Question

The upper end of a uniform open-link chain of length L, and mass per unit length λ, is released from rest at x = 0.

The lower end is fixed at point A as shown in the figure. Find the tension T(x) in the chain at point A after the upper end of the chain has dropped the distance x. Assume the free fall of the chain.

• A. $\frac{3\lambda xg}{2}$
• B. $\frac{\lambda g}{2}(L+3x)$
• C. $\frac{\lambda g}{2}(L+x)$
• D. $\frac{\lambda g}{2}(L+5x)$

Answer 1

Answer: (A)

$\frac{3\lambda xg}{2}$

Length of the left part$=L-\frac{x}{2}$

Length of the right part$=\frac{x}{2}$

Due to the falling of left part of the chain, the momentum of the left part keeps changing with time, giving rise to a force$=\frac{dp}{dt}$

$=\frac{d\left(mv\right)}{dt}=\frac{d\left(\lambda \right(L-\frac{x}{2}\left)\right)}{dt}=-\frac{\lambda v^2}{2}$

The gain in kinetic energy is due to the loss in potential energy of the chain$=\frac{1}{2}m{v}^2=mgx$(center of mass of chain falls by x)

$⟹v^2=2gx$

Tension on the string= Gravitational pull on right part of the chain+Force due to change in momentum of left part

$=\left(\lambda \frac{x}{2}\right)g+\frac{\lambda \left(2gx\right)}{2}=\frac{\lambda gx}{2}+\lambda gx=\frac{3\lambda xg}{2}$