A 21.0-kg box rests on a frictionless ramp with a 15.1° slope. The mover pulls on a rope attached to the box to pull it up the incline. If the rope makes an angle of 43.6° with the horizontal, what is the smallest force F the mover will have to exert to move the box up the ramp?

I would like to have one of these frictionless ramps.

The force in the direction of the ramp is Fcos43.6
The weight portion trying to slide the box down the ramp is 21Sin15.1

Net pulling force up the ramp
Fcos43.6-21sin15.1
set that to zero, andsolve for F

To find the minimum force required to move the box up the ramp, we need to consider the components of the forces acting on the box.

Let's break down the forces acting on the box:
- The weight of the box, acting vertically downward, can be expressed as: mg, where m is the mass of the box and g is the acceleration due to gravity (9.8 m/s^2).
- The force F, exerted by the mover pulling the box up the ramp.
- The normal force, which is the perpendicular force exerted by the ramp on the box. It is equal in magnitude and opposite in direction to the component of the box's weight perpendicular to the ramp.

Now, we will analyze the forces acting in the direction parallel to the incline. The forces in this direction can be calculated using trigonometry.

The force parallel to the incline can be expressed as: F_parallel = F * sin(43.6°), where F is the total force exerted by the mover.

The weight component parallel to the incline can be expressed as: W_parallel = mg * sin(15.1°), where m is the mass of the box and g is the acceleration due to gravity.

Since the box is in equilibrium and not moving, the total force parallel to the incline must equal the weight component parallel to the incline.

Therefore, F_parallel = W_parallel.

Substituting the values, we have:

F * sin(43.6°) = mg * sin(15.1°)

Solving for F, we get:

F = (mg * sin(15.1°)) / sin(43.6°)

Substituting the given values, m = 21.0 kg, g = 9.8 m/s^2, we have:

F = (21.0 kg * 9.8 m/s^2 * sin(15.1°)) / sin(43.6°)

Calculating this expression will give us the minimum force required to move the box up the ramp.

To find the smallest force F that the mover will have to exert to move the box up the ramp, we can break down the forces acting on the box.

First, we need to resolve the weight of the box into components parallel and perpendicular to the slope.

The weight of the box can be calculated using the formula:

W = m * g

where:
W is the weight of the box
m is the mass of the box (21.0 kg)
g is the acceleration due to gravity (9.8 m/s²)

W = 21.0 kg * 9.8 m/s²
W = 205.8 N

Next, we can calculate the component of the weight along the ramp.

W_parallel = W * sin(15.1°)

W_parallel = 205.8 N * sin(15.1°)
W_parallel ≈ 53.18 N

Now, we can find the force F required to move the box up the ramp by considering the forces acting along the ramp.

The force F can be broken down into two components: parallel and perpendicular to the ramp. The component parallel to the ramp opposes the weight component along the ramp.

Using the right triangle formed by the weight parallel component, the force F, and the angle between them (43.6° - 15.1°), we can calculate the parallel component of the force F.

F_parallel = F * sin(43.6° - 15.1°)

F_parallel = F * sin(28.5°)

Equating the force F_parallel to the weight component along the ramp, we have:

F_parallel = W_parallel

F * sin(28.5°) = 53.18 N

Finally, we can solve for the smallest force F:

F = 53.18 N / sin(28.5°)

F ≈ 108.7 N

Therefore, the smallest force F the mover will have to exert to move the box up the ramp is approximately 108.7 N.