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?
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) )
This answer is not in options
To find the reaction force between the weights, we can analyze the forces acting on each weight individually.
Let's assume that the weight with mass M is on the left, and the weight with mass m is on the right.
On the weight with mass M:
1. The force of gravity acts downwards with a magnitude of Mg, where g is the acceleration due to gravity.
2. The tension in the thread applies an upward force on this weight.
On the weight with mass m:
1. The force of gravity acts downwards with a magnitude of mg.
2. The tension in the thread applies a downward force on this weight.
Since the two weights are connected by a thread, they have the same tension acting on them. Therefore, the tension in the thread is the same for both weights, and it can be considered as the reaction force between the two.
Therefore, the reaction force between the weights is equal to the tension in the thread.
To find the tension, we can use the principle of conservation of energy:
The potential energy gained by weight m as it moves downwards is equal to the potential energy lost by weight M as it moves upwards.
mgh = Mgh
Where h is the height difference between the two weights.
Cancelling out the h and rearranging the equation gives:
m = M
Therefore, the reaction force between m and M is equal to the weight of mass m, which is mg.
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)