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A uniform chain of length L has one of its ends attached to the wall at point A, while 3L4 length of the chain is lying on the table as shown in figure. Then, minimum co-efficient of static friction between table and chain, so that the chain remains in equilibrium is


  1. 13
  2. 15
  3. 14
  4. 34


Solution

The correct option is C 14
FBD of the two parts of the chain are given below:


Let λ=mass per unit length of uniform chain
in kg/m
Now mass of the part of chain hanging from the wall, 
m1=L4λ kg

Similarly for part of the chain resting on table:
mass m2=3L4λ kg

From the FBD of m1,
Fcos37=λ L4g ....(1)
Fsin37=T

From the FBD of m2,
f=T=Fsin37
But fμsN=μs(34λLg)
i.e Fsin37μs(34λLg).....(2)

Substituting F from (1),
516λLg×35μs(34λLg)
316μs×34
μs14
μsmin=14

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