30 2 4 905 224 81
9 8124 81 9 81 2519
.
.
( ) . ( ). m s.
.. . ( )
. m sTherefore the initial velocity of the apple is
24.81m/s and the final velocity of the apple is 5.19 m/s.
Figure A 12.6
Kinematics of Particles—Rectilinear Motion 19
Example 12.11 A stone dropped into a well is heard to strike water after4 sec. Find the depth of the well if the velocity of the sound is 350 m/s.
Solution Let ‘d’ = depth of well. ( )d v t gt= +02 2
Initial velocity of stone when it is dropped; v0 0=v g0 0 9 81= =; .
∴ time taken by the stone to reach the water level in the wellt = ( . )d 4 905
Time taken by the sound wave to reach top = d/350(Since the sound wave travels with uniform velocity.)
mdd
d
dddd
dd
728.70;41.8
45.159,41.82
5600)0333.158(033.15801400033.15814000333.158
4350905.4
2
==∴
−=+±−
=
=−+=+
=+
Note: –159.45 is rejected because ‘d’ cannot be negative.Example 12.12 Priya throws a stonefrom the top of a building 30m high withan initial velocity of 25.5 m/s directedvertically upward. On its way down, thestone just misses the edge of the roof asshown in Fig. 12.22(a). Determine
(a) the velocity ‘v’ and elevation ofthe stone above the ground atany time ‘t’
(b) the time needed for the stone toreach its maximum height
(c) the maximum height above theground level.
(d) the time needed for the stone toreach the top of building and it’svelocity during downward jour-ney
(e) velocity when it reaches theground and
(f ) position and velocity whent s s= 2 5, . Figure A 12.07(a)
20 Engineering Mechanics
Solution Given data: 2
00
sm81.9sm5.25
m30
−===
avy
Equations of motion:
v v at ty y v t aty t t
= + = −= + += + −
0
0 02
2
255 9 811 2
30 255 4 905
. .( )
. .The velocity ‘v’ above the ground at any time ‘t’ is given by
v = 25.5 – 9.81t.The elevation of the stone above the ground level at any time ‘t’ is
given byy t t= + −30 255 4 905 2. .
When it reaches the maximum height (B), velocity is zero.∴ v t tB . . ; . s.= − = =255 9 81 0 2 6
Therefore the time needed to reach the maximum height is 2.6s
Figure 12.7(b)(c)
Kinematics of Particles—Rectilinear Motion 21
Figure 12.7(d)
∴ = + −=
yy
BB
. ( . ) . ( . ). m
30 255 2 6 4 905 2 663142
2
Thus the maximum height yB above the ground level is 63.142 m.
When the stone returns to point C, yc = 30y t t= + −30 255 4 905 2. .
∴ 30 30 255 4 905 2= + −. .t t∴ t t( . . )255 4 905 0− =t = 0 and t = 5.2s.
v tv
CC
. .
. . ( . ) . m s= −
∴ = − = −255 9 81255 9 81 52 255
Therefore the time required to reach the top of building during thedownward journey is 5.2s and its veloticy is – 25.5 m/s
When stone reaches the ground level at ‘D’, yD = 0
∴ O t t= + −30 255 4 905 2. .∴ 4 905 255 30 02. .t t− − =
t
t s
=+ ± − −
= −
255 255 4 4 905 302 4 905
6 2 1
2. ( . ) ( . )( )( . )
. , s.when t = 6.2 sec
v tv
= −= − = −
255 9 81255 9 81 6 2 35322
. .
. . ( . ) . m sDTherefore, when the stone reaches the ground level, its velocity is –
35.322 m/s.
22 Engineering Mechanics
t s vy
t s vy
= = − == + − =
= = − = −= + − =
2 255 9 81 2 58830 255 2 4 905 2 6138
5 255 9 81 5 235530 255 5 4 905 5 34 875
2
2
: . . ( ) . m s. ( ) . ( ) . m
: . . ( ) . m s. ( ) . ( ) . m s
It is shown in Fig. 12.22.Example 12.13 A body moving with uniform acceleration is observedto travel 33 m in eighth second and 53 m in the thirteenth second of itstravel. Calculate the velocity at start and uniform acceleration.
Solution Given: s v t at
d d
= +
= =0
2
8 13
12
33 53m; m
Distance travelled in ‘n’ secs = snDistance travelled in nth second = dnDistance travelled in ‘n’ seconds = +v n an0
21 2( )Distance travelled in (n – 1) sec = − + −v n a n0
21 2 1( ) ( )( )∴ Distance travelled in nth second
[ ] [ ][ ]
d v n an v n a n
d v n n a n n
d v a n
n
n
n
= +
− − +
−
= − − + − −
= +
−
0
2
02
02 2
0
21
21
12
1
22 1
( ) ( )
( ) ( )
The distance travelled in 8th second is denoted by d8 and given by∴ ( )[ ]d v a
v a
v a
8 0
0
0
2 2 8 1 33
215 33
7 5 33
= + − =
+
=
+ =
( )
( )
.Similarly, the distance travelled in 13th second is
( )[ ]d v av a
13 00
2 2 13 112 5 53
= + −= + =
( ).
Solving equation for d d8 13and , a v204 3m s m s.= =
Therefore velocity at start v0 3= m s and its uniform acceleration
a = 4 2m s
Kinematics of Particles—Rectilinear Motion 23
Example 12.14 The motions of two particles A and B are definedby a tA = 3 and a tB = 2 2 where ‘a’ is in m s2,’v’ is in m s, and ‘t’ is inseconds. Find xB A and vB A at t = 5s. At t = 0, xA m;= 10 xB m;= 13and v vA B m s.= =2 2
Solution The motion of particle ‘A’ is a tA = 3
∴ v t c t cA .= + = +32
152
12
1
x t c t cA.= + +153
3
1 2
x t c t cA .= + +05 31 2
At t = 0: v cA m s= ⇒ =2 21At t = 0: x cA m= ⇒ =10 102For particle A, x t t v tA A. ; . .= + + = +05 2 10 15 23 2
Motion of particle ‘B’ is a t v t cB B; ( )= = +2 2 32 33
x t c t cB ( )= + +
23 4
43 4
At t = 0: v cB m s= ⇒ =1 13At t = 0: x cB = ⇒ =13 134For particle B, x t t v tB B. ; . .= + + = +0167 13 0 667 14 3
Thus, x x x t t t
v v v t tB A B A
B A B A
. . ,
. . .
= − = − − +
= − = − +
0167 0 5 3
0 667 15 1
4 3
3 2
At t = 5s; xB A . m;= 39 875 and vB A . m s= 46875
Example 12.15 Blocks ‘A’ and ‘B’ are initially at rest when yA m= 5and yB m= 4 as shown in Fig. 12.27. Block ‘B’ is accelerated downwardat a constant rate of 0.6 m s2. Determine the relative vertical positionsand velocity of blocks 2s after the motion begins.Solution Length of the rope PQRSB is constant and also OA and CDare constants. If ‘B’ moves down ‘A’ moves up.
or, 2y yA B+ = constant( ) ( )2 y y y yA A B+ + + =∆ ∆ B constant
2 02 02 0
∆ ∆y yv va a
AA BA B
+ =+ =+ =
B
24 Engineering Mechanics
If aB . m s ( )= ↓0 6 2 then aA . m s= −0 3 2
i.e aA . m s ( )= ↑0 3 2
)(sm6.0)(sm2.1)2(6.0
A
BB
↑=↓===
vtav
m0.84.4–5.2y
m4.440.321–05
a21vyy
m5.240.62104
a21vyy
B/A
2A/OA/OA/OA
2B/OB/OB/OB
==
=××+=
++=
=××++=
++=
tt
tt
Example 12.16 A ball is thrown verti-cally upwards from 20 m level in an eleva-tor shaft with an initial velocity of 25 m s .At the same instant, an open platform el-evator passes the 5 m level with a constantvelocity of 2 m s . Determine when andwhere the ball will hit the elevator.Solution Velocity of ball,v v atB OB= +
i.e v tB .= −25 9 81Position of ball, y y v t atoB OB ( )= + + 1 2 2
y t tB .= + −20 25 4 905 2
Velocity of elevator vE m s= 2Since the elevator passes with uniform velocity, aE = 0
Figure 12.08
Figure A 12.09(a)
Kinematics of Particles—Rectilinear Motion 25
Figure 12.09
Position of elevator, y tE = +5 2
When the ball hits the elevator, y yB E=
∴20 25 4 905 5 24 905 23 15 0
2
2+ − = +
− − =t t t
t t.
.
{ }t
t
= ± − −
= ± = −
23 23 4 4 905 15 9 81
23 28 7 9 81 527 058
2( ) ( . )( ) .
. . . , .
It hits the elevator after 5.27s at an elevation ofy tE ( )( . ) . m= + = + =5 2 5 2 527 1554
(y – t) curves for ball and elevator are shown in Fig. 12.28.
26 Engineering Mechanics
let A = Initial positon of ball andlet B = ymax of ball,At B, vB = 0∴ 25 9 81 0 2 55− = =. ; .t t s
yBall ( . ) . ( . ) .= + − =20 25 2 55 4 905 2 55 51862
At A y t s vA A; . ; and m s.= = = −20 51 25At ′c the ball hits the elevator where, y yE B . m.= = 1554Just before the ball hits the elevator, velocity of ball is
v = − = −25 9 81 5 27 26 7. ( . ) . m s.Example 12.17 Parvathy drops astone from the top of a tower. When ithas fallen at a distance of 10 m,Nirmala drops another stone from apoint 38 m below the top of the tower.If both the stones reach the ground atthe same time as shown in Fig. 12.29.Calculate (a) height of the tower and(b) the velocity of the stones when theyreach the ground.
SolutionSolutionSolutionSolutionSolution Stone 1: u1 0= and
s gt= ( )1 2 2
Time to reach
‘A’: ( )10 9 81 22= . t
∴ t = 1.43sVelocity at A, v gtA . ( . ) . m s= = =9 81 143 14 028Stone (1) with vA reaches ground in ‘t’ seconds and stone (2) with
u = 0 reaches ground in ‘t’ seconds.∴ − = +
− =H t tH t
10 14 028 4 90538 4 905
2
2. .
.Solving the above two equations;∴ 28 14 028 2= =. ;t t s and
H = + =38 4 905 2 57 622. ( ) . m∴ Height of tower = 57.62 mWhen the stones reach the ground, the velocities are,
Figure 12.10
Kinematics of Particles—Rectilinear Motion 27
v gv g
1
2
2 57 62 2 9 81 57 62 33622 19 62 2 9 81 19 62 19 62
= = == = =
( . ) ( . ) ( . ) . m s( . ) ( . ) ( . ) . m s
∴ Velocity of stone 1 ( ) . m sv1 33 62= andVelocity of stone 2 ( ) . m sv2 19 62= when they reach the ground.
Example 12.18 Satheesh drops a stone from a tower 120 m high. Atthe same time, Saravanan projects another stone vertically upwards fromthe foot of the tower. If both the stones meet at 30 m above the ground,determine
(a) the velocity with which Saravanan projected the stone.(b) the velocities of stones as they cross.
Solution Given: Stone-A (dropped from top); v A0 0=
y y v t at
y tv t
A
AA
OA OA
..
= + +
= −= −
12
120 4 9059 81
2
2
Stone B (projected upwards).
v v t
y v t tB
B
OB
OB
.
.
= −
= −
9 81
4 905 2
30 120 4 905 2= − . ;t t = 4.28s
30 4 28 4 905 4 2802= −v B ( . ) . ( . ) ;
28 00=v . m sOBWhen they cross each other,
v tvA
B
. . ( . ) . m s( ). ( . ) . m s( )
= − = − = − ↓= − = − ↓
9 81 9 81 4 28 4198728 9 81 4 28 13987
After reaching the maximum point, ‘B’ returns to the ground and meets‘A’.Example 12.19 Water drops from a tap at a uniform rate of ‘n’ dropsper second. If the distance between two adjecent drops is ‘y’ when thetrailing drop has been in motion, show that )2/()( 2ngngty += where‘t’ is the time of trailing drop in motion. Neglect air resistance.Solution Consider two adjacent drops of water ‘A’ and ‘B’ where ‘A’is leading and ‘B’ is trailing.
Figure 12.11
28 Engineering Mechanics
When drop B travels ‘t’ seconds, drop ‘A’ would have travelled t + (1/n) seconds.
Initial velocity of drops = 0
∴ =
+
=
s g tn
s g tBA and ( )12
1 12
22
Distance between two drops = = −y s sA B
y g tn
t
y gn
tn
ygtn
gn
=
+
−
=
+
∴ = +
12
1
12
1 2
2
22
2
2
Example 12.20 A train is travelling with an acceleration as shown inFig. 12.32(a). Assuming that it starts from rest, determine (a) distancetravelled; (b) the time taken by the train till it comes to a complete halt.
Figure 12.13(a)
Solution At t = 0, x = 0; t = 2s, x = A; t = 5s, x = BWe know that
( )
( )
v v a dt a t
x x v dt v t
t
t
t
t
2 1
1
1
2
1
2
− = ∫ = −
− = ∫ = −
area of diagram in 1 and 2
area of diagram in 1 and 22
Figure 12.12
Kinematics of Particles—Rectilinear Motion 29
The train starts from rest. Therefore, v .0 0=
c v v a t
v
A
A
Area of ( ) in OA( . )( ) .. m s
− = −= =
∴ =
016 2 32
32v v a tv
v
B AB
B
in AB. ( . )( ) .
. m s
− = −− = =
∴ =
Area of ( ) 32 08 3 2 4
56When it comes to rest at C, vC = 0
v v a tt
t
C − = −− = −∴ =
B BC. ( . )
s
Area of ( ) in 0 5 6 14
4Thus (a – t) and (v – t) curves are shown in Fig. 12.32(b) and
Fig. 12.32 (c) respectively.
Figure 12.13(b)
Figure 12.13(c)
Distance travelled,x x v t
x
A
A
( ) OA
( )( . ) . m
− = −
∴ − =
+ =
0
012
2 0 32 32
Area of in
x x v t
x
B A
B
( ) AB
. ( ) ( . . ) . m
− = −
∴ − =
+ =
Area of in
3212
3 32 5 6 132
30 Engineering Mechanics
xx x v t
x
x
BC B
C
C
. m( ) BC
. ( ) ( . ) .
. . . m
=− = −
∴ − = + =
= + =
164
16 4 12
4 56 0 112
164 112 27 6
Area of in
Therefore total time taken for a complete halt = 9 sec. and total dis-tance travelled = 27.6m.Example 12.21 A particle moves in a straight line with a velocity shownin Fig. 12.33(a). Knowing that x = –20m and t = 0, draw (x – t), (a – t),curves for 0 < t < 18s and determine (a) total distance travelled after 12s(b) two values of ‘t’ for which particle passes through origin.
Figure 12.14(a)
Solution
At , O;, A; m s, B; m s, D; m s
ABD
t x vt x vt x vt x v
= = == = == = = −= = = −
0 08 2412 2418 24
0
a = dv/dt = slope of (v – t)
a
a
in OA m s
in AB ( ) m s
2
2
= =
= − − = −
24 8 324 24
412
‘a’ after ‘B’ = 0
Kinematics of Particles—Rectilinear Motion 31
Figure 12.14(b)
We know that
x x v dt v t2 11
2− = ∫ = −Area of curve in 1 and 2 ( )
At t = 0, x0 20= − m
x x v t
x x
A
A
( )
( ) ( )
m
− = −
=
∴ = + = − + =
0
0
24 8 96
96 20 96 76
Area of in OA
= 12
x x v t
x x
x
C A
C A
C
( )
( )( ) m
m
− = −
− = =
∴ = + =
Area of in AC12
2 24 24
76 24 100
x x v t
x x
x
B C
B C
B
( )
( )( )
m
− = −
− = − = −
∴ = − =
Area of in CB12
2 24 24
100 24 76
x x v tx x
x
D BD B
D
( )( )
.
− = −− = − = −
= − = −
Area of in BD24 6 144
76 144 68Distance travelled after 12 sec (from 0 to 12 sec) is = 20 + 76 + 24
+ 24 = 144m.It crosses origin during OA (0 < t < 8) and DB (12 < t < 18).Let E and F be the time when x = 0.
32 Engineering Mechanics
( )( )
3.65sec.40;3
321–20x
OEint–vofArea–
2
2E
OE
==
+=∴
=
tt
t
xx
x x v tx t t
F BF
( )( ) ; . s
− = −= + − = =
Area of in BF76 24 0 3172 2
i.e total time at E = 3.65s and at F = 12 + 3.17 = 15.17s.Motion curves are shown in Fig. 12.33(c).
Figure 12.14(c)
Kinematics of Particles—Rectilinear Motion 33
EXERCISES
1. The motion of a particle moving along a straight line is defined bythe equation. x t t= − +36 1083 2. .where ‘t’ is in sec ‘x’ is in ‘m’. Plot x – t, v – t and a – t curves fromt = 0 to t = 8: use ∆t = 1 second.
2. The motion of a particle is defined by the relation x = 7.5 + 22.5t– 13.5t t2 315+ .where ‘x’ is in ‘m’ and ‘t’ is in. sec. Determine (a) the positionwhere the velocity is zero (b) position and total distance travelledwhen acceleration is zero and (c) plot motion curves.
3. The velocity of a car travelling on road is expressed asv t t= + −3 4 05 2. where ‘v’ is in m sec and ‘t’ is in s. Calculate(a) average acceleration between (i) t = 3s and t = 5s (ii) t = 10s tot = 15s and (b) instantaneous acceleration when (i) t = 4s(ii) t = 12s.
4. A particle moves along z-axis according to the equationz t t( ) .= −4 8 2 + 3.2t + 4.8. Determine (a) average velocity be-tween t = 2s to t = 5s (b) instantaneous velocity at t = 4.5s. (c)average acceleration between t = 2s to t = 5s. (d) instantaneousacceleration at t = 4, 5s. (e) plot z – t, v – t and a – t curves from t= 0s to t = 6s with ∆t = 1s.
5. The position of a particle is defined by the relationship y = 8 sin 2t+ 5, where y is in metres and ‘t’ is in seconds. Calculate position,velocity and acceleration at t = 2s and 3s.
6. The velocity of a particle is expressed as v t t= − −4 5 18 152. and x= –75 when t = 6 seconds. Determine position and accelerationwhen t = 4s.
7. The acceleration of a particle is expressed as a = 10 – 10t where‘a’ is in m s2 and ‘t’ is in s. At t = 3s, x = 0 and v = –15m/s.Determine the position and velocity when t = 5s.
8. The acceleration of a particle is defined by the equation, a = 10sin t where ‘a’ is in m s2 and ‘t’ is in seconds. At time t = 0,x0 0= and v0 20= m s. Determine the first position ‘x1’ whenvelocity is maximum.
9. The acceleration of a particle is given by a x= −30 4 5 2. . The par-ticle starts with no initial velocity at the position x = 0. Determine(a) the velocity when x = 1.5m (b) the position when velocity isagain zero and (c) position where velocity is maximum.
34 Engineering Mechanics
10. The acceleration of a particle moving along a straight line is de-fined by a s= − −α 2. The particle starts from rest at s = 20 m, and itis observed that velocity is 2.828m/s when s =10 m. Determine (a)value of α and (b) velocity when s = 5 m.
11. A particle moves along a straight line with an accelerationa s= 6 1 3( ) where ‘s’ represents displacement. When t = 3s,s = 64 m and its velocity is 48 m/s. calculate velocity and accel-eration when t = 2s.
12. The acceleration of a particle is expressed as a x= −20 3 2 . Theparticle starts with no initial velocity at the position x = 0. Deter-mine (i) velocity when x = 1.0m (ii) position where the velocity isagain zero, (iii) position where velocity is maximum.
13. The retardation of a particle moving along a straight line is k v3
where k is a constant and ‘v’ is the velocity at that instant. If theinitial velocity is ‘v0’, find the velocity and time when the particletravels at a distance ‘L’.
14. The acceleration of a particle moving along a straight line isa v= 3 2 3( ) . When t = 3s, its displacement s = 37.516m and ve-locity v = 42.87m/s. Determine the displacement, velocity and ac-celeration when t = 5s.
15. A body moves along a straight line with an acceleration a t= −3 4 .After 4 seconds from the start of observation, it’s velocity is 45m sand after 6 seconds from the start of observation it’s position fromorigin is 150m.a) Determine acceleration, velocity and its position from originwhen t = 0.b) What is the time and distance travelled when velocity is zero.
16. The values of displacement-time and velocity-displacement aregiven in Table E12.1 and Table E12.2 respectively. Draw s-t curveand v-s curve and determine the acceleration at t = 3s.Table E12.1
t(sec) 0 1 2 3 4 5s(m) 1 8 27 64 125 216
Table E12.2
v(m/s) 30.23 48.07 63.0 76.32 88.57 100.02s(m) 32 64 96 128 160 192
Kinematics of Particles—Rectilinear Motion 35
17. A ball after falling 5 seconds from rest breaks a glass plate in itsdirection of motion and it loses 30% of its velocity. How far will ittravel in the next second?
18. A particle falls from rest and in the last second of its motion ittravels 98.1m. Find the height from which it fell and the time ofjourney.
19. A particle falling under gravity falls 48m in a certain second. Findthe time required to cover the next 48m. Assume g = 9 81 2. m s .
20. A particle falling under the action of gravity passes two points ‘B’and ‘C’ 65 m apart vertically in 0.5sec. From what height the par-ticle above point ‘B’ started to fall.
21. The space described by a freely falling body in the last one secondof its motion to that described in the last but one second are in theratio of 5:4. Calculate the height from which the body was droppedand the velocity with which it strikes the ground.
22. A particle freely dropped from a height ‘H’ above the ground travels60% of total distance of its fall in last one second. Calculate ‘H’.
23. Two cars A and B are travelling in the same direction at constantvelocities as shown in Fig. E12.1. Determine the distance betweenthe cars 8s after the instant when they are 400m apart.
Figure E 12.1
24. Shankari and Sharanya are sitting in cars A and B respectively,300m apart and are at rest. Shankari starts her car and moves to-wards B with an acceleration of 0.5 m s2. After three seconds,Sharanya starts her car towards A with an acceleration of 1 m s2 .Calculate the time and point at which two cars meet with respectto ‘A’. (Refere Fig E12.2).
Figure E 12.2
36 Engineering Mechanics
25. The velocities of two cars A and B are defined by v tA .= +6 152
and v tB . .= −2 4 3 2 where v is in m/s and t in seconds. Determinethe relative position xB A and relative acceleration aB A att = 4s. When t = 0, assume x xA B.=
26. The motion of two particles A and B are expressed as v t tA = +3 22
and x tB = −4 3 where x is in m, v is in m/s and t is in s. DeterminexB A, vB A and aB A at t = 6s. When t = 0 assume xA = 0.5m.
27. Two cars are moving along a straight road and are 1.5km apart, inthe same direction. The velocity of first car is 80 km/hr and thesecond is 60 km/hr. First car travels with uniform velocity whilethe second accelerates uniformly at the rate of 10 m s2 . Deter-mine the time required to overtake.
28. A stone is allowed to fall from the edge of a tower 100m high.What will be its velocity after 2 seconds? Find the distance it wouldhave travelled at the end of 2 seconds and time to reach the ground.Find also it’s velocity when it reaches ground level.
29. A stone takes 1 seconds to reach the ground when thrown downwith some velocity from the top of a tower. It takes 6.2 seconds toreach the ground when thrown up with the same velocity. Calcu-late the initial velocity of throw and the height of the tower.
30. A ball is thrown vertically upwards with a velocity of 12m/s froman elevation 50m above ground level. After two seconds, anotherball is thrown vertically upwards from an elevation of 25m with avelocity of 16m/s above ground level. Determine when and whereboth the balls would meet and their velocities of the balls whenthey meet.
31. A baseball is hit in such a way that it travels straight upward afterbeing struck by bat. Mahesh observes that it requires 4s for theball to reach its maximum height. Find its (a) initial velocity and(b) maximum height. Neglect the effects of air resistance.
32. A stone is projected vertically upwards from the top of a buildingwith a velocity of 40 m/s. Five seconds later another stone is justreleased from the same point. If it is seen that both the stonesreach the ground at the same time, what must be the height ofthe building.
33. It is observed that a particle describes 396.9m in 3 sec, 392m inthe next 4 sec and 269.5m in the next 5s. Show that this is consis-tent with a particle moving with uniform retardation and find thetime before it comes to rest.
Kinematics of Particles—Rectilinear Motion 37
34. Vivithashree throws a stone vertically upwards from a tower 30mhigh with an initial velocity of 10m/s. At the same instant, Gowthamthrows another stone vertically upwards with an initial velocity of22m/s from ground. Draw v – t and y – t curves for the two stonesand determine when and where the stones cross.
35. Priya throws a stone vertically downwards from the top of a build-ing with some initial velocity. The stone takes 3 sec to reach theground. When she throws another stone vertically upwards withsame initial velocity it takes 5 sec to reach the ground. Calculatethe initial velocity of throw and height of building.
36. Three points P, Q and R are spaced 150m apart along a straightroad. A particle start from rest at ‘0’ and accelerating uniformlypasses ‘P’ and takes 15s to cross ‘Q’ and takes another 9s to crossR. Calculate (a) acceleration (b) velocity at P, Q and R (c) dis-tance of ‘P’ from O.
37. A car starts from rest on a straight road and travels with uniformacceleration of 0.8 m s2 for the first 10s and then travels withuniform velocity for the next 30s. It then decelerates at the rate of0.5 m s2 and comes to rest. Determine(a) total time it takes to complete the trip;(b) total distance travelled by car and(c) draw motion curves.
38. Ramakrishnan travels a distance ‘AB’ with a constant velocity of54 km/hr in 6 minutes. But, Sivaraman travels the same distancein 9 minutes. Starting from rest, he moves with uniform accelera-tion attains a maximum velocity and then retards uniformly to stop.The acceleration is (2/3) of retardation numerically. Find the ac-celeration, retardation and maximum velocity.
39. The speed of a body at a certain time is given in the table. Drawspeed-time graph and find (a) the distance travelled during first 20seconds and (b) total distance travelled.
Time (s) 0 5 10 15 20 25 30speed (m/s) 0 10 13 20 25 25 22
40. The maximum possible acceleration or deceleration that a trainmay have is 10m s2 while its maximum speed is 72 km/hr. Findthe shortest interval of time in which the train can reach from onestation (A) to other station (B) if AB = 300 m.
38 Engineering Mechanics
41. A car moves with uniform velocity of 36 km/hr. in the first 90sec. Itaccelerates uniformly at 2 m s2 and attains a maximum velocity of72km/hr. It moves further with this uniform velocity for the next 5minutes and moves with uniform retartation and comes to rest inthe next 90 seconds. Find the total time of the journey and thedistance travelled.
42. Two cars A and B start from rest at the same time from two stationsspaced ‘L’ apart along the same straight road. Car ‘A’ travels firstwith an acceleration of 0.6 m s2 reaching a maximum velocity of30m/s and travels uniformly with this velocity. Car ‘B’ travels firstwith an acceleration of 0.75 m s2 reaching a maximum velocity of24m/s and travels uniformly with this velocity. Both the cars crosseach other at ‘C’ which is exactly mid-way between the two sta-tions. Determine ‘L’ and time taken by each car to cross ‘C’.
43. Two balls are projected simultaneously with the same velocity fromtop of a tower, one vertically upwards and the other vertically down-wards. If they reach the ground in times t1 and t2 respectively,show that the time for each to reach the ground if they are simplydropped from the top of the tower is t t1 2 .
44. A particle is projected vertically upwards. Show that if t1 and t2are the times at which it passes through a point at a height ‘h’above the point of projection in ascending and descending, thent t h g1 2 2= ( ).
45. A particle is projected vertically upwards with a velocity ‘u’ andafter ‘t1’ seconds another particle is projected upwards from thesame point with same velocity. Prove that the two particles will meetat a height( )4 82 2
12u g t g− after a time ( ) ( )t u g1 2 + seconds.
46. A particle projected vertically upwards is seen at ‘P’ after ‘t1’seconds. If it takes ‘t2’ seconds more from point ‘P’ to the point ofprojection. Prove that(a) Height of ‘P’ above point of projection = gt t1 2 2(b) Maximum height reached = +g t t( ) .1 2
2 847. Two particles move in the same straight line starting from the same
point and at the same moment. The first particle moves with uni-form velocity ‘u’, while the second starts from rest and moves withuniform acceleration ‘a’. During the interval of time before the sec-ond particle meets the first, show that the maximum distance be-tween them is ( )u a2 2 at the end of time (u/a) from the start.
Kinematics of Particles—Rectilinear Motion 39
ANSWERS
1. Solution 23 8.106.3 ttx +=
6.216.21
6.218.10 2
+−=
=
+−=
=
tdtdva
ttdtdxv
t x v a
0 0 0 21.61 7.2 10.8 02 14.4 86.4 –21.63 0 –32.4 –43.24 –57.6 –86.4 –64.85 –180 –162 –86.46 –388.8 –259.2 –108.07 –705.6 –378 –129.68 –1152 –518.4 –151.2
2. Solution
tdtdva
ttdtdxv
tttx
927
5.4275.22
5.15.135.225.7
2
32
+−=
=
+−=
=
+−+=
when v = 0; 4.5t2 – 27t + 22.5 = 0
( )1,5
220366
0562
=
−±=
=+−
t
t
tt
At t = 1sec, x = 7.5 + 22.5 (1) – 13.5 (1)2 + 1.5 (1)3 = 18At t = 5sec, x = 7.5 + 22.5 (5) – 13.5 (5)2 + 1.5 (5)3 = –30when a = 0; 9t – 27 = 0; t = 3 sec.At t = 3sec, x = 7.5 + 22.5 (3) – 13.5 (3)2 + 1.5 (3)3 = –6At t = 2sec, x = 7.5 + 22.5 (2) – 13.5 (2)2 + 1.5 (2)3 = –10.5At t = 0, x = 7.5At t = 1, v = 0 hence changes the direction.
40 Engineering Mechanics
Figure A 12.1
Total distance travelled = 10.5 + 7.5 + 16.5 = 34.5Answer(a) position when velocity is zero = 18 m(b) position when acceleration is zero = –6 m(c) total distance travelled = 34.5 m
3. Solution v = 3 + 4t – 0.5t2
At t = 3, v = 3 + 4 (3) – 0.5 (3)2 = 11.5At t = 5, v = 3 + 4 (5) – 0.5 (5)2 = 10.5
Average acceleration 2m/s5.0)35(
)5.115.10( =−−=
At t = 10, v = 3 + 4(10) – 0.5(10)2 = –7At t = 15, v = 3 + 4(15) – 0.5(15)2 = –49.5
Average acceleration 2m/s5.8)1015(
)7(5.49 −=−
−−−=
v = 3 + 4t – 05t2
tdtdva −== 4
At t = 4s; a = 4 – 4 = 0At t = 12s; a = 4 – 12 = –8 m/s2
Answer (a) (i) – 0.5 m/s2; (ii) – 8.5 m/s2
(b) (i) 0; (ii) – 8 m/s2
4. Solution z = –4.8t2 + 3.2t + 4.8
6.9
2.36.9
−=
=
+−=
=
dtdva
tdtdzv
At t = 2s; z = –4.8 (2)2 +3.2(2) + 4.8 = –8At t = 5s, z = –4.8 (5)2 + 3.2(5) + 4.8 = –99.2
Average velocity m/s4.30)25(
)8(2.99 −=−
−−−=
At t = 4.5s; v = –9.6 (4.5) +3.2 = 40 m/sAt t = 2s, v = –9.6 (2) + 3.2(5) = –16 m/s
Kinematics of Particles—Rectilinear Motion 41
Average acceleration 2m/s6.9)25(
)16(8.44 −=−
−−−=
At t = 4.5s; a = –9.6 m/s2
Answer (a) –30.4 m/s2
(b) –40 m/s(c) –9.6 m/s2
(d) –9.6 m/s2
5. y = 8 sin 2t + 5
tdtdva
tdtdyv
2sin32
2cos16
−=
=
=
=
At t = 2s, v = – 10.458 m/s; a = 24.218 m/s2
At t = 5s, v = – 13.425 m/s; a = 17.409 m/s2
At t = 2s; y = – 1.054 mAt t = 5s; y = 0.648 mAnswer –1.054 m; –10.458 m/s; 24.218 m/s2
0.648 m; –13.425 m/s; 17.409 m/s2
6.
ttxx
ttdtdxv
152
5.4
15185.4
2
0
2
−
+=
−−==
x = x0 + 1.5t3 – 9t2 – 15t75 = x0 + 1.5(6)3 – 9(6)2 – 15(6): x0 = 165x = 165 + 1.5 t3 – 9t2 – 15tv = 4.5t2 – 18t – 15
189 −=
= t
dtdya
At t = 5s; x = 165 + (1.5) (5)3 – 9 (5)2 – 15 (5) = 52.5 mAt t = 5s; a = 9(5) – 18 = –27 m/s2.
