Question

A box is lying on an inclined plane what is the coefficient of static friction if the box starts sliding when an angle of inclination is 60°

Answer 1

A box is lying on an inclined plane , if box starts sliding then , at this instant :
downward force along plane = upward force along plane.

downward force along plane = mgsin60° [ components of weight along plane]
upward force along plane = static friction

but we know, $f_s=\mu_s N$
where fs is static friction , $\mu_s$ is coefficient of static friction and N is normal reaction act between box and plane.
N = mgcos60° [ components of weight perpendicular to plane]

so, mgsin60° = $\mu_s$ mgcos60°
$\mu_s=tan60^{\circ}$

hence, coefficient of static friction = tan60° = √3

Answer 2

Answer:

√3 = 1.73

$fs = mus \: \times n$

Explanation:

as we know force on inclined plane will be mgsin60°

N on inclined plane will be = mgcos60°

applying above formula

mgsin60°=mu*mgcos60°

tan60°=√3

mu=1.73