Question

A body of mass 400g slides on a rough horizontal surface. If the frictional force is 3.0 N, find (a) the angle made by the contact force on the body with the vertical and (b) the magnitude of the contact force. Take g = 10$\frac{m}{s^2}$

• A. ${37}^{\circ }$, 5N
• B. ${37}^{\circ }$, 4N
• C. ${53}^{\circ }$, 4N
• D. ${53}^{\circ }$, 5N

Answer 1

The correct option is A

${37}^{\circ }$, 5N

Let the contact force on the block by the surface be F which makes an angle $\theta$ with the vertical

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

N = mg = (0.400 kg) $\frac{10m}{s^2}$ = 4.0 N.

The frictional force is f = 3.0 N.

(a)

Or, θ = ${tan}^{-1}$ ( $\frac{3}{4}$) = ${37}^{\circ }$.

(b)The magnitude of the contact force is

F = $\sqrt{N^2+f^2}$

= $\sqrt{{\left(4.0N\right)}^2+{\left(3.0N\right)}^2}$ = 5.0 N.