7.
20
2
0
510
21010
1010
ttvv
ttvv
tdtdvv
−+=
−+=
−==
At t = 3s; –15 = v0 + 10(3) – 5(3)2; v0 = 0
42 Engineering Mechanics
320
32
0
2
2
667.15
35
210
510
510
ttxx
ttxx
ttdtdxv
ttv
−+=
−
+=
−=
=
−=
At t = 3s; 0 = x0 + 5(3)2 –1.667(3)3; x0 = 0
tattv
ttx
1010510
667.152
32
−=−=
−=∴
At t = 5s; x = 5(5)2 – 1.667(5)3 = –83.375 mv = 10(5) – 5(5)2 = –75 m/s.
8. Given: a = 10 sin t; at t = 0, x0 0= and v0 20= m sThe acceleration of the particle is given by
00
00
cos10cos10]cos[cos10
][cos10sin10
sin10
sin10
0
0
0
ttvvttvv
ttdtdv
tdtdv
tdtdva
t
t
tt
v
v
+−=−−=−∴
−==
=
==
∫∫
Substituting the given data in the above equation
dttdx
tdtdxv
ttv
)cos1030(
cos1030
cos103010cos1020
−=
−==
−=+−=
Integrating ∫∫ −=t
t
x
x
dttdx00
)cos1030(
)sin(sin10)(30 000 ttttxx −−−=− .Substituting the given data, we get
ttx sin10300 −=−
Kinematics of Particles—Rectilinear Motion 43
Hence the displacement, velocity and acceleration are given by;sin1030 ttx −= tv cos1030 −= and ta sin10=
When v is maximum, 0sin100 =⇒== tadtdv∴ ...,2,,0 ππ=t
At t = π,v is maximum
sm40cos1030m248.94sin1030
max
1
=−==−=∴
πππ
vx
9. Given data: 25.430 xa −= and v = 0 at x = 0The acceleration of the particle is given by
dxxdvvdtdx
dxdv
dtdva
)5.430( 2−=∴
==
since a x= −30 4 5 2.
Integrating,
132
132
2
2360
)35.4(30)2(
)5.430(
cxxv
cxxv
dxxdvv
+−=
+−=
−= ∫∫
when x v c= = ⇒ =0 0 01,∴ v x x2 360 3= −(a) When x = 1.5, 875.79)5.1(3)5.1(60 32 =−=v
)360(
sm937.822 xxv
v
−=
=∴
(b) v = 0; x = 0 and x x m2 20 4 472= ⇒ = . m
(c) When v is maximum, dvdx
x= − =60 9 02
∴ x = 2.582 m.
10. Given data: 22 ssa αα −=−= −
Particle starts from rest, v = 0 when s = 20mWhen s = 10m, v = 2.828 m/sThe acceleration of the particle along a straight line is
dssdvvsdt
dsdsdv
dtdv
sdtdva
)( 2
2
2
−−=∴
−==
−==
α
α
α
44 Engineering Mechanics
Integrating, v s c
2 112 1
= −−
+−
α ( )
vs
c2 12 2= +α
When s v c= = ⇒ + =20 0 220
2 01, .α
∴ c1 0 05= − . α
When s v c= = ⇒ + =10 2 828210
2 81, . .α
0 2 0 05 2 8. ( . ) ( )α α+ − =∴ α = = −
= −
80 4160 8
12
; c
vs
When s v v= = ⇒ =5 24 4 92; . m s11. Given data: a s= 6 1 3( )
The acceleration of the particle along a straight line is given bya f s
advdt
dvds
dsdt
vdv a ds s ds
=
= =
= =
( )
( )6 1 3
Integrating, v s C
v s c
v s c
2 1
13
1
22 6
4 31
2 4 31
26
1
4 32
9 2
13
=
++
= +
= +
+
( )
( )
when s v c= = = +64 48 48 9 64 22 4 31; ( ) ( )
∴ =c1 0
∴ v s v s
vdsdt
dtdsv
dss
dt s ds
2 4 3 2 3
2 32 3
9 3
33
= =
= = =
= −
( ) ( );
;( )
Kinematics of Particles—Rectilinear Motion 45
Integrating, ( )31
3
3
2 3 1
23
2
1 3
1 3 2
1 3 2
t s c
t s c
t s c
=− +
+
= +
= +
− +( )
( )
( )
( )
When ( )t s c= = ⇒ = +3 64 3 64 31 32, ( )
∴
3 43
3
1
22
1 3
= +
= −
= −
c c
t s
;
or ( ) . ( ) .t s i e s t+ = = +1 11 3 3
Velocity and acceleration is given by
v s ta s t
= = += = +
3 3 16 6 1
2 3 2
1 3( )( )
When t = 2s, sva
= + == + == + =
( ) m( ) m s.( ) m s
2 1 273 2 1 276 2 1 18
3
2
2
12. Escape velocity is the minimum velocity with which a particle shouldbe projected vertically upwards from the surface of earth so that itdoes not tend to return to the earth.Let ‘O’ be the centre of earth. Consider a particle projected fromthe surface of the earth. From Fig. 12.19a,
But
r R x
a gRr
g RR x
a dvdt
dvdx
dxdt
v dvdx
= +
= − = −+
= = =
2
2
2
2( )
( )
Equating the above two equations
vdvdx
gRR x
v dv gR dxR xve
= −+
∫ = −+
∫
2
20
22
0
( )
( )
α Figure A 12.2(a)
46 Engineering Mechanics
Where ve = escape velocity
gRv
gRRgR
xRgR
v
e
e
2
12
2
0
22
=
−=−=
+=−
α
If the same particle is projected from top of atower or hill of height ‘ 0h ’ above the surfaceof earth as shown in Fig. 12.19(b).
22
2
2
)(
)(
xRdxgRdvv
xRgR
dxdvva
+−=
+−==
Integrating
( )
000
2
0
2
22
22
0
Since
12
)(
0
0
hRrr
gRhR
gR
xRgRv
xRdxgRdvv
he
hve
+=−=+
−=
+=−
+−= ∫∫
α
α
Therefore0
22rgRve =
Hence escape velocity ‘ve’when the particle is projected from sur-face of earth, gRve 2=Escape velocity ‘ve’ when the particle is projected from a point‘ 0h ’above the surface of earth, )2( 0
2 rgRve = where
00 hRr +=13. Given data: a = 8 – x
Particle starts from rest and at t = 0, x = 0We know that acceleration can be written as a = f(x) since theparticle starts from rest, at t = 0, v = 0.
∴ dxadvvvdxdva
dtdx
dxdvxf
dtdva
==
===
;
)(
Figure A 12.2(b)
Kinematics of Particles—Rectilinear Motion 47
Integrating, v dv adx x dx= = −∫ ∫ ∫ ( )8
122
1
22
216
28
2
cxxv
cxxv
+−=
+
−=
where c1 = constant of integration.
At t = 0, v = 0 and x c= ⇒ =0 01 .∴ v x x2 216= − and a x= −8(a) Velocity when x = 10m
.sm746.760)10()10(16 22
==−=
vv
Therefore velocity of the particle when x = 10 m is v = 7.746 m/s.(b) Position when velocity ‘v ’ becomes zero
v x x2 16 0= − =( )∴ x = 0 and x = 16 m
Therefore the position of the particle when velocity is zero is at x =0 and x = 16 m(c) Velocity when acceleration is zero.
a = 8 – x = 0; x = 8.
sm86464)8(16
162
22
==−=
−=∴
vv
xxv
When the acceleration is zero, the velocity of the particle is givenby v = 8 m/s.
14. Given: 32)(3 va =When t = 3, s = 37.516m and v = 42.87 m/s.The acceleration of the particle is given by
dvvvdvdt
vdtdva
vfa
)32(32
32
)(3
)(3
)(
−==∴
==
=
Integrating,
131
1
1)32(
)32(
33
1)32(3
3
cvt
cvt
dvvdt
+=
++−
=
=
+−
−∫∫
48 Engineering Mechanics
At 5.187.42,3 1 −=⇒== cvt331 )5.0(3)5.0(3 +=⇒=+ tvvt
we observe that,dtds
dsdv
dtdva ==
32)(3 vdvvv
advds ==
dvvds 313 =
Integrating, ∫ ∫= dvvds 313
24
2
34)5.(
433
)34(3 ctscvs ++
=⇒+=
when t = 3, s = 37.516; 02 =c
2
3
4
)5.0(3
)5.0(
)5.0)(41(
+=
+=
+=∴
ta
tv
ts
when t = 5s
2sm75.90
sm375.166m766.228
=
==
a
vs
15. Given data: a x= −30 4 5 2. and v = 0 at x = 0The acceleration of the particle is given by
dxxdvvdtdx
dxdv
dtdva
)5.430( 2−=∴
==
since 25.430 xa −=
Integrating,
132
132
2
2360
)35.4(30)2(
)5.430(
cxxv
cxxv
dxxdvv
+−=
+−=
−= ∫∫
when x v c= = ⇒ =0 0 01,∴ v x x2 360 3= −(a) When x = 1.5, 875.79)5.1(3)5.1(60 32 =−=v
)360(
sm937.822 xxv
v
−=
=∴
Kinematics of Particles—Rectilinear Motion 49
(b) v = 0; x = 0 and x x m2 20 4 472= ⇒ = . m
(c) When v is maximum, dvdx
x= − =60 9 02
∴ x = 2.582 m.16. s-t and v-s curves are drawn.
Figure A 12.3(a)
The acceleration is given by
dtds
dsdv
dtdva ==
From Fig. 12.20(a), when t = 3, dtds
at 444.446.3
160 ==A
50 Engineering Mechanics
From Fig. 12.20(b), when 64=s ,
dsdv
at 469.016075 ==B
2sm844.20)444.44)(469.0( ==
=∴
dtds
dsdva
17. Freely falling ball = =v0 0.By equations of motion
sm05.49)5(81.981.9
)81.9(905.421
0
220
==∴=+=
==+=
vtatvv
atattvs Q
After breaking glass plate, sm335.34)05.49(7.01 ==vDistance travelled in next one second 2
1 )21( attv +=
m24.39)1(905.4)1(335.34 2
=+=
Therefore the distance travelled in the next second is 39.24 m18. Let ‘t’ be the time of journey and ‘H’ be the height from which it
had fallen. Since it falls freely, its initial velocity v0 is 0.
2
20
905.421
tH
attvH
=
+=∴ since .00 =v
Distance travelled in (t – 1) sec is given by2)1(905.4 −= ts
∴ Distance travelled in last one second = 98.1
stttt
sH
5.10;1.98)12(905.41.98)1(905.4905.4
1.9822
==−=−−
=−∴
∴ m776.540)5.10(905.4 2 ==HHence the maximum height from which it had fallen is 540.776m andthe time taken for its journey is 10.5 seconds.
19. Let ‘ 0v ’ be the initial velocity at the beginning of certain second.
( )sm095.43;)1(905.4)1(48
905.421
02
0
20
20
=+=∴
+=+=
vv
ttvattvs
Therefore the initial velocity sm095.430 =v
Figure A 12.4
Kinematics of Particles—Rectilinear Motion 51
Velocity at the end of ‘t’ seconds
Kinetics of Particles—Newton’s Second Law 1
1 41 41 41 41 4
Kinetics of Particles —Kinetics of Particles —Kinetics of Particles —Kinetics of Particles —Kinetics of Particles —Newton’Newton’Newton’Newton’Newton’s Second Laws Second Laws Second Laws Second Laws Second Law
CONCEPTS
In this chapter, kinetics of particles involving Newton’s second law andD’Alembert’s principle is described. Application to connected bodies arealso illustrated. Analysis of motion of a particle subjected to central forcewith particular application to satellite motion is also dealt in this chapter.
• According to Newton’s second law of motion,r rF ma=
where r rF F= Σ when more than one force act on the particle, and
ra is the acceleration in the direction of resultant.
• While considering the motion of particles, rectangular compo-nents normal and tangential components and radial and trans-verse components are used.(i) In rectangular components,
ΣF max x= ; and ΣF may y=(ii) In tangential and normal components,
Σ ΣF ma F mat t n n= =; and
where and advdt
vdvds
av
t n= = =2
ρ(iii) In radial and transverse components,
Σ ΣF ma F mar r= =; and θ θ
[ ] [ ]where and a r r a r rr = − = +&& & && & &θ θ θθ2 2
• Equations of motion of a particle, can be written asΣr rF ma− = 0. where inertial force F mai = −
r. Hence inertial force
vector −mar of magnitude ‘ma’ and direction opposite to the accel-
eration of particle is added. The particle is said to be in dynamicequilibrium. Thus a dynamics problem is transformed to a staticsproblem. This principle is known as D’Alembert’s principle.
2 Engineering Mechanics
• If a particle is in static equilibrium, the resultant of external forcesis zero. However, if a particle is in dynamic equilibrium, the result-ant of external forces is not zero.
• In the analysis of lift motion, Tension in string (T) due to accel-eration (a) of the lift is,
T = W [1 + (a/g)] for upward motionT = W [1 – (a/g)] for downward motion
• Linear momentum of a particle is expressed asr r
rr r
G mv
F ma m dvdt
dGdt
=
= = =
The rate of change of linear momentum is the resultant of forcesacting on the particle.
• Angular momentum rH0 of a particle about ‘O’ is the moment
about ‘O’ of the linear momentum mvr of the particle.
r r rH r mv0 = ×The rate of change of angular momentum of a particle about ‘O’is the sum of moments about ‘O’ of the forces acting on theparticle.
• When the only force acting on a particle is a force rF directed
towards or away from a fixed point ‘O’, the particle is said to bemoving under a central force.The angular momentum of a particle moving under a central forceis constant, both in magnitude and direction.
• According to Newton’s law of gravitation,
F GMmr
W mg= =2 and
g GMR
GM gR= =22 or
• For a satellite moving under central force motion,1
12ru
GMh
C redc
= = + =
+cos
cosθ
θ and
where e chGM
d C= =2
1and ( )
If e = 0, circleIf 0 < e < 1, ellipseIf e = 1, parabolaIf e > 1, hyperbola
Kinetics of Particles—Newton’s Second Law 3
• h rv r v= = 0 0• For an elliptical path,
rA when θ = 0 is perigee; r a eA ( )= +1 ; rP when θ =180 is apo-gee; r a eP ( )= −1
where a r r= +( )( )max1 2 0
b a e b r r= − =1 20 and max
• Periodic time of a planet is the time required by the body todescribe its orbit.
T Ah
= 2
where A = Area of orbit, h rv r v= = 0 0
• Escape velocity, v GMr
gRresc = =2 2
0
2
0
represents the smallest v0 for which the particle will not return toits starting point.
• The trajectory of the satellite is determined from v0 as,If v v0 > esc HyperbolaIf v v0 = esc ParabolaIf v v0 < esc Ellipse
If vv
0 2= esc Circle
• Kepler’s laws of planetary motion:(1) Every planet moves in an elliptical orbit with the sun located
at one of the focal points.(2) The radius vector drawn from the sun to any planet sweeps
equal areas in equal time intervals.(3) The square of orbital period of any planet is proportional to
the cube of the semi-major axis of the elliptical orbit.Example 14.1 A block of mass 80 kg rests on a horizontal plane asshown in Fig. 14.12(a). Find the magnitude of force ‘P’ required to give theblock an acceleration of 2sm4=a to the right. The coefficient of frictionbetween the block and the plane is 0.30.
Figure 14.1(a)
4 Engineering Mechanics
Solution Due to the force the block moves to the right with a = 4 2m s .The block is subjected to its weight (W = mg), the force ‘P’, the verticalreaction ‘N’ and frictional force ‘F’. Applying Newton’s Second Law F =ma, as shown in Fig. 14.12(b),
Figure 14.01(b) Newton’s Second Law
Since the motion is to the right, frictional force acts in a direction op-posing the motion.
→+ ΣF max = .
P F macos .30o − =↑+ ΣFy = 0
− − + == + ⋅
784 8 30 030 784 8
. sin .sin .
P NN P
o
o
[ ]
N727.775320)8.784(3.0)30sin3.030(cos
.3208.78430sin3.030cos
).4(803.030cos
=∴=−−
=+−
=−
PP
PP
NP
oo
oo
o
The above problem is also solved using D’Alembert’s principle byapplying the inertia force ‘ma’ in a direction opposite to that of motion.The block is in dynamic equilibrium as shown in Fig. 14.12(c)
Figure 14.12(c) D’Alembert’s principle
→+ ΣF P N max = − − =0 30 0 3 0; cos . .o
P Ncos . ( ) .30 0 3 80 4 0o − − =
Kinetics of Particles—Newton’s Second Law 5
+↑ ΣF P Ny = − − + =0 30 784 8 0; sin . .o
N P= +sin . .30 784 8o
[ ]N727.775
0320)8.784(3.0)30sin3.030(cos03208.78430sin3.030cos
=∴=−−−
=−+−∴
PP
PPoo
oo
Example 14.2 A block 80 N is pulled up a smooth plane by a force ‘P’as shown. Determine the acceleration along the plane. (See Fig. 14.13)
Figure 14.02(a) Figure 14.02(b)
Solution
Mass of block = 80/9.81=8.155 kgComponent of weight parallel to the plane =W sin .α , where ‘α ’ is the
inclination of the plane. Let us assume that the block moves up the plane.
Figure 14.02(c) Figure 14.02(d)
Resolving the forces along the plane ′x and normal to the plane ′y asshown in Fig. 14.13(c) we get,
→+ ΣF F W Px i′ = − − + =0 0; sin cosα θ− − + =∴ = −
8155 80 35 30 15 02 073 2
. ( ) sin cos. m s
aa
o
6 Engineering Mechanics
The block does not move up the plane and moves down the plane witha = 2 073 2. m s .
↑+ 015sin35cos;0 =++−=Σ ′ PNWFyo
015sin3035cos80 =−= oN N768.57=∴ N
When ‘P’ is increased to 75 kN, again let us assume that the blockmoves up the plane.
ΣF P W Fx i′ = − − =0 0; cos sinθ α
75 15 80 35 8155 03257 2
cos sin . ( ). m s
o o− − =∴ =
aa
035cos8015sin75;0 =−+=Σ oo NFyN121.46=∴ N
When P = 30 N, the block moves down the plane with an accelerationof 2.073 m s2.
When P = 75N, the block moves up the plane with an acceleration of3.257 m s2.Example 14.3 A flexible chain of length ‘L’ rests on a table with length‘C’ overhanging the edge as shown in Fig. 14.14(a). The system originallyat rest is released. Describe the motion. Self weight of chain is w/unitlength.Solution Let ‘T’ be the tension in the chain when a portion of lengthhas fallen down.
Considering the freebody diagram as shown in Fig. 14.14(b)T m aT m a wxm w L x g m w x ga a a d x dt
− =+ − == − == = =
A AB B
A B
A
( ) ;
00
2 2B
T m a
T wg
L x d xdt
=
= −
A A
( )2
2
Considering the freebody diagramas shown in Fig. 14.14 (c)
T m a wx
T wx wxg
d xdt
+ − =
= −
B B 02
2
Fig. 14.03(a)
Fig. 14.03(a) Dynamic equilib-rium of chain on table
Kinetics of Particles—Newton’s Second Law 7
wg
L x d xdt
wx wxg
d xdt
−
− −
=( )
2
2
2
20
wLg
d xdt
wx
d xdt
g L x
⋅
=
=
2
2
2
2( )
Solution of this second order differential equation is,
( ) ( )x A e B e
g L t g L t= +
−( ) ( )
The constants A and B are evaluated using conditions(1) At x = c, v = 0(2) At t = 0, v = 0.
( ) ( )
teLgB
eLgAdtdxv
eBeAx
Lg
tLg
tLg tLg
)(
)(
)(
.)(
)(
−−
==
+=−
At x = c, t = 0; c = A + BAt V = 0, t = 0;
( ) ( )00
2
05
= −= −
∴ = =
= +
−
A g L B g LA B
A B c
x c e eg L t g L t
.( )
. ( ) ( )
Example 14.4 The bob of a pendulum, 3.5 m long, describes an arc ofa circle in vertical plane as shown in Fig. 14.15(a). If the tension in thestring is 2.5 times the weight of the bob for the position shown, find thevelocity and acceleration of the bob in that position.Solution Since the bob describes the arc of a circle, the acceleration innormal (an) and tangential direction (a t) are considered and inertia forcesare shown for the bob in dynamic equilibrium as shown in Fig. 14.15(b).
Fig. 14.03(c) Dynamicequilibrium of chain
overhanging
8 Engineering Mechanics
Figure 14.04(a) Figure 14.04(b) Dynamic equilibrium
→+ ΣF ma mgt t= − =0 30 0: sin o
∴ a gt = = =sin . sin . m s30 9 81 30 4 905 2o o
↑+ ΣF T ma mgN n= − − =0 30 0: cos o
But T mg= 2 5.
sm49.7)(velocity
.102.56)5.3(029.16;
sm763.16
)029.16()905.4(
sm029.1630cos5.2
22
2
22
22
2
=∴
===
=
+=
+=∴
=−=∴
v
vva
aaa
gga
n
nt
n
ρ
o
and acceleration ( ) .a = 16 029 m s2
tan ;α α= =a an t 73o
Example 14.5 A small sphere of weight ‘W’ is held as shown in Fig.14.16(a) by two wires ‘AB’ and ‘CD’. Determine the tension in the wires‘AB’ and ‘CD’. Also determine the acceleration ofthe sphere and tension in wire CD, if the wire ABis cut.Solution The sphere is at rest and is in equilib-rium.
→+ ΣF T Tx = − =0 40 40 0: cos cosCD ABo o
Fig. 14.04(c)
Fig. 14.05(a)
Kinetics of Particles—Newton’s Second Law 9
∴ T TCD = ΑΒ
↑+ ΣF T T Wy = + − =0 40 40 0: sin sinAB CDo o
∴ == =
2 400 778
T WT T WABAB CD
sin.
o
When the wire ‘AB’ is cut, the ball attached towire ‘CD’ swings with an acceleration as shown inFig. 14.16(c). Immediately after it is cut, its initial ve-locity is zero. Hence normal component of accelera-tion, a vn = =2 0ρ . The direction of motion andparticle in dynamic equilibrium are shown in Fig.14.16(c).
WWTaTmaWF
ggamamgmaWF
n
nn
t
ttt
643.050cos;0Since050cos:0
sm515.7776.050sin050sin;050sin:0
CD
CD
2
====−+=Σ
===∴=−=−=Σ
o
o
o
oo
Figure 14.05(c) Dynamic equilibrium of sphere
Therefore the acceleration of sphere when the wire AB iscut = 7.515m s2 and the tension in the string CD = 0.643 W.
Example 14.6 A particle of mass ‘m’ rests on the top of a smoothsphere of radius ‘r’ as shown in Fig. 14.17(a). Assuming that the particlestarts to move from rest, determine the point at which it leaves the sphere.
Fig. 14.05(a)
10 Engineering Mechanics
Figure 14.06(a) Figure 14.06(b) Particle indynamic equilibrium
Solution Let θ be the angular displacement at time ‘t’ during its travel.In free body diagram, the reaction ‘R’ from sphere is along the radial lineand weight of the particle, W = mg.
Inertia force mat and man are also shown.
∴ = − == + − =Σ
ΣF W ma
F R ma Wt t
n n
0 00 0
: sin: cos
θθ
But, distance travelled ‘s’ = rθvelocity ‘v’ = = =ds dt r d dt r( )θ ω
acceleration a dv dt r ddt
r d dt rt = = = =ω θ α( )2 2
normal acceleration a v r rn = =2 2ω
W ma a Wmt tsin ; sinθ θ− = =
0
a r ggrt = = =
α θ α θsin ; sin
We know, ωθ
αω
= =ddt
ddt
and
∴ = =
∴ =
dt d d
d d
θω
ωα
α θ ω ω
But α θ θ θ ω ω=
=gr
gr
d dsin , sin
Integration from 0 to ‘θ ’ ( ) sin .g r d dθ θ ω ωθ ω
0 0∫ = ∫
Kinetics of Particles—Newton’s Second Law 11
[ ] [ ][ ]
[ ]
( ) cos ( )
( ) cos cos( ) cos
g r
g rg r
− =
− + =− =
θ ω
θ ωθ ω
θ ω0
20
2
2
1 2
0 21 2
When the particle leaves the sphere, normal reaction ‘R’ will be zero.Hence from the equation,
R mama W
nn
+ − =∴ =W cos
cosθ
θ0
But a v r rr
rn = = =22
2( ) .ω ω
Substituting in man, [ ]m r mg g rω θ ω θ2 2= =cos ; ( ) cos
∴
− = =
− =
gr
g r( cos ) ( )cos
coscos
12 2
12
2θ ω θ
θ θ
∴ = = =∴ =
1 15 1 15 0 66748 2
. cos ; cos ( . ) .( . )
θ θθ o
Therefore the particle leaves the sphere when θ = ( . )48 2 o
Example 14.7 A block of mass ‘m’ mayslide freely on a frictionless arm ‘OA’ whichrotates in horizontal plane at constant rate θ0as shown in Fig. 14.18(a). Initially ‘B’ is re-leased from a distance ‘a’ from 0. Determine(a) Velocity component vr along OA as afunction of ‘r’. (b) The magnitude of hori-zontal force
rF exerted on mass ‘B’ by the arm
OA in terms of r.
Solution F ma m r rF ma m r r
r r= = −= = +
(&& ( &) )( && & &)
θθ θθ θ
2
2Since the mass can slide freely on the arm Fr = 0. It is subjected to Fθ ,
only as shown in Fig. 14.18(b).Using Newton’s second law,→+ [ ]ΣF ma m r rr r= = −: && ( &)0 2θ
↑+ [ ]ΣF ma F m r rθ θ θ θ= = +: && & &2
Fig. 14.07(a)
12 Engineering Mechanics
Since rr r r r
v r r v vr r r= + = +& &λ θ λ λ λθ θ θIn this case v rr = & and r v&θ θ=
&& &r vdvdr
drdt
vdvdrr
rr
r= = ⋅ =
Substituting this in equation,
[ ]m r r&& ( &)− =θ 2 0
∴ − =
− =
∴ =
&& ( &)
( & )
( & )
r r
vdvdr r
v dv rdr
rr
r r
θ
θ
θ
2
02
02
0
0
Integrating v dv r dr
v r
r rv
v
a
r
r
v
v
a
r
r r0
0
02
2
02
2
2 2
∫ = ∫
=
( & )
( & )
θ
θ
Since v0 0= : [ ]v r ar2
02 2 2= −( & )θ or [ ]v r ar = −&θ0
2 2
Since & &θ θ= =0 constant; &&θ = 0 and &r dr dt vr= =
[ ][ ]
F m r rF m r vr
= += +
&& & &
( ) &
θ θθ
20 2 0
{ }F mv m r ar= = −2 20 02 2
0& & ( ) &θ θ θ
∴ = −F m r a2 02 2 2( & ) ( )θ
Hence, at any position ‘r’ of block ‘B’,
v r a F m r ar = − = −& ( ); ( & ) ( )θ θ02 2
02 2 22and
Example 14.8 A lift carries a weight of 3600 N and is moving with auniform acceleration of 35 2. m s as shown in Fig. 14.21. Determine thetension in the supporting cable when
a) lift is moving upwards; andb) lift is moving downwards.
Assume g = 9 81 2. m s .
Fig. 14.07(b)
Figure 14.08(a)Upward motion
Kinetics of Particles—Newton’s Second Law 13
Solution
[ ]
[ ]
T W ma
T W ma W Wg
a
T W a g
T
T ma W
T W ma W Wg
a
T W a g
T
1
1
1
1
2
2
2
2
0
1
3600 1 359 81
4884 404
0
1
3600 1 359 81
231560
− − =
= + = +
= +
∴ = +
=
+ − =
= − = −
= −
∴ = −
=
( )..
.
( )( . )
( . ).
N
N
Example 14.9 A lift has an upward acceleration of 2.5m s2. Whatpressure will a man of weight 800 N exert on a floor of the lift? Determinethe pressure he would exert if the lift has an acceleration of 2.5m s2 down-wards. Also determine the upward acceleration to cause the weight toexert a pressure of 1200 N on the floor. Assume g = 9 81 2. m s .
Solution
(a) Upward motion: [ ]T W a g= +1 ( )
T1 800 12 59 81
1003874= +
=
..
. N
(b) Downward motion: [ ]T W a g= −1 ( )
T2 800 12 59 81
596126= −
=
..
. N
(c) If T N= 1200 , for upward motion
[ ]T W a ga
= +
= +
1
1200 800 19 81
( )
.
a = −
=9 81
1200800
1 4 905 2. . m s
Example 14.10 Two equal weights ‘W’ are connected by a light string(assume weightless) passing over a frictionless pulley as shown in
Figure 14.08(b)Downward
motion
14 Engineering Mechanics
Fig. 14.27(a). A weight ‘w’ whose magnitude is much less than ‘W’ isadded to one side causing that weight to fall. Determine the accelerationof the system, assuming that the weights start from rest.
Fig. 14.27(a) Fig. 14.27(b)
Fig. 14.27(c)
Solution m W gA ;= and m W w gB ( )= +
For block A: T W m a− − =A 0T W g a W− =( )
For block B: T m a W w+ − + =B ( ) 0
T W wg
a W w++
= +( )
Qm W wgB =+
∴+
=2W w
ga w
a g wW w
=+( )2
Kinetics of Particles—Newton’s Second Law 15
Example 14.11 Two blocks ‘A’ and ‘B’ ofmasses kg280A =m and kg420B =m arejoined by an inextensible cable as shown inFig. 14.28(a). Assume that the pulley is fric-tionless and µ = 0 30. between block ‘A’ andthe surface. The system is initially at rest.Determine (a) acceleration of block A; (b) ve-locity after it has moved 3.5 m; and (c) veloc-ity after 1.5 seconds.
Fig. 14.11(b) Fig. 14.11(c)
Solution
N2.4120N8.2746
420;280
BB
AA
BA
======
gmWgmW
kgmkgm
Block A: +↑ ΣF Ny = =0 27468: .
→+ ΣF m a Tx = − − + =0 0 3 27468 0: . ( . )A
T a− =280 824 04.
Block B: +↑ ΣF T m ay = + − =0 4120 2 0: .B
T a+ =420 4120 2.Solving the above two equations
700 3296 16 4 709280 4 709 824 04 2142 56
2a aT
= =∴ = + =
. ; . m s( . ) . . N
Hence a = 4 709 2. m s and T = 2142 56. NVelocity after it has moved a distance of 3.5 m is,
v v as
v
202
22
0 2 4 709 355741
= += +
∴ =( ) ( . )( . ). m s
Fig. 14.11(a)
16 Engineering Mechanics
Velocity after 1.5 sec,v v at
v
= += +
∴ =
00 4 709 157 064
( . )( . ). m s
Example 14.12 A body of mass 25 kg resting on a horizontal table isconnected by a string passing over a smooth pulley at the edge of thetable to another body of mass 3.75 kg and hanging vertically as shown inFig. 14.29(a). Initially, the friction between 25 kg mass and the table is justsufficient to prevent the motion. If an additional 1.25 kg is added to the3.75 kg mass, find the acceleration of the masses.
Figure 14.12(a) Figure 14.12(b)Solution
mA kg;= 25 mB . kg= 3 75
W m gW m g
F N
A AB B
. N;. N
= == ==
2452536 788
µwhere N = 245 25. ;and F T= = 36 788.
∴ µ = =
=( )
..
