First slide

Friction on inclined plane of angle more than angle of repose

Question

A box of mass 8 kg is placed on a rough inclined plane of inclination 30°. Its downward motion can be prevented by applying a horizontal force F, then value of F for which friction between the block and the incline surface is minimum, is

Moderate

A

$\frac{80}{\sqrt{3}} N$
B

$40 \sqrt{3} N$
C

$\frac{40}{\sqrt{3}} N$
D

$80 \sqrt{3} N$

Solution

For friction to be minimum,
$F cosθ = m g sinθ$
$\Rightarrow F = m g tanθ = 8 \times 10 \times \tan 30^{0} = \frac{80}{\sqrt{3}} N$

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