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A block of mas M is resting on an inclin...

A block of mas M is resting on an inclined plane as shown in the figure. The inclination of the plane to the horizontal is gradually increased. It is found that when the angle of inclination is `theta` the block just begins to 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 ?

Text Solution

Verified by Experts

The weight mg of the block has two rectangular components `mg cos theta` perpendicular to the plane and `mg sin theta` down the plane

Force of friction `f=mu R=mu mg cos theta` is also down the plane
therefore `F=mg sin theta +f`
`mg sin theta+mu mg cos theta`
But `mu = tan theta therefore F=mg sin theta+ tan theta mg cos theta`
`=2 mg sin theta`
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Knowledge Check

  • A block of mass m is resting on an inclined plane. The inclination of the plane to the horizontal is gradually increased. It is found that when angle of inclination is theta , the block just begins to 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?

    A
    `2mg sin theta`
    B
    `mg sin theta`
    C
    `mg cos theta`
    D
    `2 mg cos theta`

  • A block of mass 4kg rests on an an inclined plane. The inclination of the plane is gradually increased. It is found that when the inclination is 3 in 5 (sin=theta(3)/(5)) the block just begins to slide down the plane. The coefficient of friction between the block and the plane is.

    A
    `0.4`
    B
    `0.6`
    C
    `0.8`
    D
    `0.75`

  • A block of mass m is placed on a rough inclined plane. When the inclination of the plane is theta , the block just beging to slide down the plane under its own weight. The minimum force applied parallel to the plane, to move the block up the plane, is.

    A
    `mgsintheta`
    B
    `2mgsintheta`
    C
    `mgcostheta`
    D
    `mg tantheta`

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