A partly hanging uniform chain of length L is resting on a rough horizontal table. l is the maximum possible length that can hang in equilibrium. The coefficient of friction between the chain and table is

Question

A partly hanging uniform chain of length L is resting on a rough horizontal table. l is the maximum possible length that can hang in equilibrium. The coefficient of friction between the chain and table is (A) (l/L-l) (B) (L/l) (C) (l/L) (D) (l L/L+l)

Tardigrade Admin · Accepted Answer

If μ is the mass/length, then Weight of hanging length =μ lg Weight of chain on table =μ(L-l) g R=μ(L-l) g f=μs R=μs μ(L-l) g Equating, μs μ(L-l) g=μ l g or μs=(l/L-l)

Q. A partly hanging uniform chain of length $L$ is resting on a rough horizontal table. $l$ is the maximum possible length that can hang in equilibrium. The coefficient of friction between the chain and table is

$\frac{l}{L - l}$

$\frac{L}{l}$

$\frac{l}{L}$

$\frac{l L}{L + l}$

If $μ$ is the mass/length, then
Weight of hanging length $= μ l g$
Weight of chain on table $= μ (L - l) g$
$R = μ (L - l) g$
$f = μ_{s} R = μ_{s} μ (L - l) g$
Equating, $μ_{s} μ (L - l) g = μ l g$
or $μ_{s} = \frac{l}{L - l}$

Q. A partly hanging uniform chain of length L is resting on a rough horizontal table. l is the maximum possible length that can hang in equilibrium. The coefficient of friction between the chain and table is

L−ll​

lL​

Ll​

L+llL​