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    Block and Inclined Plane

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    A block is on an inclined plane. There is a horizontal force on the block. The block is at rest. There is a certain coefficient of friction.

    a) find force (magnitude and direction)
    b) find normal force exerted on the block by the plane

    Given: you can give the block a mass of 5.00 kg an angle of 60 degrees and the coefficient of friction 0.300

    Please see the attachment.

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    Please see attachment.
    Here is solution to this problem:

    Let the inclined plane make an angle of 60 degrees with the horizontal.
    let the value of g be 10 m/s^2.
    Weight of the block acting vertically downward = mg = 5*10 = 50 N
    Component of mg(50N) parallel and down the inclined plane = mg sin 60
    Component of mg(50N) normal to the inclined plane(pressing the plane) = mg cos 60

    Let the horizontal force F be applied to the ...

    Solution Summary

    Resove gravitational force acting on the block kept on the inclined plane in two components; one parallel to the plane and the other perpendicular to the plane.