Particles of mass 3m and 5m hang one at each end of a light inextensible string which passes over a pulley.the system is released from rest with the hanging parts taut and vertical.during the subsequent motion the resultant force exerted by the string on the pulley is of magnitude-
To find the magnitude of the resultant force exerted by the string on the pulley during the motion of the system, we need to analyze the forces acting on the system.
In this scenario, there are two forces acting on the pulley: the tension force from the lighter mass and the tension force from the heavier mass. These tensions are transmitted through the string to the pulley.
Let's assume that the lighter mass (3m) is on the left and the heavier mass (5m) is on the right. Since the system is released from rest, the heavier mass will start to descend while the lighter mass starts to rise.
As the lighter mass rises, the tension in the string connected to the left side of the pulley will decrease. Let's call this tension T1. Similarly, as the heavier mass descends, the tension in the string connected to the right side of the pulley will increase. Let's call this tension T2.
Now, let's analyze the forces acting on the pulley:
1. The downward force due to the heavier mass: This force is equal to the weight of the heavier mass, which is 5mg (m times the acceleration due to gravity g).
2. The upward force due to the lighter mass: This force is equal to the weight of the lighter mass, which is 3mg.
Since the system is in equilibrium, the net force acting on the pulley is zero. Therefore, the magnitudes of the upward and downward forces must be equal:
3mg = 5mg
Simplifying the equation:
3m = 5m
This equation cannot be satisfied, so there is no magnitude of the resultant force exerted by the string on the pulley during the subsequent motion of the system. The tension forces in the string balance out the weights of the masses, resulting in a net force of zero on the pulley.