Question

A stationary pulley carries a rope one end of which supports a ladder with a man and the other end a counter weight of mass $M$. The man of mass $m$ climbs up a distance $\ell$ w.r.t the ladder and then stops. The displacement of the centre of mass of this system is :

• A. $\frac{m\ell }{M+m}$
• B. $\frac{m\ell }{2M}$
• C. $\frac{m\ell }{M+2m}$
• D. $\frac{m\ell }{2M+m}$

Answer 1

The correct option is B $\frac{m\ell }{2M}$

Let $x$ be the displacement of ladder. So displacement of with respect to ladder is $l-x$. (As we take downward direction as negative.)

Displacement of center of mass$=\frac{Mx-x(M-m)+m(l-x)}{M+M-m+m}=\frac{ml}{2M}$