A 149 kg crate is pushed at constant speed up a frictionless inclined plane of $\theta$ = 51 $^{\circ}$ with the horizontal. Assuming one pushes with a horizontal force, what is the magnitude of the force F (in N) required to do this?

To calculate the magnitude of the force F required to push the crate up the inclined plane, we need to consider the forces acting on the crate.

First, let us draw a free-body diagram of the crate on the inclined plane.

The weight of the crate acts vertically downward and can be calculated using the formula:

Weight = mass * acceleration due to gravity

Weight = 149 kg * 9.8 m/s^2 ≈ 1458 N

The weight can be split into two components: one perpendicular to the inclined plane (mg * cos θ) and one parallel to the inclined plane (mg * sin θ), where θ is the angle of inclination.

mg * cos θ acts perpendicular to the inclined plane, and mg * sin θ acts parallel to the inclined plane.

Since the crate is moving at a constant speed, the applied force must be equal in magnitude and opposite in direction to the parallel component of the weight to counterbalance it.

Therefore, F = mg * sin θ.

F = 149 kg * 9.8 m/s^2 * sin 51° ≈ 1142 N.

So, the magnitude of the force F required to push the crate up the frictionless inclined plane is approximately 1142 N.

Wc = m * g = 149kg * 9.8N/kg = 1460 N. =

Wt. of the crate.

Fp = 1460*sin51 = 1135 N. = Force parallel to the plane.
Fv = 1460*cos51 = 919 N. = Force perpendicular to the plane.

Fap-Fp = m*a
Fap-1135 = m*0 = 0
Fap = 1135 N. = Force applied parallel to the plane.

Fh = Fap*cos51 = 1135*cos51 = 714.3 N. =
Required hor. force.