Question

A flatbed truck rolls up a 10 degree incline. There are 3 forces acting on an unsecured crate in the back. The weight is 2500 newtons straight down. The normal force has a magnitude of 2462 newtons 10 degrees to the left of vertical. The frictional force is 180 newtons 10 degrees above horizontal to the right. Find the magnitude and angle of the net force.

Answer 1

The magnitude of the net force on the crate is 2230.12 N at an angle of 179.214°.

To find the magnitude and angle of the net force, we need to add the forces acting on the crate. The weight force of 2500 N is straight down, so its horizontal component is 0 N and vertical component is -2500 N. The normal force of 2462 N has a vertical component of 2462 sin(10°) N = 427.11 N upward and a horizontal component of -2462 cos(10°) N = -2407.08 N to the left. The frictional force of 180 N has a vertical component of 180 sin(10°) N = 30.97 N upward and a horizontal component of 180 cos(10°) N = 176.96 N to the right.

To find the magnitude of the net force, we need to add the horizontal components of the forces: -2407.08 N + 176.96 N = -2230.12 N. To find the angle of the net force, we can use the tangent function: angle = atan(30.97 N / -2230.12 N) = -0.786°. Since the angle is negative, we can add 180° to make it positive. Therefore, the magnitude of the net force is 2230.12 N and the angle is 179.214°.