Question

a uniform chain of mass M and length L is lying on a horizontal table with one third of its length hanging over the edge of the table if the chain is in limiting equilibrium what is the coefficient of friction for the contact between the table and chain​

Answer 1

Answer:

0.5

Explanation:

Mass of chain on table = (2/3)*M

Friction force = uMg = u(2/3)*Mg

This will be equal to mg of hanging part

Mg/3 = u(2/3)*Mg

u = 1/2 = 0.5

Answer 2

Answer:

0.5 will be the coefficient of friction.

Explanation:

Since we get from the question that 2/3rd of the chain of length L is lying on the horizontal table so, the mass of chain which is lying on table = (2/3)*M.

Also, from the above data's we will get the Friction force = μMg = μ(2/3)*Mg  where the μ is the coefficient of static friction between the table and the chain.

And we know that this frictional force will be equal to weight or mg of hanging part  of the chain.

Mg/3 = μ(2/3)*Mg .

On solving we will get the value of coefficient of friction to be μ = 1/2 = 0.5.