Question

The wooden box lying at rest on an inclined surface of a wet wood is held at static equilibrium by a constant force $\overrightarrow{F}$ applied perpendicular to the incline. If the mass of the box is $1kg$, the angle of inclination is ${30}^o$ and the coefficient of static friction between the box and the inclined plane is $0.2$, the minimum magnitude of $\overrightarrow{F}$ is (Use $g=10m/s^2$)

• A. $0N$ as ${30}^o$ is less than angle of repose
• B. $\ge 1N$
• C. $\ge 3.3N$
• D. $\ge 16.3N$

Answer 1

Answer: (D)

$\ge 16.3N$

$mg\sin \theta =\mu (F+mg\cos \theta )$
$F=\frac{mg\sin \theta }{\mu }-mg\cos \theta =mg[\frac{\sin \theta }{\mu }-\cos \theta ]=10[\frac{1}{2\times 0.2}-\frac{\sqrt{3}}{2}]=5[5-\sqrt{3}]=5(5-1.732)=16.34$