engineering mechanics prelim question paper solution …vidyalankar.org/upload/3_mech_soln.pdf ·...

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Vidyalankar F.Y. Diploma : Sem. II

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

Prelim Question Paper Solution

Q.1 Attempt any TEN of the following : [20]Q.1(a) Difference between mass and weight? [2](A) Difference between mass and weight

Mass Weight 1. Mass is the quantity of matter

contained in a body. Weight of a body is the force with which the body is attracted by the earth towards its centre.

2. It is scalar quantity. it is a vector quantity. 3. S.I. unit of mass is ‘ky’ S.I. hit of weight is ‘N’.

Q.1(b) What is mechanics & classified it? [2](A) Mechanics is a branch of Physical science which deals with study of forces and their

effect on a body in motion and rest condition.

Q.1(c) State Newton’s Laws of motion? (First, Second & Third) [2](A) Nowtons 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.

% = 100Output

input

Q.1(e) What is meant by ideal machine? [2](A) Ideal machine : If the efficiency of the machine is 100% i.e. when the output is equal to

the input, the machine is called an ideal or perfect machine. input = output.

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.

Mechanics Statics

Dynamics Kinematics

Kinetics

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Vidyalankar : F.Y. Diploma Mechanics

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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 adjcement 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.”

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) State Lami’s theorem. [2](A) “if there 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”.

sin sin sin

P Q R

Q.1(j) What is coefficient of friction? [2](A) Coefficient of friction :

Coefficient of friction is defined as the ratio of limiting friction to the normal reaction at the surfaces of contact.

= F/R

Q.1(k) Define centre of gravity? [2](A) Centre of gravity : “The centre of gravity of a body can be defined as the point throught

which the whole weight of the body is assumed to act, irrespective of the position of a body”.

Q.1(l) After giving 100 N.m input to a machine 80 N.m output is obtained. What is its efficiency?

[2]

(A) = 100output

input

= 80

100100

= 80 %

= 80 %

R

Q

P

B

O

C

D

Q sin

R

P

Q Q

Q cos

A

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Q.2 Attempt any FOUR of the following : [16]Q.2(a) A machine has V.R. = 50. A load of 6 KN is lifted by an effort of 300 N.

Calculate efficiency and effort lost in friction. [4]

(A) V.R = 50, W = 6KN = 6000N, P = 300N

M.A = w / P = 6000

300 = 20 M.A = 20

= .. 100

M AV R =

20100

50 = 40% = 40 %

Effort lost in friction

Pf = P Pi = WP V R =

6000300

50

Pf = 180N

Q.2(b) The diameter of wheel in a differential wheel and axle is 36 cm. The axles are 9 cm and 6 cm diameter if the efficiency of the machine is 80%, Find the load lifted by an effort of 100 N.

[4]

(A) D = 36cm, d1 = 9 cm, d2 = 6 cm, = 80%, P = 100N, W = ?.

V.R = 1 2

2

D

d d

= 2 36

9 6

= 72

3 = 24

M.A = W/P = 100

w

= .

100.

M A

V R 80 =

/100100

24

W

80 = 100100

W

W = 80 100 24

100

= 1920N = 1.92KN

Q.2(c) A screw jack of pitch 8 mm a lever of 250 mm length if the efficiency of machine is 30%, find the effort required to lift a load of 1500 N.

[4]

(A) Pitch = 8 mm, L = 250 mm, = 30 %. W = 1500N find effort (P) = ?

V.R = 2LP

= 2 250

8

= 196.35

M.A. = W/P = 1500

P

= .

100.

M A

V R 30 =

1500 1100

196.35

P

P = 25.46N

Vidyala

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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 = sin

sin

F

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 = sin

sin (

F

= 8 sin 60

sin 60 30

=8 0.866

sin 90

F2 = 6.93 KN

Q.2(e) Find the orthogonal (mutually perpendicular) components of the following forced all acting away from the point : (i) 100 N acting at 30 west of south. (ii) 400 N acting the North. (iii) 250 N acting North-East. (iv) 150 N acting the west.

[4]

(A) (i) Orthogonal components of 100 N Fx = 100 cos 60, Fy = 100 sin 60 Fx = 50N, Fy = 86.6N

(ii) Orthogonal components of 400N

Fx = 400 cos 50, Fy = 400 sin 90 Fx = 0, Fy = 400N

(iii) Orthogonal components of 250N

Fx = 250 cos 45, Fy = 250 sin 45 Fx = 176.78N, Fy = 176.78N

W E

N

S 100N

150N

400N 250N

45 45

30

F = 8KN

F2

F1 = 4KN = 60

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(iv) Orthogonal components of 150N Fx = 150 cos 180, Fy = 150 sin 180 Fx = 150N, Fy = 0

Q.2(f) Three forces 4N, 8N and 12N are acting along the sites of a triangle AB, BC

and AC respectively. Calculate the moments of all the forces about point C, if AB = BC = AC = 0.8m.

