First slide

Friction on horizontal surface

Question

A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.5. The magnitude of the force acting upwards at an angle of 60° from the horizontal that will just start the block moving is

Moderate

A

5 N
B

$\frac{20}{2 + \sqrt{3}} N$
C

$\frac{20}{2 - \sqrt{3}} N$
D

10 N

Solution

$R + F \sin 60^{0} = mg$
R = mg - $\frac{\sqrt{3} F}{2}$
If block just starts moving
$F \cos 60^{0} = f_{m a x} = μR$
or $F + \frac{\sqrt{3} F}{2} = 10 or F = \frac{20}{2 + \sqrt{3}}$

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