@ @

@ @

@pa‹šb«@ÙîãbØî¾a@ñ†bß@ @

Mechanics - Dynamics @ @@òÜy‹¾a@òjÜİÛ¶ëþa@@O@ïaŠ‡Ûa@Ý—ÐÛaïãbrÛa@ @

2007@@O2008@ @

@ @

†a‡Çg@ @

@‘‡äè¾a@Zñ†ìÇ@‹‚bÏ@ãìí@ @ñ†b¾a@‘Š‡ß@ @

@ @

]1[

Dynamics Dynamics is that branch of mechanics which deals with the motion of particles , lines and bodies. So , the dynamics is divided into two parts : 1 – Kinematics : which deals with the motion of the bodies without consideration of the forces

required to produce the motion including the position , speed , velocity , acceleration … etc . 2 – Kinetics : which deals with the motion of the bodies with consideration of the forces required

to produce the motion . Motion in a straight line : We define some point in space as a reference point or origin. For a particle at P: In general : Displacement ( S ) as (vector) unit : m

dtdsVVelocity =)( unit : m/sec

2

2

)(dt

sddtdvaonaccelerati == unit : m/sec2

Example ( 1 ) : The position ( S ) of a body moving along a horizontal strait line is given by the equation : ( S = 6t2 - 4 ) , where ( S ) in ( m ) , ( t ) in ( sec ) , the body is in ( 4 m ) to the right of the origin when t = 0 . Determine : a – the displacement of the body during the time interval from t = 2 sec to t = 4 sec . b – the velocity and acceleration of the body when t = 4 sec . Solution : a - S1 = 6 t2 – 4 = 6 ( 2 ) 2 – 4 = 20 m , S2 = 6 t2 – 4 = 6 ( 4 ) 2 – 4 = 92 m q = S2 – S1 = 92 – 20 = 72 m .

b - dtdsV = = 12 t = 12 * 4 = 48 m / sec ,

dtdva = = 12 m / sec 2 .

Example ( 2 ) : A body moves on a straight line according to the equation : S = 4 t 3 + 2 t 2 – 2 t + 5 , where (s) is displacement , ( t ) is the time interval , Find out displacement , velocity , and acceleration when t = 3 sec . Solution : S = 4 t 3 + 2 t 2 – 2 t + 5 1- The displacement at t= 3 sec : S = 4 ( 3 ) 3 + 2 ( 3 ) 2 - 2 ( 3 ) + 5 = 125 m

2- The velocity at any instant : dtdsV = = 12 t 2 + 4 t - 2

The velocity when t = 3 sec : V = 12 ( 3 ) 2 + 4 ( 3 ) – 2 = 118 m / sec

3- The acceleration at any instant : 2

2

dtSd

dtdva == = 24 t + 4

The acceleration when t = 3 sec : a = 24 ( 3 ) + 4 = 76 m / sec 2

O

P

(O is the reference point)

]2[

Example ( 3 ) : The rectilinear motion of a body is given by the equation ( S = V 2 - 9 ) , When ( S ) is in ( m ) and ( V ) in ( m / sec ) , Determine : the acceleration of the body ? Solution :

2

2

sec / m 21

2VVa . 2 ....... ,

0 2dtds ....... 9

==⇒===

−=−=

aVVdtdVa

dtdSV

dtdVVVS

Example ( 4 ) : The motion of a particle is defined by the relation x= t2 – 10 t + 30 , where x is expressed in metres and t in seconds , Determine

a- The time when velocity is zero b- The position and the total distance traveled when t = 8 seconds

Solution :

ondsttt

tdtdxv

ttx

sec 51020102

102

30102

=⇒=⇒=−

−==

+−=

a-

m 14308*1083010

2

2

=+−=

+−= ttx

Example ( 5 ) :

A line rotates in a vertical plane according to the law 2 2 23 −−= ttθ , where (θ ) gives the angular position of the line in radians , (t) is the time in seconds, The line is turning clockwise when t=1 sec , Determine A- The angular acceleration when t = 3 sec . B- The value of (t) when the angular velocity is zero C- the total angle turned through during the time t=1 sec to t=3 sec. Solution :