.F N36 78824525
015
When 1.25 kg is added to 3.75 kg mass, it moves down.∴ = × =W 5 9 81 49 05. . N
Figure 14.12(d) Figure 14.12(e)
Fig. 14.12(c)
Kinetics of Particles—Newton’s Second Law 17
05.495
005.495:0788.3625
)25.245(15.025
25.245;025.245:0015.025:0
AA
A
=+
=−+=Σ=−=−∴
==−=Σ=+−−=Σ
aTaTF
aTaT
NNFTNaF
y
y
x
Solving the above two equations30 12 262 0 4087 2a a= =. ; . m s
∴ = + =T 25 0 4087 36 788 47 006( . ) . . NTherefore acceleration of masses a = 0 409 2. m s andTension in string, N006.47=T
Example 14.13 A body weighing 400N is resting on a rough plane inclined at20o to the horizontal as shown in Fig.14.30(a). It is pulled up the plane from restby means of a light flexible rope runningparallel to the plane and passing over alight frictionless pulley at the top of theplane. The portion of rope beyond the pul-ley hangs vertically and carries a weight225 N at the end. If the coefficient of fric-tion for the plane and the body is 0.15, find(a) tension in rope; (b) acceleration of thebody; and (c) distance moved by the body in 3 seconds, starting from rest.Solution kg775.4081.94001 ==m
kg936.2281.92252 ==mFor Block 1:
+→ ΣF F W F Tx i′ = − − − + =0 20 01: sin o
− − − + =m a N T1 400 20 015 0sin .o
↑+ ΣF Ny′ = − =0 400 20 0: cos ;o N = 375877.
Substituting N = 375877. in the first equation− − − + =40 775 400 20 015 375877 0. sin . ( . )a T
∴ T a− =40 775 193190. .
Fig. 14.13(a)
18 Engineering Mechanics
For Block 2:+↑ ΣF T m a Wy = + − =0 02 2:
T a+ − =22 936 225 0.∴ T a+ =22 936 225.Solving the above two equations∴ 63 711 3181. . ;a = a = 05 2. m sAcceleration of the body = 0.5 m s2
∴ T = + =19319 40 775 0 5 213578. . ( . ) . NTherefore tension in the string = 213.578N
.m25.2)3)(5.0)(21()3(0
)21(2
20
=+=
+=
s
attvs
Therefore distance covered by the body in 3 sec-onds = 2.25m.Example 14.14 Two rough planes inclined at 30oand 60o to the horizontal and the same height are placedback to back. Masses of 12 kg and 24 kg are placed onthe faces and are connected by a string passing overthe pulley on the top of planes as shown in Fig 14.31(a).If µ = 0 6. find the resulting acceleration.
mm
W m gW m g
AB
A AB B
kgkg. N. N
==
= == =
1224117 7223544
Solution Assume that block‘A’ moves up the plane and ‘B’moves down the plane.
Figure 14.14(b) Figure 14.14(c)
Figure 14.13(b)
Figure 14.13(c)
Figure 14.14(a)
Kinetics of Particles—Newton’s Second Law 19
For block A:→+ ΣF a T W Nx ′ = − + − − =0 12 30 0 6 0: sin .A A
o
− + − − =12 117 972 30 0 6 0a T N. sin . Ao
↑+ ΣF W Ny′ = − + =0 30 0: cosA Ao
NA ( . ) cos . N= =117 72 30 101949o
T aT a− = +− =
12 117 72 30 0 6 10194912 120 029
. sin . ( . )
.
o
For Block B:ΣΣ
F T a N WF W N
x
y
′
′
= + + − == − + =
0 24 0 6 60 00 60 0: . sin: cos
B B
B B
o
o
NB . cos .= =23544 60 117 72o
T a+ = − +24 0 6 117 72 23544 60. ( . ) . sin o
Solving these two equationsT aT a+ =− =
24 13326512 120 029
.
.∴ 36 13236 0 368 2a a= =. ; . m s
∴ T = + =120 029 12 0 368 124 445. ( . ) . NTherefore the resulting acceleration is 0.368 m s2
Example 14.15 Block A and B weighing 500 N and 1500 N respectivelyare connected by a weightless rope passing over a frictionless pulley asshown in Fig. 14.32(a). Assume thecoefficient of friction as 0.3 on allcontact surfaces and determinethe velocity of the system 5 s afterstarting from rest.
Solution
W N m W gW N m W g
A A AB B B
, . kg, . kg
= = == = =
500 50 9681500 152 904
W WB A ,> hence block ‘B’ moves downand Block ‘A’ moves up.Block A:→+
030sin
:0
AA
AA
=−−
−=Σ ′
FW
amTFxo
T a N− − − =50 968 500 30 0 3 0. ( ) sin .A Ao
↑+ ΣF W Ny′ = − + =0 30 0: cosA Ao
Figure 14.15(b)Dynamic
equilibrium of block A
Figure 14.15(a)
20 Engineering Mechanics
904.379968.50N013.433
A
A
=−∴=
aTN
Block B:ΣF T m a W FT a N
x ′ = + − + =+ − + =
0 60 0152 904 1500 60 0 3 0
: sin. sin .
B B B B
B B
o
o
ΣF W NN
y′ = − + ==
0 60 0750
: cosB B
B
o
038.1074904.1520)750(3.0
60sin1500904.152
B
B
=+=+−+∴
aT
aT o
Since a a aA B ,= =T a
T aaa
− =+ =
=∴ =
50 968 379 904152 904 1074 038203872 694134
3405 2
. .
. .
. .. m s
T Nv v at= + == + = + =
379 904 50 968 3405 553450 3 405 5 17 0250
. . ( . ) .. ( ) . m s
Therefore tension in rope (T) = 553.45 Nacceleration (a) = 3.405m s2 and velocity after 5sec = 17.025 m/s.
Example 14.16 A small block of weight 60 N starts from rest at ‘P’ andslides down on the plane of the roof surface inclined at 32o to the horizon-tal as shown in Fig. 14.33(a). Find the horizontal distance from Q at whichthe block will hit the ground. The coefficient of kinetic friction between theblock and roof is 0.25. Also find the velocity when the block reaches theground.
Figure 14.16(a)
Figure 14.15(c)Dynamic equilibrium
of block B
Kinetics of Particles—Newton’s Second Law 21
Solution sin ; . m32 3 5661o = =PQ PQMotion down the plane ‘PQ’
N W F N WW F ma
W W ma
= = =∴ − − =
− − =
cos ; cossin
sin cos
θ µ µ θθ
θ µ θ00
[ ]a Wm
a g
=
−
= −= − =
sin cos
(sin cos ). (sin . cos ) . m s
θ µ θ
θ µ θ9 81 32 0 25 32 3119 2o o
Consider motion of block,Velocity at P, vP = 0, since it starts from rest.
v v as
v
Q P
Q
( . )( . ). m s
2 2 20 2 3119 56615943
= += +
∴ =The block when falls from ‘Q’ it becomes a projectile with an initial
velocity v0 and angle of projection α .The equation of projectile is,
y xg
vx= −tan
secα α2
02
2
2v0
2
5 943320 625
0 8481391
== −= −==
. m s
tan .cos .
sec .
αααα
o
∴ = − −
= − −
y x x
y x x
( . ) . ( . )( . )
. .
0 625 4 905 13915943
0 625 01932
22
2
At R, y = –8,∴ + − =
+ − =01932 0 625 8 0
3235 41408 0
2
2. .
. .x x
x x
x
x
=− ± − −
∴ = −
3235 3235 4 41408
25018 8 253
2. ( . ) ( . )
. m, . mx cannot be negative. Hence x = 5.018m
Figure 14.16(b)
Figure 14.16(c)
22 Engineering Mechanics
Time taken to reach ‘R’ from Q x vx= = =( ).
. coss0
5 0185 943 32
1o
v v gt
v v v
v v
yR yQ
x y
y x
= + = + =
= + = + =
= =
( . sin ) . ( ) .
( . ) ( . ) . m s
tan ; ( . )
5943 32 9 81 1 12 959
504 12 959 13896
68 75
2 2 2 2
θ θ o
Example 14.17 A horizontal force P = 600 N is exerted on block A ofmass 120 kg as shown in Fig. 14.34(a). The coefficient of friction betweenblock A and the horizontal plane is 0.25. Block B has a mass of 30 kg andthe coefficient of friction between it and the plane is 0.40. The wire be-tween the two blocks makes 30o with the horizontal. Calculate the tensionin the wire.
Figure 14.17(a)
Solution Both the blocks move towards the right. The free body dia-grams of blocks are drawn.
W m gW m g
B BA A
. . N. . N
= = × == = × =
30 9 81 294 3120 9 81 1177 2
Figure 14.17(b) Figure 14.17(c)
Block B: →+ ΣF T F Fx i= − − =0 30 0: cos B Bo
T N acos . B30 0 4 30 0o − − =
↑+ ΣF T Ny = − + =0 30 294 3 0: sin . Bo
N TB . sin= −294 3 30o
Kinetics of Particles—Newton’s Second Law 23
∴ − − − =− − =
− =
T T aT a
T a
cos . ( . sin ). .
. .
30 0 4 294 3 30 30 01066 117 72 30 0
28143 110 432
o o
Block A: →+ ΣF F T Fx i= − − − =0 600 30 0: cosA Ao
600 0 25 30 120 0− − − =. cosAN T ao
↑+ ΣF T Ny = − − + =0 1177 2 30 0: . sin Ao
N TA . sin= +1177 2 30o
∴ − + − − =− + − =
+ =
600 0 25 1177 2 30 30 120 00 991 3057 120 0
121090 308 476
. ( . sin ) cos. .
. .
T T aT a
T a
o o
Equating the values of ‘T’∴ + = −= =
∴ = + =
110 432 28143 308 476 121090198 044 149 233 1327
110 432 28143 1327 147 778
2. . . .
( . ) ( . ) . m s. . ( . ) . N
a aa
T
Therefore tension in wire (T) = 147.778 N; and, acceleration of theblocks (a) =1327 2. m s towards the right.Example 14.18 Determine the ac-celeration of block ‘A’ for the systemif the system starts from rest. Coeffi-cient of friction between block ‘A’ andtable is 0.25 and that between blocks‘A’ and ‘B’ is 0.35 (See Fig. 14.35(a)).
W WWm W gm W gm W g
A BCA AB BC C
N; N;N
. kg. kg
. kg
= ==
∴ = == == =
1500 9001500
152 90591743152 905
Solution Let the tension in the string connecting the blocks A and Bbe T1. The tension in the string connecting Blocks A and C be T2. The freebody diagram is shown in Fig. 14.35(b).
Consider block C,+↑ ΣFy = 0; T m a WC C C2 0+ − =
T aC2 152 905 1500 0+ − =.
Figure 14.18(a)
24 Engineering Mechanics
Consider block A,→+ ΣFx = 0; T T m a2 1 315 600 0− − − − =A A
T T a2 1 152 905 915 0− − − =. AConsider block B,
→+ ΣFx = 0; − + + =T m a1 315 0B B− + + =T a1 91743 315 0. B
Since a a a aA B C ,= = =T a
T T aT a
T aT a
22 1
122
152 905 1500152 905 915
91743 315244 648 1230152 905 1500
+ =− − =− + = −
∴ − =+ =
.
.
.
.
.Solving the above two equations
∴ = = + =∴ = + =
a T aT a
0 679 1230 244 648 1396116315 91743 377 294
22
1
. m s ; . ( ) . N. . N
Hence T1 377 294= . N; and T2 1396116= . NTherefore acceleration of block A = 0 679 2. m s
Figure 14.18(b) Dynamic equilibrium of blocks A, B & C
Kinetics of Particles—Newton’s Second Law 25
Example 14.19 Two blocks A and B are held on an inclined plane asshown in Fig. 14.36(a). The coefficient of friction for blocks A and B withinclined plane are 0.3 and 0.2 respectively. If the blocks begin to slidedown simultaneously calculate the time and distance travelled by eachblock before block B touches block A. Suppose if they continue to moveas a single unit, determine the contact force exerting between them.W WA BN and N= =300 500 .
Figure 14.19(a)
Solution Assume that the block B touches block A at O. Considerdistance travelled by A is sA and distance travelled by B is sB. Since theinitial velocity is zero,
s v a t a ts v a t a tA 0A A A
B 0B B B
= + == + =
( ) .( ) .1 2 0 51 2 0 5
2 2
2 2
Component of weight perpendicular to the plane =W cosθComponent of weight parallel to the plane =W sinθ
Figure 14.19(b)
26 Engineering Mechanics
Frictional force, F N W= =µ µ θcos
mW
mW
AA
BB
gkg;
gkg= = = =30581 50 968. .
Consider block A:Σ
Σ
F W N
N WF W F m a
m W g F Na
a
y
x
′
′
= − + =
= = == − − =
= = =− − =
∴ =
0 0
300 35 2457460 0
30581300 35 0 3 245746 30581 0
3216 2
; cos
cos cos .; sin
. ;sin . ( . ) .
. m sA
A A
A AA A A AA A A A A
A
θ
θθ
µ
o
o
Consider block B:− + =
= = =− − =
− − =∴ =
W NN W
W F m aaa
B B
B BB B B B
B
B2m s
coscos cos .sin
sin . ( . ) ..
θθθ
0500 35 409 576
0500 35 0 2 409 576 50 968 0
4 020
o
o
When the blocks touch, m
mseconds
B A
B B
A A
B A
s s ss a t ts a t t
s s tt
− = == == =
∴ − = ==
10 0505 2 0105 16080 402 10 05
5
2 2
2 2
2
.. .. .. .
Distance travelled by A A= = =s 1608 5 40 202. ( ) .Distance travelled by B B= = =s 2 01 5 50 252. ( ) .After 5 seconds, they move together as a single unit with same
acceleration ‘a’.ΣF W F m a
W F m a
a
a
x ′ = − − +− − =− +−
− + =
∴ = =
00
300 35 0 3 245 746500 35 0 2 409 576
30581 50 968 0303 22281549
3 718
; ( sin )( sin )
sin . ( . )sin . ( . )
( . . )( . )( . )
.
A A AB B B
2m s
θθo
o
They move together with accelera-tion of 3.718 .sm 2 To determine contact force FC, dynamic equilibriumcondition of any one of the blocks must be considered.
Figure 14.19(c)
Kinetics of Particles—Newton’s Second Law 27
Consider block A,W F m a F
FF
A A A A C
CC N
sinsin . ( . )
. ( . ).
θ − − + =− ++ =
∴ =
0300 35 0 3 24574630581 3718
15351
o
This can also be calculated by con-sidering block B, as follows,
W F m a FF
F
B B B B C
CC N
sinsin . ( . ) . ( . )
.
θ − − − =− − =
∴ =
0500 35 0 2 409 576 50 968 3718
15374
o
Note: Contact force is a compressive force, hence it acts towards theblock.
Therefore after 5 seconds, the blocks move together as a single unitwith a = 3 718. m s2 and the contact force between them is 15.35N.Example 14.20 Determine the tensions in the string and accelerationsof the blocks A and B, shown in Fig. 14.37(a).Solution Total length of the stringis constant, i.e, 2y yA B+ = con-stant. Let ∆yA and ∆yB be thechange in position of blocks A andB respectively
2 02 0
2 0 2
∆ ∆y yv v
a a a a
A BA B
A B B A
+ =∴ + =
∴ + = = −;In other words, if block A moves
down with an acceleration ‘a’, blockB moves up with an acceleration ‘2a’.Similarly, if block A moves up withan acceleration ‘a’, block B movesdown with an acceleration ‘2a’. Inthis case, since W WA B> , let us as-sume that block A moves down withan acceleration aA. Hence block B must move up with an acceleration aB.But we know that a aB A.= −2
Let the tension in the string be T1. The free body diagram of pulley P1,block A and block B are shown in Fig. 14.37.
mm
AB
kgkg
= == =
420 9 81 42 814140 9 81 14 217
. .. .
Figure 14.19(d)
Figure 14.20(a)
28 Engineering Mechanics
From Fig. 14.37(b): 2 0 21 2 2 1T T T T− = =;From Fig. 14.37(c): T m a W2 0+ − =A A A
T a2 42 814 420 0+ − =. AFrom Fig. 14.37(d): T m a W1 0− − =B B B
T a1 14 271 140 0− − =. BBut T T a a2 12 2= =and B A
2 42 814 42028542 140
11
T aT a+ =− =
.
.AA
Solving the above two equations 99 898 140. aA =
∴ =aA2 m s1416. ( )↓
T a a a1 140 28542 180 416 2 2 832= + = = =. . ; .A B A2N and m s ( )↑
Figure 14.20(b) Figure 14.20(c) Block A inPulley P1 dynamic equilibrium
Figure 14.20(d) Block B indynamic equilibrium
Kinetics of Particles—Newton’s Second Law 29
Example 14.21 Block A and blockB of weights 750 N and 1500 N respec-tively start from rest as shown in Fig.14.38(a). The coefficient of friction be-tween block A and the surface of theinclined plane is 0.2. Determine the ac-celeration of each block and tension inthe wire. Assume the pulley to beweightless and frictionless.Solution Let us first determinewhether the block moves up the plane or slide down the slope and also themaximum frictional force that could mobilized to keep the system frommoving. Assume that the system is at rest; the forces acting on the bodyare shown in free body diagram.
Figure 14.21(b) Figure 14.21(c)
N904.129)519.649(2.02.0375;075030sin750
519.649;30cos750
max =====+−−==
NFFFNN
o
o
The body cannot remain at rest since ‘F’required to keep the body at rest is greaterthan Fmax. Therefore it should move up theplane.
The system has one degree of freedom,therefore magnitude of accelerations can beexpressed in terms of aA.
If block A moves up the plane with an ac-celeration ‘aA’, simultaneously block B movesdown with an acceleration a aB A( )= 2
Figure 14.21(a)
Figure 14.21(d) Dynamicequilibrium of ‘B’
30 Engineering Mechanics
Block B: a aB A.= 0 5 and moves downward.
kg905.15281.9
1500BB ===
gWm
Figure 14.21(e) Figure 14.21(f) DynamicPulley equilibrium of ‘A’
ΣF T aT a
y = + − =+ − =
0 152 905 1500 076 453 1500 0
1
1
; ..
B
APulleyMass of the pulley is very small; hence it is neglected. T T1 22 0− =Block A
m WgAA kg= = =750
9 8176 453
..
904.504453.761500453.760904.504453.76030sin750)519.649(2.0453.76030sin:0
519.649;030cos750:0
A2
A1
A2
2A
A2maxAA
=−=+=−−=−+−−=−+−−=Σ==+−=Σ
′
′
aTaT
aTTa
WTFamFNNF
x
y
o
o
Since T T1 22= ,3 2004 904 668 3012 2T T= =. ; . N
and aA2m s= 213.
a TB2m s N= =1065 1336 6021. ; .
Therefore tension in the wire (T N2 ) .= 668 301
Acceleration of block A, aA2m s= 213.
Acceleration of block B, aB2m s= 1065. ( )↓
Kinetics of Particles—Newton’s Second Law 31
Example 14.22 A satellite is launched in a direction parallel to thesurface of earth with a velocity of 35000 km/hr at an altitude of 600 km asshown in Fig. 14.46. Determine (a) eccentricity of orbit (b) maximum alti-tude reached by the satellite (c) periodic time. M = ×598 1024. ,kgR = 6380 km and G = × −6 67 10 11 3 2. .m kg sSolution
1
2rC
GMh
= +cosθ
where m kg sGM
= ×= ×
−6 67 10598 10
11 3 2
24.. kg
h r r r r v= = =2 & ( &)θ θ
∴ = = + ×
= ×
h r v
h
0 03
3
10
6380 600 10 35000 103600
678611 10
[( ) ]
1
1 6 67 10 598 106 78611 10
1 8 66134 10
2
11 24
10 2
8
rC GM
h
rC
rC
= +
= + × ×
×
= + ×
−
−
cos
cos( . ) ( . )
( . )
cos ( . )
θ
θ
θ
At θ = 0: r r= = ×036980 10
∴×
= + × −16980 10
8 66134 1038
( )( . )C
∴ = × −C 56653 10 8.
∴
= × + ×− −1 56653 10 8 66134 108 8
r( . ) cos ( . )θ
‘r’ is also expressed in terms of eccentricity ‘e’ as
rh GM
e=
+( )
cos
2
1 θAt θ = 0:
11
8 66134 10 6980 101654
2
08 3+ = =
× ×=−e h GM
r( )
( . ) ( ).
32 Engineering Mechanics
e = <0 654 1. , hence orbit is an ellipse.
∴ At θ = = × + ×− −180 1 56653 10 180 8 66134 108 8o o; ( . ) cos ( . )maxr∴ = =rmax .33377391 33377 391m km
max. . .( ) ( )
( ) [ . ] . km
( . ) [ ( . ) ].
maxmax
( )
altitude kmmajor axis
minor axis
= − = − =− = = +
= + =
− = − = −=
r Ra r r
a
b a eb
33377 391 6380 26997 3911 2
1 2 6980 33377 391 20178 695
1 20178 695 1 0 65415265072
0
2 2 1 2
This can also be calculated as, b r r= 0 max
period
secondshr
hr seconds
T abh
TT
= =×
×== ==
2 2 20178 695 15265 072 106 78611 10
28520475333 7 927 55 12
6
10π π( . ) ( . )
( . )
. min .min
Figure 14.22
EXERCISES
1. A particle of mass 45 kg falls from a height of 12 m and penetratesin to the ground. If the resistance of penetration is uniform and isequal to 15000 N, find the depth through which it penetrates.
Kinetics of Particles—Newton’s Second Law 33
2. A locomotive exerts a constant pull on a train of mass 300 tonnesup an incline of 1 in 200 measured along the slope and in 4 km, thespeed drops from 72 km/hr to 60 km/hr. If the frictional resistance isconstant at 50 N/tonne, find the pull of the locomotive.
3. A 20 kg mass is subjected to forces as shown in Fig. E 14.1. Find itsacceleration.
Figure E 14.1
4. A 15g bullet is shot with an initial velocity of 250 m/s into a blockof wood as shown in Fig. E 14.2. It stops after penetrating 32.5 mminto the wood. Find the force needed to stop it. Assume uniformdeceleration.
Figure E 14.2
5. A force of 640 N acts on a body having a weight of 150 N, for 12seconds. If the initial velocity is 6 m/s find (a) acceleration pro-duced in the direction of force (b) distance moved by the body in12 seconds.
6. Aravind weighing 650 N dives into a swimming pool from a towerof height 20 m. He was found to go down in water by 3 m and thenstarted rising. Calculate the average resistance of water. Neglectair resistance.
7. A box is placed in the rear side of van and the van decelerates at3.5m s2. Determine the static coefficient of friction between thebox and bed of the van required such that box will not slide.
8. A train of weight 5000 kN is moving up an inclined plane 3 in 150with an acceleration of 0.05m s2. Tractive resistance is 5N per kN.Determine the acceleration if it moves with same tractive force (a)on a level track (b) down the plane inclined at 3 in 150.
34 Engineering Mechanics
9. A box of mass 70 kg resting on a horizontalsurface is subjected to a force of 350 N appliedat o40 to the horizontal as shown in Fig. E14.3. The coefficient of friction between thebox and the surface is 0.25. Determine the ve-locity of the box after it has moved 5.5m.
10. A train weighing 1800 kN has a velocity of 60 km/hr when it is atthe beginning of 1.5% grade as shown in Fig. E 14.4. If the pullexerted by the engine is 15 kN and tractive resistance is 9 N per kNload find the distance which it will travel up the inclined planebefore coming to rest.
Figure E 14.4
11. A train has a weight of 2500 kN. The frictional resistance is 7.2 Nper kN. Determine the pull exerted by locomotive to increase thespeed from 40 km/hr to 60 km/hr within a period of 2.5 min.
12. A 25 kg box is dropped on to a body of a van moving at a speed of60 km/hr. If the coefficient of friction is 0.5, calculate how far thevan will move before the box stops slipping?
13. A car of mass ‘m’ is travelling around a curve with uniform veloc-ity of 72 km/hr. The coefficient of friction between the tyres andthe ground is 0.65. Determine the minimum radius of the curve sothat the car does not skid.
14. Consider the amusement ride shown in Fig. E 14.5. A train of smallcars with passengers travels inside the loop of tracks on a path 50m diameter. Assume that the tracks can apply only normal forces‘ Rn ’ to the cars. The weight of a car is 5000 N. (a) Determine thespeed at T at which there is no force from tracks on a car. (b) Atpoint ‘L’ if the speed of the train is 20 m/s determine normal forceon a car at ‘L’.
Figure E 14.3
Kinetics of Particles—Newton’s Second Law 35
Figure E 14.5
15. Determine the velocity of the car moving on a highway curve ofradius ρ = 100 m banked through θ = 20o. The velocity on a bankedroad is the velocity at which a car should travel, if no lateral frictionis exerted on its wheels.
16. The motion of a block of mass 2.5 kg on a horizontal plane isdefined by the relations, r t= +2 1 2( cos )π and θ π= 2 t where‘r’ is in metres, t in seconds and θ in radians. Determine the radialand transverse components of the force exerted on the block when(a) t = 0 and (b) t = 1.25 sec.
17. An experimental car of mass 1500 kg travelling on a horizontalroad is tracked by radar located at ‘O’ as shown in Fig. E 14.6. At aparticular r =120m, & m s,r = 3 && ;r = 0 and θ =50o ,
secrad25.0=θ& and && .θ = 0 Calculate horizontal force exerted onthe car.
Figure E 14.6
36 Engineering Mechanics
18. A 3 kg block ‘A’ on a rotating arm is at r = 0.5m when & , &r = =0 5θ rad/s and && .θ = 08 rad/s. Determine the net force on the block. (ReferFig. E 14.7)
Figure E 14.7
19. A 80 kg mass is resting on a smoothplane which is inclined at 30o withhorizontal as shown in Fig. E 14.8.A cable passes from this mass overa frictionless pulley to 160 kg masswhich when released will drop ver-tically. Determine the velocity of160 kg mass after 4 seconds of itsrelease from rest.
20. A 80 kg mass is resting on an in-clined plane which makes an angleof 30o with horizontal. A cablepasses from this mass over a fric-tionless pulley at the upper end ofthe plane.The free end of the cableis connected to a weight of 1569.6N. If the coefficient of friction be-tween the 80 kg mass and the planeis 0.25, calculate the time taken for the hanging weight to descendby 8.5 m (see Fig. E 14.9).
21. Find the frictional force on Block ‘A’ if the tension in the cable is600N. Assume m m kgA B= =100 for the blocks in Fig. E 14.10.
Figure E 14.8
Figure E 14.9
Kinetics of Particles—Newton’s Second Law 37
Figure E 14.10
22. Two bodies of weight W NA = 800 and W NB = 400 are connectedto the two ends of light inextensible string, passing over a smoothpulley. The 800 N weight is placed onrough horizontal surface whose coeffi-cient of friction is 0.25. The 400 N weightis hanging vertically in air. If the sys-tem is released from rest and Block ‘B’falls through a vertical distance of 2.5m, determine the velocity attained byblock ‘B’.
23. Block ‘A’ and ‘B’ are connected with abar of negligible weight as shown in Fig. E 14.11. If A and B, eachweigh 300 N, with µA = 0 25. and µB = 0 5. , Calculate the accel-eration of the system and the force in the bar.
24. Two bodies of weight N800A =W and N500B =W are con-nected to the two ends of a light inextensible string which passesover a smooth pulley. The weight 800N is placed on an inclinedplane of angle 15o. If the coefficient of friction is 0.2, determine theacceleration, tension in string and the distance moved by 500 N in5 seconds starting from rest.
25. Two smooth inclined planes whose inclinations with horizontal
are 30o and 20o, are placed back to back. Two blocks of weights,W NA = 300 is placed on 30o plane and N500B =W is placed on20o plane. They are connected by a light inextensible string pass-ing over a smooth pulley. Calculate the acceleration and tension inthe string.
26. Two rough inclined planes whose inclinations with horizontal are30o and 18o, are placed back to back. Two blocks of weights,WA = 1800 N is placed on 30o plane and NW 500B = is placed on
Figure E 14.11
38 Engineering Mechanics
18o plane and are connected by a light inextensible string passingover a smooth pulley. If the coefficient of friction is 0.35, find theacceleration of the blocks and tension in the string.
27. Find the acceleration of the blocks and the tension in the cable forthe system shown in Fig. E 14.12. (a), (b) and (c).
Figure E 14.12(a) Figure E 14.12(b)
Figure E 14.12(c)
28. In the pulley system shown inFig. E 14.13, the masses of blocks Aand B are 24 kg and 12 kg respec-tively. The masses of the pulley andrope can be neglected.(a) Determine the acceleration of
each block and tension in therope
(b) Instead of block A, if a force of235.44 N acts, determine the ac-celeration of block B and ten-sion in the rope.
Figure E 14.13
Kinetics of Particles—Newton’s Second Law 39
29. In the system of connected blocks shown in Fig. E 14.14., thecoefficient of kinetic friction under blocks A and C is 0.25. Calcu-late the acceleration of each block and tension in the connectingcable. The pulleys can be assumed to be frictionless and of negli-gible weight.
Figure E 14.14
30. Determine the acceleration of the system shown in Fig. E 14.15 andtension in cables. Assume that cables pass through frictionlesspulley. Given: µA .= 0 3 and µB .= 0 25
Figure E 14.15
31. Two blocks ‘A’ and ‘B’, connected by a thread, move along arough horizontal plane under the action of a force 490 N as shownin Fig. E 14.16. Coefficient of friction between sliding surface ofweights and the plane is 0.3. Determine the acceleration of thesystem and tension in the cable.
40 Engineering Mechanics
Figure E 14.16
32. A body weighting 600 N and lying on a smooth table top is con-nected by eans of two ropes as shown in Fig. E 14.17. If N90L =Wand N30R =W , calculate the tension in the ropes and the result-ing acceleration.
Figure E 14.17
33. Blocks ‘A’ and ‘B’ are connected with a bar of negligible weight,as shown in Fig. E 14.18. If Aand B weigh N100A =Wand N50B =W and coeffi-cient of friction µA .= 0 25and 375.0B =µ , determinethe acceleration of the sys-tem and the force in the bar.
34. Three blocks A, B and C are connected by strings that pass overfrictionless pulleys as shown in Fig. E 14.19. The acceleration ofthe system is 2.5m s2 to the right and the surfaces are rough. Findthe tension in the strings and coefficient of kinetic friction be-tween the blocks and surfaces (Assume same µ for both theblocks). N.250 and N125N,75 CBA === WWW
Figure E 14.18
Kinetics of Particles—Newton’s Second Law 41
Figure E 14.19
35. A system of weights connected by string and passing over thetwo pulleys P1 and P2 are shown in Fig. E 14.20. Find the accelera-tion of each block. Assume smooth frictionless pulleys.
N.225 and N450,N675 CBA === WWW36. A child Swathi of mass 16 kg slides down on an inclined plane in a
park as shown in Figure E 14.21. Assuming that the resistance dueto friction and air is 100 kN, calculate the time taken for her to slidedown the incline.
Figure E 14.20 Figure E 14.21
37. A man of mass 90 kg draws water from a well as shown in Fig. E14.22. The coefficient of friction between the ground and his feet is0.2. If acceleration of bucket is 0.2m s2, the determine the maxi-mum mass of water (together with the mass of bucket) he can drawwithout slipping?
42 Engineering Mechanics
Figure E 14.2238. Determine the velocity at which a rocket can permanently escape
the attraction of earth.39. A particle falls vertically from rest as shown in Fig. E 14.23 in a
medium whose resistance is proportional to the velocity of theparticle. Prove that, ( )v g e t= − −β β( )1 andy v g v g= − − −β β β( ) log ( ).2 1
40. A particle of weight ‘W’ is suspended on a wire of length ‘L’ asshown in Fig. E 14.24 Determine the period and frequency.
Figure E 14.23 Figure E 14.2441. A ball of weight ‘W’ is rotated at con-
stant speed ‘v’ in a vertical circle ofradius ‘R’ as shown in Fig. E14.25. Findthe tension in the cord when the ball isat (a) position ‘P’ and (b) position ‘Q’.
42. A particle of mass ‘m’ slides down africtionless chute and enters a loop asshown in Fig. E 14.26. Find the heightrequired at the start in order that theparticle may make a complete circuit inthe loop.
Figure E 14.25
Kinetics of Particles—Newton’s Second Law 43
43. The ball of mass ‘m’ shown in Fig. E14.27, moves in a circle ofradius ‘R’ with a speed ‘v’. Find the tension in the cord.
