# Movement of man and ladder and their center of mass

Suppose there is a massless frictionless pulley. A rope over it carries a mass $M$ and on other side carries a ladder of mass $(M-m)$ and a man on that ladder, of mass $m$. Now the man starts climbing the ladder at a velocity $v$ upwards($v$ in ground frame).

The ladder and the block also move, but I cannot happen to imagine his hands going down on the ladder legs and the ladder going down in the ground frame. I need to somehow get to the movement of center of mass.

My view is that since momentum should be conserved(he is climbing the ladder with distinct downward forces that span over very small time, say $dt\rightarrow 0$, so impulse should be $0$). This means that $v_{cm}$ should be $0$, since its expression has total momentum of system in the numerator, which was initially $0$ and hence finally will be $0$. This means the center of mass should not change, which somehow seems absurd, as the man is moving up. Could anyone make clear what goes down and whether $y_{cm}$ changes or not?

(And yes cm means center of mass.)

-
You are right in thinking that this won't end with the man drifting in one direction or the other; the forces will balance out as he accelerates and decelerates. However, the center of mass of the whole system will go up. $y_\text{cm}$ does change. Try writing out the potential energy of the man, then the ladder, and then the mass in terms of the man's displacement and the mass's displacement. Figure out how much work the pulley will do on the system. – krs013 Mar 27 '13 at 13:39
Also, are you talking about the center of mass of the man and the ladder, or the whole system together? – krs013 Mar 27 '13 at 13:58
@krs013: I'm talking about the man($m$)+ladder$(M-m)$+block($M$) system; And I think the pulley won't do any work, the force of tension is by the rope. I don't understand what should I do with potential energy.(Please be kind enough to guide further) – Ashish Gaurav Mar 27 '13 at 14:03
This is a very similar problem, but looks at displacement of cm, not velocity. It should be easy to convert from displacement to velocity. It is Problem and solution 23: books.google.com.au/… – Mew Mar 27 '13 at 14:18
Just for clarity's sake, I'll point out that the pulley does work on the entire system in that the center of mass moves up and the force that sustains it is provided by the pulley. Of course, it's a frictionless pulley, so it's not doing any work rotationally, just vertically. Good job figuring it out! – krs013 Mar 27 '13 at 14:35

It is easy to see from this website, problem/solution 23, http://books.google.com.au/books?id=8NtJLfGf94QC&pg=PA431&lpg=PA431&dq=monkey+climbing+ladder+on+pulley&source=bl&ots=2tWFRhdKME&sig=9ocddC7lk53A1_XdrVAwPm7MBw0&hl=en&sa=X&ei=Jv9SUevUEoTziAfSs4HIBg&ved=0CDAQ6AEwAA#v=onepage&q=monkey%20climbing%20ladder%20on%20pulley&f=false, that as the man moves up a distance $l'$, the centre of gravity moves up by a distance $l = \frac{ml'}{2M}$. Where $m$ is the mass of the man, and $M$ is the mass of the counter-weight.
This means that if the man is moving at a velocity of $V$ meters per second, then the centre of gravity will move up at a rate of $\frac{mV}{2M}$ meters per second as well.
I finally think something works. I don't need to imagine anything either. So, I assume the block's lower part is at height $H$ from the ground, ladder's last step at height $h$, the height of ladder being $l$ and the distance of man from the bottom being $x$ (where $x<l$ ).
Now $$y_{cm}=\frac{M(H)+(M-m)(h+\frac{l}{2})+m(h+x)}{(M)+(M-m)+(m)}$$ This solves as $$y_{cm}=\frac{H}{2}+\frac{h}{2}+\frac{l}{4}+\frac{m}{M}(\frac{x}{2}-\frac{l}{4})$$ After differentiating, I have $$v_{cm}=\frac{1}{2}(\dot H+\dot h)+\frac{m\dot x}{2M}$$ Since whatever be the velocity of block, ladder moves opposite to it by same velocity as the string/rope is inextensible. Hence $\dot H+\dot h=0$ So, $$v_{cm}=\frac{m\dot x}{2M}$$