( ) 2

2

23

rad/sec 1443 6: sec 3on when tacceleratiangular The

instantany at 4 6

4 3

2 2

=−=

=

−==

−==

−−=

α

ωα

θω

θ

tdtd

ttdtd

tt

( )sec3.13

443043

04304343

2

2

==⇒=⇒=−

=−⇒=−

−=

ttt

ttttttω

( ) ( )

( ) ( )( ) rad 1037Q

rad 72323

sec1 rad 32121

sec1

12

232

2

231

1

=−−=−==−−=

=−=−−=

=

θθθ

θθ

θ

tat

tat

]3[

Example ( 6 ) : The rectilinear motion of the a body is given by the equation ( S = r . sin wt ) , if ( r ) and ( w ) are constants , Show that the acceleration is equal to ( a = - w 2 . S ) ? Solution :

S = r . sin wt , dtdsV = = r . cos wt . w = r . w . cos wt

dtdva = = r . w . ( - sin wt . w ) = - w2 . r . sin wt = - w2 . S

Example ( 7 ) : A circular body rotates according to the equation ( S = 2 t 2 – 4 t ) , Find the radius of the body to give an angular acceleration of ( 4 rad / sec ) ? Solution : S = 2 t 2 – 4 t V = dS / dt = 4 t – 4 a = dV / dt = 4 a = r . α 4 = r * 4 …… r = 1 m Falling Objects and Gravitational Acceleration : Galileo showed that if an object falls toward the earth under the influence of gravity and all other forces are negligible then …… 1. The acceleration is the same for all objects. 2. The acceleration has a constant value. This ‘gravitational’ acceleration has a value of 9.81 m/sec2 at the Earth’s surface. Motion at constant acceleration : Equations of motion at constant acceleration : For motion in the x direction:- A particle starts from O with a velocity Vo : (S = 0 at t = 0 and V = Vo at t = 0 )

2 21 . tatVoS += ………….. ( 1 )

( Vo ) is the initial velocity of the body at t = 0

Differentiating (1) with respect to time…. taVdtdsV Of +== ………. ( 2 )

( Vf ) is the final velocity of the body

Rearranging (2) ….. aVVt Of −

= Substituting for ( t ) in (1) we get :

S a 2V 22O +=fV ……………… ( 3 )

These three are equations of motion at constant acceleration. Corresponding variation of a) position, b) velocity and c) acceleration with time :

vacuum In air

]4[

The following table explain the three equations of motion with constant acceleration according to the type of a motion :

Rectilinear motion ( horizontal motion )

Vertical motion Circular motion ( Angular motion )

Upward motion Downward motion

atVofV += gtVofV −= gtVofV += t . αωω += of

2

2

1 . tatVS o += 2

2

1 . tgtVh o −= 2

2

1 . tgtVh o += 2

2

1 . tto αωθ +=

S 222 aVoVf += h g 222 −=VoVf h g 2 22 +=VoVf θαωω 22 2 += of

Vf = final velocity ( m / sec ) , Vo = initial velocity ( m / sec ) , a = linear acceleration ( m / sec2 ) S= distance ( displacement ) ( m ) , g = gravitational acceleration ( m / sec2 ) , ω f = final angular velocity ( rad / sec ) , ω o = initial angular velocity ( rad / sec ) , α = angular acceleration ( rad / sec 2 ) , θ = angular distance ( angular displacement ) ( rad ) S = r . θ , V = r . ω , a = r . α Example ( 8 ) : A body is fall down from ( 5 m ) high , In what time does reach the earth ? Solution :

h g 222 +=VoVf = 0 – 2 * 10 * 5 = 100 = 10 m / sec

gtVofV += 10 = 0 + 10 * t ....................... t = 1 sec Example ( 9 ) : A stone is thrown vertically into the air from a tower ( 100 m ) high , at the same instant that a second stone is thrown upward from the ground . The initial velocity of the first stone is ( 50 m / sec) and that of the second stone is ( 75 m / sec ) , When and Where will the two stones be at the same height from the ground ? Solution :

