Question

A uniform chain of length L and weight W is changing vertically from its ends A and B which are close together. At a given moment the end B is released. Find the tension at A when B has fallen a distance x < L.

Answer 1

GO THROUGH THE ATTACHED IMAGE..

Answer 2

Solution :-

Initially length of chain on either side = L/2.As soon as end B is released the lower point C begins to descend. When B falls a distance X, C descends through a distance X/2.

Let m be mass per unit of chain, then a small mass dM = m dx of the chain will fall through a distance X/2 and would acquire a velocity V given by

$\frac{1}{2} (dM)u {}^{2} = (dM)g . \frac{x}{2} = > u = \sqrt{gx}$

when it reaches lowest point C, it is brought to rest.

• Force, F = rate of change of momentum.

$= \frac{(mdx)u}{dt} = m( \frac{dx}{dt} )V = mv {}^{2} \\ \\ = mgX(∵ \: u = \frac{dx}{dt} )$

$= m( \frac{L}{2} + \frac{X}{2} ) + \frac{Wx}{L} = \frac{W}{2} (1 + \frac{3x}{L} )$

$As \: mg= \frac{Mg}{ L} = \frac{W }{L} \: \: ∴ F = \frac{Wx}{L}$

• Tension A = weight of length [L/2 + W/2] of chain + F Force, F = rate of change of momentum.

• Tension A = weight of length [L/2 + X/2] of chain + F.

$= m( \frac{L}{2} + \frac{X}{2} ) + \frac{Wx}{L} = \frac{W}{2} (1 + \frac{3x}{L} )$