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Mechanics

A fixed wedge ABC is in the shape of an equilateral triangle of side l. Initially, a chain of length 2l and mass m rests on the wedge as shown. The chain is slowly being pulled down by the application of a force F as shown. Work done by gravity till the time, the chain leaves the wedge will be :

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10 Years agoGrade
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2 Answers

Profile image of Mallikarjun Maram
10 Years ago
Initial center of mass of the chain is at a height of \frac {L}{2} sin 60^o = \frac {\sqrt {3}}{4}L, from base of wedge.  Final center of mass is at a depth L below the base of wedge.  Hence total distance moved by center of mass of the chain is (\frac {\sqrt {3}}{4} + 1)L.  Work done by the gravitational force is then equal to \frac {mgL(\sqrt {3} + 4)}{4}.
 
Profile image of DHEERAJ
10 Years ago
Initial center of mass of the chain is at a height of \frac {L}{2} sin 60^o = \frac {\sqrt {3}}{4}L, from base of wedge.  Final center of mass is at a depth L below the base of wedge.  Hence total distance moved by center of mass of the chain is (\frac {\sqrt {3}}{4} + 1)L.  Work done by the gravitational force is then equal to \frac {mgL(\sqrt {3} + 4)}{4}.