Question

A fixed wedge $ABC$ is in the shaped 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. Find the work done by gravity till the time, the chain leaves the wedge.

Answer 1

Side of the equilateral triangle $=l$

Initially centre of mass of the chain $=\frac{L}{2}\sin 60=\frac{\sqrt{3}}{4}L$

Final centre of mass is at depth L

Total distance moved by the centre of mass of chain

$=\frac{\sqrt{3}}{4}L+L$

$=L(\frac{\sqrt{3}}{4}+1)$

Work done

$W=F.l$

$=mgl$

$=L(\frac{\sqrt{3}}{4}+1)mg$

$=mgl(\frac{\sqrt{3}}{4}+1)$