2

2

1 . tatVS o +=

h 1 = 50 * t - ½ * 10 * t 2 = 50 t - 5 t 2 ……… ( 1 ) h 2 = 75 * t - ½ * 10 * t 2 = 75 t - 5 t 2 ……… ( 2 ) ± h 1 = - 50 t ± 5 t 2 h 2 = + 75 t - 5 t 2 ------------------------------- adding h 2 - h 1 = 25 t Then , h 2 - h 1 = 100 m

∴ 100 = 25 t → t = 4 sec h 1 = 50 * 4 – 5 ( 4 ) 2 = 200 – 80 = - 120 m h 2 = 75 * 4 – 5 ( 4 ) 2 = 300 – 80 = 220 m

]5[

Example ( 10 ) : A stone is thrown vertically upward returns to the earth during (5 sec) , How high does it go? Solution :

2

2

1 . . tgh =

h 1 = 0.5 * 10 * t12 = 5 t1

2 h 2 = 0.5 * 10 * ( 5 - t1 ) 2 = 5 ( 25 – 10 t1 + t1

2 ) = 125 - 50t1 + 5t12

h 1 = h 2

5t12

= 125 – 50 t1 + 5 t12

5t12 – 125 + 50 t1 - 5t1

2 = 0 → 50 t1 = 125 → t1 = 2.5 sec h = ½ * 10 * ( 2.5 ) 2 = 31.25 m Example ( 11 ) : A vehicle starts motion from rest , then accelerated uniformly until its velocity reaches 20 m/sec after 30 sec . Find :

a. The acceleration of vehicle b. The displacement of the vehicle during the period c. The displacement of the vehicle during the last seconds from its motion

Solution a-

2sec6030020

sec30 sec20 0

m/.aa*atVV

t m/V,Vof

fo

=

+=+=

===

b-

m)(.*a.t.tVx o 2703060210

21 22 =+=+=

c-

m..xx

m.)*(.*a tVo.tx

ondth ng the hicle duriment of vee displaceWe find th

7173252270

32522960210

21

sec29

2930

22

=−=−

=+=+=

( )

( )

m.

*.*

ta.tVx

lutionAnother so

o

717

130260210

1221

30

=

−+=

−+=

]6[

Example ( 12 ) : A body moving according to the relation x = 20 + 4 t2

a- Determine the displacement of the body during the interval t1 = 2 sec and t2 = 5 sec b- Find average velocity during this interval c- Find the velocity at t= 2 sec

Solution : a-

m 8436120 12010020)5(420

364*420)2(420420

12

22

21

2

=−=−=∆=+=+=

=+=+=

+=

xxxmx

mxtx

b-

m/sec 283

842536120

12

12 ==−−

=−−

=ttxxv

c- ( )

sec1628080

420 2

m/*t

dttd

dtdxv

=+=+=

+==

Example ( 13 ) : A body moving with constant acceleration 4 m/sec2 , the position of the body is x = 5 m at t = 0 with initial velocity 3 m/sec .

a- Determine the position and the velocity after 2 sec b- Where the body maybe in position when its velocity 5 m/sec

Solution : a-

( )

sec/112*43.

192*4*212*35

.21.

2

20

mtaVoVf

m

tatVoxx

=+=+=

=++=

++=

b- ( )( )

( )

mx

xx

xxaVoVf

7856

5892554*235

222

022

==

−+=−+=

−+=

]7[

Example ( 14 ) : A stone is dropped down a well ( ر and ( 5 sec ) later , the sound of splash is heard , if the , ( بئvelocity of the sound is ( 332 m / sec ) , what the depth of the well ? Solution :

t 332 t . v ==⇒= hthV ……… ( 1 )

.صوت الحجر ) سماع ( هو زمن وصول ) t( حيث 2

2

1 . tgtVh o −= = 0 * t +

21 * 10 * ( 5 – t ) 2

زمن صعود الصوت الى خارج البئر + هو زمن وصول الحجر الى قاع البئر ) sec 5( اذ ان الزمن

h = 21 * 10 ( 5 – t ) 2 = 5 ( 25 – 10 t + t 2 ) = 125 – 50 t + 5 t 2 ……….. ( 2 )

