How Does a Frictionless Pulley System Affect the Force on a Ceiling Hook?

[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 :/

 

[Saitama]

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

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

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.

 

[Simon Bridge]

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. [edit]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 $w_1=m_g, \; w_2=m_2g \; w_p=70\text{N}$