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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 $\vec{F}$ applied perpendicular to the incline. If the mass of the box is 1kg, the angle of inclination is 30⁰ 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=10m/s² )

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$0 N, a s 30^{0} i s l e s s t h a n a n g l e o f r e p o s e$

$\geq 1 N$

$\geq 3.3 N$

$\geq 16.3 N$

answer is D.

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Detailed Solution

The frictional force will balance the component of weight.

$mgsinθ = μ (F + mgcosθ)$

$F = \frac{mgsinθ}{μ} - mgcosθ = mg [\frac{sinθ}{μ} - cosθ]$

$= 10 [\frac{1}{2 \times 0.2} - \frac{\sqrt{3}}{2}] = 5 [5 - \sqrt{3}] = 16.34 N$

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