:وط الحجر وصعود الصوت ألن عمق البئر هو نفسه في حالتي سق) 2( مع المعادلة ) 1( بمساواة المعادلة 332 t = 125 – 50 t + 5 t 2 5 t 2 – 282 t + 125 = 0

:المعادلة بأستخدام طريقة الدستور لحل

10227282

5*2125*5*42)282()282( ±=

−±−−=t

أن قيمة ( + ) في حالة اعتماد االشارة الموجبة ي t = 55.95 sec : هي ) t( ف زمن الكل اورد في السؤال اذ ان ال ذا خالف م وه ) t = 0.44 sec (ساوي يان الزمن حيث ، وعليه يجب اعتماد االشارة السالبة ، ) sec 5( لنزول الحجر وصعود الصوت يساوي

h = 332 * 0.44 = 148.3 m :سيكون ) h( اذن عمق البئر Example ( 15 ) : A car with ( 90 Km / hr ) velocity is suddenly stopped by breaks , Find the time interval to stop the car after ( 50 m ) distance ? Solution :

asVoVhrKmV

f 2

sec / m 2536900

60*601000*9090

22 +=

====

( 25 ) 2 = 0 + 2 * a * 50 …. → a = 625 / 100 = 6.25 m / sec2 2 t. a .

21 t . Vo S +=

50 = 0 * t + ½ * 6.25 t 2

50 = 3.125 t 2 …….. → t 2 = 16 …….→ t = 4 sec Example ( 16 ) : A sphere weighs ( 49 N ) , Joined with a wire and rotated in a horizontal plane , if the length of wire is ( 1 m ) with ( 30 r . p . m ) , Find the tension in the wire ? Solution :

Nrg

WT 3.481*2)(*1049*2*

sec / rad 602*30

===

==

πω

ππω

]8[

Example ( 17 ) : The 10 m rod in fig. moves with its ends in contact with ( x ) and ( y ) axes , the rod has an angular velocity of 5 rad / sec clockwise , and an angular acceleration of 8 rad / sec2 counterclockwise . Determine the velocity of ( G ) when the rod is in this position . ( θ = 60 ) Solution : X = 6 cos θ Y = 4 sin θ

( ) ( )

sec/2660sin5*6sin..6

sin6*sin6cos6cos6cos6

mwdtd

dtd

dtd

dtd

dtd

dtdxVx

−=−=−=

−=−=====

θ

θθθθθθθ

( ) sec/ 1060cos*5*4cos..4cos4sin4sin4 mw

dtd

dtd

dtd

dtdyVy ======= θθθθθ

( ) ( ) sec/ 9.271026 2222 mVyVxV =+=+= Example ( 18 ) : A point moves along the curve y2 = 16 x ( x and y in metres ) in such a manner that the y coordinate of its position at any time is y = t2 - 4t , where t is in seconds , Determine the velocity of the point when t = 5 sec . Solution :

( )

( ) ( ) sec/ 75.783*25235

41

sec5

tan223

41

23*214*

161

21

161

1616

168

161

16816 , 416 , 4 , 16

23

23

23

234234

2342222

mV

t tvelocity aThen, the

tat any ins velocity t thettV

tttdtdxV

ttttttx

tttxttxttyxy

=+−=

=

+−=

+−==

+−=+−=

+−=−=−==

]9[

Newton’s Laws (Empirical laws governing motion) : Newton’s First Law : Every body continues in its state of rest or of motion at a constant velocity unless acted on by an unbalanced force (F ) . i.e. if F = 0 the a = 0 ( Force is something which changes the state of motion of a body ) The total force F can be the sum of several forces Newton’s Second Law : If an unbalanced force ( F ) acts on a body it produces an acceleration ( a ) where : F = m . a F = resultant force , m = mass of the body , a = linear acceleration Mass : defined by this law is a measure of a body’s resistance to motion (inertia) and is called the inertial mass. At a fixed place on the earth’s surface: F = W = m . g ( W ) is called the weight of a body and ( g ) is the gravitational acceleration at the earth’s surface. Newton’s Third Law : “To every action there is an equal and opposite

reaction”. FAB = F BA Example ( 19 ) : What forces act on a 2 kg tin of chocolates on a table? If an Engineering student decides to test the strength of the tin and pushes down on the lid with a force of 40N, what changes will occur in the forces acting on the box of chocolates? Solution : The reaction force R increases to 138 N , The weight of the box of chocolates remains the same. Example ( 20 ) : Determine the weight of the body (A) to give the body (B) of ( 20 N ) weight a downward acceleration of 0.5 m/sec2 ? Solution : For the body (A) : F = m . a

