1

Vidyalanakar F.Y. Diploma : Sem. II

[AE/CE/CH/CR/CS/CV/EE/EP/FE/ME/MH/MI/PG/PT/PS] Engineering Mechanics

Time : 3 Hrs.] Prelim Question Paper Solution [Marks : 100

Q.1 Attempt any TEN of the following : [20]Q.1(a) Define funicular polygon. [2](A) Funicular Polygon : The polygon is so constructed by drawing the lines in the respective spaces of space diagram;

parallel to the rays of polar diagram by maintaining the order is called as ‘funicular polygon’. Q.1(b) State any two advantages and any two disadvantages of friction. [2](A) Advantages of friction (1) Man cause can easily walk on horizontal surface due to friction. (2) There is safe movement of vehicles by applying breaks. (3) One can hammer the nail into wall due to friction. Q.1(c) State Newton’s Laws of motion? (First, Second & Third) [2](A) Newton first law : “ Everybody continues to be in its state of rest or of uniform motion in a

straight line, unless it is acted upon by some external agency”. Second Law : “The rate of change of momentum is directly proportional to applied forces”. Third Law : To every action there is equal and opposite reaction.

Q.1(d) What is efficiency of a machine? [2](A) Efficiency () : The efficiency of a machine is the ratio of output to the input of a machine and

is generally expressed as a percentage.

% = Output

100input

Q.1(e) What is Bow's notation ? Explain with a sketch. [2](A) Bow’s notation Bow’s notation is used designate a force as per this notation, each

force is designated or named by two spaces one on each side of the line of action of a force. This space are generally named by capital letter’s as A, B, C serially.

Explanation A force say ‘F’ acting on rigid body divided space above or below it

into two parts, say A and B hence the force ‘F’ is named as AB.

Q.1(f) Define Resolution of force [2](A) Resolution of force :

The way of representing a single force into number of forces without changing the effect of force on body is called resolution of force.

Q.1(g) State Parallelogram Law of forced with neat sketch. [2](A) Parallelogram Law of forces :

“If two forces acting at and away from the point be represented in magnitude and direction by the two adjacent sides of a parallelogram, then the diagonal of the parallelogram passing through the point of intersection of two forces, represents the resultant in magnitude and direction.”

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Vidyalankar : F.Y. Diploma Mechanics

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Q.1(h) State principle of transmissibility of force. [2](A) Principle of transmissibility of force : If force acts at a point on a rigid body, it is assumed to act at any other point on line of action of

force within the body. Q.1(i) Define Lami’s theorem. [2](A) Lami’s theorem states that, if three forces acting at a point

on a body keep it at rest, then each force is proportional to the sine of the angle between the other two forces.

As per Lami’s theorem,

31 2 FF F

sin sin sin

Q.1(j) State any two types of beams with diagram of each. [2](A) Following are the different types of beams –

(1) Simply supported beam

(2) Cantilever beam

(3) Over hanging beam

(4) Fixed Beam

(5) Continuous beam

Q.1(k) Explain meaning of self locking machine. State the condition for it. [2](A) SelfLocking Machine A machine which is not capable of doing work in reverse direction even on removal of effort,

then the machine is called as selflocking or Nonreversible machine. Condition for SelfLocking Machine Efficiency < 50 % < 50 %

B

O

C

D

Q sin

R

P

Q Q

Q cos

A

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Q.1(l) Define cone of friction. [2](A) Cone of friction It is an imaginary cone generated by revolving resultant reaction about the normal reaction,

when body is in limiting equilibrium condition. OR

When body placed on horizontal surface is in limiting equilibrium by a horizontal force P, then the resultant reaction makes an angle with normal reaction.

If we gradually change the angle of force P through 360, then imaginary cone is generated as resultant reaction revolves around normal reactions. This imaginary cone is called cone of friction.

Q.2 Attempt any FOUR of the following : [16]Q.2(a) Resolve the force 19 MN along 22 and 32° on either side of it. [4](A) Resolve the force 19 MN along 22 and 32 on either side of it. Let 1 = 22 and 2 = 32.

