Rotational Dynamics and pulley

[J89]

Homework Statement

Two weights are connected by a very light flexible cord that passes over a 50 N frictionless pulley of radius .300 m. Pulley is a solid uniform disk and is supported by a hook connected to a ceiling. What force does the ceiling exert on the hook? There are two weights around each side of the disk, one is 75 N and the other is 125 N. Answer is 249 N.

Homework Equations

I = 1/2 MR^2
F=ma

The Attempt at a Solution

 

[tiny-tim]

Two weights are connected by a very light flexible cord that passes over a 50 N frictionless pulley of radius .300 m. Pulley is a solid uniform disk and is supported by a hook connected to a ceiling. What force does the ceiling exert on the hook? There are two weights around each side of the disk, one is 75 N and the other is 125 N. Answer is 249 N.

Hi J89!

(I assume that "frictionless pulley" means that there is friction between the pulley and the rope, with no slipping)

The tension in the two sections of rope will be different … call them T_m and T_M, and call the linear acceleration a …

then apply good ol' Newton's second law three times, ie to each of the weights and the pulley (separately), to get T_m + T_M

 

[nlightner]

Based on the given information, we can use the equation F=ma to determine the acceleration of the system. Since the pulley is frictionless, the net force acting on the system is equal to the sum of the tensions in the cord on either side of the pulley. We can set up two equations, one for each weight, as follows:

For the 75 N weight:
F - T = ma

For the 125 N weight:
T - F = ma

Where F is the force exerted by the ceiling, T is the tension in the cord, m is the mass of each weight (assuming they are equal), and a is the acceleration of the system.

Since the weights are connected by the same cord, the tension on either side of the pulley must be equal. Therefore, we can set the two equations equal to each other and solve for F:

F - T = T - F
2F = 2T
F = T

Now, we can substitute this value of T into either of the original equations to solve for F:

F - T = ma
F - F = ma
0 = ma

Since the acceleration of the system is 0 (due to the weights being balanced by the tension in the cord), the force exerted by the ceiling, F, must also be 0. This means that the ceiling is not exerting any force on the hook.

In order for the system to have an acceleration of 0, the sum of the torques on the pulley must also be 0. The torque on the pulley is equal to the force applied (F) multiplied by the radius of the pulley (0.300 m). Since the force is 0, the torque must also be 0. This means that the force exerted by the ceiling is not necessary for the system to remain in equilibrium, and the pulley can support the weights on its own.

Therefore, the answer of 249 N is not correct. The ceiling does not exert any force on the hook, and the pulley can support the weights on its own due to its rotational dynamics.