(1) ........ *1010

*10

11

11

aaT

AA

AA

WWT

WW

=−

=−

For the body (B) : F = m . a

N 191201205.0*220

*220

*102020

222

2

22

22

=⇒=−⇒=−=−=−

=−

TTTTT

T

aa

For the pulley : T2 = 2 T1 → 19 = 2 T1 → T1 = 9.5 N a2 = 0.5a1 a1=1 Subst. In (1) :

AW = 8.63 N

]10[

Example ( 21 ) : Determine the acceleration of each block , and the tension in the cord , if the fixed drum is smooth ? Solution : For the first block (10 N ) weight : F = m . a

) 1 ( ....... 10

. 101010

aT

aT

=−

=−

For the second block ( 40 N ) weight : F = m . a

) 2 ( ....... 440

. 104040

aT

aT

=−

=−

Subst. (1) in (2) : 40 – T = 4 ( T – 10 ) → …. 40 – T = 4 T – 40 → ….. T = 16 N Subst. In (1) : 16 – 10 = a ……… a = 6 m / sec2 Example ( 22 ) : Block ( A) weighs ( 8 N ) , Block ( B) weighs ( 16 N ) , The horizontal slot is smooth , the length of the slot is ( 7.2 m ) , Determine : a – the tension in the cord . b – the acceleration of each block . c – the required time for the block ( B ) to complete its motion

along the slot . ( g = 10 m / sec2 ) Solution : a – For the block ( A ) : F = m . a

T – 8 = 108 . a ……→ 10 T – 80 = 8 . a ------- ( 1 )

For the block ( B ) : F = m . a

10 - 54 T =

1016 . a ……→ 500 – 40 T = 80 a -------- ( 2 )

From ( 2 ) : 80T 40500 −

=a -------- ( 3 )

Subst. ( 3 ) in ( 1 ) : 10 T – 80 = 8 ⎟⎠⎞

⎜⎝⎛ −

80 40500 T

100 T – 800 = 500 – 40 T ….. → T = 9.2 N

b- Subst. In ( 3 ) : 2sec / m 6.180

2.9*40500=

−=a

c- 2

21 tat.VS o +=

S= the length of slot

7.2 = 0 * t + ½ ( 1.6 ) t2 ……→t = 3 sec

]11[

Example ( 23 ) : Determine the weight of the body ( A) to give the block ( 100 N ) an acceleration of ( 0.5 m / sec2 ) , µ = 0.1 ? Solution : For the block ( 100 N ) : Ff = µ . N = 0.1 * 100 cos 60 = 5 N F = m . a T – 100 sin 60 - Ff = ( 100 / 10 ) . a T – 91.6 = 10 . a T – 91.6 = 10 . * 0.5 T = 96.6 N For the block ( A ) : F = m . a WA – T = (WA / 10 ) . a WA – 96.6 = (WA / 10 ) * 0.5 10 WA - 960.6 = 0.5 * WA 10 WA - 0.5 WA = 960.6 9.5 WA = 960.6 WA = 960.6 / 9.5 = 193.2 N Example ( 24 ) : A man wants to slide the homogeneous ( 100 N ) box shown in fig . across the floor by pushing on it with the force ( P ) , the coefficient of friction between the box and the floor is ( 0.2 ) . Determine the force ( P ) to give the box an acceleration of ( 8 m / sec2 ) ? Solution :

∑ Fy = 0 → N – 100 = 0 → N = 100 Ff = µ . N = 0.2 * 100 = 20 N F = m . a P - Ff = ( 100 / 10 ) * 8 P – 20 = 10 * 8 P = 80 + 20 = 100 N

60 o

T

]12[

Problems : 1- A particle moves along a plane curve having the parametric equations x = 2t , y = 2t2 – 4 ,

where the rectangular coordinates x and y are measured in feet and t is the time in seconds. Determine (a) the displacement of the particle during the time interval from t = 0 to t = 2 sec; (b) the velocity of the particle when t = 2 sec .

