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18 January, 04:42

A 50.0-kg box is resting on a horizontal floor. a force of 250 n directed at an angle of 32.0° below the horizontal is applied to the box. what is the minimum coefficient of static friction between the box and the surface required for the box to remain stationary?

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  1. 18 January, 07:29
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    The minimum value friction is equal to the horizontal force component of 250 N force.

    The horizontal force component can be found by multiplying the 250N force by cosine of 32 degrees. = 212N

    F = μ N

    Hence, divide the horizontal component of the force by the Normal Force.

    Normal Force equals to the Vertical component of the 250N force plus the weight of the box. = 490.5N

    The answer is: 0.432
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