A block of mass M is resting on an inclined plane as shown in Fig. The inclination of the plane to the horizontal is gradually increased. It is found that when the angle of inclination is $θ$ the block just begins to the slide down the plane. What is the minimum force F applied parallel to the plane that would just make the block move up the plane?

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$M g \sin θ$

$M g \cos θ$

$2 M g \cos θ$

$2 M g \sin θ$

answer is D.

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Detailed Solution

The component of weight Mg of the block along the inclined plane $= M g \sin θ$ .

The minimum frictional force to be overcome is also $M g \sin θ$ . To make the block just move up the plane the minimum force applied must overcome the component $M g \sin θ$ of gravitational force as well as the frictional force $M g \sin θ$ .

$\Rightarrow F = 2 M g \sin θ$

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