Question

A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25 s, then the maximum fraction of the length of the chain that can hang over one edge of the table is

• A. 0.2
• B. 0.25
• C. 0.35
• D. 0.15

Answer 1

Answer: (A)

0.2

Answer 2

Let M is the mass of the chain of length L. If y is the maximum length of chain which can hang outside the table without sliding, then for equilibrium of the chain, the weight of hanging part must be balanced by the force of friction on the portion on the table

W=$f_L$......(i)
But from figure
$W=\frac{M}{L}yg$ and $R=W'=\frac{M}{L}(L-y)g$
So that
$f_L=\mu R=\mu\frac{R}{L}(L-y)g$
Substituting these values of W and $f_L$ in eqn.(i),
we get $\mu\frac{M}{L}(L-y)g=\frac{M}{L}yg$
or $\mu(L-y)=y$ or $y=\frac{\mu L}{\mu +1}=\frac{0.25 L}{1.25}=\frac{L}{5}$
or $\frac{y}{L}=\frac{1}{5}=\frac{1}{5}\times100$
$=20\%$