Figure E 14.26 Figure E 14.27
44. A body starts with a velocity ‘v0’ and moves with a retardation
equal to β times the distance travelled ( ).a x= −β Show that the
distance travelled by the particle before it comes to rest is v0 β .45. A body moving along a straight line is subjected to a resistance
which produces a retardation βv3, where ‘v’ is the velocity at anyinstant ‘t’. Show that the velocity and distance travelled are ex-pressed as,
v vsv
t s sv
=+
=
+
0
0
2
0112β
βand
where s = displacement and v0 = initial velocity.46. If the planes are smooth and W WB A> , find the acceleration, of the
blocks and the tension in the string for the system shown inFig. E 14.28.
Figure E 14.28
44 Engineering Mechanics
47. A satellite describes an elliptic orbit about a planet of mass ‘M’. Ifr0 and r1 represent the minimum and maximum values of the dis-tance ‘r’ from the satellite to the centre of planet, derive
1 1 2
0 12r r
GMh
+
=
where ‘h’ is the angular momentum per unit mass of satellite.
ANSWERS
1. M = 45 kg; h = 12 m; R = 1500 NVelocity at ground level m/s344.15)12)(81.9(22 === ghWhen it penerates the ground, let the acceleration be ‘a’ W = 45 ×9.81 = 441.45 NR – W + ma = 0(1500 – 441.45 + 45a) = 0a = 323.523 m/s2 (declaration)
m364.0)523.323(2)344.15(
22
20
2
=∴−+=
+=
SSO
asVV
depth of penetration = 0.364 m2. m = 300 t = 300 × 103 kg
anlgles allfor 200
1tansin
== θθ
S = 4 km = 4000 m
222
20
2 m/s015.0)4000(2
)20()667.16(;2
m/s667.163600
)1000(60km/hr60
m/s203600
)1000(72km/hr72
−=−=∴+=
===
===
aasVV
V
V
f
f
o
If P = pull,
Frictional resistance tonne300tonne50 ×= NF
F = 15000 NComponent of weight parallel to plane = W sinθ
147152001)81.910300( 3 =
××=
Figure A 14.1
Figure A 14.2
Kinetics of Particles—Newton’s Second Law 45
N2521501500014715)015.0)(10300(
0sin3
==−−−×−
=−−−
PP
FWmaP θ
Pull of the locomotive (P) = 25215 N3. →+ 631.1530sin15025cos100 =−=∑ xF
↑+ 166.17230cos15025sin100 =+=∑ yF
2
22
m/s644.820
874.172;
81.84;tan
874.172)166.172()631.15(
===
=∑∑
=
=+=
amaR
FF
R
x
y αα
4. Vo = 250 m/s; Vf = 0; S = 32.5 mm = 0.0325 m
2
2
22
m/s46.961538
);0325.0(2)250(
2
−=
+=
+=
a
aO
sVV of α
If ‘R’ is resistance– R – ma = 0; R = – maR = –(0.015) (–961538.46)R = 14423.076
5. F = 640 N; t = 12 SW = 150 N; Vo = 6 m/s
kg291.1581.9
150 ==m
F = ma
m56.3085
)12)(855.41(21)12(6
21
m/s855.41)291.15(
)640(
2
2
2
=
+=
+=
==
=∴
S
attVS
mFa
o
a) Acceleration = 41.855 m/s2
b) distance travelled = 3085.56 m
Figure A 14.3
Figure A 14.4
46 Engineering Mechanics
kg259.6681.9
650;N650 =
== mW
Velocity at water level, )20)(81.9(22 == ghVo
m3
0m/s809.19
=
==
S
VV
f
o
ion)(deceleratm/s399.65
)3(2)809.19(
2
2
2
22
−=∴
+=
+=∴
a
aO
asVV of
If ‘R’ is resistance,R + ma – W = 0R + (66.259) (–65.399) – 640 = 0∴ R = 4973.272 NAverage resistance of water = 4973.272 N
7. – F – ma = 0F = –ma
357.081.9
)5.3(
ondecelerati9
9
=−
−=∴
−=
−=
µ
µ
µ
a
aWW
Coefficient of static friction = 0.3578. W = 5000 kN = 5000 × 103 N
N25000
kN5000kN1N5
N10100
501)105000(sin
kg50968481.9
)105000(
3
3
3
=
×=
×=
×=
=×=
F
W
m
θ
–Wsinθ – F – ma + P = 0P = (100 × 103) + (25 × 103) + (509684 × 0.05) = 150.484 × 103N
Figure A 14.5
Figure A 14.6
Figure A 14.7
Kinetics of Particles—Newton’s Second Law 47
(a) on a level trackP – F – ma = 0150484 – 25000 – 509684 a = 0∴ a = 0.246 m/s2
(b) down the plane inclined at 3 in 150P – ma – F – Wsinθ = 0150484 – 509684a – 25000 + 100000 = 0a = 0.443 m/s2
Pull = 150.484 × 103 Nacceleration on level track = 0.246 m/s2
acceleration down the plane = 0.443 m/s2
9.
Figure A 14.8→+ 0)(25.0º40cos350;0 =−−=∑ maNFy
↑+ 07.686º40sin350;0 =+−−=∑ NFy
∴ N = 911.676∴ 350 cos 40º – 0.25 (911.676) – 70a = 0∴ a = 0.574 m/s2
asvv 220
2 +=when the box starts from rest, v0 = 0∴ v2 = 0 + 2 (0.574) (55)∴ v = 2.513 m/s
10.
Figure A 14.9For small values of α, sin α = tan α = (1.5/100)
Figure A 14.7(b)
Figure A 14.4
48 Engineering Mechanics
Weight component opposing motion = W sin atractive resistance = 9 N per kN
Tractive resistance ‘F’ N16200kN1800kN1N9 =×
=
kg10)486.183(81.9
)101800(;kN1800 33
=×===g
WmW
0sin;0 =−−−=∑ ix FWFPF α
0)10486.183(10015)1800(200.1615)10( 33 =×−
−− a
∴ 183.486 × 103a = –28.2 × 103
a = –0.154 m/s2
v = (v0)2 + 2as
When it comes to complete halt, v = 0
m/s667.163600
1060 3
0 =×=v
0 = (16.667)2 + 2(– 0.154)s∴ s = 901.912 mIt travels 901.912 m up the inclined plane before coming to rest.
11. Frictional force = 7.2 N per kN
∴ Total frictional force ‘F’ N18000)kN2500()kN1(
N2.7 =×=
m/s667.163600
1060km/hr60
m/s111.113600
1040km/hr40
3
3
0
=×==
=×==
v
v
kg10842.25481.9
)102500(/
kN2500m/s037.0150/)111.11667.16(
33
2
×=×
==∴
==−=∴
gWm
Wa
→+ 0;0 =−+−=∑ FPmaFx P = ma + F = (254.84 × 103) (0.037) + 18000 = 27429.154 N
∴ Pull = 27.429 kN12. Let us assume that the van is moving from left to right. When the
box is dropped, box tries to move to the left. It moves to the left tillit reaches the velocity of van. During that time van would havetravelled a distance ‘s’.
Kinetics of Particles—Newton’s Second Law 49
W = 25 × 9.81 = 245.25 N
Figure A 14.10∴ 0.5 (N) = ma
2m/s905.425)25.245(5.0 =⇒= aaTime required for the box to reach the speed of van is ‘t’.
sec4.3)(905.40667.16
m/s667.16)6060(
100060km/hr60
0
=⇒+=+=
=×
×==
ttatvv
v
Distance travelled by van,2
0 21 attvs
+=
Since the van travels with uniform speed of 60 km/hr, a = 0∴ s = v0t = 16.667(3.4) = 56.668 m
13. →+ N65.0;0N65.0;0 ==−=∑ nnx mamaF
where r
van
2=
+↑ mgNNmgFy ==+−=∑ ;0;0substituting N = mg in the firstequation
3765.6)81.9(65.02
==
∴
rv
Given m/s20)6060()1000(72km/hr72 =
×==v
m730.623765.6
)20(3765.6
22===∴ vr
∴ If man is greater than the frictional force, the car will not skid.Therefore the minimum radius (r) for which the car does not skid =62.730 m.
Figure A 14.11
50 Engineering Mechanics
14. The normal acceleration is acting towards centre. The dynamicequilibrium is shown in Fig. 14.12.Consider the position T, where there is no force in the track i.e.,Rn = 0.man – W – Rn = 0
m/s661.15)25(81.9
0;0)(;022
===∴
=−
=−=
gRv
mgRvmW
RvmR
T
TTn
Consider the position, L
N944.815425
)20(
81.95000
0
2
2
2
==
=
=
=
−
RvmR
RvmR
Ln
Ln
Therefore at position T, speed, vT = 15.661 m/s and at position L,normal reaction Rn = 8154.944 N
15.
m/s317.1820sin)81.9()100(sin9
sin9
sin
2
2
===
=
=
θ
θ
θ
rV
rVmm
rVmW
16.
]2[
])([ 2
θθ
θ
θ&&&&
&&&
rrmF
rrmFr
+=
−=
0;2cos8
2;2sin4
2;2cos22
2
22
2
2==−=
=
==−=
=
=+=
dtdt
dtrdr
dtdt
dtdrr
ttr
θθππ
πθθππ
πθπ
&&&&
&&
substituting in the eqn. Fr
]82cos16[
])2)(2cos22()2cos8[(22
22
πππ
ππππ
−−=∴
+−−=
tmF
ttmF
r
r
Figure A 14.12(a)
Figure A 14.12(b)
Figure A 14.13
Kinetics of Particles—Newton’s Second Law 51
]82cos16[
)]2)(2sin4(2)0()cos22[(]2[and
22 πππ
ππππθθ
θ
θ
−−=∴
−++=+=
tmF
ttmrrmF &&&&
(a) At t = 0; Fr = 2.5 (– 236.871) = –592.176 N; and Fθ = 2.5(0) = 0
(b) At t = 1.25; Fr = 2.5 [–16π2 cos (2.5π) – 8π2] = –197.392 N; and
Fθ = 2.5 [ – 16π2 sin (2.5π)] = –157.914 N
17.
Figure A 14.14 Figure A 14.14(a)
]2[
])([ 2
θθ
θ
θθ&&&&
&&&
rrmmaF
rrmmaF rr
+==
−==
Substituting the given values in the above eqn.,Fr = 1500 [0 – 120(0.25)2] = –11250 N = –11.25 kNFθ = 1500 [120 (0) + 2(3)(0.25)]
= 2250 N = 2.25 kN
)º31.11(;tan
kN473.1122
=
=
=+=
αα θ
θ
r
r
FF
FFR
172.769.38sin473.11sin955.869.38cos473.11cos
−=−==−=−=−=
ββ
RRRR
y
x
)º69.38(31.1150
=−=
−=
β
αθβ
jiRrrr
172.7955.8 −−=∴18. Fr = m ar
])([ 2θ&&& rrmFr −=
θθ maFFr
=−=−= 5.37])5(5.00[3 2
Figure A 14.14(b)
52 Engineering Mechanics
N519.37)2.1()5.37(
2.1)]5)(0(2)8.0)(5.0[(3]2[
22 =+=
=+=+=
F
FrrmF
θ
θ θθ &&&&
19. mA = 80 kg; mB = 160 kgWA = mA g = 784.8 NWB = mB g = 1569 N
Figure A 14.16(a)
The inclined plane is smooth, and block B moves down when it isreleased.Block A
4.39280080º30sin7840º30sin;0
=−=−−=−−=′∑
A
A
AAA
aTaT
WamTxF
Block B
6.156916006.1569160
0;0
=+=−+
=−+=∑
B
B
BBBy
aTaT
WamTF
Since aA = aB = a, equations of dynamic equi-librium of block A and B are given byT – 80a = 392.4T + 160a = 1569.6Solving the above two equations3T = 2354.4∴ T = 784.8 N; and a = 4.905 m/s2
∴ v = v0 + at = 0 + 4.905(4) = 19.62 m/sTherefore velocity of 160 kg after 4 seconds of its release= 19.62 m/s
20. N = 784.8 cos 30ºN = 679.657
Figure A 14.15
Figure A 14.16(b)
Kinetics of Particles—Newton’s Second Law 53
Fmax = 0.25 N = 169.914Block A;
314.562800º30sin8.784914.16980
0º30sin;0 max
=−=−−−
=−−−=∑ ′
A
A
AAx
aTaT
WFamTF
Block B
06.1569160
0;0
=−+
=−+=∑
B
BBBy
aT
WamTF
Since aA = aB = a,T – 80a = 562.314T + 160a = 1569.6Solving the above two equations3T = 2694.528T = 898.176 N; a = 4.198 m/s2
sec01.2
)198.4(21)0(5.8
21
2
20
=∴
+=
=
t
tt
attvs
Therefore hanging weight B takes 2.01 sec-onds to descend by 8.5 m.
21. T + mB a – 981 = 0600 + 100 a – 981 = 0a = 3.81 m/s2
Figure A 14.18 Figure A 14.18(a)
T –FA– mB a = 0600 – FA – 100 (3.81) = 0∴ F = 219 NFrictional force on block ‘A’ = 219 N.
22. FA = 0.25 (800) = 200
kg775.4081.9
400;kg549.8181.9
800 ==== BA MM
Figure A 14.17
Figure A 14.17(a)
54 Engineering Mechanics
Figure A 14.19 Figure A 14.19(a)
T – 200 – 81.549 a = 0; T – 81.549 a = 200T + 40.775 a – 400 = 0; T + 40.775 a = 400Solving T = 333.333a = 1.635
m/s859.2);5.2)(635.1(2220
2 === VasVV23. )º07.28(;158tan == θθ
The blocks move down the planeFor Block A:
0)º07.28sin(30025.0581.300sin300581.30;0
=−++=−++=∑ ′
A
Ax
NaTFaTF θ
Figure 14.20(a)
712.264)07.28cos(300
0cos300;0
==
=−=∑ ′
A
Ay
N
NF θ
Kinetics of Particles—Newton’s Second Law 55
T + 30.581a + 0.25 (264.712) – 300 sin (28.07)º = 0T + 30.581a = 74.987For Block B:
712.264
0)07.28cos(300;00)º07.28sin(3005.0581.30
0sin300581.30;0
=
=−=∑=−++−
=−++−=∑
′
′
B
By
B
Bx
N
NFNaTFaTF θ
–T + 30.581a + 0.5(264.712) – 300 sin (28.07) = 0–T + 30.581a = 8.809T + 30.581a = 74987Solving the above two equations
Figure 14.20(b)
61.162a = 83.796; a = 1.37 m/s2
∴ T = 74.987 – 30.581 (1.37) = 33.091Therefore acceleration of the system is a = 1.37 m/s2; and force inthe bar T = 33.091 N.To check: For the total system
0sin5.025.0;0 =−+++=∑ ′ θABABAx WNNamamF(30.581 + 30.581)a + 0.25 (264.712) + 0.5(264.712) – (2) 300(sin 28.07) = 0∴ a = 1.37 m/s2
Hence it is correct.24. WA = 800 N; mA = 81.549 kg
WB = 500 N; mB = 50.968 kgWA sinθ = 207.055 NWA cosθ = 772.741 NFA = 0.2 (772.741) = 154.548 N
Figure A 14.21
56 Engineering Mechanics
Figure A 14.21(a)(b)
T – 207.055 – 154.548 – 81.549 a = 0T + 50.968 a – 500 = 0T – 81.549 a = 361.603T + 50.968 a = 500
m05.13)5)(044.1(21
21
N789.446
m/s044.1)132517()397.138(
2
2
2
=
=
+=
=
==∴
S
attVS
T
a
o
25. WA = 300 NWB = 500 N
Figure A 14.22
kg968.5081.9
500
kg581.3081.9
300
=
=
=
=∴
B
A
m
m
WA sin 30 = 300 sin 30 =150WB sin 20 = 500 sin 20 = 171.010Planes are smooth and aA = aB = a
Kinetics of Particles—Newton’s Second Law 57
Figure A 14.22(a)(b)
For Block A, T – 150 – mA a = 0T – 30.581 a = 150
For Block B, – T –50.968+171.01 = 0T + 50. 968 a = 171.01
Solving, 81.549 a = 21.01 ∴ a = 0.258 m/s2
T = 171.01 – 50.968(0.258) = 157.860 NTension in the string (T) = 157.860 NAcceleration of the blocks (a) = 0.258 m/s2
26. WA = 1800 NWB = 500 N
kg968.5081.9
500
kg486.18381.9
1800
=
=
=
=∴
B
A
m
m
WA sin θ = 1800 sin 30 = 900WA cos θ = 1800 cos 30 = 1558.846 NWB sin θ = 500 sin 18 = 154.509 NFA = µ (WA cos θ) = 0.35 (1558.846) = 545.596 NFB = µ (WB cos θ) = 0.35 (475.528) = 166.435 N
Figure A 14.23(a)(b)
For Block A, T + mA a + FA – 900 = 0T + 183.486 a + 545.596 – 900 = 0T +183.486 a = 354.404
Figure A 14.23
58 Engineering Mechanics
For Block B, T – mB a – FB – 475.528 = 0T – 50. 968 a –166.435 – 475.528 = 0T – 50.968 a = 641.963∴ 234.454 a = –287.559∴ a = –1.227 m/s2
and T = 50.968 (–1.227) + 641.963 T = 579.425 N
Since ‘a’ is negative, Block ‘B’ moves down; Block ‘A’ moves up.Tension (T) = 579.425 NAcceleration (a) = 1.227 m/s2
27. Distance moved block ‘B’ is half of the distance moved by block ‘A’.WA = 190 NWB = 150 N
N57)190(3.0
5.02
===
=
=∴
AA
AA
B
WF
aaa
µBlock B: TBaB – WB = 0
TB + 15.291 aB – 150 = 0TB + 15.291 aB = 150
Block A: TA – mA aA – FA = 0TA + 19.368 aA – 57 = 0TA + 19.368 aA = 57
TB + 15.291 aB = 150TA – 19.368 aA = 57Since TB = 2TA and aB = 0.5 aA,2TA + 7.646 aA = 150TA – 19.368 aA = 57∴ 46.386 aA = 36 aA = 0.776 m/s2
TA = 57 + 19.368 (0.776) = 72.030Tension in cable A, TA = 72.030Tension in cable B, TB = 144.060acceleration of A, aA = 0.776 m/s2
acceleration of B, aB = 0.388 m/s2
b. WA = 200 NWB = 300 N
kg581.30
81.9300
kg387.2081.9
200
=
=
=
=
B
A
m
m
Figure A 14.24(a)
Figure A 14.24(b)
Figure A 14.23(c)
Kinetics of Particles—Newton’s Second Law 59
WA cos θ = 200 cos 35 = 163.830 NWA sin θ = 200 sin 35 = 114.715 NFA = µ (WA cos θ) = 0.25 (163.830) = 40.958 NIf acceleration of block A is aA, acceleration of block B is aB = 0.5 aATB = 2TABlock A:TA – 114.715 – 20.387 aA – FA = 0TA – 20.387 aA = 114.715 + 40.958 = 155.673TB + 30.581 aB – 300 = 0TB + 30.581 aB = 300∴ TA – 20.387 aA = 155.6732TA + 15.291 aA = 300∴ 56.065 aA = –11.346aA = –0.202 m/s2
∴ TA = 155.673 + 20.387 aA=155.673 + 20.387 (–0.202)
TA = 151.555 NTension in cable A, TA = 151.555 NTension in cable B, TB = 303.110 NBlock A moves down and Block B moves up,
aA = 0.202 m/s2
aB = 0.101 m/s2
c. WA = 600 NWB = 400 N2aA = aB TA = 2TB
kg775.40
81.9400
kg162.6181.9
600
=
=
=
=
B
A
m
m
WA sin θ = 600 sin 40 = 385.673 NWA cos θ = 600 cos 40 = 459.627 NFA = µ (WA cos θ) = 0.25 (459.627) = 114.907 NFor block B: TB + MB aB – 400 = 0
TA + 40.775 aB = 400TA – mA aA – 385.673 – 114.907 = 0TA – 61.162 aA = 500.58
Hence, the two equations are,TB + 40.775 aB = 400
Figure A 14.24(d)
Figure A 14.24(e)
Figure A 14.24(f)
Figure A 14.24(g)
60 Engineering Mechanics
TA – 61.162 aA = 500.58Since 2aA = aB and TA = 2TB∴ 56.065 aA = –11.3460.5TA +81.55 aA = 400TA – 61.162 aA = 500.58224.262 aA= 299.42∴ aA = 1.335 m/s2
TA = 500.58 + 61.162 (1.335) = 582.231 NTension in cable A, TA = 151.555 NTension in cable B, TB = 303.110 Nacceleration of A, aA = 1.335 m/s2
acceleration of B, aB = 2.67028. Total length of the rope is constant
i.e., yA + 2yB + d = constant and
020202
=+=+∴=∆+∆
BA
BA
BA
aavvyy
If block A moves downwards with anacceleration aA, simultaneously block -B moves upwards with an accelerationaB. (aB = 0.5 aA)For block A+↑ 0;0 =−+=∑ Aiy WFTFT + 24aA – 235.44 = 0T + 24 aA = 235.44For block B
↑+ 02;0 =−−=∑ Biy WFTF2T – 12aB – 117.72 = 0;2T – 12aB = 117.72Substitutng aB = 0.5 aA2T – 12 (0.5aA) = 117.722T – 6aA = 117.72T + 24aA = 235.44Solving the above two equations9T = 4(117.72) + 235.44 = 706.32T = 78.48 NaA = (2T – 117.72)/6 = 6.54 m/s2
∴ aB = 0.5 aA = 3.27 m/s2
Figure A 14.25(a)
Figure A 14.25(b)
Kinetics of Particles—Newton’s Second Law 61
If a force 235.44 N is applied instead of blockA, then tension in the rope is constant and isequal to 235.44 N.From Fig. 14.25(c): T – 235.44 = 0;
T = 235.44From Fig. 14.25(d)
↑+ 02;0 =−−=∑ BBBy WamTF2(235.44) – 12 aB – 117.72 = 0aB = 29.43 m/s2
Therefore acceleration of each block andtension in the rope are given as (a) aA = 6.54 m/s2 )(↓ ; aB = 3.27 m/s2 )(↑ ;and T = 78.48 N respectively. (b) If a forceA, 235.44 N is applied, the accelerationof block B aB and tension (T) are givenas T = 235.44 N; aB = 29.43 m/s2 ).(↑In both the cases, magnitude of force applied at ‘A’ is same. But aBis different. In case (a), tension in the rope is reduced by inertia ofblock A.
29. In the case of connected particles as shownin Fig. 14.58, the directions of motions mustbe determined. We observe that the length ofconnecting cable is constant. Hence down-ward motion of Block ‘A’ will raise block ‘B’upwards, and upward motion of block ‘C’ willlower the block ‘B’ downwards. The actualvalue of tension in the cable will determinethe motion. Weights of blocks when resolvedinto parallel and perpendicular to theplane.WA cos 50º = 750 cos 50º = 482.091WA sin 50º = 750 sin 50º = 574.533WC cos 50º = 300 cos 50º = 192.836WC sin 50º = 300 sin 50º = 229.813mass of blocks are also calculatedas,mA = WA/g = 76.453 kgmB = WB/g = 91.743 kgmC=WC/g = 30.581 kgLet us assume the block ‘B’ is at rest
Figure A 14.25(c)
Figure A 14.25(d)
Figure A 14.26(a)
Figure A 14.26(b)
62 Engineering Mechanics
Let us consider block ‘A’. Components of weights are shown inFig. 14.26(d).
Fig. A 14.26(c) Block A Fig. A 14.26(d) Block C
→+ 011.4522.120450533.574 −=++−=∑ ′xFi.e., Block ‘A’ is not in equilibrium hence it moves downwards.Similarly, let us consider block C. Components of weights are shownin Fig. 14.26(e).→+ 011.4209.48813.229450 −=−−=∑ ′xF∴ Block ‘C’ is not in equilibrium and it should move up the plane.Hence the entire system cannot remain at rest. The motion of blocksare assumed as follows.
Block A — moves down the planeBlock B — moves downBlock C — moves up the plane.
Since total length of the cable is constant,sA + 2sB + sC = 0sC = –(sA + 2sB)
i.e. displacement of C occurs in the upward direction, displace-ment of A and B in the downward direction.
∴ cC = cA + 2vB; aC = aA + 2aBDynamic equilibrium of each block is considered
Figure 14.26(e) Dynamic equilibrium of ‘A’
Kinetics of Particles—Newton’s Second Law 63
→+ 0522.120533.574:0 =+++−=∑ ′ AAx amTF
011.454453.76 =+∴ AaT
+↑ 02:0 =−+=∑ BBBy WamTF 900743.912 =+ AaT
022.278581.300209.48813.229:0
=−=−−−=∑ ′
C
CCx
cTamTF
Figure 14.26(f) Figure 14.26(g) Dynamic equilibrium of block’B’ Dynamic equilibrium of block ‘C’
Since aC = aA + 2aBT – 30.581(aA + 2aB) = 278.022T – 30.581 aA – 61.162 aB = 278.022The three equations are,T + 76.453 aA = 454.0112T + 91.743 aB = 900T – 30581 aA – 61.162 aB = 278.022
]2900[473.911
]011.454[76453
1
Ta
Ta
B
A
−
=
−
=∴
)900(743.97162.61
)011.454(453.76581.30022.278
743.91324.122
453.76581.301
022.278)2900(743.91162.61)011.454(
453.76581.30
+
+=
++
=−
−−
−∴
T
TTT
64 Engineering Mechanics
2
2
m/s856.32
m/s930.0]011.454[453.761
N896.382;625.1059)7674.2(
=+=
=−
=∴
==∴
BAC
A
aaa
Ta
TT
Therefore tension in cable, T = 382.896 Nacceleration of A, aA = 0.930 m/s2
acceleration of B, aB = 1.463 m/s2
acceleration of C, aC = 3.856 m/s2
30. WA = mAg = 98.1 N; WB = mBg = 294.3 N; and WC = mCg = 490.05 N
866.0cos,5.0sin;º30
6.0cos,8.0sin;13.53;34tan
222
1111
===
====
θθθ
θθθθ
Components of weights A and B parallel and perpendicular to theplane are;
N871.254cos;N15.147sinN86.58cos;N48.78sin
22
11
====
θθθθ
BB
AA
WWWW
when block ‘C’ moves down; block ‘B’ moves to the right andblock ‘A’ moves up.FA = 0.3 NA = 0.3 (58.86) = 17.658 NFB = 0.25 NB = 0.25 (254.871) = 63.718 NBlock A: T1 – WA sin θ1 – FA – mA aA = 0
T1 – 78.48 – 17.658 – 10a = 0T1 – 10a = 96.138
Block B: T2 – T1 + WB sin θ2 – FB – mB aB = 0T2 – T1 + 147.15 – 63.718 – 30 a = 0T2 – T1 – 30a = –83.432
Block C: T2 + mC aC – WC = 0T2 + 50a = 490.05T2 – T1 – 30a = –83.432T2 – (96.138 + 10a) – 30a = –83.432T2 – 40a = 12.706
∴ 90a = 477.844; a = 5.309 m/s2
Hence T2 = 490.05 – 50a = 224.6 N; and T1 = 96.138 + 10a = 149.228 N
Kinetics of Particles—Newton’s Second Law 65
Figure 14.27 Dynamic equilibrium of blocks A, and B and C
Therefore, tension in cable connecting B and C (T2) = 224.6 Ntension in cable connecting A and B (T1) = 149.228 Nacceleration of blocks (aA = aB = aC) = 5.309 m/s2.
31 . WA = 245 NWB = 980 N
kg898.9981.9
980
kg975.2481.9
245
=
=
=
=
B
A
m
m
0.3 WA = 73.5; 0.3 WB = 294
Figure 14.28
T – 73.5 – 24.975 a = 0– T – 99.898 a + 490 – 294 = 0T – 24.975 a = 73.5
66 Engineering Mechanics
T + 99.898 a = 196124.873 a = 122.5a = 0.981 m/s2
T = 73.5 + 24.975(0.981) = 98 Nacceleration (a) = 0.981 m/s2
Tension (T) = 98 N
32. kg174.981.9
90;N90 =
== LL MW
kg058.381.9
30;N30 =
== RL MW
kg162.6181.9
600;N600 =
== MW
Assume that WL moves down:TL – WL + ML a = 0TL + 9.174 a = 90TR – WR – MR a = 0TR – 3.058 a = 30–TL + ma + TR = 0–TL + 61.162 a + TR = 0∴ TL = TR + 61.162 a∴ (TR + 61.162a) + 9.174a = 90TR + 70.336 a = 90TR – 3.058a = 30∴ 73.394a = 60∴ a = 0.818 m/s2
TR = 30 + 3.058(0.818) = 32.501 NTL = 32.501 + 61.162(0.818) = 82.532 NTension, TL = 82.532 NTension, TR = 32.501 Nacceleration, a = 0.818 m/s2
33. WA = 100; WB = 50WA cos 40º = 76.604WA sin 40º = 64.279WB cos 40º = 38.302WB sin 40º = 32.149FA = 0.25 (76.604) = 19.151FB = 0.375 (38.302) = 14.363
Figure A 14.29
Figure A 14.29(a)
Figure A 14.29(b)
Kinetics of Particles—Newton’s Second Law 67
kg097.581.9
50;kg194.1081.9
100 ==== BA MM
Figure A 14.30
T + 10.194a + 19.151 – 64.279 = 0T + 10.194a = 45.128– T – 32.140 + 5.097a + 14.363 = 0– T + 5.097a =17.777
N192.3777.17)114.4(097.5
m/s114.4291.15905.62 2
=−=
==∴
T
a
34. kg645.781.9
75;N75 =
== AA MW
kg484.2581.9
250;N250
kg742.1281.9
125;N125
=
==
=
==
CC
BB
MW
MW
)125()()952.64()cos(
952.6430cos75cos5.3730sin75sin
µµµθµ
θθ
====
====
BB
AA
A
A
WFWF
WW
Block C, moves down; with a = 2.5 m/s2; hence block moves to theright and block A moves up. Unknowns are T1, T2, & µ.T2 + MC a – WC = 0T2 + 25.484(2.5) – 250 = 0T2 = 186.29T2 – T1 – MBaB – FB = 0T1 + 125 µ = 186.29 – 12.742 (2.5)T1 + 125 µ = 154.435T1 – MA aA – WA sinθ – FA = 0T1 – 7.645(2.5) – 37.5 – µ(64.952) = 0
Figure A 14.31(a)
Figure A 14.31(b)
68 Engineering Mechanics
T1 – 64.952µ = 56.613Solving, 189.952 µ = 97.822µ = 0.515∴ T1 = 56.613 + 64.952(0.515) = 90.063 T1 = 90.063 N; T2 = 186.29 N; µ = 0.515
35. Let block ‘A’ be moving down with an ac-celeration aA. Hence Pulley P2 must alsomove up with aA. Let the tension in thestring be T1.Let block ‘B’ move down with an acceleration aB. Hence block Cmust move up with same acceleration aB. Since they are connectedto Pulley P2, moving with an acceleration aA upwards.Absolute acceleration for B, = aB – aAAbsolute acceleration for C, = aB + aAT1 – 2T2 = 0; T1 = 2T2WA = 675 N; MA = 68.807 kgWB = 450 N; MB = 45.872 kgWC = 225 N; MC = 22.936 kgT1 + mA aA – WA = 0T1 + 68.807 aA – 675 = 0T1 + 68.807 aA = 675T2 + mB (aB – aA) – WB = 0T2 + 45.872 (aB – aA) – 450 = 0T2 + 45.872 (aB – aA) = 450T2 – mC (aB – aA) – WC = 0T2 – 22.936 (aB – aA) = 225T1 + 68.807aA = 675 (1)T2 + 45.872 (aB – aA) = 450 (2)T2 – 22.936 (aB – aA) = 225 (3)Since T1 = 2T22T2 + 68.807 aA = 675(2) + 2(3) gives,3T2 – 91.744 aA = 900T2 + 34.404 aA = 300∴ 64.985 aA = 37.5aA = 0.577 m/s2
T1 = 675 – 688.807(0.577) = 635.298 N∴ T2 = (T1/2) = 317.649 N
Figure A 14.31(c)
Figure A 14.32
Figure A 14.32(a)
Figure A 14.23(b)
Kinetics of Particles—Newton’s Second Law 69
45.872 aB = 450 – T2 + 45.872 aA = 450 – 317.649 + 45.872 (0.577)
∴ aB = 3.462 m/s2
Absolute acceleration of B= aB – aA = 2.885 m/s2
Absolute acceleration of C= aB + aA = 4.039 m/s2
T1 = 635.298 N T2 = 317.649 N aA = 0.577 m/s2
Absolute aB = 2.885 m/s2
Absolute aC = 4.039 m/s2
36. m = 16 kg
868.0sin26.60
75.125.3tan
==
=
=
θθ
θ
W = 16 × 9.81 = 156.96 NF = 100 NWsin θ = 156.968 sin 60.26 = 136.241 NWsin θ – F – ma = 0136.241 – 100 – 16 a = 0∴ a = 2.265 m/s2
m301.45.25.3 22 =+=SInitial velocity Vo = 0
sec95.1
)265.2(21
04301
21
2
2
=∴
+=
+=
t
t
attVS o
37. m = 90 kgW = 90 × 9.81 = 882.9 NW sinθ = 882.9 sin 15 = 228.511Wcosθ = 882.9 cos 15 = 825.816F = µ W cosθ = 0.2 (825.816) = 165.163T – Wsinθ – F = 0
Figure A 14.32(c)
Figure A 14.33
Figure A 14.34
70 Engineering Mechanics
T – 228.511 – 165.163 = 0T = 393.674 NT – ma – W = 0393.674 – m(0.2 + 9.81) = 0m = 39.328 kgmass of water = 39.328 kg
38. The minimum velocity with which a rocket should be projected toescape from the gravitational influence of the earth is known asescape velocity.
