Question

A uniform chain hangs partly from the edge of a table.As a result the remaining pan on the table is in equilibrium.If 1/n th of the total length of the chain hangs down,coefficient of static friction between chain and table is​

Answer 1

Answer:

μ = 1 / ( n - 1 )

Explanation:

Given----> 1/nth part of uniform chain hangs down and remaining part of chain lies on the table at rest

To find-----> Coefficient of static friction between chain and table

Concept used ---->

1) When body about to move in other words ready to move over a surface then frictional force is static

Static friction = μ R

2) If a body is in rest or equilibrium than resultant force on body is zero

3) Weight of a body

= mass of body × gravitational acceleration

Solution----> Let length of whole chain be l , then

Length of part of chain hanging = l / n

Length of part of chain resting on table = ( l - l/n )

Let mass of per unit length of chain be m

Mass of hanging part of chain = ( l/n ) m

Mass of chain rest on table = ( l - l/n ) m

Weight of hanging part = ( l/n ) mg

Weight of part of chain rest on table = ( l - l/n )mg

Let , Tension in chain be T and normal reaction of table on part of chain rest on table be R.

Now for equilibtium of hanging part,

T = ( l/n ) mg .....................( 1 )

Now , for equilibrium of part of chain resting on table

R = ( l - l/n ) m g

T = μ R

Putting value of R in it , we get,

T = μ ( l - l / n ) m g ...............( 2 )

Equating value of T from ( 1 ) and ( 2 ) , we get

( l/n ) m g = μ ( l - l/n ) m g

m g is cancel out from each side and we get ,

l/n = ( l - l/n ) μ

l / n = l ( 1 - 1/n ) μ

l is cancel out from both sides and we get,

1 / n = ( n - 1 ) μ / n

n is cancel out from each side

1 = ( n - 1 ) μ

μ = 1 / ( n - 1 )