Question

A block of mass $m=2kg$ is resting on a rough inclined plane of inclination ${30}^o$ as shown in figure. The coefficient of friction between the block and the plane is $\mu =0.5$. What minimum force $F$ should be applied perpendicular to the plane on the block, so that block does not slip on the plane.? $(g=10m/s^2)$

• A. zero
• B. $6.24N$
• C. $2.68N$
• D. $4.34N$

Answer 1

The correct option is C $2.68N$

$mg\sin \theta =\mu (mg\sin \theta +f)$
$\frac{mg}{2}=\frac{1}{2}(mg\frac{\sqrt{3}}{2}+F)$
$mg(1-\frac{\sqrt{3}}{2})=F$
$20\left(0.13\right)=2.68N$