θθθ cos1)/(
cos1/
cos1
222
eMGvr
eMGh
eedr
+=
+=
+=
I e < 1, it is an ellipse, for rocket to escape, the orbit should be aparabola therefore e = 1.
km/hr40255m/s966.11181
)106380()1067.6()1098.5(2
12
)11(
3
1124
22
==
×××==∴
+=
−
esc
esc
esc
vv
MGv
MGvrr
39. Let, v = Velocity andy = displacement at time ‘t’.
+↑ 0;0 =−+=∑ WmaRFy
Since ‘R’ is proportional to velocity, R = kvWe know that ma = W – R
dtvg
dvvgdtdv
vgvmkg
mRg
mR
mWa
=−
−=∴
−=
−=−=−=
)(;
ββ
β
Integrating, 1)log(1 ctvg +=−
− β
β
When t = 0, v = 0; gc log11
−=
β
)]log([log1
log1)log(1
vggt
gtvg
ββ
ββ
β
−−
=
−=−
−
Figure A 14.34(a)
Kinetics of Particles—Newton’s Second Law 71
[ ]t
t
t
egv
egvg
eg
vggvgt
gvgt
β
β
β
β
β
βββββ
−
−
−
−=∴
=−
=−∴
−=−−−=−
1
)(]/log[log)log(
The above relationship is used to find ‘v’ at any time ‘t’. As timeincreases, .0=− te β
=∴
βgvmax
At this velocity, acceleration is zero, hence this is known as limit-ing velocity. But v = (dy/dt).
[ ]dtegdtvdy tβ
β−−
== 1
Integrating the above equation,
21 cetgy t +
+
= −β
ββ
At t = 0, y = 0; 22 βgc −=
[ ]t
t
egtgy
gegtgy
β
β
ββ
βββ
−
−
−
+
=
−
+
=∴
12
22
But
−=−=
−
=
−−
−
gvee
gv
egv
tt
t
ββ
β
ββ
β
1;1
)1(
taking log on both sides,
−=−
gvt ββ 1log
−
−=
gvt β
β1log1
72 Engineering Mechanics
substituting in the eqn. for y,
[ ]
( )t
t
egvvgvg
egtgy
β
β
βββ
ββ
ββ
−
−
−=∴
−
−
−
=
−
−
=
11log1
12
−
−
−=∴
ββ
βv
gvgy 1log2
40.
θθ
θθθ
cos;0cos;0
sinsin;0sin;0
WmaTmaTF
gmWamaWF
nnn
ttt
=−=−=∑
===+=∑
Figure A 14.35
Distance travelled = s = Lθ
==
==∴ 2
2
2
2and/
dtdL
dtsda
dtdLdtdsv t
θθ
substituting ‘at’ in the first eqn. θ is positive, anticlockwise
=− 2
2sin
dtdLg θθ
)sin, of valuessmallfor (0or
0sin
2
2
2
2
θθθθθ
θθ
==+
=+
Lg
dtd
Lg
dtd
∴ The solution of differential equation is
( ) ( )tLgBtLgA )(cos)(sin +=θ
Kinetics of Particles—Newton’s Second Law 73
The values of A and B can be calculated using boundary condi-tions.The angular velocity ( )Lg=ω in rad/s.
Therefore period (T) = time required to complete one cycle.
=
ωπ2T
and frequency (f) = number of cycles per second
=
=
=
Lg
Tf
ππω
21
21
41. The ball of weight W is subjected to tension in the cord. Centrip-etal force required to hold the particle in circular path acts towardsthe centre. Hene inertia force acts along the radial direction butacts onwards (away from centre). Free body diagram of ball indynamic equilibrium is shown in Fig. 14. 25.(a) Tension in cord at ‘P’ (Fig. 14.36(b))
↑+ 0;0 =−−=∑ WTmaF ppy
mgWRvap == and
2
substituting in the above equation
pTgRvm =
−
2
It must be noticed that tension in the cord becomes zero when
.2
gr
v =
But in that case, gravitational attraction of earth pro-
duces the required radial or centripetal acceleration at ‘A’ since
the centripetal acceleration
r
v2
is then equal to gravitational
acceleration ‘g’. If ‘v’ becomes very small ,2
gr
v <
the pull of
gravity exceeds the required centripetal force and hence the objectwill fall out from the circular path before reaching A. The studentscan try this experiment with very small velocity.(b) Tension in cord at ‘Q’ (Fig. 14.36(b))
74 Engineering Mechanics
Figure A 14.36(a) Dynamic Figure A 14.36(a) Dynamic equilibrium at ‘P’ equilibrium at ‘Q’
↑+ 0;0 =−−=∑ QQy maWTF
+= g
RvaQ
2
42. Considering the motion down the plane,
θθθ
sin;0sin0sin
gamamgmaW
==−=−
If the distance travelled along theplane is ‘s’,
gHvsH
gsv
sgv
asvv
B
B
B
AB
2;/sin
sin2
sin20
2
2
2
2
22
==
=
+=
+=
θ
θ
θ
Let us consider the movement alongthe circleAt point B: N – W – maB= 0
+=+=
===
DvmmgmaWN
Dv
Dv
rva
BB
BBBB
2
222
2
2)2(
Figure A 14.37(a)
Kinetics of Particles—Newton’s Second Law 75
Figure A 14.37(b)
Figure A 14.38
At point B, gHvB 2=
When the particle is at the topposition‘C’, mac – W – Nc = 0But Nc = 0; mac = W
2; 2
2 gDvmgr
vm cC ==
But gDvgsvv BBC 22 222 −=−=
)(2222 DHggDgHvC −=−=
DDH
DHDDHgDDHg
25.14
554;44
2/)(2
=
=∴
==−=−∴
Hence the particle should leave at a height H = 1.25 D to completethe whole circuit.
43. If the ball of mass ‘m’ moves in a circle of radius R, then it calledconical pendulum.The tension in the cord ‘T’ must satisfy two conditions: Verticalcomponent must balance the weight of the ball and horizontalcomponent the radial force.∴ T cos θ = W = mg
==
RvmmaT n
2sinθ
Dividing the above two equations
Rgv
mgRvm
TT
2
2
tan
cossin
=∴
=∴
θ
θθ
The above equation is independent of mass ‘m’.Hence the angle ‘θ’ at watch he ball hangs to follow the circularpath is not influenced by mass of the ball ‘m’.
76 Engineering Mechanics
Tension in the cord, gRvmT +
=
2
44.45.46.47.
Kinetics of Particles—Work and Energy 1
1515151515
Kinetics of Particles —Kinetics of Particles —Kinetics of Particles —Kinetics of Particles —Kinetics of Particles —WWWWWork and Energyork and Energyork and Energyork and Energyork and Energy
CONCEPTS
• The work done by a constant force rF acting on a particle is the
product of the component of force in the direction of displace-ment and the magnitude of displacement.
U F ds F ds
F dx F dy F dzx y z
121
2
1
2
1
2
= ∫ ⋅ = ∫
= ∫ + +
r rcos
( )
θ
• The work done by a varying force acting on a particle movingalong x-axis from x x x x= =1 2to is,
U F dxx
x
x= ∫1
2
where Fx is the component of force in x-direciton.• Kinetic energy of a particle of mass ‘m’ moving with speed ‘v’ is
T m v= ( ) .1 2 2
• According to the principle of work and energy, U T T12 2 1= −Where U12 = work of a force on the particle from position ‘1’ to‘2’.
T1 = Kinetic energy at initial position ‘1’T2 = Kinetic energy at final position ‘2’
• Work of a weight ‘W’ moving from position 1 to 2 is
U W dy Wy Wy121
2
1 2= − ∫ = −
• Work of a force Fs of spring during a finite displacement fromx x= 1 to x x= 2 is,
2 Engineering Mechanics
( ) [ ]dU F dx kx dx U kx dx kx kxsx
x= = − = − ∫ = −; and 12 1
222
1
2 12
Work done is positive if it returns to undeformed position.• Work done by the gravitational force
rF exerted by a particle of
mass ‘M’ located at ‘O’ on a particle of mass ‘m’ moving from 1 to2 is
U GMmr
dr GMmr r12
1
2
2 2 11 1= − ∫ =
−
• Gravitational potential energy of a particle of mass ‘m’ which is atan elevation ‘y’ is, U mgyg = .
• Elastic potential energy stored in a spring of stiffness ‘k’ is,U kxs = ( ) .1 2 2
• A force is conservative if the work done by that force on a par-ticle is independent of the path taken by the particle when itmoves from position (1) to position (2). In other words, a force issaid to be conservative if the work done by the particle is zerowhen it moves through an arbitrary closed path and returns to itsinitial position. A force which does not satisfy this condition isnon-conservative.
• A potential energy funciton (V) is associated with the conserva-tive force. If a conservative force acts on a particle which movesalong the x-axis from x x1 2to , then.
U V V V V12 1 2 2 1= − = − −( )where ‘V’ is the potential energy.For a closed path, U F ds= ⋅ =∫
r r0
r r r r r r rF F i F j F k V
x i Vy j V
z kx y z= + + = −
+
+
∂∂
∂∂
∂∂
• Total mechanical energy of a system is the sum of the kineticenergy and potential energy i.e., E = T + V
• When a particle moves under the action of conservative forces,E T V T V= + = + =( ) ( )1 1 2 2 constant
• When a particle is subjected to non-conservative and conserva-tive forces, work done by the non-conservative force is equal tothe change in the total mechanical energy of the system (loss inenergy).
• Positions of stable equilibrium correspond to those points forwhich potential energy is minimum.
Kinetics of Particles—Work and Energy 3
• Positions of unstable equilibrium correspond to those points forwhich the potential energy is maximum. Position of neutral equi-librium correspond to those points for which potential energy isconstant.
• Power is the rate at which work is done.
.W7461sNm111
Power
===
⋅=⋅==
hpsJW
vFdt
rdFdtdU rr
rr
Example 15.1 A particle moving in the xy-plane undergoes a displace-ment r
r rs i j= +( )4 6 m with a constant force
r r rF i j= − +( )5 10 N acting on
it. Calculate the work done, magnitude of force and magnitude of displace-ment.
Solutionr r rF i j= − +5 10
m211.7)6()4(
''nt displaceme of Magnitude
N180.11)10()5(
'' force of Magnitude
2222
2222
=+=+=
=+−=+=
yx
s
FF
F
yx
U F s
i j i j= ⋅= − + ⋅ +
r rr r r r
( ) ( )5 10 4 6= − ⋅ − ⋅ + ⋅ + ⋅= − − + +
∴ =
5 4 5 6 10 4 10 620 0 0 60
40
r r r r r r r ri i i j j i j j
U NmNote:
r r r r r r r ri i j j i j j i⋅ = ⋅ = ⋅ = ⋅ =1 0and
cos| | | | ( . ) ( . )
( . )
α
α
= ⋅ =
∴ =
r rr r
o
F sF s
4011180 7 211
60 3Example 15.2 An object weighing 80 N is pulled up, on the smoothplane by 75N force as shown in Fig. 15.26. Determine the velocity of theobject after it has moved 4m.
4 Engineering Mechanics
Figure 15.1
Solution Component of weight normal to the plane = W cos q = 80cos 35 = 65.532
Component of weight parallel to the plane = W sin θ = 80 sin 35° = 45.866 N
Component of force parallel to the plane = P cos 15° = 75 cos 15° = 72.444 N
Component of force normal to the plane = P sin 15° = 75 sin 15° = 19.411 N
∴ = − =Fnet 72 444 45866 26 578. . .∴ Work done = change in kinetic energy
U
T m vi i
= =
=
=
26578 4 106 31212
02
. ( ) . Nm
( )T mv v v
v v
f f f f
f f
=
=
=
∴ = =
12
12
80 9 81 4 078
106 312 4 078 5106
2 2 2
2
. .
. . ; . m sTherefore velocity of the block = 5.106 m/s.
Example 15.3 Two blocks A and B ofmasses, mA = 280 kg and mB = 420 kg arejoined by an inextensible cable as shown inFig. 15.27(a). Assume that the pulley is fric-tionless and µ = 0.30 between block ‘A’ andthe surface. The system is initially at rest. De-termine the velocity of block after it hasmoved 3.5 m.
Figure 15.2(a)
Kinetics of Particles—Work and Energy 5
Solution m ms v
A BA
kg; kg;m and
= == = =
280 4200 3 35 00µ . ; . ;
∴ = == =
W m gW m g
A AB B
NN
274684120 2
..
Normal reaction at ‘A’ and tension in cable do no work. When block ‘B’moves down, block ‘A’ moves to the right.
N NN
N
A AA AA
= == ==
WFF
274680 3 2746 8
824 04
.. ( . )
.µ
Figure 15.2(b) Forces which do work are represented
Work doneNm
B A= − = −=
( ) ( . . ) ( . ).
W F sUif
4120 2 824 04 351153656
Initial kinetic energy of system, = Ti
T m v m vi i i=
+
12
12
2 2A A B B( ) ( )
Since blocks start from rest, initial velocity ( ) ( )v vi iA B= = 0
Let the final velocity be v; ( ) ( )v v vf fA B= =
∴ =
+
=T m v m v vf12
12
3502 2 2A B
According to principle of work energyU T T
vv
if f i= −
= −∴ =
11536 56 350 05741
2.. m s
Therefore, velocity of the blocks after they move at adistance of 3.5 m is 5.741 m/s. Figure 15.2(c)
6 Engineering Mechanics
Note: If the tension in the string is to be determined, we must useNewton’s second law or D’Alemberts principle.
Since,
( . ) ( . )
v v as
af2
02
2
2
5741 0 2 35
= +
= +
∴ =∴ + − =∴ = −
=
aT aTT
4 708420 4120 2 04120 2 420 4 7082142 84
2..
. ( . )
.
m s
NExample 15.4 In the pulley system shown in Fig. 15.28, the masses ofblocks A and B are 24 kg and 12 kg respectively. Neglect the masses ofpulleys and rope. Find the velocity of the blocks when block A moves by1.5 m assuming that the blocks start from rest.
Figure 15.3
Solution m mA Bkg kg= =24 12;
W m g W m gA A B BN and N= = = =23544 117 72. ; .Total length of rope is constant.
∴ + + =+ =+ =
y y dy yv v
A BA BA B
constant22 02 0
∆ ∆.
If ‘A’ moves down, ‘B’ moves up. Since block ‘A’ moves down by 1.5m, ‘B’ moves up by (1.5/2) = 0.75 m. Therefore work done by ‘A’ is positivebut by ‘B’ is negative.
Kinetics of Particles—Work and Energy 7
∴ = −= − =
U W WU
if
if
A BNm
( . ) ( . ). ( . ) . ( . ) .15 0 75
23544 15 117 72 0 75 264 87Initially, the blocks start from rest, therefore initial velocity
( ) ( )v vT
i ii
A B= =∴ =
00
But v vB A= 05. and block A moves down and block B moves up.
T m v m v
v
T v
f
f
=
+
=
+
∴ =
12
12
12
24 05 12
135
2 2
2 2
2
A A B B
A
A
( ) ( )
( ) [ ( . ) ( )]
.According to principle of work and energy,
U T T vv
if f i= − = −∴ =
; . ..
264 87 135 04 429
2A
A m sVelocity of block ‘B’, v vB A m s= =( ) . .2 2 215Therefore Velocity of block B when block A moves by 1.5 m is 2.215 m/s.
Example 15.5 The acceleration due to gravity of a particle falling to-wards earth is, a g r= − R2 2 where ‘r’ is the distance from the centre ofthe earth to the particle; ‘R’ is the radius of earth and ‘g’ is the accelerationdue to gravity at the surface of earth. Determine the escape velocity usingthe principle of work and energy.Solution Escape velocity defines the minimum velocity with which aparticle should be projected vertically upward from the surface of earthsuch that it does not return to the earth. i.e., v = 0 when r = ¥.
Force, F when the particle is at a distance x,F W x= − R2 2
∴ work done
( ) ( )
= ∫
= − ∫
= −
R
RR
R
x
x
F dx
U W dxx
U Wx R
( )22
2 1 1
When x U W= ∞ = −, RAccording to the principle of work and energy, U T Tf i= − and
Figure 15.4
8 Engineering Mechanics
( )T m v v
WR Wg v
v v gR
f f f
i
i
=
= =
∴ − = −
= =
12
0 0
0 12
2
2
2
; .QAt maximum height,
escExample 15.6 A block of weight 12N falls at a distance of 0.75 m on topof the spring. Determine the spring constant if it is compressed by 150 mmto bring the weight momentarily to rest.Solution Initial velocity = 0. Work donefrom ‘A’ to ‘B’
U WhAB Nm= = =12 0 75 9( . )kinetic energy at ‘A’ = 0; and kinetic
energy at ‘B’
=
=
=
12
12
129 81
0 612
2 2
2
m v v
v
B B
B
..
Work done = change in kinetic energy∴ =
∴ =9 0 612
3835
2..
vvB
Bm s
This is same as the velocity of the freely falling bodyv ghB m s= = × × =2 2 9 81 0 75 3835. . .
Work done from ‘B’ to ‘C’ = 12 (0.15) = 1.8 NmWork done by spring ( ) ( )= − = − = −kx k k2 22 015 2 0 01125( . ) . Nm
∴ Total work done U kBC = −18 0 01125. .Since it comes to rest momentarily at C, vC = 0∴ Change in kinetic energy = −T TC B
T m v
k
k
B B
N m
=
=
=
∴ − = −
∴ = =
12
12
129 81
3835 8 995
18 0 01125 8 99510 7950 01125
959 556
2 2.
( . ) .
. . .( . )
..
The above problem can also be done by considering the positions ‘A’and ‘C’ directly. Since, it is at rest at both ‘A’ and ‘C’, velocities are zero.
Hence v vA Cand= =0 0; .∴ change in kinetic energy = 0
Figure 15.5
Kinetics of Particles—Work and Energy 9
∴ work done from A to C = 0; ( )U W h x k xAC = + − =( ) 2 2 0
∴ + − =
∴ =
12 0 75 015 0152
0
960
2( . . ) ( . )k
k N mTherefore the spring constant is k = 960 N/m.
Example 15.7 A 50N block is released form rest on an inclined planewhich is making an angle of 35° to the horizontal. The block starts from ‘A’,slides down a distance of 1.2 m and strikes a spring with a stiffness of 8 kN/m. The coefficient of friction between the inclined plane and the block is0.25. Determine (a) the amount the spring gets compressed and (b) distancethe block will rebound up the plane from the compressed position.
Figure 15.6
Solution Component of weight pallelled to plane =W sinθComponent of weight perpendicular to plane = W cosθ
WW
sin sin .cos cos .
θθ
= == = =
50 35 28 67950 35 40 958
o
o
NN NA
Frictional force, F = = =µN NA 0 25 40 958 10 240. ( . ) .Since the block starts from rest, initial velocity is zero. When the spring
compresses to the maximum level, final velocity will be zero.∴ Work done = change in kinetic energy = 0The component of weight along the plane, frictional force and spring
force are the forces which do the work. Assume that the deformation ofspring as ‘x’. Then
( )( )
∴ − + − =
− + − =
− − =
( . ) ( . )
( . . ) ( . ) ( )
. .
28 679 10 24 12 12
0
28 679 10 24 12 12
8000 0
4000 18 439 221268 0
2
2
2
. x kx
x x
x x
10 Engineering Mechanics
x x2 3 34 6098 10 55317 10 0− × − × =− −. .
x
x
=× ± × + ×
= × ± = −
∴ =
− − −
−
( . ) ( . ) ( . )
. . . , .
.
4 6098 10 4 6098 10 4 55317 102
4 6098 10 014882
0 077 0 072
0 077
3 3 2 3
3
m∴ Energy stored in spring= = =( ) ( ) ( )( . ) .1 2 1 2 8000 0 077 237162 2kx NmWhen the block rebounds up and travels to a maximum distance along
the plane, the component of weight and frictional force act in the samedirection opposing the motion. Let ‘x1’ be the distance it rebounds,
23 716 28 679 10 240 609
11
. ( . . ).
= +∴ =
xx m
Therefore(a) When the block hits the spring, the spring is compressed to
0.077 m(b) It rebounds up the plane by 0.609 m.
Example 15.8 A spring is used to stop a 10 kg package which is mov-ing down on an inclined plane and makes an angle of 25° with horizontal.The spring constant is k = 30 kN/m and is held by cables so that it isinitially compressed by 80 mm. If the velocity of the package is 8 m/s whenit is at 15.5m from the spring, determine the maximum additional deforma-tion of the spring in bringing the package to the rest position. Assumeµ = 0.30.
Figure 15.7
Solution When the package comes to rest, the final velocity v2 0= ;Let the additional deformation be ‘ ∆x ’.
Kinetics of Particles—Work and Energy 11
N673.26)25cos1.98(3.0N
N909.8825cos1.98cosN
N459.4125sin1.98sin
N1.9881.910
A
A
===
===
==
=×==
o
o
o
µ
θ
θ
F
W
W
mgW
Work done by package )5.12(786.14)5.12()673.26459.41( xx ∆+=∆+−=Work done by spring from 0.080 to ( . )0 080 + ∆x is given by
Figure 15.7(d)
PP
minmax
( . )( . )
= == +
30000 0 08 240030000 0 08 ∆x
Work done by spring P P= −
+
= −
+ +
= − +
1212
2400 2400 30000
2400 15000 2
( )
[ ]
[ ].
min max ∆
∆ ∆
∆ ∆
x
x x
x x∴ = + − +
=
=
=
Net work done
kinetic energy at
Nm
U x x x
T mv
T
122
1 12
12
14 786 12 5 2400 15000
1 1212
10 8 320
. ( . ) ( )
' ' ,
( )( )
∆ ∆ ∆
Kinetic energy at ‘2’ T mv2 221 2 0= =( )
320)150002400()5.12(786.14 21212
−=∆+∆−∆+
−=∴
xxx
TTU
Figure 15.7(c)
12 Engineering Mechanics
.m121.0
2)0337.0(4)159.0(159.0
00337.0159.0
0825.504214.238515000
2
2
2
=∆∴
−−±−=∆
=−∆+∆
=+∆−∆−
x
x
xx
xx
The package is brought to rest when the spring deforms additionallyby 0.121 m.Method 2: This can also be calculated by considering the initial deforma-tion as zero.
Distance between the block and the spring is 12.5 – 0.08 = 12.42 m∴ let the total deformation of the spring be ‘x’.
( ) [ ]14 786 12 42 12 30000 0 08 0 08 320
0121 0 014 0
0121 0121 4 0 014 2 0193
2
2
. ( . ) ( ) ( . ) ( . )
. .
. ( . ) ( . ) .
+ − + − − = −
− − =
= ± − −
=
x x x x
x x
x m
Initial deformation of spring xi = 0 080. mTherefore additional deformation, ∆x x xi= − = 0113. m
Example 15.9 The mass ‘m’ of a pendulum is connected to a string oflength ‘L’ which is released from the horizontal position of the string asshown in Fig. 15.37. Show that, at the lowest position of the mass, thetension in the string is three times the value when the mass is simplesuspended in that position.Solution At any intermediate position ‘C’, W and T are acting. But ‘T’ isalways normal to the path and towards the centre of curvature. Hence, ‘T’does not do any work and work is done only by ‘W’.
Work done U WLAB =
At A, kinetic energy T m vA A= =( )1 2 02
At B, kinetic energy T m vB B= =( )1 2 02
( )∴ = −
=
= = =
U T T
WL m v v WLm
gL v gL
AB B A
B B B12
2 2 22 2; ;
Kinetics of Particles—Work and Energy 13
Figure 15.8
To find the tension at ‘B’, dynamic equilibrium of pendulum is consid-ered as shown in Fig. 15.37.
a v gLL
g
T W maT mg m g mgT W
n
n
BB
B BBB
R= = =
∴ − − == + ==
2 2 2
02 3
3( )
When the mass is simply suspended, T WB =∴ =TB 3 (Tension due to simply suspended).
Velocity at intermediate positon C:Work done from A to C U WLAC = sinθKinetic energy at A, TA = 0Kinetic energy at C, T m vC C= ( )1 2 2
∴ = −
=
=
U T T
WL m v
v gL
AC C A
C
C
sin
sin
θ
θ
122
2
Writing the dynamic equilibrium equation along the string AC,T ma Wa v R gL L gT m g W W
n
n
AC C
C CAC
− − − == = == + =
cos( )sin sin
( sin ) sin sin
90 02 2
2 3
2θ
θ θθ θ θ
The variation of v TC Cand are shown in Table 15.1
14 Engineering Mechanics
Table 15.1 )sin3();sin2( ACC θθ WTgLv ==
θθθθθ 0 10° 20° 30° 40° 50° 60° 70° 80° 90°v
gLC
2 0 0.417 0.585 0.707 0.802 0.875 0.931 0.969 0.992 1
TWAC
3 0 0.174 0.342 0.5 0.643 0.766 0.866 0.940 0.985 1
For the pendulum, law of conservation of mechanical energy can alsobe verified.
∴ = −= −
=
= =
Potential energy ( )and
Kinetic energy ( )
V W L LmgL
T mv
m gL mgL
( sin )( sin )
( sin ) sin
θθ
θ θ
11212 2
2
∴ Total energy for any value of ‘θ ’ (E)= Potential energy + Kinetic energy= − +=
mgL mgLmgL
( sin ) sin1 θ θ
Table 15.2 ( ( sin ); sin ; )V mgL T mgL E mgL= − = =1 θ θ
θθθθθ 0 10° 20° 30° 40° 50° 60° 70° 80° 90°V
mgL 1 0.826 0.658 0.5 0.358 0.234 0.134 0.060 0.015 0T
mgL 0 0.174 0.342 0.5 0.642 0.766 0.866 0.940 0.985 1E
mgL 1 1 1 1 1 1 1 1 1 1
Example 15.10 A particle of weight‘W’ is released from rest at ‘P’ on thesurface of a smooth circular cylinder ofradius ‘R’. The particle slides from point‘P’ to point ‘Q’ where it leaves the sur-face of cylinder. Determine (a) At whatangle ‘θ ’ does it leave the surface and(b) with what velocity, will it leave the surface.Solution At ‘P’ the particle is at rest. Hence kinetic energy TP = 0 andpotential energy V W RP = sin .α
Figure 15.9(a)
Kinetics of Particles—Work and Energy 15
At ‘Q’, particle leaves the cylinder. As the particleslides down from ‘P’ to ‘Q’, smooth surface exerts aradial reaction on the particle. As the particle movesalong the surface, work done by this reaction is zero.Component of weight ‘W’ perpendicular to surface (alongradial line) is W Wcos ( ) sin .90 − =θ θ
ma Wmv W
n + − =
+ − =
N
R N
sin
sin
θ
θ
0
02
mv W mv WR2 2
2 2 2
2
RN and NR
(KE) (PE)2
NR
= − = −
= −
sin sinθ θ
At Q, normal reaction is zero; hence (KE) = (PE)/2Note: This equation is valid only at the point where it leaves the cylin-der and at all other points, normal reaction (N) will be present.
When it leaves, N = 0 KE PE= 2 at any point ‘Q’
∴
=
=
=
=
At Q Q
Q
Q
, sin
sin sin
sin
m v WR
v Wm
R g R
v g R
2 22
2
θ
θ θ
θAt Q, potential energy V WQ R= sinθKinetic energy, T m vQ Q= ( )1 2 2
( )( )
T mg W
V T
W W
E W
Q
Q Q
Q
R R
Total energy at Q
R R
R
= =
∴ = +
= +
=
12 2
215
sin sin
sin sin
. sin
θ θ
θ θ
θ∴ By conservation of mechanical energy,
E EW W
v gR gR
v gR R
P Q
Q
Q
R RHence
andsince
=∴ =
= =∴ = =
= = =
= =
−
sin . sinsin ( . ) sin . sinsin . sin sin [ . sin ]
sin ( . ) sin sin . sin
. ( ) sin . sin
α θθ α αθ α θ α
θ α θ α
α α
151 15 0 667
0 667 0 6670 667 0 667
0 667 2 558
1
Figure 15.9(b)
16 Engineering Mechanics
∴ When the particle starts sliding from angleα it leaves the surface atθ = −sin 1 [ . sin ]0 667 α with a velocity of 2 558. R sin .α
Table 15.3 ( sin . sin ; . sinθ α α= =0 667 2 558v RQ )
ααααα sinα θθθθθ vQ
90° 1 41.84 2.558 R80° 0.985 41.06 2.539 R70° 0.940 38.81 2.480 R60° 0.866 35.29 2.380 R
The values of θ and velocity for some values of ‘α ’ are shown inTable 15.3.Example 15.11 A 5 kg block slides from rest at point ‘A’ along africtionless inclined plane making an angle of 25° with horizontal. Deter-mine the speed of the block at ‘B’ at a distance of 3m from A.
Figure 15.10(a)
Solution Using the principle of work and energy
U T TAB A= −Bwhere UAB = work done from A to BThe component of weight of the block acting parallel to the plane is
WUsin ( . ) sin .
. ( ) .θ = × =
∴ = =5 9 81 25 20 729
20 729 3 62187
o NNmAB
Kinetic energy at A A A= = =T m v( ) ( )1 2 02
Since block starts from rest at A, vA = 0Kinetic energy at B B B= =T m v( ) ( )1 2 2
T v v
U T T vv
B B B
AB A BB m s
=
=
∴ = − = −∴ =
12
5 2 5
62187 2 5 04 988
2 2
2
( ) .
; . ..B
Kinetics of Particles—Work and Energy 17
The problem can also be analysed using the principle of conservationof mechanical energy. The block is subjected togravitational force which is conservative. Hence,total energy V + T = constant
i e V T V T. ., A A B B+ = +Potential energy at A, V W yA A=
= ×=
( . ) ( sin ). .
5 9 81 3 2562 187VA Nm
Kinetic energy at A, T m vA A= =( )1 2 02
∴ Total energy at A is E V TA A A= + = 62 187.Potential energy at B, V W yB B= = × =( . ) ( )5 9 81 0 0Kinetic energy at B, T m v vB B B= =( ) .1 2 2 52 2
∴ Total energy at B, E V T vB B B B= + =2 5 2.
sincem s
A B BB
E E vv
= =∴ =
, . ..
62 187 2 54 988
2
Therefore speed of the block ‘B’ at a distance of 3m from A isvB m s= 4 988. .Example 15.12 A uniform chainof length ‘L’ lies on a smooth tablewith length ‘a’ overhanging from theedge of a table as shown in Fig. 15.40.If the system is released from rest, findthe velocity with which the last linkleaves the table.Solution Let the weight of thechain be w/unit length and the centreof gravity of segments of chains ‘AO’and ‘OB’ are at the mid-points of thesegments. The only force acting on them is the self-weight. Selecting‘AO’ as the datum
V wa a wai = − = −[ ( )] ( )2 22
Initially, it is at rest, hence Ti = 0
∴ = + = −E T V wai i i ( )2 2
Figure 15.10(b)
Figure 15.11
18 Engineering Mechanics
When the last link leaves the table,
V wL L wL
T m v wLg
v
E T V wLg
v wL
f
f f f
f f f f
= − = −
= =
∴ = + = −
( )2 2
12
12
2 2
2
2 2
2 2
Using the principle of conservation of energy,
− =
−
− =
− =
= −
wa wLg
v wL
w L a wLg
v
L a Lg
v
v g L L a
f
f
f
f
2 2 2
2 2 2
2 2 2
2 2
2 2 2
2 2( )
( )
( ) ( )VelocityExample 15.13 A satellite is launched in a direction parallel to thesurface of the earth with a velocity of 35000 km/hr, from an altitude of 600km. Determine the maximum altitude reached by the satellite using prin-ciple of work and energy.