[4]

(A) Find mc = ? In CDB, CB2 = CD2 + BD2

(0.8)2 = CD2 + (0.4)2

CD = 0.48 = 0.693 m in ACD,

sin 60 = CD

AC =

0.8

CD

CD = 0.8 sin 60 CD = 0.693 m

Mc = 4 CD = 4 0.693 = 2.77 N.m

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:

= 1tan

X

y

F

F = 1 55 49

tan5.85

= 83.98 with horizontal.

60

60 60

12N 8N

4N B A

C

0.4 m 0.4 m

0.8m

D

15 cos60

25 sin30

30N

20N

15 sin6015 N

25N

60 30 25 cos30

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Q.3(b) Calculate the magnitude and direction of resultant for concurrent force system as shown in Figure.

[4]

(A)

Fx = 50 cos 30 80 cos 75 100 + 60 cos 45 = 34.98 N

Fy = 50 sin 30 + 80 sin 75 60 sin 45 = 59.85 N

R = 2 2 Fx Fy = 2 234.98 59.85

= 69.32 N

tan = Fy

Fx =

59.85

34.98 = 1.71

= tan1 (1.71) = 59.69 = 59.7 with horizontal

Q.3(c) Forces of 1, 2, 3 and 4 KN respectively act at one of the angular points of a regular pentagon towards the other four angular points taken in order. Find the resultant in magnitude and direction.

[4]

(A) Exterior angle = 360

5

= 72

interior angle = 180 72 = 108 BAC = CAD = DAE

= 108

3 = 36

Fx = 1 + 2 cos36 + 3 cos72 4 cos72

= 2.310 KN Fy = 2 sin36 + 3 sin72 + 4 sin72

= 7.833 KN

R = 2 2 Fx Fy = 2 22.310 7.833

= 8.17 KN

tan = Fy

Fx =

7.833

2.310 = 3.39

= tan1 (3.39) = 73.57

60 N

50 N

36

36 36 72

1 KN A B

C

D

E 3 KN

2 KN 4 KN

80 N

100 N45

30

15

60 N

50 N

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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

= 1tan Fy

Fx = tan1 5/4 = 51.34

Q.3(e) Six forced 10N, 20N, 10N, 10N, 20N, 20N are concurrent and pull in sense and

make equal Angle with each other. Calculate graphically to resultant of the forces.

[4]

(A)

(a) Space diagram

(b) vector diagram (Scale 1cm = 5N)

A B

D C

6N

4N 1N

2N

Length Ag = 2 cm R = ag scale R = 2 5 R = 10N = Angle mode by R withhorizontal = 60 Vidyala

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Q.3(f) Solve graphically and show resultant in Space diagram. [4]

(A)

(a) Space Diagram

(b) Vector Diagram

Q4 Attempt any FOUR of the following : [16]Q.4(a) Find the reaction at the Base A and inclination , for the telephone pole to be

stable. (Refer to Figure) [4]

R = ( ) af scale

= 3.8 200 = 760 N = 46 with horizontal

30 50

600 N

1200 N 800 N

300 N

400 N

65

600 N

400 N

A ∙

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(A)

MA = 0 400 h + 600 cos 0 600 sin h + RAH 0 + RAQ 0 = 0

400h = 600 sin h 400 = 600 sin

sin = 400

600 =

2

3

= 1sin 2 / 3 = 41.81 Fx = 0 RAH + 400 600 sin = 0

RAH + 400 600 23 = 0 sin = 2/3

RAH + 400 400 = 0 RAH = 0

Fy = 0 RAQ 600 cos = 0 RAQ = 600 cos = 600 cos 41.81 RAQ = 447.215N

RA = 22AH AQR R = 2 20 447.215

RA = 447.215N

Q.4(b) A sphere of weight 400 N rests in a groove of smooth inclined surfaces which are making 60 and 30 inclination to the horizontal. Find the reactions at the contact surfaces.

[4]

(A) Apply Lami’s theorem :

sin 30 AR =

sin 60 BR =

400

sin(60 30 ) =

40

sin 90

RA = 0.5 400 = 200N RB = 0.866 400 = 346.4N

600 sin

600 N

400 N

A RAH

RAQ

h 600 cos

RA

A

60 30

30 B

O

RB

60

400 H

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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 :

1

sin105 T

= 2

sin130 T

= 800

sin125

1

0.966

T = 2

0.766

T = 976.62

taking, 1

0.966

T = 976.62

T1 = 0.966 976.62 = 943.34N > 800N

taking 2

0.760

T = 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

Q.4(d) A beam AB of 9m span is simply supported at ends. The beam carries point loadof 2KN upwards at 2m from A and uniformly distributed load of 1000 N/m downwards on a length of 6m from B. Determine support reactions analytically.