Referring figure,

F1 = 2

1 2

Fsin

sin

= 19sin32

sin 22 32

F1 = 12.44 MN

F2 = 1

1 2

Fsin

sin

= 19sin 22

sin 22 32

F2 = 8.79 MN. Q.2(b) In a simple axle and wheel, the diameter of wheel is 180 mm and that of axle 30 mm. If

the efficiency of the machine is 80%, find the effort required to lift a load of 100 N. [4]

(A) Given: D = diameter of wheel = 180 mm, d = diameter of axle = 30 mm = efficiency of the machine = 80% W = load = 100 N To find : V.R: ?

Formula: V.R = D

d

Solution:

V.R. = D

d=

180

30= 6

Effort (p) required to lift a load of 100N:

= M.A

100V.R.

= W

100P V.R.

= W.A

100V.R.

MA =W

P

80 = 100

100P 6

P = 100 100

80 6

= 20.83 N

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Vidyalankar : F.Y. Diploma Mechanics

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Q.2(c) The diameters of bigger and smaller pulleys of Weston's differential pulley block are 250 mm and 100 mm respectively. Determine the effort required to lift a load of 3 kN with 80% efficiency.

[4]

(A) Given data : D = 250 mm W = 3 kN = 3000 N d = 100 mm % = 80%

(i) V.R. = 2D

D d =

2 250

250 100´

= 3.33

(ii) To find effort ‘D’

% = M.A.

V.R. 100

% = W / P

V.R. 100

80 = 3000 / P

3.33 100

80 = 3000

3.33 P 100

P = 3000 100

3.33 80

´´

P = 11.26.12 N

Q.2(d) The resultant of two forces is 8KN and its direction is inclined at 60 to one of force

whose magnitude is 4KN . Find magnitude and direction of other force. [4]

(A) F1 = 4 KN F = 8 KN = 60 F2 = ? = ?

We know that,

F1 = Fsin

sin

4 = 8 sin

sin 60

4 sin(60+) = 8 sin 4(sin60 cos + cos60sin) = 8 sin 3464 cos + 2sin = 8 sin 3464 cos = 6 sin

sin

cos

=3.464

6

tan = 0.58 = 30

also,

F2 = F sin

sin (

= 8 sin 60

sin 60 30

= 8 0.866

sin 90

F2 = 6.93 KN

F = 8KN

F2

F1 = 4KN = 60

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Q.2(e) A screw jack has an effort wheel diameter of 20 cm and pitch is 5 mm. Find velocity ratio. If a load of 1000 N is lifted by an effort of 150 N. Find the efficiency of the machine.

[4]

(A) (1) VR of simple screw jack is given by –

VR = D

p

VR = x 200

5

VR = 125.66

(2) MA = W 1000

6.66P 150

(3) % = M.A. 6.66

x100 x 100V.R. 125.66

= 5.31%

Q.2(f) Write the different types of force system. [4](A) Classification of force system : Based on Line of Action, Force system may be classified as following :

(i) Collinear forces System : The forces acting in same line of

action is called collinear forces. A collinear system is necessarily coplanar

Weight W and tension T are in same line of action. (ii) Concurrent forces : The system in which all the forces act at

same point is called as concurrent forces. A concurrent force system may be either coplanar or non-coplanar provided that there are more than two forces.

All the forces F1, F2, F3, F4, F5, F6 are meeting at point ‘O’. (iii)Nonconcurrent forces : The system in which the forces act

at different points is called nonconcurrent forces. A non-concurrent system may be either coplanar or non-coplanar

In the diagram F1, F2, F3, F4 are acting at different point. Parallel forces The system of in which line of action are parallel to each other are called

as parallel forces. A parallel force system may be either coplanar or non-coplanar.

(i) Like parallel forces : Like forces acting in same direction are called as like parallel forces. (ii) Unlike parallel forces :

Parallel forces acting in opposite direction are called as unlike parallel forces.

W

T

F6

‘O’

F1

F2

F3

F4

F5

C

A B

D

F1

F2

F3

F4

P Q R S

F1

F2

F3

F4

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Vidyalankar : F.Y. Diploma Mechanics

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Q.3. Attempt any FOUR of the following : [16]Q.3(a) Four forces 20N, 15N, 30N, & 25N are acting at 0, 60, 90 & 150 from xaxis taken

in order. Find resultant by analytical method. [4]

(A) Fx = 20 + 15cos 60 25 cos 30 `FX = 5.85 N Fy = 15 sin60 + 30 + 25 sin30 Fy = 55.49 N

Resultant force is given by,

R = 2 2x yF F

= 2 25.85 55.49

R = 55.79 N Direction:

= 1 X

y

Ftan

F

= 1 55 49tan

5.85

= 83.98 with horizontal. Q.3(b) Determine the magnitude of resultant and position of it wrt point A for the force

system shown in Figure. Solve it graphically.