Given:
.kgm1067.6
andkm6380,kg1098.52311
24
sG
RM−×=
=×=
Solution Let ‘A’ be the point of the orbit at r1 distance farthest fromthe centre of the earth. Consider the motion from point ‘O’ to ‘A’, fromwhich (Potential energy + Kinetic energy) at ‘O’ and ‘A’ are equal.
Potential energy due to gravitational force = −GMm r
( ) ( )
sm222.97223600)1000(35000hrkm35000
m1069806980)6006380(where
21
21
0
30
1
21
0
20
===
×==+=
−=−∴
v
kmr
rGMmvmr
GMmvm
(A)
Kinetics of Particles—Work and Energy 19
Figure 15.12
Since the only force acting on the satellite is the force of gravity, angu-lar momentum about the centre of the earth is conserved.
( ) ( )( )
r m v r m vv v r r
v GMr
v GMr
v v GMr r
0 0 1 11 0 0 1
02
012
1
02
12
0 1
12
12
12
1 1
==
− = −
− = −
( )
[ ]
Substituting for v1 in Eqn (A),
( )12
1 1
21
1 2
1 2 6 67 10 598 106980 10 9722 222
02 0
1
2
001
02
01 0
01 0 0
2
0
1
11 24
3 2
( )
( . ) ( . )( ) [ . ]
v rr
GMr
rr
v rr
GMr
rr
GMr v
rr
−
= −
∴ +
=
+
=
+
= × ××
−
1 1209
0 20969800 209
33397129
01
10
+
=
∴ = = =
rr
r r
.
. .. km
20 Engineering Mechanics
maximum altitude kmaltitude m
( ) . .max
h r R= − = − ==
1 33397 129 6380 27017 12927017129
EXERCISES
1. A block of mass 15 kg is pushed 3 m along a frictionless horizontaltable by a constant 50 N force directed 35° below the horizontal.Determine the work done by (a) applied force (b) normal forceexerted by the table (c) force of gravity (d) net force on the block.
2. A person lifts a 30 kg bucket from a well and does 8.0 kJ of work.Determine the depth of the well if the speed of bucket remainsconstant when it is lifted.
3. A horizontal force of 480 N is used to push a 150 kg box by 4.00mon a rough horizontal surface. If the box moves at a constantspeed, find (a) work done by 480 N force (b) energy lost due tofriction and (c) coefficient of kinetic friction.
4. A particle is subjected to a force Fx that varies with position asshown in Fig. E 15.1. Find the total work done by the force over thedistance x = 0 to x = 12m.
Figure E 15.1
5. The figure E 15.2 shows thetwo horizontal forces pullingthe box along x direction. Cal-culate the work done by eachforce as the box is displacedby 3.5m along x-axis. Alsodetermine the total work doneby the two forces in pullingthe box at this distance.
6. The magnitude of the force ( )Fx along x-axis acting on an objectvaries as a function of ‘x’ as shown in Fig. E 15.3. Find the workdone by the force in the interval (a) 0 0 3≤ ≤x . m (b)0 3 05. .≤ ≤x m (c) 0 1≤ ≤x m.
Figure E 15.2
Kinetics of Particles—Work and Energy 21
Figure E 15.3
7. A box of mass 70 kg resting on ahorizontal surface is subjected toa force of 350 N applied at 40° tothe horizontal as shown in Fig. E15.4. If the coefficient of frictionbetween the box and the surfaceis 0.25 determine the velocity ofthe box after it has moved 5.5 m.
8. A train weighing 1800 kN has a velocity of 60 km/hr when it is atthe beginning of 1.5% grade as shown in Fig. E 15.5. If the pullexerted by the engine is 15 kN and tractive resistance is 9 N per kNload, find the distance which it will travel up the inclined planebefore coming to rest.
Figure E 15.5
9. A train has a weight of 2500 kN. If the frictional resistance is 7.2 Nper kN, determine the pull exerted by the locomotive to increasethe speed from 40 km/hr to 60 km/hr within a distance of 2085 m.
10. A train of weight 5000 kN is pulled up by an engine on a level trackat a constant speed of 50 km/hr. The resistance due to friction is 15N per kN of weight of train. Calculate the power of engine.
11. A 25 kg package slides at a speed of 12 m/s from point A onsloping board. Determine the speed at point B if the coefficient ofkinetic friction is 0.3 as shown in Fig. E 15.6.
Figure 15.4
22 Engineering Mechanics
12. A 40 kg box is pushed alongthe plane by a constant force Fas shown in Fig. E 15.7. The boxstarts from rest at ‘A’ and at-tains, a speed of 4m/s after trav-elling 5m to point B. Determinethe force ‘F’ (a) if the surface issmooth (b) if the coefficient ofkinetic friction is 0.25.
Figure E 15.7
13. Find the power of a locomative drawing a train whose weight is 600kN up an incline of 1 in 100 at a steady speed of 4.5 km/hr. Assumefrictional resistance to be 8N per kN.
14. A small box of mass 25 kg starts from rest at ‘A’ and slides downthe inclined plane as shown in Fig. E 15.8. Determine the distanceit travels along the horizontal plane before it comes to rest. As-sume that the velocity at ‘B’ for the motion along ‘BC’ is the sameas it has gained during travel from A to B and also assume that thecoefficient of kinetic friction is 0.35 for the surfaces AB and BC.
Figure E 15.8
15. In the above problem, if ‘BC’is also inclined as shown inFig. E 15.9, determine the dis-tance it will move along theincline BC. Also determine thevelocity at B when it returnsto B.
16. A body weighing 400 N is resting on a rough plane inclined at 20°to the horizontal. It is pulled up the plane from rest by means of
Figure E 15.6
Figure E 15.9
Kinetics of Particles—Work and Energy 23
light flexible rope run-ning parallel to the planeand passing over a lightfrictionless pulley at thetop of the plane. Theportion of the rope be-yond the pulley, hangsvertically and carries aweight of 225 N at the end. If the coefficient of friction between theblock and the plane is 0.15, find the velocity of the system after ithas moved 2.25 m.
17. Two rough planes inclined at 30° and 60° to the horizontal and ofthe same height are placed back to back. Masses of 12 kg and24 kg are placed on the faces and connected by a string passingover the top of the planes. If µ = 0.6 find the velocity of the blockswhen they travel a distance of 10 m, starting from rest. (Refer Fig.E 15.10).
18. A small block of weight 60 N starts from rest at ‘P’ and slides downthe plane of the roof surface which is inclined at 32° to the horizon-tal as shown in Fig. E 15.11. Find the horizontal distance from Q, atwhich the block will hit the ground. The coefficient of kinetic fric-tion between the block and the roof is 0.25. Also find the velocitywhen the block reaches the ground.
Figure E 15.11
19. The system shown in Fig. E 15.12 is initially at rest, has two blocksA and B, weighing 750 N and 1500 N respectively. If coefficient offriction between block A and the surface of inclined plane is 0.2,determine the velocity of block A when it moves along the planeby 1.5m. Assume pulley to be weightless and frictionless.
Figure E 15.10
24 Engineering Mechanics
Figure E 15.12
20. Determine the velocity of blocks A and B when block ‘C’ movesdown by 1.5 m. Assume that cables pass through frictionless pul-ley. µA = 0 3. and µB = 0 25. . Also find the tension in the cables.m mA Bkg kg= =10 30; and mC kg= 50 . (Refer Fig. E 15.13).
Figure E 15.13
21. Two bodies of weight W WA BN and N= =800 400 are con-nected to the two ends of light inextensible string, passing oversmooth pulley. The weight WA is placed on rough horizontal sur-face whose coefficient of friction is 0.25 and WB is hanging verti-cally in air. If the system is released from rest and block ‘B’ fallsthrough a vertical distance of 2.5m, determine the velocity attainedby ‘B’.
22. Two bodies of weight W WA BN and N= =800 500 are con-nected to the two ends of light inextensible string which passesover a smooth pulley. The weight 800 N is placed on an inclined
Kinetics of Particles—Work and Energy 25
plane of angle 15° and block ‘B’ is vertically hanging in the air. Ifthe coefficient of friction is 0.2, determine the velocity of block ‘B’if it falls through a vertical distance of 2.0 m.
23. Find the velocities of blocks, if block B shown in Fig. E 15.14(a), (b)and (c) falls vertically at a distance of 1.5m.
Figure E 15.14(a) Figure E 15.14(b)
Figure E 15.14(c)
24. Consider the amusement ride as shown in Fig. E 15.15. Assumethat mass of train m = 5000 kg, the height is 40 m and the minimumspeed in the loop is vB m s= 35. . Calculate the required take offspeed vA.
Figure E 15.15
26 Engineering Mechanics
25. A car weighing 2500 N starts from rest at point A and moveswithout friction, down the track as shown in Fig. E 15.16. Deter-mine the force exerted by the car at point B where the radius ofcurvature of the track is 18 m. Also determine the minimum safevalue of radius of curvature at point C.
Figure E 15.1626. A 3 kg mass starts from rest
and slides a distance ‘d’ downa frictionless 30° inclinedplane and strikes an un-stressed spring of negligiblemass as shown in Fig. E 15.17.The mass slides an additionaldistance of 0.25m as it isbrought momentarily to restby compressing a spring ofstiffness 500 N/m. Find the initial distance ‘d’ between the massand the spring.
27. A spring is used to stop a 10 kg package which is moving down on25° incline as shown in Fig. E 15.18. The spring constant k = 30 kN/m is held by the cables so that it is initially compressed by 120 mm.Knowing that the velocity of package is 3 m/s when it is 8m from thespring, determine the maximum additional deformation of the springin bringing the package to rest. Assume the surface to be smooth.
Figure E 15.18
Figure E 15.17
Kinetics of Particles—Work and Energy 27
28. At a mine, the end of a side track is to be provided with a springbumper. The spring must be capable of stopping a 60 kN car. Thevelocity of the car at ‘P’ is 1.5 m/s. From ‘P’ it travels along the 3%grade and then to the bumper as shown in Fig. E 15.19.Determine the stiffness of spring ‘k’ if it is to stop the car afterbeing compressed by 300 mm. At what distance ‘x’, the car will rollback along the level track after rebounding from the point of maxi-mum compression? Assume that the track resistance is contant ofmagnitude 600N.
Figure E 15.19
29. A spring gun having a spring constant 5000 N/m is held in a verti-cal position. If the original compression of the spring is 250mm,determine the projection of a ball of weight 25 N from the com-pressed position. Also determine the velocity of the ball at a heightof 1.6 m.
30. Consider the swing shown in Fig. E 15.20 whose string length is 4m. A child, Amitesh, of mass 50 kg swings on it. Calculate themaximum speed of Amitesh if the maximum angle of the supportingstrings with the vertical is 65°. Neglect frictional loss.
Figure E 15.20
28 Engineering Mechanics
31. A particle of mass ‘m’ = 20 kg is released from point ‘P’ on africtionless surface shown in Fig. E 15.21. Determine (a) particle’sspeed at ‘Q’ and ‘R’ (b) Net work done by the force of gravity inmoving from P to R.
Figure E 15.21
32. A 25 kg block is released from a point ‘P’ as shown in Fig. E 15.22.The path is smooth except for the region ‘QR’ of length 6m. Theblock travels down the path strikes a spring of stiffness, k = 2.25kN/m and compresses the spring by 0.30 m after which it comesmomentarily to rest. Determine the coefficient of kinetic frictionbetween the surface ‘QR’ and the block.
Figure E 15.22
33. A 10 kg collar slides with negligible friction along a vertical rod asshown in Fig. E 15.23. The spring is undeformed when collar ‘A’ isat the same elevation as point 0. If it is released from rest aty1 0 6= . m , determine the velocities of collar at y2 0= andy3 0 5= − . .m
Kinetics of Particles—Work and Energy 29
Figure E 15.23
34. If a system has two masses m mA Band which are released fromrest, find the velocity of mB after it has fallen a vertical distance of0.8 m. Neglect the inertia of pulleys and assume thatm mA B= = 25 kg.
Figure E 15.24
35. A particle of mass ‘m’ is attached to one end of a thin rod ofnegligible mass and length ‘L’ as shown in Fig. E 15.25. The otherend is attached to a frictionless pin. Determine the minimum initial
30 Engineering Mechanics
velocity ‘ v0 ’ to be given to the particle at thelowest position so that the particle reaches thetop position. Show that the minimum velocityis 2 gL .
36. In the above problem, if the mass is attached toan inextensible wire instead of a rod of negli-gible weight, determine the minimum velocity‘ v0 ’ so that the particle reaches the top posi-tion and also show that the minimum velocityis 5gL.
37. A particle of mass ‘m’ starts from rest and slidesdown on a frictionless track as shown in Fig. E 15.26. It leaves thetrack horizontally and strikes the ground as shown in the Fig. E15.26. Determine ‘h’.
Figure E 15.26
38. A block of mass 0.250 kg is placed on top of a vertical spring ofconstant k = 8000 N/m and pushed downward, compressing thespring 0.150 m. After the block is released, it travels upward andthen leaves the spring. Determine the maximum height it rises abovethe point of release.
39. A 5 kg mass is suspended on a rigid rod ofnegligible mass and length L = 600 mm asshown in Fig. E 15.27. The rod has a hingeat ‘O’. Pendulum is released from rest atθ = 0. Determine the speed whenθ = 90o and θ = 135o neglecting friction.
Figure E 15.25
Figure E 15.27
Kinetics of Particles—Work and Energy 31
ANSWERS
1. W = 15 × 9.81 = 147.15 NNormal reaction = 147.15+50 sin 35 = 175.829 NThe block moves horizontally by 3 ma) Work done by applied force = (50 cos 35)3 = 122.873 Nmb) work done by normal force = 0c) work done by force of gravity = 0d) Net work done = 122.803 Nm
2. W = 30 × 9.81 = 294.3 NU = 8 kJ = 8000 J = 8000 NmIf ‘d’ is depth of well,U = wd; 8000 = 294.3 d; d = 27.183 m
3. W = 150 × 9.81 = 1471.5a) work done by 480 N force = 480(4)
= 1920 Nmwork done = change in kinetic energysince the box moves with constantspeed, change in kinetic energy = 0;Hence net work done = 0(480 – F) 4 = 0; ∴ F = 480∴ Energy lost due to friction = 1920 Nm
326.05.1471
480 ===NF
kµ
4. Workdone = area of (F – x) diagram
Total work done
+
++
×+
=
)2.1)(20(218.4)2040(
21
]6.320[)4.2)(20(21
= 252 Nm5. Work done by 80 N force = (80 cos 30°) x
Work done by 70 N force = (70 cos 50°) x∴ When the box moves 3.5 m along x-axis,
Work done by 80 N force = (80 cos 30°) 3.5 = 242.487 NmWork done by 70 N force = (70 cos 50°) 3.5 = 157.483 Nm
∴ Total work done = (80 cos 30° + 70 cos 50°) 3.5 = 399.97 Nm = 399.97 Joule
6. ∫=2
1
12
x
xx dxFU
Figure A 15.1
Figure A 15.2
32 Engineering Mechanics
Nm5.18)15()25(6
Nm15)5.0)(30((c)
Nm5.2)2.0)(25(21(b)
Nm6)3.0)(40(21(a)
3.0
0
1
5.0
3.0
3.0
3.0
0
=+−+==
===
−=
==
=
==
∫
∫
∫
∫
dxFU
dxFU
dxFU
dxFU
xtotal
xIII
xII
xI
7. Weight of box (W) = mg = 686.7N.∴ Normal reaction at surface = N = 350 sin 40º + 686.7 N = 911.676∴ Frictional force (F) = µN = 0.25 (911.676) = 227.919 NWhen the box moves to the right by5.5 m, weight, Normal reaction andvertical component do no work.∴ Fnet = 350 cos 40º – 227.919 = 40.197 N∴ Work done = Fnet (s) = 40.197 (5.5) = 221.084 NmFrom work-energy principle,Uif = Tf – Ti
222
20
35)70(5.021
021
ffff
i
vvmvT
vmT
==
=
=
=
m/s513.235084.221 2 =⇒=∴ ff vv∴ Velocity of box = 2.513 m/s.
8. For small values of α = sin α = tα = (15/100) tractive resistance= 9 N per kN.∴ Total tractive resistance F = (9N/1kN)× 1800 kN = 16200 N
kg10486.18381.9
101800
N1800
33
×=×==∴
=
gWm
W
Figure A 15.3
Figure A 15.4
Kinetics of Particles—Work and Energy 33
Component of weight opposing motion = W sin α = 1800 × 0.015 = 27 kN
∴ Pull (P) = 15 kNTotal force opposing motion = 16.200 + 27 = 43.200 kN.∴ Net force = 15 – 43.200 = – 28.200 kN.Hence when the train moves up, there will be a loss in kineticenergy. From work energy principle,
sU
TTU
if
ifif
)10200.28( 3×−=
−=
m/s667.16
3600)1060(
hrkm60
)0rest, tocomesIt (021
3
2
=×==
==
=
i
fff
v
vmvT Q
∴ – (28.200×103) s = – 25485186∴ s = 903.73 m
Therefore train travels at a distance of 903.73 m up the plane beforecoming to rest.
9. Total frictional resistance F = (7.2 N/1kN) 2500 kN = 18000 N.Let the pull be ‘P’ N. Therefore effective force is (P – 18000) N.Since the velocity changes from 40 km/hr to 60 km/hr
m/s667.163600)1060(km/hr60
m/s111.113600)1040(km/hr403
3
=×==
=×==
f
i
v
v
Change in kinetic energy = Tf – Ti
kg10842.25481.9
102500 33
×=×==g
Wm
According principle of work-energy,Work done (Uif) = change in kinetic energy (Tf – Ti).
N91.27431
])111.11()667.16)[(10842.254(212085)1800( 223
=∴
−×
=−
P
P
Therefore total pull required is, P = 27.432 kN.10. Work done = change in kinetic energy.
21
22 2
121)( mvmvsFP
−
=−
Since v1 = v2 =50 km/hr, P = F
Frictional force ‘F’ N75000kN5000kN1
N15 =×=
34 Engineering Mechanics
∴ Pull required = 75000 N∴ Power = 75000 N × Velocity
Since velocity = 50 km/hr m/s889.13360050000 ==
∴ Power = 75000 N × 13.889 m/s = 1041675 W = 1041.675 kWTherefore power of the engine is 1041.675 kW.
11. W = 25(9.81) = 245.25 NComponent of weight normal to the plane = Wcos α = 245.25 cos 35 = 200. 897 N∴ Friction = µ.N = 0.3 (200.897) = 60.269Component of weight parallel to the plane
= W sin α = 245.25 sin 35 = 140.670
UAB = (140.670 – 60.269)6 = 482.406velocity at A, UA = 12 m/s
kinetic energy at A, 1800)12(2521
21 22 =
=
mv
kinetic energy at B, 22 5.12
21
BB Vmv =
According to Principle of work and energy. UAB = TB – TA
m/s513.1318005.12406.482 2
=∴−=
B
B
VV
12. W = mg = 392.4 NW sin α = 392.4 sin 12 = 81.585W cos α = 392.4 cos 12 = 383.825
320)4)(40(
21
21
021
22
2
=
=
=
=
=
BB
AA
mVT
mVT
a) If surface is smooth,UAB = (F – 81.585) 5∴ UAB = TB – TA(F – 81.585)5 = 320∴ F = 145.585 N
b) If coefficient of friction is 0.25,frictional force = 0.25 (383.825) = 95.956
Figure A 15.5
Kinetics of Particles—Work and Energy 35
∴ UAB = (F – 81.585 – 95.956)5∴ UAB = TB – TA(F – 81.585 – 95.956)5 = 320F = 241.541
13. Friction = 8 N per kN
Total friction N4800)kN600(kN1N8 =
=
N6000kN6100
1600sin ==
=θW
Since it moves with steady speed, change in kinetic energy is zero.Hence net workdone must be zero.Hence, P – F – W sin θ = 0P – 4800 – 6000 = 0P = 10800 NPower of locomotive VF
rr⋅=
kN5.13W13500PowerNm/s1350025.1)10800(Power
m/s25.13600
)10(5.4km/hr5.4Velocity3
====
===
14. m = 2.5 kg; W = 2.5(9.81) = 24.525 kNWsinθ = 24.525 sin 30 = 12.263 NWcosθ = 24.525 cos30 = 21.239 NFAB = µ Wcosθ = 0.35 (21.139) = 7.4 NUAB = (Wsinθ – FAB) os
= (12.263 –7.4) 40 = 214.52 Nvelocity at A = 0; Velocity at B = VBAccording to principle of work and energy,UAB = TB – TA
kinetic energy at B, 22 25.1
21
BB VmVTB =
=
kinetic energy at C, 021 2 =
= AA VmT
m/s1.1325.152.214 2
=∴=∴
B
B
VV
In BC, Block travels along horizontal. Hence frictional forceFBC = 0.µW = 0.35 (24.525) = 8.584 NIf block comes to rest at C, thenUBC = TC – TB
Figure A 15.6
36 Engineering Mechanics
m990.24
)1.13()5.2(210)(585.8 2
=∴
−=−
x
x
distance travelled along ‘BC’ = 24.990 m15. W = 24.525 N
m = 2.5 kgµ = 0.35velocity at A = 0velocity at B = 13.1 m/s.W sin θ = 24.525 sin 40 = 15.764FBC = µ(W cosθ) = 0.35 (24.525 cos 40) = 6.576 Nvelocity at C = 0
UBC = TC – TB
m602.9
)1.13()5.2(210)576.6764.15( 2
=∴
−=−−
x
x
When it returns to B from C, motion is downwards,
22
2
25.121
021
BBB
CC
VVmT
mVT
=
=
=
=
m/s401.825.1223.88 2
=∴=
−=∴
B
B
VV
TCTBUCB
distance travelled up along BC = 9.602 mvelocity at B when it returns to B = 8.401 m/s
16. WA = 400 N; WB = 225 N;mA = WA/g = 40.775 kg; and mB = WB/g = 22.936 kgComponent Weight ‘WA’ parallel to the plane = WA sin θ
= 400 sin 20º = 136.808 NComponent of Weight ‘WA’ normal to the plane = WA cos θ
= 400 cos 20º = 375.877 Nfrictional force, FA = µ NA = 0.15(375.877) = 56.382 NBlock ‘B’ moves down and Block ‘A’ moves up the plane. Theforces which do work are shown in Fig. 15.32 (b). The componentof WA normal to the plane and tension in the string do no work.Work done, Uif = (225 – 136.808 – 56.382) 2.25 = 71.573 Nm
Figure A 15.7
Figure A 15.7(a)
Kinetics of Particles—Work and Energy 37
Initial kinetic energy, 22 )(
21)(
21
iBBiAAi vmvmT
+
=
(vA)i = (vB)i = 0 since they start from rest.(vA)f = (vB)f = v
Figure A 15.8
Final kinetic energy 22 )(21)(
21
fBBfAAf vmvmT
+
=
22 856.31]936.22775.40[
21 vvT f =+
=
According to principle of work and energy,Uif = Tf – Ti71.573 = 31.856 v2 – 0∴ v = 1.5 m/s
Therefore velocity of blocks after they move 2.25 m is 1.5 m/s.17. mA = 12 kg; WA = mA g = 117.72 N
mB = 24 kg; WB = mB g = 235.44 NAssume that block ‘A’ moves up and block B moves down.Component of weight parallel to plane = WA sin θComponent of weight normal to plane = WA cos θFor block A; WA sin θ = 117.72 sin 30º = 58.86 N
WA cos θ = 117.72 cos 30º = 101.949 NFor block B; WB sin θ = 235.44 sin 60º = 203.897 N
WB cos θ = 235.44 cos 60º = 117.72 NFrictional force F = µNComponent of weights WA and WB parallel to the planes and fric-tional forces do work and they are shown in Fig. 15.9.
38 Engineering Mechanics
Figure A 15.9
Work done, Uif = (– 58.86 – 61.169 – 70.632 + 203.897)10 = 132.36 NmInitial kinetic energy is zero, since the blocks start from rest; andtherefore vA = vB = v.According to principle of work and energy
Uif = Tf – Ti132.36 = 18 v2
∴ v = 2.712 m/sTherefore velocity of blocks after they move 10 m is 2.712 m/s.
18. sin 32º = (3/PQ); PQ = 5.661 mMotion down the plane ‘PQ’
N795.31º32sin60sin
N721.12N25.0N883.50º32cos60cos
====∴
====
θ
θµ
WF
WNNF
Work done from P to Q= UPQ = (31.795 – 12.721) (5.661)
UPQ = 107.978 NmInitial Velocity at P = vP = 0
m/s942.5;058.3978.107
058.3)81.9/60(21
)(21;0)(
21
2
22
22
==
−=∴
=
=
==
=∴
PQPQ
QQQ
QQPP
vv
TTU
vvT
vmTvmT
From Q to R, Let us apply the principle of conservation of energy.At Q, potential energy = mgh = 60(8) = 480 Nm
Figure A 15.10
Kinetics of Particles—Work and Energy 39
kinetic energy 22 )942.5)(81.9/60(21
21
=
= mv
= 107.974 NmTotal energy = 587.974 NmAt ‘R’ potential energy is zero,
∴ kinetic energy 22 058.3)81.9/60(
21
RR vv =
=
and total energy 2058.3 Rv=m/s866.13;974.587058.3 2 ==∴ RR vv
When it falls freely from Q to R, it is subjected to acceleration dueto gravity only. x component of velocity must be same.At Q, vxQ = vQ cos 32º = 5.942 cos 32º = 5.039 m/s
vyQ = vQ sin 32º = 5.942 sin 32º = 3.149 m/sAt R, vxR = vR cos α = 13.866 cos α
vyR = vR sin α = 13.866 sin α13.866 cos α = 5.039; α = (68.70)ºvyR = 13.866 sin (68.7) = 12.919 m/s
Time taken to reach ‘R’ from ‘Q’ is, vRy = vQy + gt12.919 = 3.149 + 9.81(t); t = 1 sec
∴ horizontal distance m039.5===′ tvxQxRQ19. Let us assume that block ‘A’ moves up the plane by 1.5 m from
rest. If block A travels up by ‘x’ m, then block ‘B’, moves down by(x/2) m.
AAB vvv 5.0)2( ==∴component of weight A parallel to plane = WA sin 30º = 375 N
NA = WA cos30º = 649.519 NFA = µ NA = 0.2 (649.519) = 129.904 N
The forces which do work are : (1) component of WA along plane:(2) FA: and (3) WB.Work done
U = (– 375 – 129.904) (1.5) + 1500 (15/2)U = 367.644 Nm
Since blocks start from rest, Ti = 0Let vA = v
22
21
21
5.05.0
BBAAf
AB
vmvmT
vvv
+
=
==∴
40 Engineering Mechanics
2
22
340.57
])5.0(905.152453.76[21
vT
v
f =∴
+
=
Using the principle of work and energy,
m/s266.1m/s;532.2m/s532.2
034.57644.367; 2
==∴=∴
−=−=
BA
ifif
vvv
vTTU
20. WA = mA g = 9.81 N; WB = mB g = 294.3 N; WC = mC g = 490.05 N
6.0cos;8.0sin13.53);34(tan
11
11
====
θθθθ
N871.254cos;N15.147sin
N86.58cos;N48.78sin866.0cos;5.0sin;º30when
22
11
222
====
===
θθθθ
θθθ
BA
AA
WWWW
Block ‘C’ moves down, block ‘B’ moves to the right and henceblock ‘A’ moves up.
N718.63)871.254(25.0N658.17)86.58(3.0
======
BBB
AAA
NFNF
µµ
The following forces which do work: (1) WC (2) components paral-lel to the plane of weights WA, WB; and (3) FA, FB.
vA = vB = vC = vSince they start from rest, Ti = 0Work doneU = 490.05(1.5) – 78.48(1.5) – 17.658(1.5) + 147.15(1.5) – 63.718(1.5)∴ U = 716.016 Nm
m/s989.3
045016.716;
45]503010[21
)(21
2
22
2
=∴
−=−=
=++
=∴
++
=
vvTTU
vvT
vmmmT
if
f
CBAf
To find the tension in the cable ‘D’ Alembert’s principle is used.aA = aB = aC = a
For Block C,
2
22
22
m/s304.5
)5.1(2)0()989.3(
2
=∴
+=
+=
a
a
asvv if
Kinetics of Particles—Work and Energy 41
Dynamic equilibrium of block ‘C’
N.85.224005.490)304.5(50
0
2
2
2
=∴=−+
=−+
TT
WamT CCC
Dynamic equilibrium of ‘A’
N.178.1490658.1748.78)304.5(10
0sin
1
1
11
==−−−
=−−−
TT
FWamT AAAA θ
Therefore velocity of blocks = 3.989 m/s and tensionT1 = 149.178 N; T2 = 224.85 N
To check: The result can be verified by considering the dynamicequilibrium of ‘B’.
Figure 15.11
0718.63)304.5(30178.14915.14785.2240sin 122
=−−−+=−−−+ BBBB FamTWT θ
Hence it is correct.21. WA = 800 N
F = µWA = 0.25(800) = 200 N
kg775.4081.9
400
kg549.8181.9
800
==
==
B
A
M
M
Work is done by WB and F.∴ U12 = (400 – 200) 2.5 = 500Initial kinetic energy
21
21
21 )(
21)(
21)(
21
BBAAAA VmVmVm
+
+
=
since they start from rest, (VA)1 = (VB)1 = 0Final velocity, (VA)2 = (VB)2 = V
Figure A 15.11
Figure A 15.12
42 Engineering Mechanics
∴ Final kinetic energy ])775.40549.81[(21 2V+
=
T2 = 61.162 V2
∴ U12 = T2 – T1500 = 61.162 V2
∴ V = 2.859 m/s
22.
kg968.5081.9
500;N500
kg549.8181.9
800;N800
===
===
BB
AA
MW
MW
WA cosα = 800 cos15 = 772.741 NWA sinα = 800 sin15 = 207.055 NFrictional force = 0.2 (772.741) =154.548 N∴ U12 = (500 – 207.055 – 154.548)2 = 276.794 NInitial kinetic energy = 0
Final kinetic energy 2)968.50549.81(
21 V+
=
= 66.259 V2∴ U12 = T2 – T1 276.794 = 66.259 V2
∴ V = 2.044 m/s
23.
