# Centre of mass help!

1. Dec 7, 2012

### SK_JEE

1. A stationary pulley carries a rope,one end of which supports a ladder with a man and the other a counter weight mass M.The man of mass 'm' climbs up a distance "L" wrt the ladder and then stops.The displacement of CM of the system is?

2. Relevant equations
How do i approach this question

3. The attempt at a solution
As the man moves up,the mass M moves up too..so ml+ML/m+M ,something like this..?

Last edited by a moderator: Dec 7, 2012
2. Dec 7, 2012

### Staff: Mentor

Welcome to the PF.

If the counterweight has the same mass as the person, and they are connected by a rope over a pulley, as the man moves up the counterweight should move down, no?

3. Dec 7, 2012

### SK_JEE

Aren't they in equilibrium,so I guess it shouldn't move at all..

4. Dec 7, 2012

### Staff: Mentor

Yeah, but the problem says the man uses the ladder to move himself. Kind of like an astronaut pulling himself along inside the Space Station...

5. Dec 7, 2012

### SK_JEE

So,if the man moves up,the ladder moves down..and the mass M moves up?

6. Dec 7, 2012

### Staff: Mentor

Oh, sorry, I misread your first post with the question. So the counterweight supports both the man and the ladder? I was assuming that the ladder was fixed, but re-reading your post it says that the man and ladder are both on one end of the rope, balanced by the counterweight.

Hmm. So it's more like the man is holding the ladder, and can move it up and down with respect to his position? Is there a figure showing the problem?

7. Dec 7, 2012

### SK_JEE

There isn't any diagram,but I assumed and made a quick one,here
http://gyazo.com/ee2fda8cfbe22eae4fc549534b053db4

8. Dec 7, 2012

### Staff: Mentor

Okay, so it's more like the man is holding the ladder. Instead of really climbing up the ladder, it's more like the man is lowering the ladder, because there is nothing supporting the ladder, so the man can't really "climb up".

If the man is lowering the ladder and that does not cause him to move anywhere himself, what happens to the CM of the system?

9. Dec 7, 2012

### SK_JEE

Not too sure about the ques,but ans isnt 0
its
ml/2m

10. Dec 7, 2012

### haruspex

I can't make sense of the set-up. No mass is mentioned for the ladder, so I assume that's to be ignored. So, before the man steps on the ladder the counterweight must be on the floor. Presumably the counterweight weighs less than the man.
As soon as the man steps on the ladder, the ladder is not going to be vertical. But now you get into awkward questions about the angle the rope makes to the vertical. OK, maybe we can assume a very long rope and short ladder, so we get to treat the rope as always vertical. But now, below a certain critical height of the man, the counterweight remains on the floor; above that height, the counterweight rises and doesn't stop till the man reaches the floor. I see no reason why equilibrium would be reached in between.
On top of all that, it's unclear whether the distance climbed and CoM displacement are supposed to be measured vertically or in line of movement.

11. Dec 8, 2012

### vela

Staff Emeritus
I think, as SK_JEE has assumed, the system is in equilibrium initially, so the mass of the ladder has to be M-m.

You might approach this problem using energy methods. As the man climbs, he does work. That work goes into changing the potential energy of the man, the ladder, and the counterweight.

So the m's cancel, leaving you with l/2, right? What is l supposed to represent? M and m are different variables as are L and l. Don't use them interchangeably.

Last edited: Dec 8, 2012
12. Dec 8, 2012

### haruspex

OK, but I still don't have the picture. Is the lower end of the ladder on the ground? If so, the ladder will be upright and remain so as the man ascends. Neither the ladder nor the counterweight will move.
If the ladder is suspended, then why does it not say so, rather than merely 'supported'? But supposing it is, I get mL/(2M-m).