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1. Equations of Equilibrium and Statics
Q.1 A smooth hollow cylinder of radius r open at both ends rests on
Hollow
a smooth horizontal plane. Two smooth spheres of weights W1
cylinder
and W2 and radii r1 and r2, respectively are placed inside the
cylinder, with the larger sphere (radius r2) resting on the
horizontal plane as shown in figure.
Determine the minimum weight Wof the cylinder that will prevent
the cylinder from tipping over.
[CSE (Mains) 2009: 12 Marks]
Solution:
Let weight of the cylinder that will prevent the cylinder from tipping over be w.
Force diagrams:
h
C
Fn., Fn L = 2r— (ri + r2)
From AABC, h2 = (r1 + r2)2 - [2r- (r1 + r2)]2
h2 = (r1 + r2)2 - 4r2- (r1 + r2)2 + 4r(r1 + r2)
h2 = 4r(r1 + r2)- 4r2 = 4 r[ri + r2- 11
h = V4r(r1+ r2 - r)
F„ (L )
Also,
w1 =
(h
[ 2r -(r1+r2)
Fr, = w1
\14r(r1+ r2 - r)
Taking equilibrium of opposite torques about the lower left corner of cylinder,
Ti = w(r + t)
L
T2 -— Fn r2 = wilh-jr2
T3 = Fn(r2+ h)
T4 = Torque supplied by normal force of the floor on sphere (radius r2)
T4 = 0 (when cylinder starts to tip)
Since, Ti + T2 = T3
w(r + t)+ F„ r2 = Fn(r2+ h)
Pozevioad
2 Civil Services Mains Mechanical Engineering • Paper-I
Solved Papers
w(r + t) = Fnx h = L
w1[2r -(ri + r2)]
Wrnin (r
Q.2 A homogeneous ball of weight Q and radius a as well as a weight Pare suspended
by cords from a point 0 as shown in figure. The distance OM is b. Find the
inclination 4) of OM with the vertical when the system is in equilibrium.
[CSE (Mains) 2009 : 8 Marks]
Solution:
Given: ball weight = 0, radius = a, weight = P, OM = b.
Let inclination of OMwith vertical when the system is in equilibrium be (0).
OM = b
MN = b sin (1) and ON = b cos (1)
MX = a
NX=MX-MN=a-bsin 4'
Taking moments about point 0.
P x (WO = 0 x (MN)
P(a- b sin (I)) = 0 (b sin (p)
Pa- Pb sin (1) = Qbsin 4'
Pa= Pb sin + Ob sin (1) = (Pb + 0b) sin (I)
sin =
(PbP+ a0b)
= sin Pa
Pb+Qb
Q.3 A body A of weight 10 kN is connected to an another body B of weight
50 kN, resting on a smooth table of weight 200 kN through an
inextensible thread, passing over a freely rotating pulley mounted on a
corner of the table. Find the vertical component of the reaction of the
ground on the table when the bodies A and B are in motion. Does the
reaction change with time? The system is shown in figure.
[CSE (Mains) 2009: 12 Marks]
Solution:
10 3
Given: WA = 10 kN; m - 101 =1000 kg
A ()
50 x103
50 kN; m = 5000 kg
8 10
200 x103
We = 200 kN; me 20000 kg
10 a
Let The the tension in string and N be the normal force. T
T= mBa [a = acceleration of system]
Net We
T. (5000 a) N
Also,
MAg - MA a
104- (5000) a = 103 a 1T
104
a = 1.666 m/sec2
mAg a
5x103 +10'
MIME ERSY Engineering Mechanics 1 3
Considering equilibrium of forces in vertical direction on block.
2 Tsin 0 + Nc = NB + Wc
1\10 = NB + Wc — 2T sin 0
N0 = (200 + 50) — 2 T sin 0
and T = 5000 a = 8333.3 N
Alc = 250 — 16666.67 sin 6N
This reaction is constant and will not change with time.
Q.4 A body A weighing P1 descends down an inclined plane D which
makes an angle a with the horizontal and pulls a load B that weighs
P2 by means of a weightless and inextensible string passing over a
pulley C as shown in figure. Determine the horizontal component of
the pressure with which the inclined plane D acts on the floor rib E.
[CSE (Mains) 2010: 20 Marks]
Solution:
Let the tension in a string that is inextensible and weightless string be T.
And a be the common acceleration of system
T— P2 g =
P2 a
T=
P2(a +
and Pig sin a — T =
Pia
From (i) and (ii) Pea P2g = Pig sin a — Pia
(P2 + Pi)a = Pig sin a— P2g
Pigsina—P2g
a —
P1+ P2
[Pigsina—P2g
and T p + g
. 2
Pi+ P2 ]
[Pigsina—P2g +Pig+ P2g]
P2
P1+ P2
Posina +Pig]
T = P1P2 [gsina + g]
[ P1+ P2
P1 + P2
Let the horizontal component of the pressure with which the inclined plane D acts on the floor rib E be FH.
FH = 2 T cos a
2P1P2
(gsina+g)cosa
F
H = +P2
Q.5 The crane shown in figure below is supported by cables BD and BE. The distances marked are in m.
What are the tensions in the cables BC, BD and BE, when BC is parallel to the x-axis?
50 kN
Ball and socket joint
[CSE (Mains) 2010: 8 Marks]
4 Civil Services Mains Mechanical Engineering • Paper-I
► Solved Papers
Solution:
Given:
When BC is parallel to x-axis.
Let tension in BC; TBD = tension in BD; TBE = tension in BE.
TBC =
AE = AD = ,I82 +102 =12.80 m
AE 12.80
L ABD = L ABE = ai tan a1 = AB = 12
a1 = 46 86°
•
LDBA = al = 46.86°
TBC
Equilibrium of forces at joint C.
12
tan 0 = —
12 1
0 _ 45.
TAccos0 = 50
50 TBD
T — = 70.71kN
Ac cos45°
TBC = TAC sin 0 = 70.71 sin 45° = 50 kN Tap
Considering equilibrium of forces at joint B.
TBD = THE TBE
Component of TBD and TBE along AD and AE respectively are:
TBD = (TBD)Along AD = 0.729 TBD
E (TBE) Along AE = TBE sin = TBE sin 46.86° = 0.729 TBE
TBE (TBE )along x axis = E COS a2
where, tan a2 = — = 0.8 = 38.659°
10
TBE = TBE cos 38.659° = 0.5692 TBE
Similarly, TBD = 0.5692 TBD
and TBE = 50 kN
BD= 43.92 kN
TBE = T
2. Friction
Q.6 A stuntman drives a motorcycle around a circular vertical wall 30 m in diameter. The coefficient of
friction between the tyre and wall is 0.6. Determine the minimum speed that will prevent sliding down
the wall. Determine the angle also by which the motorcycle is inclined to the horizontal.
[CSE (Mains) 2012: 12 Marks]
Solution:
Given d = 30 m, µ = coefficient of friction = 0.6
Let the minimum speed that will prevent sliding down the wall be v.
my
N = Normal force =