2- A point moves along the curve y2 = 36x ( x and y in feet ) in such a manner that the y coordinate of its position at any time is y=6t2 . Determine the acceleration of the point when t =2 sec .

3- The velocity of a point moving in the xy plane is given by the equations Vx = 4t – 1 and Vx =2. If the point is at ( 3 , 4 ) measured in feet , when t =1 sec , Determine the equation of the path in terms of x and y .

4- Block A in Fig . weighs 20 lb and the coefficient of friction between A and the plane is 0.30 . The initial velocity of A is 5.0 fps to the right, and during the next 3 sec the block is displaced 30 ft to thr right. Determine the weight of block B.

5- In Fig. , the 2.00-lb A is connected by a cord to the 12.00-lb block B . The coefficient of friction between A and the plane is 0.20 , and that between B and the plane is 0.30 . The blocks are moving to the right with an acceleration of 6.00 fps2 to the right . Determine the force P .

6- A man wants to slide the homogeneous 100-lb box in Fig . (a) across the floor by pushing on it with a force P as indicated. The coefficient of friction between the box and the floor is 0.20. Can an acceleration of 8.0 fps2 to the right to produced as indicated without tipping the box ? If so , determine magitude of the force P .

]13[

WORK , POWER & ENERGY WORK : The work done by a constant force ( F ) during a displacement ( S ) of the force is defind as the product of the magnitude of the force , the magnitude of the displacement , and the cosine of the angle between the force and the displacement , that is : W : work done , F : Applied force , S : displacement , W = F . S . cos θ POWER : The rate of work done by the body per time . P : Power , F : Applied force , S : displacement , V : velocity

P = VF

tSF

tW . .

==

ENERGY The ability to do the work . Divided into two parts : 1 – Potential Energy ( P.E ) : The energy which stored in or losed from the body . P.E = W . h = m . g . h W : the weight of the body , h : the high of the body 2 – Kinetic Energy (K.E ) : The ability of the body to do the work due to its velocity . K.E = ½ . m . V 2 m= mass of the body , V = velocity Example ( 1 ) : A body of ( 50 N ) weight , falls down from ( 500 m ) hight , Show that the potential energy is changing to kinetic energy ? Solution : P.E = m.g.h = 50 X 10 X 500 = 250 000 J V f 2 = 2 . g . h = 2 X 10 X 50 = 10 000 K.E = ½ . m . V f 2 = ½ X 50 X 10 000 = 250 000 J YES , the potential energy is changing to kinetic energy . Example ( 2 ) : Find the work done by a body of ( 5 N ) weight which is falled down from ( 10 m ) hight ? Solution : Work done = P.E = m . g . h = ( 5/10) X 10 X 10 = 50 J Example ( 3 ) : Determine the total energy of ( 10 tonnes ) airoplane mass when it is flying with ( 15 m/sec ) and of ( 2500 m ) high? Solution : P.E = m . g . h = 10 X 1000 X 10 X 2500 = 250 000 000 J K.E = ½ . m . V f 2 = ½ X 10 000 X ( 15 ) 2 = 1125000 J E T = P.E + K.E = 250 000 000 + 1125 000 = 26125 000 J Problems : 1 - Each of the three balls has a mass mand is welded to rigid angular

frame of negligible mass. The assembly rests on a smooth horizontal surface. If a force Fis applied to one bar as shown, determine (a) the acceleration of the point O and the (b) the angular acceleration of the frame.

2 - The figure shows three forces applied to a 0.46 kg crate that

moves leftward by 3.00 m over a frictionless floor. The forces magnitudes are F1 =5 N, F2 =4.5 N, and F3 =9 N. a) Find the work done on the crate by each of the three forces. b) Find the change in the crate’s kinetic energy. c) If the crate is stationary after 3.00 m, find the crate’s initial velocity.