# Statics/Equilibrium Physics question-- Two weights and a single pulley

1. Sep 16, 2014

### zubb999

1. The problem statement, all variables and given/known data
The 60 lb collar "A" is on a friction-less vertical rod and is connect to a 65 lb counterweight "C". Determine the value "h" (which is the height between "A" and "B") for which the system is in equilibrium. Also, find the horizontal force "N" acting on collar "A".
Here is a rough typed up picture of what the picture looks like, the zeros (0) are empty space

<--15in-->
00000000B0 ^
0000000/0|0 |
000000/00|0 |
00000/000|0 |
0000/000 C0 h
000/0000000 |
00/00000000 |
0/000000000 |
A0000000000v

The horizontal distance between A and B is 15 in. There is a pulley at B and a rope/string going from a A to B, and B to C. From what I can tell, the rope is continuous from points A to C. I have no idea where to get started on this or how to ultimately find h or the horizontal force acting on point A.
A is a 60 lb weight.
C is a 65 lb weight.

2. Relevant equations
All I have so far are these few things:
∑Fx = 0
∑Fy = 0
TBC = 65 lb
TBA = 65 lb

3. The attempt at a solution
I really don't know where to begin with this problem. I'm thoroughly confused as to how I should go about finding "h" or the horizontal force at "A", and I haven't been able to find similar problems to this one. If anyone can help me out, it'd be much appreciated.

2. Sep 16, 2014

### BvU

Hello Zubb, and welcome to PF.

Nice picture..
As a first step it might be nice to consider only A and the section AB of the rope. What vertical force is needed to keep A from moving up (or down) ?
The rope can only exercise a force in a direction along the rope. With one component known and the tangent of the angle expressed in h and the 15 in, you have an expression for the total tension in the rope. Then it's time to bring B and C back in the considerations.

3. Sep 17, 2014

### zubb999

Ok...

So according to what you just said, I should have this at my disposal:
TBAy = 60 lb.
tanθ = h/15 in. --> θ=tan-1(h/15 in.)
If TBA = TBAy x sinθ
then TBA = TBAy x sin(tan-1(h/15 in.))
If TBA = TBAx x cosθ
then TBA = TBAx x cos(tan-1(h/15 in.))

Right?

And then when I bring TBC to put all this into the ∑Fx and ∑Fy equations, and I end up with this:
∑Fx = TBA/(cos(tan-1(h/15 in.))) = 0? [substituted TBAx out]
∑Fy = A[-60 lb.] + C[-65 lb.] + TBA/(sin(tan-1(h/15 in.))) + TBC[65 lb.] = 0 [substituted TBAy out]

I'm terribly sorry, but I'm not seeing how this helps me...
I really do appreciate this help though.

4. Sep 17, 2014

### BvU

Well, you have TBA,y, you have TBA and you have a relationship with one unknown, h.
What can be easier ?