Atwood Machine + Torque - need a nudge

[glossolalia]

Homework Statement

Two weights, 75N (mass 1) & 125N (mass 2) are connected by a very light flexible cord that passes over a 70.0 N frictionless pulley of radius 0.200 m. The pulley is a solid uniform disk and is supported by a hook connected to the ceiling. What force does the ceiling exert on the hook?

Homework Equations

I_cylinder=mr^2/2
tau = I(alpha)
alpha= a(r)
F=ma

The Attempt at a Solution

I'm realizing there're going to be two tensions on either side of the pulley and they are generating torque. What I've come up with:

7.653kg(a) = T_1 - 75N
12.755(a) = 125N - T_2
or
a = (T_1 - 75N)/7.653kg = (125N - T_2)/12.755

tau = 1.4kgm^2(.2m)(a)

(The angular acceleration (alpha) = .2*a)

I'm having a hard time substituting to get my values for my tension. i realize this is probably basic algebra that I'm just not seeing, but a push would really help. thanks!

 

[gneill]

One torque will be positive, the other negative. Or, if you will, one is trying to make the pulley turn clockwise while the other opposes it. So in this case your net torque will be (T2 - T1)r.

Take another look at your value for the moment of inertia for the pulley; it looks like an order of magnitude slipped.

 

[glossolalia]

got sloppy and used the N instead of the kg for the inertia. right. so i get T_2 - T_1 = Ia/r

or .715kgm(a). I'm still staring at three variables and my eyes are starting to bleed.

T_2 - T_1 = .715kgm(a)
7.653kg(a) = T_1 - 75N
12.755(a) = 125N - T_2

sure, let's factor out a and get
(T_1 - 75N)/7.653kg = (125N - T_2)/12.755 = .715kgm/(T_2 - T_1)

now what? can you tell the 17 years between high school algebra and college physics is taking its toll?

 

[gneill]

Suppose we just consider masses for now, rather than weights. Also, let's avoid dragging actual numbers through the manipulations until the end game. Then for the tensions,

T1 = (a + g)m1 and T2 = (g - a)m2 (for suitable choice of direction for a).

Further, if we looks at the situation with the pulley,

α = (τ2 - τ1)/I

You can convert the tensions T1 and T2 to torques τ1 and τ2 by a brisk massage with r. And the angular acceleration is related to the linear acceleration by a = αr.

So take the expressions for the tensions, make them over into torques, and plug them into the angular acceleration equation. Convert α to a. Solve for a.

With a you can go back and solve for the two tensions, which you should be able to do something clever about to resolve the problem posed.

 

[glossolalia]

i knew it was something simple i was missing in substitution - if you say

T_2 = T_1 + 1/2m_pulley(a)
then
T_1 + 1/2m_pulley(a) = m_2(g-a)
then
T_1 = m_2(g) - (1/2m_pulley + m_2)a
then
m_2(g) - (1/2m_pulley + m_2)a = m_1(a-g)
and we get values

Thanks very much.

For anyone else working this problem, googling along, realize your final answer is going to be the sum of the tensions and the weight of the pulley. Huzzah!