SSC JE Mechanical Engineering 2014 Set 1 Paper

Question : 116 of 200

Marks: +1, -0

A body of weight 30 N rests on a horizontal floor. A gradually increasing horizontal for applied to the body which just starts moving when the force is 9 N. The coefficient of friction between the body and the floor will be

10/3
3/10
1/3
1/9

Validate

Solution:

F = µ.N or

µ= F/N

. Here, Friction Force (limiting, F) = 9 N and Reaction (N) = 30 N.
So,

µ= 9/30 = 3/10

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