Engineering Mechanics

(^640) „„„„„ A Textbook of Engineering Mechanics
Example 31.13. Two bodies A and B of masses 30 kg and 10 kg are tied to the two ends of
a light string passing over a composite pulley of radius of gyration as 70 mm and mass 4 kg as
shown in Fig. 31.12.
Fig. 31.12.
Find the pulls in the two parts of the string and the angular acceleration of the pulley.
Solution. Given: Mass of the body A (m 1 ) = 30 kg; Mass of the body B (m 2 ) = 10 kg;
Radius of gyration of the pulley (k) = 70 mm = 0.07 m; Mass of the pulley (M) = 4 kg; Internal
diameter of the pulley (d 1 ) = 100 mm = 0.1 m or radius (r 1 ) = 0.05 m and external diameter of
pulley (d 2 ) = 200 mm = 0.2 m or radius (r 2 ) = 0.1 m.
Angular acceleration of the pulley
Let P 1 = Pull in the string carrying 30 kg mass,
P 2 = Pull in the string carrying 10 kg mass, and
α= Angular acceleration of the body.
From the geometry of the masses, we find that turning moment of the mass 30 kg
(i.e., 30 × 0.05 = 1.5 kg-m) is more than that of the mass 10 kg (i.e., 10 × 0.1 = 1 kg-m). It is thus
obvious, that the 30 kg mass will come downwards and the 10 kg mass will go upwards, when the
system is released.
Let a 1 = Acceleration of the mass 30 kg, and
a 2 = Acceleration of the mass 10 kg,
We know that the mass moment of inertia of the pulley,
I=Mk^2 = 4 (0.07)^2 = 0.02 kg-m^2
First of all consider the motion of mass 30 kg, which is coming down. We know that the
forces acting on it are m 1 g = 30 × 9.8 = 294 newtons (downwards) and P 1 newtons (upwards). As the
mass is moving downwards therefore, resultant force
= 294 – P 1 ...(i)
Since the mass is moving downwards with an acceleration (a 1 ), therefore force acting on the
body
= 30 a 1 ...(ii)
Equating equations (i) and (ii),
294 – P 1 = 30 a 1 ...(iii)