Question

A heavy uniform chain lies on a horizontal top of table.
If the coefficient of friction between the chain and the
table is 0.25, then the maximum percentage of the
length of the chain that can hang over one edge of the
table is
(a) 20%
(b) 25% (c) 35%
(d) 15%​

Answer 1

Answer:

Let x be the length of rope on table and (L−x) is the length hanging. Let total mass be M.  

Therefore mass of hanged part & on table part are  

L M  (L−x)&  L M   x respectivelyMaximum Friction force: f  max  

=μN=μ(  L M   x)g=  L μMgx ​    Weight of hanging chain: W=m  ′  g= L M (L−x)g= L M(L−xZ

For system to be stable:       L M(L−x)g =  L μMgx   (L−x)=  4 1 ​ x  x=  5 4 ​   L Therefore hanging part is   5 L .(20%)

Explanation:

Answer 2

Answer:

Explanation:

Let M = mass of chain and L = length of chain. LI = length of the hanging part.

Chain will be at rest if :

ML'g=μ

M

L(L-L')g⇒

L'

L=

μ

1+μ=

0⋅25

1+0⋅25×100=20%