Question

The coefficient of static friction between a block of mass m and an incline is ${\mu }_s$ = 0.3. What can be the maximum angle θ of the incline with the horizontal so that the block does not slip on the plane? (Also called angle of repose)

• A. $\theta$ = ${tan}^{-1}$ (0.3)
• B. $\theta$ = ${cot}^{-1}$ (0.3)
• C. $\theta$ = ${sin}^{-1}$ (0.3)
• D. $\theta$ = ${cos}^{-1}$ (0.3)

Answer 1

The correct option is A

$\theta$ = ${tan}^{-1}$ (0.3)

The situation is shown in figure

The forces on the block are

(i) (1) the weight mg downward by the earth,

(ii) the normal contact force N by the incline, and

(iii) the friction f parallel to the incline up the plane, by the incline

Taking components perpendicular to the incline,

N - mg cos $\theta$ = 0

or, N = mg cos$\theta$. .....(i)

Taking components parallel to the incline,

f - mg sin$\theta$ = 0

or, f = mg sin$\theta$

or, ${\mu }_s$N = mg sin$\theta$. .....(ii)

Dividing (ii) by (i) ${\mu }_s$ = tan$\theta$

or, $\theta$= ${tan}^{-1}$ ${\mu }_s$= ${tan}^{-1}$ (0.3)