Two masses M and m are connected to the two ends of an inextensible st

1.	Two masses M and m are connected to the two ends of an inextensible string. The string passes over a smooth frictionless pulley. Calculate the acceleration of the masses and the tension in the string. (Given M > m)
Answer» Let a = acceleration with which the mass "M" moves downwards and mass "m" moves upwards. T = tension in the string Net downward force acting on mass M is F = Mg - T But F = Ma Hence, Ma = Mg - T ...(i) Net upward force acting on mass 'm' is ma = T - mg ....(ii) Adding (i) and (ii), we get (M + m)a = g(M - m) ⇒ a = ({M - m}/{M + m}) x g Putting this value in equation (ii), we have m({M - m}/{M + m})g = T - mg ⇒ T = mg[{M - m}/{M + m} + 1] = [{2M}/{M + m}] x g

Two masses M and m are connected to the two ends of an inextensible string. The string passes over a smooth frictionless pulley. Calculate the acceleration of the masses and the tension in the string. (Given M > m)

Answer»

Let a = acceleration with which the mass "M" moves downwards and mass "m" moves upwards.

T = tension in the string

Net downward force acting on mass M is

F = Mg - T

But F = Ma

Hence, Ma = Mg - T ...(i)

Net upward force acting on mass 'm' is

ma = T - mg ....(ii)

Adding (i) and (ii), we get

(M + m)a = g(M - m)

⇒ a = ({M - m}/{M + m}) x g

Putting this value in equation (ii), we have

m({M - m}/{M + m})g = T - mg

⇒ T = mg[{M - m}/{M + m} + 1]

= [{2M}/{M + m}] x g