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A wooden box lying at rest on an inclined surface of a wet wood is held at static equilibrium by a constant force $\mathbf{F}$ applied perpendicular to the incline. If the mass of the box is $1 \mathrm{~kg}$, 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 $\mathbf{F}$ is (Use $g=10 \mathrm{~m} / \mathrm{s}^2$ )
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Verified Answer
The correct answer is:
$\geq 16.3 \mathrm{~N}$
According to question, we can draw the following diagram.
$$
\begin{aligned}
m g \sin \theta & =\mu(F+m g \cos \theta) \\
F & =\frac{m g \sin \theta}{\mu}-m g \cos \theta \\
& =m g\left[\frac{\sin \theta}{\mu}-\cos \theta\right]
\end{aligned}
$$
Here,
$$
\begin{aligned}
m & =1 \mathrm{~kg}, g=10 \mathrm{~m} / \mathrm{s}^2, \theta=30^{\circ}, \mu=02 \\
F & =1 \times 10\left[\frac{\sin 30}{0.2}-\cos 30^{\circ}\right] \\
& =10\left[\frac{1}{2 \times 02}-\frac{\sqrt{3}}{2}\right] \\
& =10\left[\frac{5}{2}-\frac{\sqrt{3}}{2}\right]=5[5-\sqrt{3}] \\
& =5[5-1.732]=16.34 \mathrm{~N}
\end{aligned}
$$
$$
\begin{aligned}
m g \sin \theta & =\mu(F+m g \cos \theta) \\
F & =\frac{m g \sin \theta}{\mu}-m g \cos \theta \\
& =m g\left[\frac{\sin \theta}{\mu}-\cos \theta\right]
\end{aligned}
$$
Here,
$$
\begin{aligned}
m & =1 \mathrm{~kg}, g=10 \mathrm{~m} / \mathrm{s}^2, \theta=30^{\circ}, \mu=02 \\
F & =1 \times 10\left[\frac{\sin 30}{0.2}-\cos 30^{\circ}\right] \\
& =10\left[\frac{1}{2 \times 02}-\frac{\sqrt{3}}{2}\right] \\
& =10\left[\frac{5}{2}-\frac{\sqrt{3}}{2}\right]=5[5-\sqrt{3}] \\
& =5[5-1.732]=16.34 \mathrm{~N}
\end{aligned}
$$
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