Engineering Mechanics

(^642) „„„„„ A Textbook of Engineering Mechanics
Example 31.14. Two bodies A and B, of mass 150 kg and 75 kg respectively are supported
by a string of negligible mass and pass over a composite pulley. The bodies rest on two smooth
inclined planes as shown in Fig. 31.13.
Fig. 31.13.
If the pulley has a mass of 75 kg and radius of gyration of 100 mm, find the accelerations of
the masses A and B and pulls in the string. Neglect friction in the bearings.
Solution. Given: Mass of the body A (m 1 ) = 150 kg; Mass of the body B (m 2 ) = 75 kg; Mass
of the pulley (M) = 75 kg; Radius of gyration (k) = 100 mm = 0.1 m; External radius of the pully (r 1 )
= 250 mm = 0.25 m and internal radius of the pulley (r 2 ) = 125 mm = 0.125 m.
Pulls in the strings
Let P 1 = Pull in the string carrying 150 kg mass, and
P 2 = Pull in the string carrying 75 kg mass.
From the geometry of the masses, we find that turning moment of mass 150 kg (i.e., 150 sin
45° × 0.25 = 150 × 0.707 × 0.25 = 26.5 kg-m) is more than that of the mass 75 kg (i.e., 75 sin 30° ×
0.125 = 75 × 0.5 × 0.125 = 4.7 kg-m). It is thus obvious that the 150 kg mass will come downwards
and the 75 kg mass will go upwards, when the system is released.
Let a 1 = Acceleration of the 150 kg mass,
a 2 = Acceleration of the 75 kg mass, and
α= Angular acceleration of the pulley.
We know that mass moment of inertia of the pulley,
I=Mk^2 = 75 (0.1)^2 = 0.75 kg-m^2
First of all, consider the motion of 150 kg mass, which is coming down. We know that the
force acting on it, along the plane, = m 1 g sin θ 1 = 150 × 9.8 sin 45° = 150 × 9.8 × 0.707 = 1039 N.
(downwards) and P 1 (upwards). As the mass is moving downwards, therefore resultant force
= 1039 – P 1 ...(i)
Since the mass is moving downwards with an acceleration (a 1 ), therefore force acting on
the body
= 150 a 1 ...(ii)
Equating equations (i) and (ii),
1039 – P 1 = 150 a 1 ...(iii)
Now consider the motion of 75 kg mass, which is going upwards. We know that the force
acting on it, along the plane, = m 2 g sin θ 2 = 75 × 9.8 sin 30° = 75 × 9.8 × 0.5 = 367.5 N (downwards)
and P 2 (upwards). As the mass is moving upwards, therefore resultant force
=P 2 – 367.5 ...(iv)