A 50.0-kg box is resting on a horizontal floor. A force of 250 N directed at an angle of 26.0° 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?
A) 0.441
B) 0.654
C) 0.375
D) 0.866
E) 0.406

Question

Quizplus · Accepted Answer

First, we need to find the normal force acting on the box. This force is equal to the weight of the box, which is (50.0 kg)(9.81 m/s^2) = 490.5 N.

Next, we need to find the force of friction required to keep the box from sliding. Using trigonometry, we can find that the horizontal component of the applied force is (250 N)cos(26.0°) = 221.3 N. Therefore, the force of friction must also be 221.3 N in the opposite direction to balance out the horizontal forces.

Finally, we can use the formula for static friction to find the minimum coefficient of static friction required:

f_static = μ_static * N

Where f_static is the force of friction, N is the normal force, and μ_static is the coefficient of static friction. Plugging in our values, we get:

221.3 N = μ_static * (490.5 N)

μ_static = 0.451

This value is closest to choice C, 0.375, so that is our answer.

A 500-Kg Box Is Resting on a Horizontal Floor