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?

Question

Jonathan Henderson

Physics

3 May, 10:14

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?

+4

Answers (1)

Know the Answer?

Madelyn Fuentes · 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