3.0
291.1581.9
150;N150
368.1981.9
190;N190
=
=
==
=
==
µ
BB
AA
MW
MW
Assume that Block ‘B’ moves down; hence Block ‘A’ moves to theright.xA + 2yB = constant,displacement of A = 2 (displacement of B)Hence, velocity of A = 2 (velocity of B)work is done by FA and WB
FA = µ WA = 0.3 (190) = 57 NU12 = – 57(3) + 150 (1.5) = 54
Initial kinetic energy, T1 = 0
Final kinetic energy, 222 2
121
BBAA VmVmT
+
=
Figure A 15.13
Kinetics of Particles—Work and Energy 43
22
222
382.46
]))(291.15()2)(368.19[(21
B
BB
VT
VVT
=
+=
According to principle of work and energy,U12 = T2 – T1
2382.4654 BV=∴ VB = 1.079 m/s VA = 2.158 m/sVelocity of A, VA = 2.158 m/sVelocity of B, VB = 1.079 m/s
b. WA = 200 NWB = 300 N
kg581.3081.9
300
kg387.2081.9
200
=
=
=
=
B
A
M
M
WA cos θ = 200 cos 35 = 163.830 NWA sin θ = 200 sin 35 = 114.715 NFA µ (WA cos θ) = 0.25(163.83) = 40.95 NIf block ‘B’ moves down by ‘yB’ then block A moves up by (2yB)workdone due to movemnt of block B by 1.5m,U12 = (– WA sinθ – FA) (3) + WB (1.5) = (114.715 + 40.958) (3) (300) (1.5)U12 = –17.019Initial kinetic energy, T1 = 0
Final kinetic energy, 222 2
121
BBAA VmVmT
+
=
21212
22
222
065.56019.17
065.56
]581.30)2(387.20[21
B
B
BB
V
TTUVT
VVT
=−
−=∴=
+
=
Block B moves up, VB = 0.551 m/sBlock A moves down, ∴ VA = 1.102 m/sVelocity of A, VA = 1.102 m/sVelocity of B, VB = 0.551 m/s
44 Engineering Mechanics
c. WA = 600 NWB = 400 N2 VA = VB
kg775.4081.9
400
kg162.6181.9
600
=
=
=
=
B
A
m
m
WA sinθ = 600 sin 40 = 385.673 NWA cosθ = 600 cos 40 = 459.627 NFA = µ (WA cosθ ) = 0.25 (459.627) = 114.907 Nworkdone = U12; If block moves down by 1.5 m, block A moves upby 0.75U12 = (–WA sin θ – FA) 0.75 + WB (1.5) = – (385.673 + 114.907) 0.75 + 400 (1.5)U12 = 224.565Initial kinetic energy, T1 = 0
Final kinetic energy, 222 2
121
BBAA VmVmT +=
22
22
2
033.28
)775.40(2
)162.61(21
B
BB
VT
VVT
=
+
=
Using principle of work and energy,U12 = T2 – T1
2033.28565.224 BV= ∴ VB = 2.830 m/s VA = 1.415 m/svelocity of A, VA = 1.415 m/svelocity of B, VB = 2.830 m/s
24. UAB = TB – TA
222
22
2500500021)(
21
30625)5.3(500021
)(21
AAAA
BB
vvvmT
vmT
=
=
=
=
=
=
m/s232.282500306251962000
1962000)40(81.950002
=∴−=−
−=×−=−=
A
A
AB
vv
mghU
Kinetics of Particles—Work and Energy 45
25. At point B:At point C:
Figure 15.14
Work done from A to B: Weight (gravitational force) of the cardoes work. Since it is conservative force, work done UAB = Wh =2500 (30) = 75000 Nm.
kg842.25481.9
2500 ===g
Wm
change in kinetic energy ][21 22
AB vvm −
=
starting from rest at A, vA = 0
222
2
m/s7.3218
)261.24(
m/s261.24
))(842.254(2175000
====
==
=
Rvaa
v
v
BBn
B
A
∴ considering the dynamic equilibrium at B in Fig. 15.39b,
N333.10833)7.32)(842.254(25000
=+=∴=−−
B
nB
NmaWN
Similarly, apply work-energy principle from A to C,UAC = Wh = 2500 (30–12)45000 Nm
change in kinetic energy ))(842.254(21 22
AC vv −
=
Considering the dynamic equilibrium at C in Fig. 15.39c,NC + m aC – W = 0
In the limiting case, NC = 0∴ (254.842)aC = 2500∴ aC = 9.810 m/s2
46 Engineering Mechanics
m00.3681.9
)793.18(;81.922
===∴ RRvC
Note: When R > 36.00, is not equal to zero.Therefore Velocity at B = vB = 24.261 m/s
Velocity at C = vC = 18.793 m/sForce exerted at B = 10833.333 N and safe value of radius ofcurvature = 36 m.
26. W = 3 × 9.81 = 29.43 N W sin30º = 14.715 N
Block starts from rest and finally comes to rest.Hence, change in kinetic energy = 0
Workdone 021)25.0()715.14( 2 =
−+ kxd
.m812.0
0)25.0)(500(21)25.0()715.14(
021))(sin(
2
212
=
=
−+
=
−+=∴
d
d
kxxdWU θ
27. When the package comes to rest, final velocity V2 = 0Let additional deformation be ‘∆x’W = 10 × 9.81 = 98.1 NWsinθ = 98.1 sin25 = 41.459 NSurface is smooth; hence no frictional force.work done by package = 41.459 (8 + ∆x)work done by spring from 0.12 m to (0.12 + ∆x) is,Pmin = 30000 (0.12) = 3600Pmax = 30000 (0.12 + ∆x )Workdone by spring
xPP ∆+
−= )(
21
maxmin
]150003600[
]3000036003600[21
2xx
xx
∆+∆−=
∆∆++
−=
∴ Net work done = 41.459(8 + ∆x)–(3600 ∆x + 15000 ∆x2)
Kinetic energy at ‘1’, 2
11 21 mVT
=
Figure A 15.15
Kinetics of Particles—Work and Energy 47
Nm45)3()10(21 2
1 =
=T
Kinetic energy at ‘2’, T2 = 0∴ U12 = T2 – T1(41.459) (8 + ∆x) – (3600 ∆x + 15000 ∆x2) = –45–15000 ∆x2 – 3558.541 ∆x + 376.672 = 0∆x2 + 0.237 ∆x – 0.0251 = 0
m079.02
)0251.0(4)237.0(237.0 2
=∆
−−±−=∆
x
x
The package is brought to rest when the spring deforms addition-ally by 0.079 m.Method 2: This can also be calculated by considering the initialdeformation as zero.Distance between block and pring = 8 – 0.12 = 7.88 mLet the total deformation be ‘x’
00248.010764.2
0697.371459.4115000
453000021)88.7(459.41
32
2
2
=−×−
=++−
−=
−+
− xx
xx
xx
m159.0
)0248.0(4)10764.2(10764.2 233
=
−×±×+= −−
x
x
Initial deformation = 0.12 m∴ additional deformation = 0.039 m
28. W = 60 kN = 60000 N; m = (60000/9.81) = 6116.208 kgFor small values of α, 03.0%3tansin === αα
N1800)03.0(60000sin ==∴ αWInitial elevation of ‘P’ above Q = 50 sin α = 50(0.03) = 15 mInitial kinetic energy of car
Nm734.6880)5.1()208.6116(2121
2
2
=
=
= vm
Work done by graivty = 60000 × 1.5 = 90000.00 NmTotal distance travelled by car = 50 + 80 + 0.3 = 130.3 m∴ Work done against friction = 600 (130.3) = 78180 Nm
48 Engineering Mechanics
Work done in compressing the spring
kkkx 045.0)3.0)((21
21 22 =
=
=
Finally when the car comes to rest its kinetic energy is zero.Work done = change in kinetic energy
N/m86.415571734.6880045.0781809000
=∴−=−−∴
kk
Energy stored in the spring is used to push the car to the left by adistance ‘x’ against friciton.
mx
x
xk
168.31
0600)3.0()86.415571(21
0600)3.0(21
2
2
=∴
=−
=−
∴
Therefore (i) stiffness of spring k = 415571.86 N/m (ii) it rolls backs along the level ground by 31.168 m.
29. Energy stored in spring = (1/2) kx2
= (1\2) (500)(0.25)2
U = 156.25 NmThis energy is converted into potential energy and kinetic energyof the ball. At maximum height, velocity of the ball is zero.
m25.62525.156;0,At
81.925
212525.156
max
maxmax
2
=∴==
+=∴
hhvh
vh
At a height of 1.6 m,
m/s552.981.9
2521)6.1(2525.156 2
=∴
+=
v
v
Therefore, velocity of the ball at 1.6 m is v = 9.552 m/s.30. The system is subjected to conservative forces.
Total energy at A, EA = TA + VAwhere potential energy VA= WyA = (50 × 9.81) yA
Nm055.1133)31.2()81.950(m31.2)º65cos1(4
º65cos44
=×=∴=−=
−=
A
A
A
Vyy
Kinetics of Particles—Work and Energy 49
Kinetic energy at A, 2
21
AA mvT
=
Since it starts from rest at A, VA = 0; TA = 0∴ EA = 1133.055Since total energy is constant, energy at any other point in thepath is E = T + V.
2
21)(055.1133
055.1133
mvWy
TV
+=
+=∴
Velocity ‘v’ is maximum when potential energy ‘V’ is minimum. Aty = 0 ‘V’ is minimum at ‘B’.
732.6
))(50(21055.1133
max
2max
=∴
=∴
v
v
31. Potential energy at ‘P’ = mgh = 20 × 9.81 × 8 = 1569.6 Nm; W = mg= 196.2 N.
Kinetic energy at ‘P’ 021 2 =
= Pmv (since it starts from rest).
Potential energy at Q = 196.2 (4.8) = 941.76 NmAccording to the principle of conservation of energy,
m/s924.7
))(20(2176.9416.1569i.e., 2
=∴
+=
Q
Q
v
v
Potential energy at R = 196.2 (3.6) = 706.32 Nm
Since 2))(20(
2132.7066.1569; RQP vEE
+==
∴ vR = 9.291 m/sWork done from P to R = change in potential energy
UPR = –(VR – VP) = –[706.32 – 1569.6] = 863.28 Nm(Qgravitational force is a conservative force).
32. Kinetic energy at 0)(21 2 =
= Pvm and
Potential energy at P = mgh = (25 × 9.81) (3) = 735.75 Nm∴ Total energy at P = 735.75 NmAt S, final velocity of block is zero and the spring is compressedby 0.30 m.
50 Engineering Mechanics
Potential energy stored in spring 2
21 kx
=
Total energy at P = Total energy at S + Loss due to friction in ‘QR’ 735.75 = 101.25 + Loss in energy
Loss in energy in ‘QR’ = 634.5 NmSince frictional force opposes the motion there is a loss in energy.Normal reaction from surface ‘QR’ = W = 245.25 N.
431.05.6346)25.245(
=∴=∴
µµ
Therefore the coefficient of kinetic friction between the surfaceQR and the block is 0.431.
33. Spring is undeformed when it is at the same level as ‘O’.
Figure A 15.16
∴ Undeformed length = 0.45 mPotential energy of collar = mgy andpotential energy of spring
2
21 kx
= where ‘x’ is the elongation
of the spring.
Kinetic energy of collar 2
21 mv
=
Position 1: y1 = 0.6 mElongation of spring x = 0.75 – 0.45 = 0.30 m.
Nm86.78)3.0)(400(21)6.0)(81.9)(10( 2
1 =
+=∴ V
Kinetic energy T1 = 0 Since v1 = 0Total energy E = T1 + V1 = 78.86 + 0 = 78.86 NmPostion 2: y2 = 0; x = 0
222 2
1 kxymgV
+=
Figure A 15.16(c)
Kinetics of Particles—Work and Energy 51
0)0)(400(21)0()81.9()10( 2 =
+=
Kinetic energy, 22
222 5
21 vvmT =
=
m/s972.3;Nm86.785 222 ±==∴ vv
Postion 3: y3 = – 0.5; x = 0.223 m
Nm104.39
)223.0)(400(21)5.0)(81.9()10(
21
3
2
233
−=
+−=
+=
V
kxymgV
Kinetic energy, 23
233 5
21 vvmT =
=
m/s857.4;104.39586.78 323 ±=−=∴ vv
It is downwards.Speed of collar at y2 = 3.972 m/sSpeed of collar at y3 = 4.857 m/s
34. When mass ‘mB’ falls down by 0.8 m, the pulley P2 is lowered by
0.4 m and mass A moves by 0.4 m. Hence .2
= B
Avv
At initial position: Potential energy = 0 ; Kinetic energy = 0 ∴ Total energy = 0
At final position: When ‘B’ falls by 0.8m,Potential energy
= (mA g) (0.4) + (mB g) (– 0.8) = (25 × 9.81) [ –0.4] = –98.1 Nm
Kinetic energy 22
21
21
BBAA vmvm
+
=
2
22
2813.153
22)81.925(
21
B
BAB
B
v
vvvv
=
=
+
×
= Q
∴ Total energy 22813.1531.98 Bv+−=According to the principle of conservation of energy,(Total energy)Initial = (Total energy)Final
m/s4.0m/s8.0
2813.1531.980 2
=∴=
+−=
A
B
B
vv
v
52 Engineering Mechanics
Therefore, velocity of B = 0.8 m/s (downwards); and velocity ofA = 0.4 m/s (upwards).
35. ?36. ?37. 14. At A if velocity is VA (horizontal) time taken to reach ground,
sec55.0
81.9)5.1(2
5.121 2
==
=
t
gt
m743.1
)182.2(21)5.1(81.9
21)5.1(
m/s182.2)55.0(2.1;
2
2
=∴
=−
=−
=∴==∴
h
h
Vmhm
VVtVx
Ag
A
AA
38. Potential energy stored in spring 2
21 kx
=
90)15.0)(8000(21 2 =
=
When the block reaches maximum height v = 0; hence kinetic en-ergy is zero.
m697.36)81.925(.
9090
=×
=
=∴
h
mgh
39. Potential energy at A, = mgh = 5(9.81)(0.6) = 29.43 NmSince it starts from rest, kinetic energy at A = 0At B, potential energy is zero.
m/s431.343.295.2
5.2)(521
21
2
222
==∴
=
=
B
B
BBB
VV
VVVm
Potential energy at C = mg (L – L cos θ) = 5(9.81)[0.6] (1 – cos45) = 8.62
Figure A 15.17
Kinetics of Particles—Work and Energy 53
m/s885.2
43.29)5(2162.8 2
=∴
=
+∴
C
C
V
V
When θ = 90º, VB = 3.431 m/sWhen θ = 135º, VC = 2.885 m/s Figure A 15.17(a)
Kinetics of Rigid Bodies 1
1 91 91 91 91 9
Kinetics of Rigid BodiesKinetics of Rigid BodiesKinetics of Rigid BodiesKinetics of Rigid BodiesKinetics of Rigid Bodies
CONCEPTS
• Equation of translational motion is camFrr
=Σ where, car is theacceleration of mass centre regardless of where the resultantforce acts on the body. It is independent of rotation of the body.
• Equation of rotutional motion isαrr
cc IM =Σwhere α is the angular acceleration of the rigid body indepen-dent of translation of the body.
• Three scalar equations of motion can be written for a rigid bodyin plane motion
αcccyycxx IMmaFmaF =Σ=Σ=Σ ;;• Using D’Alembert’s principle dynamic problem is converted to
static problem and the equations of equilibrium can written as:0and;0;0 =Σ=Σ=Σ cyx MFF
• In case of rolling motion, the following three cases are consid-ered.Rolling, no sliding αµ raNF s =≤ ;Rolling, sliding impending αµ raNF s == ;Rolling and sliding αµ and;aNF k≤ are independent
• Work of a force on a rigid body
dtvFrdFUt
t
t
t∫∫ ==2
1
2
1
..12rrrr
When rd infinitesimal displacement of force F along the pathfrom 1 to 2 and dtdrv /=Work of a moment
∫=−
2
1
21
θ
θθMdU
where θd is the infinitesimal angular displacement
2 Engineering Mechanics
• Kinetic energy of the rigid body is defined as
22rottran 2
121 ωcc ImvTTT +=+=
where C denotes the centre of the mass.For rotation about an arbitrary fixed axis through O:
202
1 ωIT =
• Principle of work and energy.2112 −+= UTT
The equation is valid for an internally conservative system ofrigid bodies.
• In a conservative system, T1 + V1 = T2 + V2where V = Total potential energy = Vg + Ve
• In translation, power vFdtdUrr
.=• In rotational, power ωMdtdU =• Impulse and Momentum principle of a rigid body
mvc1
mvc2
Fdt
mdt
C C+ =C
O
Ic 1
rc
Ic 2
Figures S 19.1
Linear Impulse and Momentum:
∫ −=Σ2
1
)( 12
t
tcc vvmFdtrr
Angular Impulse and Momentum:
∫ −=Σ2
1
12
t
tccc HHdtM
r
For plane motion, Mr
Σ is parallel toωr ,
∫
∫
=+Σ
=+Σ
2
1
2
1
20100
21
t
t
t
tccc
IIdtM
IIdtM
ωω
ωω
Kinetics of Rigid Bodies 3
• Conservation of linear momentum principle:
21
0,0If
cc vvGF
rr
rr
==∆=Σ
• Conservation of angular momentum principle:
21
0,0Ifωω =
=∆=Σ cHMrr
• For a system of rigid bodies use the same fixed reference point Ofor all parts of the system
00 =∆Hr
Example 19.1 A truck moving with a velocity of 10 m/s experiences asuddenly applied brake. It was observed that truck shown in Fig. 19.4skidded to rest in 7.5 m. If the mass of the truck is 3000 kg, determine themagnitude of normal reaction and frictional force on each wheel when thetruck skidded to rest.(see Fig. 19.4)
Figure 19.1
Solution
2
20
20
m/s67.6
5.7210002
m/s10
−=
×+=+=
=
a
aasvv
v
i.e., acceleration is from right to left.Hence inertia force is from left to right
kN010.20N2001067.63000
=×=ma
W = mg = 3000 × 9.81 = 29430 N = 29.430 kN
00
BA
BA
=+−−=−+
maFFWNN
Figure 19.1 (a)
4 Engineering Mechanics
[ ]
[ ] 010.20430.29
0
BA
BA
BA
=+=+∴=++−
NNNNmaNN
µ
µ
Taking moment about A,
+
04)6.1(430.29)2.1(010.200)4()6.1()2.1(
B
B
=+−−=+×−×−
NNmgma
kN65.11kN78.17
A
B
==∴
NN
[ ]
680.001.2078.1765.11
==+
µµ
kN92.7)65.11(68.0kN09.12)78.17(68.0
AA
BB
======
NFNF
µµ
The values computed above represents the sum of reactions at the twowheels in front and two at the back.
kN96.392.721)2(
kN045.609.1221)2(
kN83.565.1121
21
kN89.8278.17
21
Arear
Bfront
Arear
Bfront
=×==
=×==
=×==
===
FF
FF
NN
NN
Example 19.2 Thin plate ABCD has a mass of 10 kg and is held inposition by the wire BH and two links AE and DF. as shown in Fig 19.6.Neglect the mass of the wire and links and determine immediately afterwire BH has cut (a) the acceleration of the plate (b) the force in each link.
Figure 19.2
Kinetics of Rigid Bodies 5
Solution After the wire BH has been cut we observe the corners A andD move along parallel circles of radius 150 mm centered respectively at Eand F. At the instant when BH is cut, the velocity of the plate is zero. Thusthe acceleration a of the mass centre C of the plate is tangent to thecircular path.
Figure 19.3
W = 10 × 9.81 = 98.1 N(a) Acceleration of the plate
2m/s50.8;010º30cos1.98
030cos0
==−
=−=Σ
aa
maWFt
Therefore acceleration of the plate 8.5 m/s2.(b) Forces in each Link 030sinDFAE =−+=Σ WFFFn (a)Taking moment of all the forces at C,
0100)30cos()250)(30sin()100)(30cos()250)(30sin(
DF
AEAE
=++−
FFFF
DF
0603.2114.38 DFAE =+ FF (b)Solving the above two equations, we get
N87.101.981109.01109.0N92.591.986109.06109.0
DF
AE
=×−=−==×==
WFWF
Example 19.3 A 100 kg pulley having a radius of gyration of 0.4 m isconnected to two cylinders as shown in Fig 19.8. Assume no axle frictionand determine the angular acceleration of the pulley and the accelerationof each cylinder.
6 Engineering Mechanics
Solution The pulley will rotate in the anticlockwise direction since massof B is greater than mass A. Assume α as the angular acceleration. Hence
αααα 3.0and5.0 BBAA ==== rara
Figure 19.3
αα
102.1965.02081.920
2081.920
A
AA
+=×+×=
+×=
T
aT
B comes down
αα
124.3923.04081.940
B
B
−=
××−×=
TT
Taking moment at C,αITT =×−× 5.03.0 AB
ααα I=+−− 5.0)102.196(3.0)124.392(222 kgm16)4.0(100 === mkI
ααααααα6.2456.31662.19
1651.986.372.117=++==−−−
rad/sec797.06.24
62.19 ==∴ α
2BB
2AA
m/s2391.0)797.0(3.0
m/s398.0)797.0(5.0
===
===
α
α
ra
ra
Figure 19.3 (d)
Kinetics of Rigid Bodies 7
Example 19.4 A 20 kg rod of length 1m is supported at a distance of 0.25 mfrom one end as shown in Fig 19.15. If therod is released at the horizontal position,determine (a) angular velocity of the rodwhen it is rotated through an angle 60ºand (b) reaction at the support.Solution From the free body diagramshown in Fig. 19.15b.
0cos;0 =−+=Σ θα mgmrRF tt (i)0sin;0 2 =−+=Σ θω mgrmRF Nn (ii)
0))(cos(;0 20 =−=Σ ααθ ImrrmgM (iii)
For the bar, I = mL2/12from eqn (iii)
( )12cos
)()cos(
222 Lrgr
mrIrmg
+=
+= θθα
substituting, r = 0.25 m; L=1m θα cos8.16=
2srad4.8218.16,º60at =×== αθ
To find ω at º60=θ
θωωθ
θωωα
dd
dtd
dd
dtd ===
θω
θθωωθω
sin8.162
cos8.16
200
=
= ∫∫ dd
θω
θω
sin6.33
sin6.332
=
=
at º60=θ srad39.5866.06.33 =×=ω To find reactions at 0
N56)4.8)(25.0(20)5.0(81.920
cos
=−×=
−= αθ mrmgRt
Figure 19.4(a)
Figure 19.4(b)
Figure 19.4(c)
8 Engineering Mechanics
N5.157
)39.5(25.020)866.0(81.920
sin2
2
=××+××=
+= ωθ mrmgRn
The magnitude of total reaction
l tangentia thetoº9.79;56
5.157tan
N160)5.157()56( 22
===
=+=
φφt
nRR
R
or 79.9 + 30 = 109.9º to the horizontal.Example 19.5 A solid spherical ball of mass m and radius r isprojected horizontally on a rough floor as shown in Fig. 19.16. The ballcontacts the floor with a velocity v0 and no angular velocity. Thecoefficient of friction between the floor and the ball is µ . Determine (a)the time and distance at which the ball will start rolling without slidingand (b) final linear and angular velocities of the ball.
Figure 19.5Solution Consider free body diagram shown in Fig. 19.16c
;0
;0;0
=−=Σ
=−=Σ−==+=Σ
αµ
µµ
IRrM
mgRFgaamRF
y
xr
For spherical base, 2)5/2( mrI =
rg
mrmgr
mrRr
25
25
25
22
µα
µµα
=
==
The linear and angular velocity may be obtained as
rgtt
gtvvatvv
25;
;
00
00
µωωαωω
µ
+=+=
−=+=
Kinetics of Rigid Bodies 9
From the above equations it is seen that linear velocity of the balldecreases continuously with time and angular velocity increases continu-ously. The velocity of the point of contact of the ball with the floor be-comes zero.
0=− rv ω
Denoting time 02
5)(; 1101 =
−−= t
rgrgtvtt µµ
gv
t
tggv
µ
µµ
72
25
01
10
=
+=
At this instant onwards, the ball starts rolling without sliding and thevalue of F is less Rµ . We can calculate velocity at that instant
rv
gv
rgt
rg
vg
vgvv
75
72
25
25
)7/5(72
00101
00
01
=
=+=
=
−=
µµµωω
µµ
Thus after time ),)(72( 01 gvt µ= the ball rolls with uniform linear andangular velocities. It can be seen that v and ω are independent of µ andt1 depends on µ . If µ is very large (i.e., ground is very rough, is verysmall.)
Example 19.6 The extremities of 2.4 m rod of mass 50 kg move freelyand with no friction along two straight tracks as shown in Fig. 19.17. If therod is released with no velocity from the position shown, determine (a)angular acceleration of the rod and (b) the reactions at A and B.
Figur 19.6(a) Figur 19.6(b)
Solution Since the motion is constrained, the acceleration of mass cen-tre C must be related to angular accelerationα . To obtain the relation, wemust first determine aA at the point A in terms of α. Assumeα is in theanticlockwise direction.
10 Engineering Mechanics
α4.2B/A =ar
Draw the vector diagram of accelerations.
αα
αα
990.230sin45sin
4.2
278.345sin
105sin4.2
30sin45sin105sin
B
A
BB/AA
=×=
=×=
==
a
a
aaa
r
r
rrr
The acceleration of C is now obtained by writing
]2.1[278.3C/AA
αα +→=+==
aaaaa c
r
rrrr
↓+==
→=−=−=
ααααα
αα
04.160sin2.1678.26.0278.3
60cos2.1278.3
y
x
a
a
r
r
Figure 19.6 Equivalent static forces
αααα
5204.1509.133678.250
244.21250
21 2222
=×==×=
=×==
y
x
mama
kgmkgmmlI
writing equations of equilibrium09.133707.0;045cos BB =−=−=Σ αRmaRF xx (i)
052490707.0
049045sin
AB
AB
=++−
=++−=Σ
RR
RmaRF yy
α (ii)
Figure 19.6(c)
60
Kinetics of Rigid Bodies 11
166.1932.4366.509
02468.29.13304.15204.1490
===−×
−×−×=Σ
αα
αααEM
(iii)
Substituting in eqn. (i)
N83.220707.0
9.133B == αR
Substituting in eqn. (ii)N24.273;0)166.152(490)83.220707.0( AA ==+×+−× RR
Example 19.7 The pendulum of an impact testing machine hinged at Oas shown in Fig 19.24 is allowed to drop from the horizontal position. Themass of the pendulum is 50 kg and that centre of mass is at a distance of 0.9from O. The radius of gyration about O is 1.1m and the hinge has a fric-tional resisting couple of 20 Nm. Determine the angular velocity the pen-dulum and linear velocity of the mass.
Solution When is reaches the vertical position
Figure 19.7
At position 1: v1 = 0; W1 = 0 Hence T1 = 0
22
22
22
222
22
2222
25.3025
)1.1(502150
21
21
21:2
ω
ω
ω
+=
×+×=
+=
v
v
mkmvTpositionAt
12 Engineering Mechanics
Assume datum is passing through position 2, then45.4419.081.9509.01 =××=×= gmV
The reactions do not do work.Work done by frictional couple MR is
22
2222
22
22
222111
21
81.0;9.0
25.3025)2/(5.4410
)2/(
ωω
ωπ
πθ
=×=
+=−+
+=++−=−=
−
−
vv
vM
VTUVTMMU
R
RR
m/s5641.29.0rad/s849.2
5.501.410
)25.3081.025(2/205.441
22
22
22
22
===
=
+×=×−
ωω
ω
ωπ
vExample 19.8 A road roller of mass = 1200 kg and radius 0.25 m ispulled with a force of P = 1500 N on a rough ground. If the roller starts fromrest and rolls without slipping, determine the distance travelled by thecentre of the roller C at which roller acquires a velocity of 3 m/s.
Figure 19.8
Solution Starting position is at rest. Therefore 011 == vω at the in-stant when v2= 3 m/s since there is not slipping
rad/s12)25.0/3(/22
=== rvω
5.3725.0120021
21cylinder 22 =××== mrI
81001445.372131200
21 2
2 =××+×==T
Kinetics of Rigid Bodies 13
m4.5810015002211
=⇒=×==+ −
xxPTUT x
Example 19.9 A high speed tester for leaf springs shown in Fig 19.26has a falling mass mA = 10 kg which has a downward speed of 5 m/secwhen it contacts the undeformed spring. Assume spring constant k = 400kN/m. Determine (a) the speed of A when the spring deflection is 0.01 m (b)the reactions at support B and C at that instant. The supports can freelymove horizontally. Neglect friction and mass of the spring.
Figure 19.9
Solution 2211 VTVT +=+
AA2
A22AA
21AA 2
1)(210)(
21 gYmkYvmvm −+=+
01.081.910)01.0(100040021)(10
212510
21 22
2A ××−×××+×=×× v
m/sec6039.4)(9810.1059810.020125)(5
9810.020)(5125
2A
22A
22A
==+−=
−+=
vv
v
Energy method cannot be used to find the reactions. Hence writing thedynamic equations of equilibrium,
0spsp =−−+=Σ amPRRFy
P causes deflection in the spring
kN2kN41001.0400 3
A
==××==
RkYP
Example 19.10 A sphere, a cylinder and a hoop, each having the samemass and radius are released from rest on an incline. Determine the veloc-
14 Engineering Mechanics
ity of each body after it has rolled through a distance corresponding to achange elevation h.
Solution The instantaneous centre of rotation is located at O
( ) 22
22
222
1
21
21
21
21
210
vrIm
rvIvm
IvmT
T
r
rr
rr
+=
+=
+=
=
ω
Figure 19.10
Frictional force does not do work. Hence
( )
( )
22
22
22
2211
21
22
221
mrIghv
mgWrIm
Whv
vrImWh
TUTWhU
+=
=+
=
+=
=+=
−
−
r
r
r
Sphere: ghvmrI 2845.0;21 2 ==
Cylinder: ghvmrI 2816.0;21 2 ==
Kinetics of Rigid Bodies 15
Hoop: ghvmrI 2707.0;2 ==
when a frictionless block slides to the same distance ghv 2,0 ==ω
222
2
2 rkmrmk
mrI ==
A hoop which has largest k attains smallest velocity. While the slidingblock which does not rotate attains largest velocity.Example 19.11 Each ofthe slender rods shown in Fig19.28 is 1.5m long and has amass of 12 kg. If the system isreleased from rest when
º60=β , determine (a) angu-lar velocity of the rod ABwhen º20=β . (b) the veloc-ity of the point D at that in-stant.Solution
Figure 19.11
When º20=β . Since vB is perpendicular toAB, vD is horizontal and the instantaneous cen-tre of rotation is at C.
BC = 1.5 m, CD = 1.026 m.Denoting ω as angle velocity of AB
ω5.1B =V
ED = 0.75; CD = 1.026 ; EDC = 70°, EC = 1.044
Figure 19.11(a)
Figure 19.11(b)
16 Engineering Mechanics
Velocity at mass centre of .75.0AB AB ω== v BD seems to rotate atpoint C.
ωωωωωωω
)044.1()(;5.1;5.1)BC(
BDBD
BDBDBD
=====
ECvPosition 1: Mass of each rod = 12 kg
restat is system thesince 0
03.15365.072.11722VN72.11781.912
1
11
==××==
=×=
TJWy
W
Position 2:
JWyV 39.604026.172.11722 22 =××==
To find T2 at 2,
[ ] ( )2BD
2ABAB
2BD
2AB2
222BDAB
21)(2
1
kgm25.25.112121
121
ωω +++=
=××===
IvvmT
mlII
2
2
22
222
414.14
5.4914.9
))(25.221
)044.1()75.0(1221
ω
ωω
ωωω
=
+=
×++×=
T
By conservation of energy T1 + V1 = T2 + V2
m/s.6.2026.1rad/s535.2
64.92414.14
39.60414.1403.153
D
BDAB
2
2
==∴===
=
+=
ωωωω
ω
ω
v
Example 19.12 A road roller of mass = 1200 kg and radius 0.25 m ispulled with a force of P = 1500 N on a rough ground. If the roller starts fromrest and rolls without slipping, determine the distance travelled by thecentre of the roller C at which roller acquires a velocity of 3 m/s.Solution The free body diagram of the roller is shown in Fig 19.25.Applying the principle of impulse and momentum,
∫Σ=−2
1
12 )()(t
txxx dtFGG
Figure 19.11(c)
Kinetics of Rigid Bodies 17
∫=−
−=−2
1
12 )()(
)1500()03(1200t
tccc dtMHH
tF
0;rad/s12)25.0/3(
5.37)25.0(120021
212
22
===
=××==
ωω
mrIc
sec6.35400180036003600150015003600
180025.0450
25.0)012(5.37
==⇒+=+=−=
==
××=−
tFttFtt
Ft
tF
833.06.3
312
==
+=
a
tvv ω
m4.5
)6.3(833.021
21 travelleddistance
2
21
=
××=
+=
x
attvx
Example 19.13 Gear A has a mass of 20 kg and radius of gyration of200 mm while gear B has a mass of 6 kg with radius of gyration 80 mm. Thesystem is at rest when a couple of magnitude M = 12 Nm applied to gear B.Neglecting friction, determine (a) the time required for the angular velocityof gear to reach 600 rev/min and (b) the tangential force which gear Bexerts on gear A.
Figure 19.12
18 Engineering Mechanics
Solution
rad/s1.2525010083.62
rad/s83.62602600
kgm0384.008.06
kgm8.0)02.0(20
A2
B2
222B
222A
=×=
=×=
=×==
=×==
ω
πω
mrI
mrI
Figure 19.12(b)For gear A 2AAA1AA )(ωω ItFrI =+
Ns32.8025.0
1.258.01.258.025.0
=×=
×=×
Ft
tF
For gear B 2BBBB1B )(ωω ItFrMtI =++
Figure 19.12(c)
s87.044.10032.883.620384.012
==+×=
tt
Thus force exerted by gear A or B N32.9287.032.80 ==
Example 19.14 A new pole of street light is designed to shear off by arelatively small force when an automobile crashes on it. For modellingsuch an event assume that car is a uniform rectangular plate of 1200 kgmass travelling at v0 = 100j km/hr as shown in Fig. 19.36 prior to the impact.