Fig.

[4]

(A)

Resultant (R) = (ae) scale

= 2 100 = 200 kN x = 5.5 1 = 5.5 m Position (x) = 5.5 m

15 cos60

25 sin30

30N

20N

15 sin60 15 N

25N

60 30 25 cos30

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Q.3(c) Calculate the moment about point ‘B’ for the force system as shown in Figure. [4]

(A) Taking moment @ point B MB = (15 x 0) + (10 x 3) (20 x 2) + (30 x 3) + (40 x 2) = 0 + 30 40 + 90 + 80 = +160 N- m ( ) = 160 N-m (Clockwise moment) Q.3(d) Four forces 6N, 4N, 2N and 1N are acting along the sides of a square AB, BC, CD

and AD respectively. Find the resultant in magnitude and direction only. [4]

(A) Fx = 6 2 = 4N Fy = 1 4 = 5N

R = 22 FyFx

= 224 5 = 6.4N

= 1 Fytan

Fx

= tan1 5/4 = 51.34

Q.3(e) Calculate the magnitude and direction of resultant for concurrent force system as

shown in Figure. Use analytical method. [4]

(A) (1) Resolving all forces ∑Fx = +(50 cos 30) (70 cos 45) + (100 cos 180) + (60 cos 70) = + 43.30 49.50 100 + 20.52 = - 85.68 ∑Fy = +(50 sin 30) + (70 sin 45) + (100 sin 180) – (60 sin 70) = + 25 + 49.50 + 0 – 56.38 = + 18.12 N (2) Magnitude of Resultant

R = 2 2Fx Fy

R = 2 2( 85.68) (18.12)

R = 87.58N (3) Direction and position of resultant As ∑ Fx is –ve and ∑ Fy is +ve, resultant lies in 2nd quadrant.

θ = tan11Fy 18.12

tanFx 85.68

θ = 11.94º

A B

D C

6N

4N 1N

2N

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Vidyalankar : F.Y. Diploma Mechanics

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Q.3(f) ABCD is a rectangle such that AB = 3m and BC =2 m. Along sides AB, CB CD and AD, the forces of 100 kN, 200 kN, 250 kN and 150 kN are acting respectively. Find the magnitude, direction and position of the resultant of the forces from C. Use analytical method only.

[4]

(A) Drawing figure from given data. Resolving the given system of force horizontally. We get, ⇌ ± ; Fx = 100 250 = 150 KN Resolving the given system of forces vertically

we get Fx = 150 200 = 50 KN

Magnitude of resultant (R):

R = 2 2Fx Fy = 2 2

150 50

R = 158.11 KN Magnitude of resultant is 158.11 KN

Q.4 Attempt any FOUR of the following : [16]Q.4(a) Two men carry a weight 670 N by means of ropes fixed to the weight. One rope is

inclined at 40 and other at 50° with the vertical. Find the tension in each rope analytically.

[4]

(A) Drawing the F.B.D in fig as follows: Let T1 and T2 be the tension in ropes. Applying Lami’s theorem, we get.

1T

sin190 = 2T

sin140=

67

sin 90

T1 =670

sin130sin 90

= 513.249

T2 = 670

sin140sin 90

= 430.66 N Q.4(b) Forces of 3, 6, 9, and 12 kN respectively acts on a regular pentagon as shown in

Figure. Find the resultant in magnitude and direction. Use analytical method only. [4]

(A)

+ ↑ ↓

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(A) (1) Exterior angle = 360/No. of angular points = 360/5 = 72º Interior angle = 180º 72º = 108º Angle BAC = Angle CAD = Angle DAE = 108º/3 = 36º (2) Magnitude of Resultant Resolving all forces ∑Fx = (12 cos 0) + (9 cos 36) + (6 cos 72) (3 cos 72) = +12 + 7.28 + 1.85 0.93 = +20.2 N ∑Fy = (12 sin 0) + (9 sin 36) + (6 sin 72) + (3 sin 72) = 0 + 5.29 + 5.71 + 2.85 = +13.85 N (3) Magnitude of resultant

R = 2 2Fx Fy

R = 2 2(20.2) (13.85)

R = 24.49N (4) Direction and position of resultant As ∑Fx = +ve and ∑Fy = +ve, R lies in 1st quadrant

θ = tan1 1Fy 13.85tan

Fx 20.2

θ = 34.44º Q.4(c) Check whether a wire having capacity of 600 N can lift

a load of 800N if it is attached as shown in Figure.

