Engineering Mechanics

(^636) „„„„„ A Textbook of Engineering Mechanics
(ii) Tension of the rope
Substituting the value of a in equation (iii),
P – 29 400 = 3000 × 8.71 = 26 130
P= 26 130 + 29 400 = 55 530 N = 55.53 kN Ans.
31.19.MOTION OF TWO BODIES CONNECTED BY A STRING AND
PASSING OVER A PULLEY
Consider two bodies connected by an inextensible light string and
passing over a simple pulley as shown in Fig. 31.10.
Let
m 1 and m 2 = Masses of the two bodies,
M= Mass of the pulley,
I= Moment of inertia of the pulley,
r= Radius of the pulley,
k= Radius of gyration of the pulley,
a= Acceleration of the two bodies,
α= Angular acceleration of the pulley, and
P 1 and P 2 = Pulls in the strings.
First of all, consider the motion of body 1 of mass m 1 , which is coming down. We know that
the forces acting on it are m 1 g (downwards) and P 1 (upwards). As the body is moving downwards,
therefore, resultant force
=m 1 g – P 1 ...(i)
Since the body is moving downwards with an acceleration (a), therefore force acting on it
=m 1 a ...(ii)
Equating equations (i) and (ii),
m 1 g – P 1 =m 1 a ...(iii)
Now consider the motion of body 2 of mass m 2 , which is going upwards. We know that the
forces acting on it are m 2 g (downwards) and P 2 (upwards). As the body in moving upwards, there-
fore, resultant force
=P 2 – m 2 g ...(iv)
Since the body is moving upwards with an acceleration (a) therefore force acting on it
=m 2 a ...(v)
Equating equations (iv) and (v),
P 2 – m 2 g=m 2 a ...(vi)
Now consider motion of the pulley, which is rotating about its axis due to downward motion
of the body of mass m 1 tied to the string. We know that linear acceleration of the body 1 is equal to
the angular acceleration of the pulley.
∴ a=rα ...(vii)
and torque, T= Net tension in the string × Radius of pulley
=(P 1 – P 2 ) r ...(viii)
We also know that torque on the pulley,
T=Iα ...(ix)
Fig. 31.10. Motion of two
bodies