Solving Dynamics Homework Problem: Ceiling Force on Hook

[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

clueless..help!

 

[CaptainEvil]

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

clueless..help!

As in any dynamics problems, start by drawing out your free body diagrams for each body - your two weights and the pulley, taking into acount all the forces acting on the weight, and the sources of torque for the pulley.

Then you can apply F = ma to each body, for each axis of movement, but don't forget your appropriate constraints.

 

[nlightner]

I would approach this problem by first understanding the concept of forces and how they act on objects. In this case, we have two weights connected by a cord, passing over a pulley and supported by a hook. The question is asking for the force exerted by the ceiling on the hook. To solve this, we need to consider the forces acting on the system and apply Newton's laws of motion.

First, we need to understand that the pulley is frictionless, meaning there is no friction force acting on it. This means that the only forces acting on the pulley are the tension forces in the cord on either side. The pulley itself does not contribute to the force exerted by the ceiling on the hook.

Next, we need to consider the forces acting on each weight. The weight of 75 N is being pulled down by gravity, while the weight of 125 N is being pulled up by the tension force in the cord. Using Newton's second law, F=ma, we can calculate the acceleration of each weight. We know that the acceleration of the system must be the same, as the cord is light and does not contribute to the overall mass.

Now, we can apply Newton's third law, which states that for every action, there is an equal and opposite reaction. This means that the tension force on one side of the pulley must be equal and opposite to the tension force on the other side. Therefore, the tension force in the cord on the side of the 75 N weight is 75 N, while the tension force on the side of the 125 N weight is 125 N.

Finally, we can calculate the force exerted by the ceiling on the hook by considering the forces acting on the pulley. As the pulley is in equilibrium, the sum of forces in the vertical direction must be zero. This means that the force exerted by the ceiling on the hook must be equal and opposite to the sum of the tension forces on either side of the pulley. Therefore, the force exerted by the ceiling on the hook is 75 N + 125 N = 200 N.

It is important to note that the given answer of 249 N is incorrect. This may be due to an error in calculation or misunderstanding of the problem. As a scientist, it is important to double check our calculations and make sure they align with the given information and physical principles.