Kinetics of Rigid Bodies 19
Assume that the force applied on the car by the pole is kN5 jF −= act-ing for a time of 0.1 second. Determine the motion of the car after impact.
Figure 19.13
Solution Initial momentum of the car
5001.05000
7.2712001200
m/s7.273600
1000100Velocity
11
1
−=×−=
×=×=
=×==
Fdt
vmv
v
yy
y
Final momentum 22 1200 yy vmv =
kmh.208.98m/s28.274166.07.27
12005007.271200
2
2
21
==−=
×=−×
=+ ∫
y
y
yy
v
v
mvFdtmv
Angular momentum ∫ =∆+ 21 ωω ccc ItMI
rad/s25.0
2000)42)(1200(121
11.05000 22
==
=+=××
f
ff
ω
ωω
Example 19.15 An elevator system is modelled as m1 = 2000 kg, m2 =1800 kg and a uniform disk of m3 = 200 kg. Calculate the velocity of m1 attime t =10s after the system starts from rest.
20 Engineering Mechanics
Figure 19.14(a)
Solution Linear momentum Assume m1 moves with a velocity of v1
11 200010)2000( vgT =−− (1)Now, m2 moves with a velocity of v1
12 180010)1800( vgT =− (2)Angular momentum
( ) 12112221
212
2212
212
12
10;10)(10)(100
1002
2002
10)(110)(1
11
vTTrvTTTT
mrI
TTrTTdtM
=−==−−=
===
−=×−=
+=
ωωω
ωω
ωω &
Adding (1) and (2) we get,
m/sec.03.5390
81.920020039038020010380200)(
1
1
11
112
=×=
==+−=+−
v
gvvgvvgTT
Figure 19.14(b)
Figure 19.14(c)
Figure 19.14(c)
Kinetics of Rigid Bodies 21
EXERCISES
Use D’Alembert’s principle to solve problems 1 to 12:1. A uniform rod of mass
‘W’ as shown in Fig. E19.1 supported by a pinconnection at A and a wireat B. What is the force ofpin A at the instant whenthe wire is released. Also determine the force at A when the rodhas rotated to 45°.
2. A stepped cylinder shown in Fig.E 19.2 having a mass of 100 kg and aradius of gyration k of 0.3 m. Theradii R1 and R2 respectively are 0.3 mand 0.6 m. A pull T equals to 200 N isexerted on the rope attached to theinner cylinder. The coefficients ofstatic and dynamic friction betweencylinder and ground are 0.1 and 0.08respectively. Find the linear and angular acceleration.
3. A rigid rod AB slides against a frictionless wall and floor as shownin Fig. E 19.3. The rod has a weight w and is made to move to theright in the plane of the paper by a force P as shown at B. What isthe angular acceleration of the rod if the initial inclination of therod is 0θ .
Figure E 19.3
Figure E 19.1
Figure E 19.2
22 Engineering Mechanics
4. A semi cylinder A shown in Fig. E 19.4 whosediameter is 0.3 m and the mass is 50 kg. What isthe angular acceleration of A at the positionshown if at this instant it has angular velocityof 2 rad/s.
5. A 2000 kg car is travelling down the hill. Whenthe brakes are applied, locks all the wheels. Cal-culate the normal and tangential force on the wheels 05=kµ as-suming symmetry between left and right wheels.
Figure E 19.56. The structure on the seismic simulator is modelled on a two dimen-
sional rigid body of weightless members and four concentratedmasses m. The hinge support at A is accelerated to the right at therate of ax.The leg at B is on frictionless surface. Determine theforces acting on legs A and B (see Fig E 19.6).
Figure E 19.6
Figure E 19.4
Kinetics of Rigid Bodies 23
7. A motor cycle v and its rider have a total mass of 200 kg. Determinethe maximum safe speed and the required angle θ of leaning onhorizontal curved road to prevent slipping or tipping (θ º is zero).
Figure E 19.7
8. The base structure of a rotating restaurant is modelled into threeuniform disks that rotate as a unit about vertical axis y. The mass mis 8000 kg and the diameter d is 7 m. Calculate the required momentto stop the disks in a time of 10 seconds from an angular speed of0.1 rpm (see Fig. E 19.8).
Figure E 19.8
9. The toggle mechanism OAB moves to the horizontal xy-plane.Each uniform bar has a mass of 12 kg and the block B has a mass of6 kg. Calculate x and y components of the force acting on pin Bfor °= 30θ F = 700 N. Friction is negligible. The system is releasedfrom rest at .30°=θ
Figure E 19.9
10. A conveyor belt starts from rest with an acceleration of 1 m/s2.Determine the angular acceleration of 2 kg sphere and 3 kg cylin-der. For both bodies, 4.0=sµ and .3.0=kµ
24 Engineering Mechanics
Figure E 19.10
11. A uniform rod AB of mass 15 kg and length 0.9 m is attached to 20kg Cart C. Neglecting friction, determine immediately after the sys-tem has been released from rest, (a) the acceleration of the cart (b)the angular acceleration of the rod (see Fig. E 19.11)
Figure E 19.11
12. Two slender uniform bars AB and BCeach of length L and mass m are pinnedtogether at B as shown in Fig. E 19.12. Ifboth the bars are simultaneously releasedat the position shown determine (a) theinitial angular accelerations of the bars and (b) the reaction atsupport A.Use energy method to solve problems 13-19
13. A solid cylinder of mass m and radius r rolls from rest down a planeinclined at an angle b to the horizontal. Determine the velocity ofcentre of cylinder after it has rolled down a distanced.
14. The bar B of mass 6 kg and length 1 m is pinned to a fixed supportO at one end and to a disc D of mass 12 kg and radius 0.2 m on theother end. (see Fig. E 19.13). The disc rolls without slipping on thecurved surface s. If the bar has an angular velocity of 2 rad/s whenit is in the horizontal position, determine the angular velocity ofthe bar when it is in vertical position.
Figure E 19.12
Kinetics of Rigid Bodies 25
Figure E 19.13
15. A 400 kg flywheel of a punching machine has a radius of gyrationof 750 mm. Each punching operation requires 2700 Joules of work.(a)knowing that a speed of flywheel 200 rev/min before punching,determine the speed immediately after punching. (b) If a constant35 N-m couple is applied to the shaft of the flywheel determine thenumber of revolutions extended before speed is again 200 rev/min.
16. A 3 kg slender rod shown in Fig. E 19.14 rotates in a vertical planeabout a point B. A spring constant k = 300 N/m and of unstretchedlength 120 mm is attached to the rod as shown, knowing that in thepositions shown the rod, has an angular velocity of 4 rad/s, clock-wise, determine the angular velocity of the rod after it has rotatesthrough (a) 90° (b) 180°.
Figure E 19.14
17. Neglecting the mass of the cable in Fig. E 19.15 find the speed of 50kg mass after it has moved to 1.6 m along the incline from theposition of rest. The coefficient of friction along the incline is 0.3.
26 Engineering Mechanics
Figure E 19.15
18. A proposed conveyor system is mod-elled as a plate of mass M moving with-out slipping on small cylindrical roll-ers of mass m and radius r. Determinethe speed of the plate after it hadmoved to a distance l. Starting fromrest along an inclined plane. Consider(a) free rollers (b) rollers freely rotat-ing about their fixed axes. Assume plate ison three rollers at all times as it moves to adistance of ‘l’ (see Fig. E 19.16).
19. In Fig. E 19.17 the linkage system rests on africtionless plane. The segment AB, BF, etceach of length 0.6m and the bars all of thesame stock, have a mass of 7.5 kg/m. Aforce F of 500 N is applied to D. What is thespeed of D after it moves 0.3 m. The systemis stretching at the configuration shown.Use impulse and momentum method to solveproblem 20 – 26.
20. A box 3 m by 4 m slides along a horizontalfloor with a velocity of 4 m/s and strikes asmall step (see Fig. E 19.18) (a) Determinethe angular velocity of the box after the im-
Figure E 19.16
Figure E 19.17
Kinetics of Rigid Bodies 27
pact. (b) will the box tip over? (c) determine the percentage loss ofenergy in the impact.
Figure E 19.18
21. A thin walled pipe of mass m and radius R is released from rest onan inclined plane making an angle of b with horizontal. Using theprinciple of angular momentum impulse, determine the velocity ofthe centre of the pipe after time t.
22. A 30 kg boy Amitesh, is in a sitting position on a swing when heis released from rest at position A at a height of 1.75 m above theground. (see Fig. E 19.19). The centre of mass of the boy is at adistance of 4 m from the support of swing. When the swing reachesthe lowest position B, the boy instantaneously stands up and hiscentre of mass is raised by 0.75 m. Determine the length ‘L’ of thecentre of mass from the ground when the swing reaches extremeposition ‘C.’
Figure E 19.19
28 Engineering Mechanics
23. Consider a rigid body initially at rest and sub-jected to an impulsive force F contained in theplane of the slab. We define centre of percus-sion P as the point of intersection of the line ofaction of F with the perpendicular drawn from‘G’. (a) show that the instantaneous centre ofrotation c of the slab is located on the line GPat a distance GC = k2/CP on the opposite side of G. (b) show thatif the centre of percussion were located at C the instantaneouscentre of rotation would be located at p (see Fig. E 19.20).
24. A computer tape moves over the two drums shown in Figure E19.21. Drum A has a mass of 0.6 kg and a radius of gyration of 19mm, which drum B has C mass of 1.6 kg and radius of gyration of31mm. In the lower portion of the tape the tension is constant andis equal to TA = 3.8 N. Knowing that tape is internally at rest deter-mine (a) The required tension constant TB if the velocity of thetape is to be V = 3 m/s after 0.24 s (b) the corresponding tension inthe portion of the tape between the drums.
Figure E 19.21
25. A disc of radius R and mass M is rotatingwith a speed of 0ω rad/s as shown in Fig.E 19.22. Another disc of radius r and massm which is at rest is dropped over thedisc.(a) Prove that trial angular velocity of
the disc
220
2
mrMRMR
+ωω
Figure E 19.20
Figure E 19.22
Kinetics of Rigid Bodies 29
(b) Show that in this problem though momentum is conserved,the energy is not.
(c) Show that if discs are identical R = r; M = m half the energy ofthe system is dissipated.
26. A tractor having a mass (with the driver) 1500 kg as shown in Fig.E 19.23. If a torque of 330 Nm is developed on the drive wheels bythe motor, what is the speed of the tractor after it moves to 3.3 m.The large drive wheels each has a mass of 100 kg and dia is 1m andradius of gyration of 0.6m. The small wheels each has a mass of 20kg and has a diameter of 0.3 with a radius of gyration of 0.24m.
Figure E 19.23
ANSWERS
1. Considering dynamic equlibrium of the ωd.
0givenat2323
1312
2
2
==
=
=
=
tLg
wLgwL
gwLI
wLI
A
A
θ
θ
θ
&&
&&
&&
at any θ
θθ
θθ
cos2331cos
22
Cg
t
gWCwL
=
=
&&
&&
Figure A 19.1
Figure A 19.1(a)
30 Engineering Mechanics
To find tθ&
θθθθ
θθθθ
θθθθ
dd
Lg
dd
dtd
dd
dtd
&&
&&
&&&&
=
=⋅==
cos23
θθ
θθ
θθθθ
θθ
θ
sin3
2sin
23
cos23
20
2
0
0
g
cg
ddcg
=
=
= ∫∫
&
&
&&
&
for θ = 45Lg
L12.238707.02 =×=θ&
cg
cg
cgan
0605.12121.2
)707.0(23/
==
=θ&&
Since total mass is concentrate on centred
068.1
707.02
3
sin32
2
53.0
)0605.1(2
2
2
22
1
1
=
×=
=
=
=
=
=
a
gCgC
Ca
gacgC
Ca
θ
θ
θ
&
&&
Considering free body
045cos45cos 21 =++ agwa
gwAx
Figure A 19.1(b)
Figure A 19.1(c)
Kinetics of Rigid Bodies 31
W125.1)59.1(707.0
)06.153.0(707.0
−=×−=
+×−=
x
x
Aw
ggwA
W375.1W125.1
W375.1
)77.006.177.053.01(
045cos45cos 21
=−=
=
×+×−=
=−+−
y
x
y
y
y
AA
A
wA
agwa
gwwA
2. Given m = 100 kgrk = 0.3 mR1 = 0.3R2 = 0.6T = 200 Nµs = 0.1µ∆ = 0.08
Assume there is no slipping on the contact surface rotation as-sume 0.
( )
45)36.009.0(100)6.03.0(100
0
22
22
2
220
=+=+=
+=
+=
I
Rrm
mRII
n
g
Taking initial at 0
33.14560
45)3.0(200
45)( 12
==
=
=−
θθ
θ
θ
&&&&
&&
&&RRT
Now employ Neton’s Law
2702007020033.16.0100
6.01002002
=+=+××+=
×−=+−
−=+−
fff
mRfT
θ
θ&&
&&
Thus for no slipping we must be able to develop friction for 270 N.
Figure A 19.2
32 Engineering Mechanics
The mex friction force their we can have is according to coulumblaw
N2.981.0100force
=××== gsg µµ
using µa = 0.08 use
2
212
rad/sec33.1
1296098
09.01003.2006.080
−=
−==−
×=×−×
−=−
θ
θθ
θµ
&&
&&
&&
&&nTRRf
Now equlibrium
2sec/m2.1
12080200100
−=
−=+−==+−
x
xxfT
&&
&&
&&µ
Thus role cylinder has a linear acceleration of 1.2 m/sec2 and angu-lar acceleration of 1.33 raw/sec2 in the clockwise direction. This isvalued for all time.
2
2
sec/row33.1
sec/m2.1
=
−=
θ
α&&
&&
3. Taking
θµθθθρ &&12
sin2
cos2
sin2
2llFlFlAB =+− (1)
we have three θ&&,, AB FFncA aFx µρ =−=∑ & (2)ycB awFy µ=−=∑ (3)
To find axc and ayc we use the related0)( =×+×+= BBcBcBc aaa ρωωρω&
+××+
+×=
jlilhk
iljlkac
sin2
cos2
cos2
sin2
θθθ
θθθ
&&
&&
−×+
+−=+
iljlk
jliljaia cycx
θθθθ
θθθ
sin2
cos2
cos2
sin2
)(
&&
&&
Kinetics of Rigid Bodies 33
→+=
↑−=
←−=
θθθθ
θθθθ
θθθθ
cos2
sin2
or
sin2
cos2
cos2
sin2
2
2
2
&&&
&&&
&&&
lla
lla
lla
cx
cy
cx
+=− θθθθρ cos
2sin
2
2&&& llMFA
( )θθθθρ cossin2
or 2&&& +=−g
WlFA (1)
( )θθθθ sincos2
2&&& −=−g
WlWFB (2)
θθθθρ &&g
WlFF AB 6sincossin =+=− (3)
Solving )cossin2(23 θθρθ W
wlg −=&&
Substituting θ = --- we get internal acceleration
)cossin2(23
000 θθθ WpWlg −=&&
4. w = 2 rad/secm = 50 kgdA = 0.3 m
2
52 Rµ=
32083
320)45128(
649
52
22
22
mRMR
RmmRIcg
=−=
−=
I above D 22
85
32083
+= RmmR
320
208320
)12583( 22 mRmRI =+=
X distance from D )30sin(8
330sin θ+−= RR
R))30(sin(375.05.0( θ+−=
Figure A 19.3
Figure A 19.3(a)
34 Engineering Mechanics
θ
θ
&&2
320208
))30sin(375.05.0(
RgW
RW
=
+−
+−=
6sin375.05.0
208320 θφθ
Rg&&
∫∫ =
==
θθθθ
θθθθ
θθθ
dd
dd
dtd
dd
&&&
&&
&&&
2
26cos375.0
6)(cos375.032475.05.0
208320
6)(cos375.05.0
208320
26)(sin375.05.0
208320
2
0
2
=
=Γ−
+Γ+−
+Γ+
=
+Γ−
∫
∫
∫
θ
θφ
θφθ
θφθ
θθθφ
θ
&
&
&
RgRg
dRg
01987.081.93202
3.04166
)(cos375.032475.05.0
26
)(cos375.032475.05.0208320
=××
×=+Γ+−
=
+Γ+−∫θφθ
θφθRg
for θ very small0.50 = 0.01987θ = 0.03974 rad/sec
substitute in θ&&
[ ]2secrad/70.1
)54346sin(375.05.015.020881.9320
=
−−××=
θ
θ
&&
&&
5. Writing equilibrium equations15cos81.9200015cos ××==+=∑ WNNv BA
46.01895=+ BA NN (1)
15sinWmaFFH BA +=+=∑ (2)
Figure A 19.3(b)
Kinetics of Rigid Bodies 35
Figure A 19.4
2sec/m1988.2
02.50782000)46.18951(5.015sin81.920002000)(
=
+=×+=+
a
aaNN BAµ
Taking at A
3.82124.0)947402(19.28427
4.0)15sin81.92000198.22000(5.115cos81.920003
035.115cos4.0)15sin(
=×−=
×+×−×××=
=+×−+
B
F
B
W
NNWWma
NA = 18951.46 – 8212.3 =10738.7FA = µ NA = 0.5 × NA = 5369.35FB = µ NB = 0.5 × NB = 4106.15NA = 10738.7 NNB = 8212.3 NFA = 5369.35 NFB = 4106.15 N
6. xA amH23
=
Taking at B
hma
V
hmaV
lVhma
lVhma
ham
ham
xB
xA
Ax
Axx
x
25
25
25
034
224
3
=
−=
−=
=×+++
Figure A 19.5
36 Engineering Mechanics
→=
↑=
↓=
xA
xB
xA
maH
hma
V
hmaV
23
25
25
7. normal acceleration r
v2=
r
mvF2
=
N196281.9200
450
200 22
=×=
==
W
vNFx
Taking at A
θ
θθ
tan1sin8.0cos8.0
=×=×
FxW
FxW
θtan4
19622 =
vassume µ is the of friction µ = 0.5
º43.6366.154
1962tan
hkm3.56m/sec66.154
9819814
N98119625.0
2
2
2
==×
=
==
=
=
=×==
θθ
µ
v
v
v
WFx
8.
Figure A 19.7
Figure A 19.6
Kinetics of Rigid Bodies 37
0rad/sec01047.0
6021.0rpm1.0
29}2781{
8
42)3(3
42)4(2
2422
222
==
×==
=++=
×+
×+
××=
f
i
w
mdmd
dmdmdmJ
πω
2
22
rad/sec001047.0
1001047.00
2
−=∞
∞+=
∞+=
∞==
tww
sww
if
if
Required moment of stop the disk
001047.02
4980009
)001047.0(2
9 2
×××=
==mdJα
Required moment = 1846.908 w stop in disk
9.
Figure A 19.8
mass of OA = mass of AB = 12 rpm mB = 6 kg
WOA = WAB = 12×9.81 = 117.72 N
JWyV 6.58230sin
172.11722 11 =××==
due to F JF 30021
70021
=×=×=
position as θ = 20from kinematics θωsin=AV
38 Engineering Mechanics
θωθω
sin2cos
=−=
B
A
VV
V2 = 2 wy2 = 2 × 117.72 × 0.171 = 40.26 Jdue to force F = 700 × 0342 = 239.4 JVc = w oc = 0.5 w; VD = 0.5 w
( ) 222222 2
1)(
21
1221
BBD vmwwJVVT ++++×=
1112121
121 2 =××== lmJ
22222
2 )684.0(621)(1
21
446 wwwwwV ××++×+
+=
=5.403 w2
T1 + V1 = T2 + V2408.6 = 5.403 w2 + 279.66 w = 9.885 rad/sec VB = 200 sin θ
= 9.770 × sin 20 = 3.3415 m/sec10. acceleration of sphere = 2 kg
mass of cylinder = 3 kg
101.062.19
0024.004.0321
21
002.005.0252
52
22
22
==
=××==
=××==
f
mI
mI
cy
s
µ
µ
µ
2 × 0.05 = 0.002 ∞cy
2raw/sec50002.0
05.02 =×=cyαCylinder
101.043.29
3
rad/sec500024.0
04.03
0024.004.03
2
==
=×=
=×
µ
α
α
cy
cy
angular acceleration of sphere and cylinder = 50 rad/sec2
Figure A 19.9
Figure A 19.9(a)
Kinetics of Rigid Bodies 39
11. mAB = 15 kglAB = 0.9
05.49.05
9.0153131
2
2
2
=×=
××=
= mlJ A
1cos34.16
sin34.16
sin21.6605.4
sin45.081.915
0sin45.0
C
J
mgJ
A
A
+−=
=
=
××=
=×−
θθ
θθθθ
θθ
θα
&
&&
&&
&&
rad/sec90.625sin34.16
sin34.16
80.14cos1634
80.14906.34.160
025When
1
1
==
=
+−=
=+×−=
==
θθθ
θθ
θθ
&&
&&
&
&
CC
2sec/m11.220
25cos9.645.01502025cos45.0
−=
+=××
−
=+×
a
a
am θ&&
acceleration of the cart = 2.11 m/sec2
angular acceleration of the rod = 6.90 rad/sec2
12.
Figure A 19.11
Figure A 19.10
Figure A 19.10(a)
40 Engineering Mechanics
Taking at A
02
322
33
523 1
21
2=
−−+
+ LmgLmgLmLLmLmL θ
θθ &&
&&&&
022
365
311
022
365
3
1
2
212
1
2
22
1
2
=−++
=−++
mgLmLmL
mgLmLmLmL
θθθ
θθθ
&&&&&&
&&&&&&
Lgmg 1262522 21 =×=+ θθ &&&& (1)
Taking at B
222 1
2
2
2 LmgmLmL =+ θθ &&&&
Lg
323 21 =+ θθ &&&& (2)
Lg
LgLg
929
151015
241044
1
21
21
=
=+
=+
θ
θθ
θθ
&&
&&&&
&&&&
Lg
Lg
Lg
Lg
2960
2927872;32
2927
22 =−==+ θθ &&&&
LgLg
2960299
2
1
=
=
θ
θ
&&
&&
25829
58871162
5887
25860
5827
2960
2299
23
223
22
21
211
mL
mLmLmg
mLmgmg
LgmL
LgmL
mLmL
mLmLmR LA
−=−=
−=−=
−+=
+
=
−+=
−++=
θθ
θθθ
&&&&
&&&&&&
Kinetics of Rigid Bodies 41
2
2960299
2
1
mLR
LgLg
A −=
=
=
θ
θ
&&
&&
13.
Figure A 19.12
The instanteneous centre of rotation constant at 0.T1 = 0
22
2
22
222
21
2121cylinderfor
21
locityangular vevelocity21
21
vmmT
mrI
vrIm
v
IvmT
+=
=
+=
==
+=
ω
ω
22 4
3 vmT =
ghv
ghv
ghm
mghv
vmmgh
TvTmghwhU
153.1
33.1
34
3443
2
2
2
2211
21
=
=
==
=
=+==
−
−
βtan153.1 dgv =
42 Engineering Mechanics
14. J of the bar
24.0204.012
2discofJ
21
126
122
2
=×=
==
===
d
B
Jmr
mLJ
If θ the angle by which moves curved discroll moved 1.2 θ
angular velocity of disc θθω &&
62.0
2.1 ==d
angular velocity of bar ωω == bPosition 1
{ }
56.5456.3614424.02
)26(24.0421
181181)126(
11
2
221
1
=+=×+=
×+×=
+=
=×=+=
vT
JJT
gggV
dDbB ωω
Position 2 V2 = 0
rad/sec2.46rad/sec467.2
56.5514.9
14.964.82
)6(24.021
2
222
222
===
=+=
+=
ωω
ω
ωωω
ωωT
15. Mass of flywheel = 400 kgradius of gyration ru = 0.75 m
2112
21
1
211
22
2700
rad/sec94.2060
220021
5.1122
75.04002
−
−
+==
=×=
=
=×==
UTTU
JT
mrJ u
πω
ω
Figure A 19.13
Kinetics of Rigid Bodies 43
rad/sec22
70.273645.11221
270094.205.11221
2
22
2
=
=×
+××=
ω
ω
no. of revolution 2294.20
200 ×=
a) spoud immorability after punching = 210 revol/min
135
352425.11221
35205.11221225.112
21
21
22
233322
=
=×××
+××=××
−==+ −
θ
θ
θ
µθUTUT
of revelution revol48.212135 ==
π16. mrod = 3 kg
{ } 2925.0)225.0(046875.0
)225.0(
)375.06.0()(
1275.03)(
2
22
2
2
2
2
2
2
=+=
+=
−+=
×==
m
ml
mL
ml
mLJ
lmLJ
BAB
cgAB
Length m39.015.036.0 22 =+=unstrectched length = 0. 12 mks = 300 N/m
39.25940.10162925.021
21
36.081.93of935.10
)12.039.0(30021)(
21
21
1
221
==××==
××==
−×==
θ&JT
V
rksV
1
111
870.23594.10275.13
EVTE
=+=+=
Figure A 19.14
44 Engineering Mechanics
If it rotates 90V2 for = 3 × 9.81(0.375 + 0.21) = 17.21
V2 spry 275.1)09.0(30021 2 =×=
22
22 2
1215.121.17
21 θθ && JEJT ++==
rad/sec4
º180
rad/sec13.6
25.37
445.5425.1887.232925.021
87.23292521425.18
2
2
2
=
=
=
=
=−=×
=×+
θθ
θθ
θ
θ
&
&
&
&
&
&
angular velocity at θ = 90º is 6.13 rad/secangular velocity at θ = 180º is 4 rad/sec
17. i = Initial positionf = final positionµif = 0.3 × 50 cos 30 × 1.6 = 20.784T1 = 0 kinetic energy at1
45.02
3.010;3.0
2150
21
222
222
=×==
+×=
Jv
JvT
θ
θ
&
&
Total length of cable = c + 2l
202
ciic&
&−
==+
velcocity of pulley is 22vv −=−=
3.0v=ω
2
222
2
75.28
1021
09.045.0
2150
21
v
yvvvT
=
××+
×+×=
784.2075.28 2
12
=
+=
v
UTT if
Figure A 19.14(a)
Figure A 19.15
Kinetics of Rigid Bodies 45
m/sec85.075.28
784.202
=
=
v
v
final speed of 50 kg mass = 0.85 m/sec18. Angle of inclined plane = β
no friction freely rotatingMass of the plate = Mmass of roller = m
2
22
22
2
2222
1
21
23
21
213
21
213
210
)sin(
vmm
rrvmvV
rv
mrMvV
VlMgU
+=
+=
=
+=
==−
µ
ω
ω
β
+=
+=
+=
=+−
3mMMlgsinv
mMMlgv
vmMlmg
vvU
β
β
β
2
3sin2
23
21)sin(
2
2
2121
19. U1–2 = F × 0.3= 500 × 0.3
U1–2 = 150 N.mT1 = 0
22
22
222
2.736.020
6.05.738
318
θθ
θ
θ
&&
&
&
=×=
××=
××= lmT
Figure A 19.16
46 Engineering Mechanics
rad/sec56.42.7
15
2.7150 2
==
=
vθ
θ
&
&
θθ
θ&& )sin(4.2
cos6.04
−=
×=
F
F
y
y
m/sec73.7
56.4)707.0(4.2º45
=×−=
=
Fy&θ
Spud of D = 7.37 m/sec20. It will be assumed that
the edge of the step Utips like on the corner.The height of the step isassumed to be egligiblecompared to dimensionsof the block during im-pact only force which ex-acts c moment at 0.5 mg.The angular momentumfirst before impact
mmmvvm 842224 =×==×=
The velocity of the centre of mass after impact v=
angular velocity ==rv
momentum after impact
mmw
mmdbmJ
8325
325)916(
3)(
322
=
=+=+=
2524) =ωa
resting = mg × 1.5 = m (14.715)b) The box by overc) % loss of energy
2
22
21
21
21
mv
Jwmv
EE −
=∆
Figure A 19.17
Kinetics of Rigid Bodies 47
%52or52.08
2524
6258
2
=
−
=m
mm
%52)
)2524)
==
=
cb
a
21.
Figure A 19.18J of pipe = mR2 = mR2
Initial angular momentum H1 = 0final angular momentum H2 = θ&J
θ&2mR=Moment at centre = (mg sin β – mg m cos β) R
Rmgdtm )cos(sin βµβ −=∫∫ =+ 21 HmdtH
)cos(sin
)cos(sin 2
βµβθ
θβµβ
−=
=−
Rgt
mRtmgR
&
&
22.
Figure A 19.19
48 Engineering Mechanics
V1 = 30 × 9.81 × 1.75 = 515.025T1 = 0
{ }
22
2
2
3021
81.930)50.1cos4(81.93025.3)75.1cos4(
vT
VV
×=
×−=×−+=
θθ
)cos2.177(475.95615
025.515)5.1cos4(3.2943021
2
2
1122
θ
θ
−=
=−+
+−+
v
v
vTvT
Assume 60=θ15v2 = 956.475 – 588.6v = 4.95 m/sec
position above the ground c2125.375.12 ×−+=
assume θ = 60º = 3.75 – 1.625 = 2.125 m23. α = angular acceleration
mrea = forceg = groundc = Instentaneous centre of rotationH1 = 0 α = constant accelerationFinal H2 = Ic ω = Ic α t
tmRtItcpmdtm crc ∞=∞=∞=∫ 2)(R = rad/sec of gyration at c
cpRr
Rcpr
tmRtcpmr
c
c
c
2
2
2
)(
)(
=
=
∞=∞
Similerly if centre of percussion located at c Ic located at ‘c’.24. JA = 0.6 (0.019)2
= 2.166 × 10–4
JB = 1.6 (0.031)2
=1.5376 × 10–3
B rotates clockwiseA rotates anticlockwiseH1 = 0
Figure A 19.20
Kinetics of Rigid Bodies 49
H2 = JB ωB
36.136022.0333
9.78038.033105376.1038.0)(
038.0)(
3
====
===
×=−
×−=
−
∫
AAA
BB
BB
rr
r
tTT
tTTBmdt
ω
ω
ω
(TB – T) 0.038 × 0.24 = 1.376 × 10–3 × 78.9 (1)(T – TA) × 0.022 × 0.24 = 2.166 × 10–4 × 136.36 (2)
N39.98.359.559.5
N59.524.022.0
36.136410166.2)(
=+=+=
=×
×−×=−
TAT
TAT
Substitute is eqn. 1
N39.9N69.22N69.22
39.93.1330.13
24.0038.09.783105376.1)39.9(
9.783105376.124.0038.0)39.9(
===
+==
××−×=−
×−×=×−
TT
T
T
T
B
B
B
B
25. Angular momentum before impact 01 ωθ JJ == &
0
2
1 2ωmRH =
When the smaller disc impact in larger disc no moment
2
22
2 22ω
+= mrmRH
Momentum is centered
a)
)(
22)(
22
2
02
0
2
2
22
mrmRmR
mRmrmR
+=
=+
ωω
ωω
Figure A 19.20(b)
Figure A 19.20(a)
50 Engineering Mechanics
b)
)costantnot isEnergy (
)(21
21
21
21
22
222
20
22011
TT
mrmRT
mRJT
≠
+=
==
ω
ωω
c)
442
21
2
2
202
202
2
20
2
1
021
ωω
ω
ωω
mRmRT
mRT
Rrmm
=×=
=
===
Hence ==− 121 21 TTT half the energy is dissipated
26. T = 330 Nmdistance moved = 3.3 mlarge wheel each mass = 100 kg
rg = 0.6 IB = 100(0.6)2 = 36
These are two wheel = 72 rB = 0.5 m
small wheel each mass = 20 kg rg = 0.24
Is = 20(0.24)2 = 1.152rs = 0.15 m
Two wheels = 2 × 1.152 = 2.304acceleration of = a; velocity = v
acceleration of big wheel vVaaB 2;2
15.0===
acceleration of small wheel vvaas 667.6667.6
15.0===
H2 = 72 × vB + 2.304 vs + 1500 × v = 330 t72 aB + 2.304 as + 1500 a = 33072 × 2a + 2.304 × 6.667 a + 1500 a = 3301659.36 a = 330
a = 0.19887 m/sec2
v2 = 2 × 0.19887 × s = 2 × 0.19887 × 3.3 v = 1.145 m/sec
speed of tractor = 1.145 m/sec