Hooks and pulleys (torque and force)

[Niles]

[SOLVED] Hooks and pulleys (torque and force)

Homework Statement

Two weights are connected by a very light flexible cord that passes over a 50.0N frictionless pulley of radius .300m. The pulley is a solid uniform disk and is supported by a hook connected to the ceiling. What force does the hook exert on the ceiling. The weight on the left is 75.0N, and the weight on the right is 125N.

The Attempt at a Solution

I can find the acceleration of the entire system, but I can't see how this would change the weight of the entire system?

 

[Shooting Star]

Why don't you start by finding the tensions in both parts of the string? Are they equal in magnitude?

 

[Niles]

No, they are are not - else the system wouldn't accelerate. But still, what do I do when I've found the tensions, the acc. and all?

 

[Shooting Star]

No, they are are not - else the system wouldn't accelerate. But still, what do I do when I've found the tensions, the acc. and all?

Then we can calculate the reaction on the pulley. But first answer the question we've asked.

 

[Niles]

Ok, the torque on the pulley is:

(T_2-T_1) * R = ½ * M * R * a (T_2 is tension for heavy block, T_1 tension for small block and a is linear acceleration of blocks and pulley)

For heavy block: m_h * a = m_h * g - T_2

For small block: m_s * a = T_1 - m_s*g

The tension for small block is: T_1 = m_h*(a-g) + ½*m_pulley*a

The tension for heavy block is: T_2 = m_s(a+g) + ½*m_pulley*a.

From here, what do I do?

 

[Niles]

Btw, I get that T_1 = 91,67 N, T_2 = 97,22 N and the acc. is 2,18 m/s^2

 

[Shooting Star]

In a massless string, which does not have any frictional force acting on it, the tension is constant through out its length. So, there is only one tension T.

For the 75 N weight, whose mass I'll call m1, and which is going up, m1a = T - m1g.

Find a similar eqn for m2.

Then find T.

The pulley has two strings hanging from it, and the tension in both is T. What will be the net reaction on the pulley?

 

[Niles]

What? The tension is the same?! How does the pulley accelerate then? And in my book they operate with two tensions.

 

[Shooting Star]

Why should the pulley accelerate? The chord just slips over it.

 

[Niles]

I'm not sure about that. The pulley has an angular acceleration.

 

[Shooting Star]

Then rewrite the original problem clearly.

 

[Niles]

The problem statement is as I wrote in the first post beginning. It's for an "Advanced Classical Mechanics"-course, so we have to operate with angular acceleration of pulley, torque and all that..

 

[Shooting Star]

Is that your logic for including angular accn of the pulley, in a situation where it possibly cannot have any?

Follow the method I've outlined in post no.7, and find the value of T.

 

[Niles]

I've found T = 93,75 N.

But still, since the two blocks are acc., then surely the pulley must have an ang. acc. as well?

 

[Shooting Star]

Please tell me why the pulley must have angular accn?

 

[Niles]

First I want to thank you, because you showed me that the net force on the ceiling is T_1 + T_2 + weight of pulley.

Second, since the two blocks are accelerating, the string is accelerating and the string turns the pulley - hence the pulley must have an angular acceleration.

This method gave me the correct answer, which is 239 N; T_1 + T_2 + weight, pulley = 239 N.

Again, thank you.

 

[Shooting Star]

Hold on a sec here. In your 1st post, you have written that it's a frictionless pulley, and confirmed it later when I asked you! If there is friction between string and pulley, T1 and T2 would be different, and the pulley will definitely undergo ang accn.

If you had been a bit more careful, we'd solved it long ago. Anyway, better late than never.

 

[Niles]

When you said frictionless, I thought you meant that there is no friction between the pulley and it's axis of rotation - I didn't think you meant the string and the pulley. But still, we solved it and I will be more careful next time.

 

[Shooting Star]

That's all right. In fact, you did the problem completely by yourself because of the misunderstanding. Next time, formulate problems carefully.