Calculating Work: Uniform Chain on Table with Hanging Part - IIT 1985

[prateek_34gem]

Homework Statement

Uniform chain of length 'L' and mass of 'M' is lying on smooth table and one third of it's length is hanging vertically down over the edge of the table. If 'g' is acceleration due to gravity , then work required to pull the hanging part onto the table is :

A)MgL
B)MgL/3
C)MgL/9
D)MgL/18
(I.I.T :- 1985)

The Attempt at a Solution

mass of hanging part will be m/3. so it will be exerting mg/3 force downward.
so this amount of force is required to pull it up.when the chain will be up its displacement along table will be l/3 because now hanging part is on table.
so

F=Mg3
S=L/3
Work = F*S= MgL/9

But it is the wrong answer.

 

[art.t]

Try lifting the center of mass of the hanging chain up on to the table and using the formula (work done against gravity = m*g*h).

The center of mass is halfway down the part that's hanging off, which was L/3. So the center of mass is L/6 below the table top. The mass of the hanging part is 1/3 of the total mass...Go for it.

 

[prateek_34gem]

ya i got it . thnx a ton