1
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
Vidyala
nkar
Vidyalankar : F.Y. Diploma Mechanics
2
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
Vidyala
nkar
Prelim Question Paper Solution
3
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
nkar
Vidyalankar : F.Y. Diploma Mechanics
4
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
Vidyala
nkar
Prelim Question Paper Solution
5
(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
Vidyala
nkar
Vidyalankar : F.Y. Diploma Mechanics
6
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
Vidyala
nkar
Prelim Question Paper Solution
7
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
nkar
Vidyalankar : F.Y. Diploma Mechanics
8
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 ∙
Vidyala
nkar
Prelim Question Paper Solution
9
(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
Vidyala
nkar
Vidyalankar : F.Y. Diploma Mechanics
10
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
Vidyala
nkar
Prelim Question Paper Solution
11
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
nkar
Vidyalankar : F.Y. Diploma Mechanics
12
(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
nkar
Prelim Question Paper Solution
13
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
Vidyala
nkar
Vidyalankar : F.Y. Diploma Mechanics
14
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
Vidyala
nkar
Prelim Question Paper Solution
15
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
nkar
Vidyalankar : F.Y. Diploma Mechanics
16
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
nkar
Prelim Question Paper Solution
17
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
nkar
Vidyalankar : F.Y. Diploma Mechanics
18
(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
nkar
Prelim Question Paper Solution
19
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
nkar