[4]

(A) Converting u.d. l to its equivalent load;

taking moments about A,

MA = 0

(RA 0) (2 2) + (6 6) (RB 9) = 0

W = 800 N

50

130

105

125

T2 T1 50

0

1000 N/M= 1 KN/M

A

6m

RB

2KN

D C

RA

1m 2m

A

RA RB

2m 1m 6m

3m 3mC D

2KN 6KN

B

(1000 6) = 6000 N

W

15

W = 800 N

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4 + 36 9RB = 0 9RB = 32 RB = 3.55KN also,

Fy = 0

RA + RB + 2 6 = 0 RA + RB = 4

RA = 4 3.55 RA = 0.45 KN.

Q.4(e) Using analytical method. Calculation the support reactions for the beam loaded as shown in Figure.

[4]

(A)

MA = 0

1200 2 + RA 0 + 2000 2 + 1200 (2 + 2) RB 6 = 0

2400 + 0 + 4000 + 4800 = 6RB 6RB = 6400 RB = 1066.67N

Fy = 0

RA + RB 1200 2000 1200 = 0

RA + 1066.67 4400 = 0 RA = 3333.33N

Q.4(f) Calculate graphically the reactions of beam at the support as shown in Figure.

[4]

1200N 2000N 300 4 = 1200N

A

RA

D E B

RB 2m 2m

2m 2m 4m

2m 2m 4m

1200N 2000N 300N/m

2m 1m 1m

A

50kN 10 kN/m

B C

Vidyala

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(A)

Space diagram (1cm = 0.5m)

(b) Vector diagram (Scale 1cm = 10 KN)

Q.5 Attempt any FOUR of the following : [16]Q.5(a) A body weighing 12 KN is lying on horizontal plane for which = 0.70.

Determine normal reaction, limiting force of friction, horizontal force required to move it and angle of friction.

[4]

(A) Using conditions of equilibrium, FX = 0 P F = 0 P = F = R P = 0.7 R

also, Fy = 0 R w = 0 R 12 = 0 R = 12 KN P = 0.7R = 0.7 12 P = 8.4 KN Horizontal force = 8.4 KN. also, = tan = 1tan = 1tan 0.70

= 34.99

P, Q, R, S Bow’s Notation Funicular Polygon

(sp) = 2.8 cm (rs) = 4.2 cm RA = (sp) scale = 2.8 10 = 28 KN RB = (rs) scale = 4.2 10 = 42 KN RA = 28 KN RB = 42 KN

F

R

P

W = 12KN

Motion

= 0.70 Vidyala

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Q.5(b) A body weighing 200 N is resting on rough horizontal plane. A pull of 30 N applied at 30 up the horizontal just moves the body. Find coefficient of friction.

[4]

(A) Fy = 0 R + 30 sin30 200 = 0 R 185 = 0 R = 185 N FX = 0 30 cos30 R = 0 R = 25.98

=25.98

185

= 0.140

Q.5(c) A block of weight 500 N is placed on a inclined plane at an angle of 20 with the horizontal. If coefficient of friction is 0.14, find the force ‘P’ applied Parallel to the plane, just to move the body up the plane.

[4]

(A) Fy = 0 R 500 cos 20 = 0 R = 469.85 N Fx = 0 P 500 sin 20 R = 0 P = 500 sin 20 + R = 171.01 + 0.14 469.85 P = 236.789N

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

500N

500 cos 20

motion

500 sin 20 F = R

R

P

20 20

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

F = R

R

30 cos30

30N

200N

30 sin30

30

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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

Q.5(e) The following readings have been taken on a machine with V.R = 40.

Load KN 1 KN 2 KN 3 KN Effort KN 48 KN 78 KN 108 KN

Find law of machine by plotting the graph.

[4]

(A) Scale x-axis 1cm = 250N y-axis 1cm = 10N

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Slope = m = 2 1

2 1

y y

x x =

108

3000 2000

78 = 0.03

Law of Machine P = (0.03W + 18)N

Q.5(f) The velocity ratio of a certain machine is 72. The law of machine is

P =

1 W + 3048

N. Find the maximum mechanical advantage and maximum

efficiency. State also whether the machine is reversible or not.

[4]

(A) VR = 72

P = 1

3048

W N

m = 1

48 and c = 30N

(i) Maximum M.A. = ? (ii) Maximum = ? (iii)To decide whether the machine is reversible or not.

(i) Maximum M.A. = 1

m =

1

1/ 48= 48

(ii) Maximum = 1

100. .

m V R

= 1

1001/ 48 72

= 66.67% > 50%

(iii)Since the maximum efficiency is more than 50%, the machine is reversible.

