# Pulley Problem - Please give me hand here!

1. Apr 16, 2006

### freecorp777

Here's a pulley problem that has got me stopped; I'd appreciate any help that's offered.

There are two blocks, A (44 N) and B (22 N) connected via a rope that stretches over a pulley. Block A is resting on a table, and block B is hanging over the edge. Block C is positioned on top of Block A.

The question reads: "Given that the coefficient of static friction between Block A and the tabel is 0.20, what is the minimum weight of Block C to keep Block A from sliding off the table?"

(The answer that is given is 66 N, but I can't seem to get this)

I tried starting off with:
Block A=M=44N Block B=m=22N
(let F stand for the friction force)
T-F=Ma and mg-T=ma
Solving for Tension and setting them equal gives:
mg-ma=F+Ma
Rearranging some more and solving for a gives:
a= (mg-F)/(m + M)
The Tension on Block A then becomes M(mg-F)/(m + M) = 8.8 N
So then in order for Block A to resist motion, the Friction between it and the table must be equal to T, so F=T=8.8 .20 * (4.489 + C) * 9.8 = 8.8 but then the mass of C ends up being zero. Where did I go wrong?

Last edited: Apr 16, 2006
2. Apr 16, 2006

### Staff: Mentor

Several problems here. One, you are trying to prevent block A from sliding, so the acceleration is zero. Two, you are given the weights of the blocks, not their masses. (22N is a force, not a mass.)

Hint: Since the entire affair is in equilibrium, what must be the tension in the rope?

3. Apr 16, 2006

### freecorp777

Well, in order for Block A *not* to move, the the frictional force on Block A must be equal to the tension right?
I redid the work for the acceleration and tension.

T = (Mmg + mF) / (M + m) M=4.489 kg m=2.245 kg

The tension (not taking Block C into account yet) is 17.6 N.
Set this equal to u(M + C) g (the frictional force) and solve for C, and I get a mass of 8.97 kg, so a weight of 88 N, which is still off.

Last edited: Apr 16, 2006
4. Apr 16, 2006

### Staff: Mentor

This is true.

Now answer my question: What's the tension in the rope? If you are looking at the situation correctly, you should be able to answer without doing any calculations.

Hint: The hanging mass is in equilibrium.

5. Apr 17, 2006

### freecorp777

Yes, I think I got it. The tension is equal to the weight of the hanging block (22 N), since hte other block is resting on the table.
It's in equilibrium, so a=0, so the whole term on that side of the equal sign becomes 0. Then T=mu * m (of Block A and C) * 9.8, then solve for the mass of Block C.

6. Apr 17, 2006

### Staff: Mentor

Exactly. But don't waste time finding the masses; you are given the weights (w = mg) and that's all you need.