Question

A body of mass 400 g slides on a rough horizontal surface. If the frictional force is 3.0 N, find the magnitude of the contact force.

(a) 5 N

(b) 10 N

(c) 15 N

(d) 20 N

I got this question in an interview for job.

The query is from Laws of Solid Friction and Limiting Friction topic in section Friction of Machine Kinematics

Answer 1

The correct answer is (a) 5 N

The explanation is: Let the contact force on the block by the surface be F which makes an angle ϴ with the vertical.

The component of F perpendicular to the contact surface is the normal force N and the component of F parallel to the surface is the friction f As the surface is horizontal, N is vertically upward. For vertical equilibrium,

N = mg = (0.400 kg) (10 m/s^2) = 4.0 N.

The frictional force is f = 3.0 N.

 F = √N^2

   = √4^2 + 3^2 = 5 N.