A 5.0-kg crate is on an incline that makes an angle 30° with the horizontal.If the coefficient of static friction is 0.5, the maximum force that can be applied parallel to the plane without moving the crate is:
A)0 N
B)3.3 N
C)21 N
D)46 N
E)55 N

Question

Quizplus · Accepted Answer

The maximum force that can be applied without moving the crate can be found by calculating the force of static friction, which is the product of the normal force and the coefficient of static friction. The normal force on an incline is given by $N = mg \cos(\theta)$, where $m$ is the mass, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $\theta$ is the angle of the incline. For a 5.0-kg crate on a 30° incline, $N = 5.0 \times 9.8 \times \cos(30°) = 42.4$ N. The force of static friction is then $f_s = \mu_s N = 0.5 \times 42.4 = 21.2$ N. However, this force acts to oppose sliding down the incline. The maximum force that can be applied parallel to the plane without moving the crate also includes the component of gravitational force parallel to the incline, which is $mg \sin(\theta) = 5.0 \times 9.8 \times \sin(30°) = 24.5$ N. Therefore, the total maximum force is the sum of the gravitational force component and the static friction force, which is $21.2 + 24.5 = 45.7$ N, which rounds to 46 N, making D the correct answer.

A 50-Kg Crate Is on an Incline That Makes an Angle