[4]

(A) Apply Lami’s theorem :

1T

sin105 = 2T

sin130 =

800

sin125

1T

0.966 = 2T

0.766 = 976.62

taking, 1T

0.966 = 976.62

T1 = 0.966 976.62 = 943.34N > 800N

taking 2T

0.760 = 976.62

T2 = 0.766 976.62 = 748.1N < 800N Since wire has a capacity of 800N. But tension in one part is 943.34N, it cannot lift a load of

800N

W

15

W = 800 N

W = 800 N

50

130

105

125

T2 T1 50

0 Vidyala

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Vidyalankar : F.Y. Diploma Mechanics

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Q.4(d) Explain the different types of beam with neat sketches. [4](A) (i) Simply supported beam A beam which is freely supported on the walls or columns at its both the ends is called as a

simply supported beam.

(ii) Cantilever beam A beam fixed at one end and free at other is called as a cantilever beam.

(iii)Overhanging beam If the end portion of the beam extends beyond the supports it is called as an overhanging

beam. A beam may be overhanging on one side or both sides.

Overhung on right side Overhung on both sides Overhung on Let side (iv) Fixed beam A beam whose both the ends are rigidly fixed in walls is called a fixed beam, constrained

beam, built-in beam or an encastre beam.

(v) Continuous beam A beam which is supported on more than two supports is called a continuous beam.

Two span continuous beam

Three span continuous beam Q.4(e) A sphere weights 1200N. It is supported by two planes at 35 and 50 to the

horizontal respectively. Calculate the support reactions. [4]

(A)

Using Lami’s theorem,

W RA RB

sin85 sin145 sin130

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1200 RA RB

sin85 sin145 sin130(2) (3)(1)

Using term (1) and (2)

1200 RA

sin85 sin145

RA = 1200 x sin145

sin85

RA = 690.92 N Using term (1) and (3)

1200 RB

sin85 sin130

RB = sin130

1200 xsin85

RB = 922.77 N Q.4(f) Find the beam reactions for the beam loaded and supported as shown in Figure. [4]

(A)

Applying condition of equilibrium Fy = 0 RA (1 4) + RB 3 = 0 RA + RB = 7 kN (i)

M@A = 0

4

1 42

´ ´ + (3 × 4) 8RB = 0

8 + 12 = 8RB RB = 2.5 kN From equation (i) RA + RB = 7 RA = 7 2.5 RA = 4.5 kN RA = 4.5 kN RB = 2.5 kN

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Vidyalankar : F.Y. Diploma Mechanics

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Q.5 Attempt any FOUR of the following : [16]Q.5(a) A block weighing 250 N is lying on a horizontal table for which the coefficient of

friction is 0.40. Determine limiting force of friction, normal reaction, resultant reaction and angle of friction

[4]

(A)

(i) Fy = 0 Normal reaction R 250 = 0 R = 250 N (ii) Fx = 0 Force of friction P F = 0 P R = 0 P 0.4 250 = 0 P = 100 N F = 100 N (iii) Angle of friction = tan tan1() = = tan1(0.4) = 21.80 (iv) Resultant reaction

2 2 2 2s R F (250) (100)

s = 269.258 N Q.5(b) A block weighing 40 kN resting on a rough horizontal plane can be moved by a

force of 20 kN applied at an angle of 40° with horizontal. Find coefficient of friction.[4]

(A) Drawing Fig from given data

Resolving the force horizontally, we get Fx = 20 cos 40 F …. (By sign convention ⇄ ±) Fx = 20 cos 40 R … ( F = R) 0 = 20 cos 40 40 … (From equilibrium Fx = 0) 0 = R 27.145 R = 27.145 N Putting the value of R in equation (i), we get

