Question

A body of mass 400 g slides on a rough horizontal surface. If the frictional force is 3.0 N, find the angle made by the contact force on the body with the vertical.

(a) 35^0

(b) 36^0

(c) 37^0

(d) 38^0

This question was addressed to me during an online exam.

Question is from Laws of Solid Friction and Limiting Friction topic in chapter Friction of Machine Kinematics

Answer 1

The correct option is (c) 37^0

For explanation I would say: 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.

tan ϴ = f/N = 3/4

or, ϴ = tan^-1 (3/4) = 37^0.