First slide

Friction on inclined plane of angle more than angle of repose

Question

The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. The coefficient of friction when the angle of inclination of the plane is 60° is

Moderate

A

$\frac{1}{\sqrt{2}}$
B

$\frac{1}{\sqrt{3}}$
C

$\frac{1}{2}$
D

$\frac{1}{3}$

Solution

$\begin{array}{l} F_{Up} = mg (\sin θ + μ_{k} \cos θ) \\ F_{down} = mg (\sin θ - μ_{k} \cos θ) \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App