Question

The chain of length L and mass per unit length rho is released from rest on the smooth horizontal surface with a negligibly small overhang x to initiate motion. Determine: (a) the acceleration of the chain as a function of x, (b) the tension T in the chain at the smooth corner as a function of x, (c) the velocity v of the last link A as it reaches the corner

Answer 1

Answer:

Part a)

$a = \frac{x}{L} g$

Part b)

$T = \rho x g(1 - \frac{x}{L})$

Part c)

$v = \sqrt{gL}$

Explanation:

Part a)

Net pulling force on the chain is due to weight of the part of the chain which is over hanging

So we know that mass of overhanging part of chain is given as

$m = \rho x$

now net pulling force on the chain is given as

$F = \rho x g$

now acceleration is given as

$F = Ma$

$\rho x g = \rho L a$

$a = \frac{x}{L} g$

Part b)

Tension force in the part of the chain is given as

$mg - T = ma$

$\rho x g - T = \rho x a$

$\rho x(g - a) = T$

$\rho x (g - \frac{x}{L} g) = T$

$T = \rho x g(1 - \frac{x}{L})$

Part c)

velocity of the last link of the chain is given as

$a = \frac{x}{L} g$

$v\frac{dv}{dx} = \frac{x}{L} g$

now integrate both sides

$\int v dv = \frac{g}{L} \int x dx$

$\frac{v^2}{2} = \frac{gL}{2}$

$v = \sqrt{gL}$