A Block of mass m is placed on a rough inclined plane. The coefficient of friction between the block and the inclined plane $μ$ . If the direction of force F can be varied, mark the correct option:

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F required to move the block up the inclined plane will be minimum when $ϕ = \tan^{- 1} (μ)$

The minimum value of F required so that the block will tend to slide up is $F_{\min} = \frac{m g (\sin θ + μ \cos θ)}{\sqrt{1 + μ^{2}}}$

The minimum value of F required so that the block will slide up is $F_{\min} = \frac{m g (\sin θ - μ \cos θ)}{\sqrt{1 + μ^{2}}}$

None of these

answer is A, B.

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Detailed Solution

For the block just to slip up the incline plane:
$F \cos ϕ = f + m g \sin θ$ and $N + F \sin ϕ = m g \cos θ$
$∴ F = \frac{m g (\sin θ + μ \cos θ)}{\cos ϕ + μ \sin ϕ}$
For F to be minimum $\cos ϕ + μ \sin ϕ$ is maximum
$∴ \frac{d}{d ϕ} (\cos ϕ + μ \sin ϕ) = 0$
$∴ F_{\min} = \frac{m g (\sin θ + μ \cos θ)}{\sqrt{1 + μ^{2}}}$