=20cos 40

27.145

= 0.56

Fig: Force are acting on a block Vidyala

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Prelim Question Paper

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Q.5(c) A body of weight 300 N, resting on an inclined plane inclined at an angle of 30 with the horizontal, just started to move down the plane. Calculate : (i) coefficient of friction (ii) angle of friction (iii)angle of repose

[4]

(A) Given : W = 300 N Find : (i) (ii) (iii)

(i) Fx = 0 F 300sin30 F = 150 N R = 150 N ( F = R)

= 150

R (1)

(ii) Fy = 0 R 300csos30 = 0 R = 259.807 N Put ‘R’ in equation (i)

= 150

259.807

= 0.577

(iii)tan = = tan1 () = tan1 (0.577) = 29.98 say 30 (iv) Angle of friction = Angle of repose = = 30

Q.5(d) A Ladder of weight 400N and length 10m is supported on smooth well with its lower end 4m from the wall. The coefficient of friction between the flower and the ladder is 0.3. Show the forces acting on the ladder and find frictional force at floor.

[4]

(A) sin = 4/10 = sin1 (4/10) = 23.58 Fx = 0 RW EF = 0 RW = FF

Fy = 0 FW + RF W = 0 0 + RF 400 = 0 RF = 400 N MA = 0 FF L cos RF 4 + 400 2 + RW 0 = 0 FF 10 cos 23.58 400 4 + 800 = 0 FF 10 cos 23.58 = 1600 800 = 800

FF = 800

10 cos 23.58 = 87.25 N

Ans. : (i) RW = FF = 87.29N (ii) RF = 400 N (iii)FW = 0

A RW

B 2m 2m

5m

5m

smooth wall W = 0

FW = RW W = 0

G

D

RF

Ff = FRF = 0.3 RF

W = 400N

C

L c

os

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Q.5(e) For a certain machine an effort of 100 N and 150 N can lift a load of 1 kN and 2kN respectively. Find the law of machine. Also calculate maximum efficiency if VR is 20.

[4]

(A) Effort (P) = 100 N and W = 1 kN = 1000 N Effort (P) = 150 N and W = 2 kN = 2000 N (i) Law of machine P = mW + C 100 = m 1000 + C (1) 150 = m 2000 + C (2) Multiplying equation (1) by 2 1000 m + C = 100 2 2000 m + C = 150 Subtracting equation (2) from equation (1) 2000 m + 2C = 200 (1) 2000 m + C = 150 (2) C = 50 N put in equation (2) 150 = m 2000 + 50 150 50 = 2000 m 100 = 2000 m m = 0.05 (ii) Law of machine P = (0.05W + 50) N

(iii) Max.M.A. = 1

m =

1

0.05

Max M.A. = 20

(iv) Max. = Max.M.A. 20

100 100V.R. 20

Max. = 100 % Q.5(f) A heavy stone of mass 450 kg is on a hill slope of 40° incline. If the between the

ground and the stone is 0.65, is the stone stable? [4]

(A) By Given : W = mg = 450 x 9.81 = 4414.5 N Weight of stone = 4414.5 N Resolving the forces perpendicular to plane, We get Fy = R 4414.5 cos 40 0 = R 3381.70 … (Fy = 0 For the stone in limiting equilibrium) R = 3381.70 N Considering the stone is just on the print of moving down the plane. We know, F = R = 0.65 3381.70 = 2193.105N ( R = 3381.70 N) F = Frictional Force = 2198.105 N.

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Prelim Question Paper

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Downward component of weight of stone = 4414.5 sin 40 = 2837.58N 2837.58 N > 2198.105 N, stone will not be stable

Q.6 Attempt any FOUR of the following : [16]Q.6(a) Locate the centroid of angle section 90 mm 100 mm 10 mm. (90 mm side is

vertical.) [4]

(A) (i) a1 = 80 10 = 800 mm2 a2 = 100 10 = 1000 mm2

(ii) x1 = 10

2 = 5 mm

x2 = 100

2 = 50 mm

y1 = 10 + 80

2 = 10 + 40 = 50 mm

y2 = 10

2 = 5 mm

1x = 1 1 2 2

1 2

a x a x

a a

=

800 5 1000 50

800 1000

´ ´

x = 30 mm

y = 1 1 2 2

1 2

a y a y

a a

=

800 50 1000 5

800 1000

´ ´´

y = 25 mm Q.6(b) Find the centroid of an inverted T-section with flange 200 mm 10 mm and a web of

