Question

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

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

Answer 1

Answer: (D)

$\geq \, 16.3 N$

Answer 2

According to question, we can draw the following diagram

$m g \sin \theta =\mu(F+m g \cos \theta)$
$F =\frac{m g \sin \theta}{\mu}-m g \cos \theta$
$=m g\left[\frac{\sin \theta}{\mu}-\cos \theta\right]$
Here, $m =1 \,kg , g=10 m / s ^{2}, \theta=30^{\circ}, \mu=0.2$
$F \left.=1 \times 10 \frac{[\sin 30}{02}-\cos 30^{\circ}\right]$
$=10\left[\frac{1}{2 \times 0.2}-\frac{\sqrt{3}}{2}\right]$
$=10\left[\frac{5}{2}-\frac{\sqrt{3}}{2}\right]=5[5-\sqrt{3}]$
$=5[5-1.732]=16.34 N$