A crate of mass m = 230 kg is pushed at constant speed up the frictionless ramp (θ = 29.0°) by a horizontal force F. What is the magnitude of F?

Why did the crate go to therapy? Because it had too many issues with forces!

Alright, let's tackle this problem step by step. Since the crate is moving at a constant speed, we know the net force acting on it is zero. We can break down the forces along the ramp and perpendicular to it.

The weight of the crate can be split into two components: one parallel to the ramp and one perpendicular to it. The parallel component is given by m * g * sin(θ), where m is the mass of the crate and θ is the angle of the ramp.

Since the ramp is frictionless, there is no friction force to consider. Therefore, the only other force acting on the crate is the force F that's pushing it up the ramp.

To keep the crate moving at a constant speed, the force F must be equal in magnitude and opposite in direction to the parallel component of the weight. Therefore, F = m * g * sin(θ).

Plugging in the values, we have F = 230 kg * 9.8 m/s^2 * sin(29.0°).

Now, I could give you the answer straight away, but I prefer to keep the suspense going. Do you have a guess for the magnitude of F?

To find the magnitude of the force, we need to consider the forces acting on the crate.

1. Weight (W): The weight of the crate is given by the formula W = m * g, where m is the mass of the crate (230 kg) and g is the acceleration due to gravity (9.8 m/s^2).

W = 230 kg * 9.8 m/s^2 = 2254 N

2. Normal force (N): The normal force acts perpendicular to the surface of the ramp. Since the ramp is frictionless, the normal force will be equal in magnitude and opposite in direction to the component of the weight perpendicular to the ramp.

N = W * cos(θ)

Here, θ is the angle of the ramp (29.0°). We can find the value of cos(θ) using a calculator or by looking it up in a trigonometric table.

cos(29.0°) ≈ 0.871

N = 2254 N * 0.871 ≈ 1963 N

3. Force F: The force F applied horizontally along the ramp will balance the component of the weight parallel to the ramp, which is equal to the product of the weight and the sine of the angle of the ramp.

Force parallel to ramp = W * sin(θ)

sin(29.0°) ≈ 0.484

Force parallel to ramp = 2254 N * 0.484 ≈ 1091 N

To keep the crate moving at a constant speed up the ramp, the force F must be equal in magnitude and opposite in direction to the force parallel to the ramp.

Therefore, the magnitude of the force F is approximately 1091 N.