Question

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?

Answer 1

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