# Pulley Tension Problem with Bosun

1. Sep 28, 2008

### Phan

1. The problem statement, all variables and given/known data
A man sitting in a bosun's chair that dangles from a massless rope, which runs over a massless, frictionless pulley, and back down to the man's hand, wants to pull himself up. The mass of the man and the chair combined is 95kg, with what force magnitude must the man pull on the rope if he wants to:
a] move up at a constant velocity
b] move up at 1.3m/s^2

2. Relevant equations
Irrelevant at the moment.

3. The attempt at a solution
I think I have found out the answer or the key to solving this problem, but there are still several points that I don't understand. I know that since the man is holding on to the rope, there must be two tensions of equal magnitudes on each side, I call that T1 & T1. Using this, I did

2T1 - mg = ma = 0

My question is, how would this change if the man was no longer holding on to the rope? I'm slightly confused, as I know that if the man lets go, he would experience mg down, since there is no longer any tension in the rope, so he would have a downward force of 931N.
So using this, if we go back to him holding the rope, is it simply because that there are two tensions, that this total force is divided in half for each side for him to be in equilibrium? Is this the right way of thinking?

Part C also asks to find the same things if a man at the bottom were to hold the rope. Am I correct in assuming that this will turn it into a normal pulley problem where for the bosun side:
T1-mg= 0?

2. Sep 29, 2008

### tiny-tim

Hi Phan!

Yes that's all correct!

"the right way of thinking"?

You can think of it any way you like, so long as it's consistent with the equations.

One way is to think of it as gearing, and to look at the work done (which is the same as energy gained) …

in the first case, pulling the rope one metre only lifts the chair half a metre … so the gearing is 2 (or is it 1/2? … i can't remember which way round it is ) … the "h" in the PE changes only half as fast as the "d" in the F.d = work done, so the F must be twice as much.