You have been asked to keep a 35.1 kilogram crate from sliding down a 50.03° incline. The coefficient of friction between the crate and the incline is 0.26, and it's on the surface of Mars (gravitational acceleration at the surface of Mars is 3.75 m/s^2). How much force must you apply to the crate?

Can someone please explain the steps to this one?

To calculate the force required to keep the crate from sliding down the incline, we need to consider the gravitational force acting on the crate and the force of friction opposing the motion.

Step 1: Calculate the gravitational force:
The gravitational force can be calculated using the formula Fg = m * g, where Fg is the gravitational force, m is the mass of the crate, and g is the gravitational acceleration on Mars (3.75 m/s^2).
Fg = (35.1 kg) * (3.75 m/s^2)
Fg = 131.625 N

Step 2: Calculate the normal force:
The normal force is the perpendicular force exerted by the incline on the crate and is equal to the gravitational force acting perpendicular to the incline. Since the incline is at an angle of 50.03°, the normal force can be calculated as follows:
Normal force = Fg * Cos(angle)
Normal force = 131.625 N * Cos(50.03°)
Normal force ≈ 131.625 N * 0.643
Normal force ≈ 84.689 N

Step 3: Calculate the force of friction:
The force of friction can be calculated using the formula Ff = μ * N, where Ff is the force of friction, μ is the coefficient of friction (0.26), and N is the normal force.
Ff = 0.26 * 84.689 N
Ff ≈ 22.034 N

Step 4: Calculate the force applied to the crate:
The force applied to the crate is equal to the force of friction since it needs to counteract the sliding force. Therefore, the force applied to the crate is approximately 22.034 N.