A uniform chain of mass M and length L is lying on a smooth table and a
part of it is hanging from the table. The work required to pull the hanging
part of the chain on to the table is mgl/32 . Find the part of the chain in terms of its
length which is hanging from the table:
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1
Answer:
The weight of hanging part ( 3/L) of chain is ( 3/1 Mg). This weight acts at centre of gravity of the hanging part, which is at a distance of ( 6/L ) from the table.
As work done =force×distance
∴W= 3Mg × 6L = 18MgL .
Answered by
0
Answer:
1/4th part of the chain is hanging from the table .
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