Friction - NEET Physics Questions
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Friction

Question 11:

A body of mass m is kept on a rough horizontal surface (coefficient of friction = μ). A horizontal force is applied on the body, but it does not move. The resultant of normal reaction and the frictional force acting on the object is given by F, where F is:

1. F = mg
2. F = μmg
3. F = mg + μmg
4. F ≤ mg(1 + μ2) 1/2
View Answer

The forces acting on the body are: Normal reaction (N=mg) (upward) and Friction force f≤μmg (opposing applied force)

The resultant force is:

F=N2+f2=mg2+(μmg)2=mg1+μ2F = \sqrt{N^2 + f^2} = \sqrt{mg^2 + (\mu mg)^2} = mg \sqrt{1 + \mu^2}

F \leq mg(1 + \mu^2)^{1/2}