A 25-N crate is held at rest on a frictionless incline by a force that is parallel to the incline.If the incline is 25°above the horizontal the magnitude of the normal force of the incline on the crate is:
A)4.1 N
B)4.6 N
C)8.9 N
D)11 N
E)23 N

Question

Quizplus · Accepted Answer

The normal force is the component of the weight perpendicular to the incline. Given the incline angle (25 degrees) and the weight of the crate (25 N), we can calculate the normal force using trigonometry. The weight of the crate acts directly downward, and we can decompose this force into two components: one parallel to the incline and one perpendicular (normal) to it. The normal force $N$ can be found using the cosine of the angle: $N = mg \cos(\theta)$, where $m$ is the mass of the crate, $g$ is the acceleration due to gravity, and $\theta$ is the angle of the incline. Since $mg = 25 N$ and $\theta = 25^\circ$, we find that $N$ is close to the weight of the crate because the angle is relatively small, and the cosine of 25 degrees is close to 1. The exact value might slightly differ due to rounding, but among the given options, 23 N (E) is the closest to the expected value, indicating that most of the weight's effect is perpendicular to the incline.

A 25-N Crate Is Held at Rest on a Frictionless