First slide

Friction on inclined plane of angle more than angle of repose

Question

A mass of 1 kg is just able to slide down the slope of an inclined rough surface when the angle of inclination is 60°. The minimum force necessary to pull the mass up the inclined plane is : (g = l0 ms^-2)

Easy

A

14.14 N
B

17.32 N
C

10 N
D

16.66 N

Solution

$\begin{array}{l} F_{min} = mgsin 60^{\circ} + μmgcos 60^{\circ} \\ = mgsin 60^{\circ} + (\tan 60^{\circ}) mgcos 60^{\circ} = 2 mgsin 60^{\circ} \\ = 2 (1) (10) (\frac{\sqrt{3}}{2}) = 17 .32 N \end{array}$

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