Q.6 Attempt any FOUR of the following : [16]Q.6(a) Find centroid of ISA 90 60 8 (L section) [4](A) a1 = 90 8 = 720,

a2 = 52 8 = 416 A = 720 + 416 = 1136

x1 = 8

2= 4

x2 = 52

82

= 34

y1 = 90

2 = 45

y2 = 8

2 = 4

x = 1 1 2 2a x a x

A =

720 4 416 34

1136

= 14.98

y = 1 1 2 2a y a y

A =

720 45 416 4

1136

= 29.58 G x y = (14.98, 29.98)

1

2

y

x

8

90

52

60

8Vidyala

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Q.6(b) A T-section has a flange 200 200mm and a web of 15x 240 mm. Find the position of centroid.

[4]

(A) x = x1 = x2 = 100 mm a1 = area of flange = 200 20 = 4000 mm2 a2 = area of web = 15 240 = 3600 mm2 A = a1 + a2 = 4000 + 3600 = 7600 mm2 y1 = C.G of area (1) y1 = 240 + 20/2 = 250 mm y2 = C.G of area (2) y2 = 240/2 = 120 mm

y = 1 1 2 2a y a y

A

= 4000 250 3600 120

7600

y = 1432000

7600

= 188.42 mm ,G x y = (100 mm, 188.42 mm).

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 3 a 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)

1

2

3

200 mm

250 mm 15 mm

20 mm

100 mm

10 mm

y

2

1

240 mm

20 mm

15mm

x

200 mm

Vidyala

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Q.6(d) A retaining wall of height 5.2 m has are side vertical the top width is 0.8m and bottom width is 3.2m. Find centroid.

[4]

(A) Split up the whole section into two ports (1) and (2) as shown in Fig. a1 = area of rectangle = 0.8 5.2 = 4.16 m2 a2 = area of triangle = 1 2 2.4 5.2

= 6.24 m2

x1 = 0.8/2 = 0.4 m x2 = 0.8 + 1/3 2.4 = 0.8 + 0.8 = 1.6 m y1 = 5.2/2 = 2.6 m y2 = 1/3 5.2 = 1.73 m

Let G ,x y

x = 1 1 2 2

1 2

a x a x

a a =

4.16 0.4 6.24 1.6

4.16 6.24

= 1.664 9.984

10.40

=

11.648

10.40

x = 1.1 m

y = 1 1 2 2

1 2

a y a y

a a =

4.16 2.6 6.24

4.16 6.24

1.73

y = 10.816 10.795

10.40

= 21.611 / 10.40 = 2.08 m G x y = (1.1 m, 2.08 m)

Q.6(e) Find the centroid of the shaded area as shown in Figure. [4]

A

1 2

0.8 m

G

B

5.2 m

y

0 0.8m 2.4 m

x

3.2 m

100 mm

100 mm

x

yVidyala

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(A) Main area a1 = area of rectangle = 100 100 = 1 104 mm2 a2 = area of quarter circle

= 2

4

r =

2100

4

= 7853.98 mm2

Shaded area, A = a1 a2 = 1 104 7853.98 = 2146.02 mm2

Distances of centroid from yaxis is :

x1 = 100

2 = 50 mm

x2 = 4

3r

= 4 100

3

= 42.44 mm

Distances centroid from x-axis

y1 = 100

2= 50 mm,

y2 = 4

3r

= 4 100

3

= 42.44 mm

x1 = y1, x2 = y2, x = y

x = y = 1 1 2 2a x a x

A

= 1 104 50 7853.98 42.44

2146.02

x = y = 77.67 mm

G x y = (77.67 mm, 77.67 mm)

Q.6(f) For the lamina shown in Fig. locate its centroid y .

[4]

(A)

100 mm

100 mm

x

y

200 mm

350 mm

360 mm

150 mm

360 mm

R = 100mm

1

3

2

150 mm

100 mm 710 mm

360 mm Vidyala

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a1 = Area of rectangle = 360 350 = 12600 mm2

a2 = Area of triangle = 1

360 3602 = 64800 mm2

a3 = Area of semi-Circle = 2

2

r =

2100

2

= 15707.963 mm2

y1 = 350

2 = 175 mm

y2 = 1

350 3603

= 470 mm

y3 = 150 + R = 150 + 100 = 250 mm

y = 1 1 2 2 3 3

1 2 3

a y a y a y

a a a

y = 126000 175 64800 470 15707.963 250

126000 64800 15707.963

= 48579009

175092.04

y = 277.448 mm y = 277.45 mm from ox

Vidyala

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engineering mechanics prelim question paper solution …vidyalankar.org/upload/3_mech_soln.pdf ·...

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