Question

Two identical weights of mass M are linked by a thread wrapped around a frictionless pulley with
a fixed axis. A small weight of mass ‘m’ is placed on one of the weights. What is reaction force
between m and M?

Answer 1

mass on light side = M
tension = T

T-Mg = Ma
T = Mg + Ma

mass on heavy side = M+m
(M+m)a = (M+m)g - T
so

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

Ma + Ma + ma = m g

(2 M+m)a = m g

a = m g/(2M+m) which we could have guessed in the beginning :)

Forces on m
Reaction force up = F
force down = m g

a down = a = m g/(2M+m)
so
mg - F = m^2 g/(2 M+m)

F = m g (1 - m/(2M+m) )

Answer 2

We can say that

a=mg/2M+m.....(equń1)

For block mg acts downward and F upward.

Assuming blocks move down, mg-F=ma
From equn 1 put value of 'a'
So F=2Mg/2M+m...(simple calculation)

Answer 3

This answer is not in options