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` `A uniform chain of mass m and length l overhangs a table. Find the work to be done by the person to put the hanging part back on the table

6 years ago

The Question says that initially the chain was hanging from the table to the ground which means that initially N(normal force)=mg .......... Now mass of length l of the chain is m.

so mass of 1 unit length of the chain will be m/l

so mass of elemental part is m/l * dx

no when we pull the chain back to the table it goes up by a distance 'x' onto the table,

length of chain on the table is also 'x'

so the normal force is m*g

because it is on the table the normal force on every point is the same i.e mg

dx part is pulled up at a particular rate

so N = m/l*g * dx

integrating from 0 to x we get

N = m/l*g * ∫dx

N = mgx / l

work done is N*g

Work is mg

^{2}x/ l

my apologies if the answer is incorrect but this is the procedure . If i have made some calculation error please excuse.

Best Of luck

Please approve !

6 years ago

mass of 1/3rd section of chain will be M/3

now consider the center of mass of hanging part will be at center of hanging part...

workdone in pulling hanging part above the table is equal to workdone in pulling center of mass of hanging part to the table...

displacement of center of mass (h)= l/6

workdone = (dm)gh

=Mgl/18

6 years ago

Dear student,

A hanging flexible chain suspended from its ends is subject to tension. The tension depends on the horizontal and vertical distances between its ends and on the length of the chain. Use the slider to move the point M to see how this tension varies along the chain.Note that the tension is calculated for a linear weight of the chain of 1 N/m (the tension varies linearly with this linear weight). The color of the chain gives the following qualitative information: green corresponds to the minimum and magenta to the maximum of the tension.

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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

6 years ago

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