# Pulley and hook problem.

Aguilar2393

## Homework Statement

Two weights are connected--Weight 1 (75.0N) and Weight 2 (125.0N)--by a very light flexible cord that passes over a 70.0N-frictionless pulley of radius 0.400m. The pulley is a solid uniform disk and is supported by a hook connected to the ceiling.

I calculated the mass of the first weight to be 7.64kg and the second weight to be 12.74kg

## Homework Equations

What force does the ceiling exert on the hook?

## The Attempt at a Solution

At first I tried adding the two weights and the pulley to get 270.0N. (A little too easy, but it doesn't hurt too try). I wasn't surprised to see that it was wrong. So, since there two different weights and a pulley, I tried using Newton's 2nd law to answer the problem.

Here's what I have:

T_1 - (m_1)(g) = (m_1)(a)
T_2 - (m_2)(g) = (m_2)(a)

and since the pulley is 70.0N and has a radius of 0.4m, I used (r x F) to get 28 N*m.

I don't know where to go from her unfortunately :/

## Answers and Replies

Saitama
T_2 - (m_2)(g) = (m_2)(a)

Check this one again. It should (m_2)g-T_2=(m_2)(a).

Aguilar2393 said:
and since the pulley is 70.0N and has a radius of 0.4m, I used (r x F) to get 28 N*m.
That's wrong. You have to take moment about the CM of pulley. You will notice that the moment of weight equals to zero. From here, you get a relation between T_1 and T_2.

Homework Helper
70.0N-frictionless pulley
What does this mean? The pulley adds an additional 70N in some direction?
Here's what I have:

T_1 - (m_1)(g) = (m_1)(a)
T_2 - (m_2)(g) = (m_2)(a)

and since the pulley is 70.0N and has a radius of 0.4m, I used (r x F) to get 28 N*m.
Your free body equations fail to account for the contribution of the 70N in the pulley. Redo. Oh wait - I think I see now... the 70N is the weight of the pulley.

If the cord does not stretch, then ##a_1=a_2##.
There is nothing wrong with the directions of ##T_1## and ##T_2## provided you are careful when you translate this to the rotational motion of the pulley.

If the two weights over the pulley were stationary - then the force on the hook would just be the sum of the weights hanging off it.
Since the pulley is not massless, it has a moment of inertia.

What you are missing is the expression for the total force on the hook - in terms of the tensions and the weight of the pulley.
Put ##w_1=m_g, \; w_2=m_2g \; w_p=70\text{N}##

Last edited: