Question

A uniform chain of length L is lying partly on a table the remaining part hanging down from the edge of the table If the coefficient of friction between the chain and table is 0.5 what is the minimum length of chain that should lie on the table to prevent the chain from slipping down the table.

Answer 1

Answer: $a = 2L / 3$

Given that the coefficient of friction is 0.5, assume that the length of chain on table be a and therefore length of chain which is left hanging be L – a.

Now to keep them equilibrium the force of friction be equal to force acted due to gravity upon the chain which will make it slide.

So,

                  $F_f = F_g\\ => \mu N = mg\\ => \frac{1}{2} m^{'} g = m g\\ =>  as\quad the\quad linear\quad mass\quad density\quad equals\quad to\quad L / m\\ => \frac{1}{2} a \lambda = ( L -a) \times \lambda => a + \frac{1}{2} a = L\\=> a = 2L / 3$

Answer 2

Answer:

answer is 1/2

for these type of questions such formula is used......