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A pulley fixed to a rigid support carries a rope whose one end is tied to a ladder and a man and the other end to the counterweight of mass $M$. The man of mass $m$ climbs up a distance of $h$ with respect to the ladder and then stops. If the mass of the rope and the friction in the pulley axle are negligible find the displacement of the center of mass of the system.

My question is that the net external force on the system is 0 then why is there any displacement of centre of mass?

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There is a net force on the system of the mass and the man in the form of the normal force provided by the pulley on the string. Similar questions have already been asked here, here and here.

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  • $\begingroup$ what is the effect of that normal force on the displacement of centre of mass. Can you please elaborate? $\endgroup$ – Physics freak Sep 28 '18 at 7:20
  • $\begingroup$ @Ritu Khanna: I believe that it has already been mentioned very clearly in the linked posts in my answer. $\endgroup$ – user7777777 Sep 28 '18 at 7:21
  • $\begingroup$ I have already gone through the links but can't see anywhere about normal force of pulley. Yes there is a mention of work done by pulley somewhere in link no 1 but it's not clear $\endgroup$ – Physics freak Sep 28 '18 at 7:42
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    $\begingroup$ @Ritu Khanna: Consider our system to be the mass and the man connected by the string. The external forces are gravity and the normal force, which act all the time. However, if the man exerts a force on the string, the normal force increases, but gravity doesn't. That's why the net force on the system is now upwards and the center of mass moves upwards accordingly. $\endgroup$ – user7777777 Sep 28 '18 at 8:01
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    $\begingroup$ Okay, I get ut now. It was very generous of you to patiently answer my query. Thanks $\endgroup$ – Physics freak Sep 28 '18 at 8:04

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