300 mm 10 mm. [4]

(A) (1) Figure is symmetric @ y-y axis and hence, x = Maximum horizontal dimension/2 = 200/2 = 100mm (2) Area calculation A1 = 200 x 10 = 2000mm2

A2 = 300 x 10 = 3000mm2

A = A1 + A2 = 5000mm2 ) Location of y y1 = 10/2 = 5mm y2 = 10 + (300/2) = 160mm

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Vidyalankar : F.Y. Diploma Mechanics

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y = 1 1 2 2A y A y

A

y = (2000 x 5) (3000 x160)

5000

y = 98mm Hence, centroid (G) for given section lies at G ( x, y ) = (100mm from OB and 98mm from OA) Q.6(c) Find the centroid of the I-section with following details.

(i) Top flange = 200 mmx 10mm (ii) Bottom flange = 100 mm 20 mm (iii)Web thickness = 15 mm (iv) over all depth = 250 mm

[4]

(A) x = x1 = x2 = 200

2 = 100 mm

a1 = 200 10 = 2000 mm2

a2 = 15 220 = 3300 mm2

a3 = 100 20 = 2000 mm2

A = a1 + a2 + a3 = 2000 + 3300 + 2000 = 7300 mm2

y1 = 20 + 220 + 10/2 = 245 mm

y2 = 220

202

= 130mm,

y3 = 20/2 = 10 mm

y = 1 1 2 2 3 3a y a y a y

A

=

2000 235 3300 130 2000 10

7300

y = 91900

7300 = 125.89 mm.

a x, y = (100 mm, 125.89 mm)

Q.6(d) Locate the position of centre of gravity for a composite body in which a cone having

400 mm diameter and 800 mm height is placed on a cube of 400 mm side coaxially. [4]

(A) (i) 2 2 6 31

1 1V r h 200 800 33.51 10 mm

3 3

3 6 32V 400 64 10 mm

1 2V V V = 97.51 106 mm3

V = 97.51 106 mm3

(ii) Distance of C.G. from base

1800

y 400 600 mm4

2400

y 200mm2

Let 1 1 2 2v y v yy

v

y = 6 6

6

33.51 10 600 64 10 200

97.51 10

´ ´ ´ ´

´

6

6

(33.51 600 64 200)y

97.51 10

y 337.462 mm

(iii)400

x 200 mm2

x 200 mm

1

2

3

200 mm

250 mm 15 mm

20 mm

100 mm

10 mm

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Prelim Question Paper

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Q.6(e) A sphere of 300 mm diameter is placed on a cube of 500 mm side coaxially. Locatethe position of centre of gravity of the assembly.

[4]

(A) (i) V1 = Volume of cube V1 = 5003 = 125 106 mm3

V2 = 34 4R

3 3 1503

V2 = 14.136 106 mm3 V = 139.136 106 mm3 (ii) Since the body is symmetrical about vertical line

500

x 250 mm2

x 250 mm (iii)Distance of c.g. from x1axis

1500

y 250 mm2

2y 500 150 650 mm

6 6

1 1 2 26

1 2

v y v y 125 10 250 14.136 10 650y

v v 139.136 10

y 290.639 mm

Q.6(f) A wall of height 6m has one side vertical and other inclined. The top thickness is 1 m

and bottom thickness is 4 m. Find its centroid. [4]

(A) From given data, Divide the section of retaining wall into rectangle (1) and triangle (2) and taking the complete

section of retaining wall in first quadrant A1 = 1 6 = 6m2

A2 = 1

3 62 = 9m2

X1 =1

2= 0.5m

X2 =1

13

= 2

= 2 m … wrt to OY

Y2 = 1

63 = 2m …. wrt to OX.

X = 1 1 2 2

1 2

A X A X

A A

=

6 0.5 9 2

6 9

= 1.4m

Y = 1 1 2 2

1 2

A Y A Y

A A

=

6 3 9 2

6 9

= 2.4m

G x, y = G 1.4m, 2.4m

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