The coefficient of static friction between the 3.78 kg crate and the 38° incline is 0.306. What minimum force must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline? The acceleration of gravity is 9.8 m/s^2. Answer in units of N.

To find the minimum force required to prevent the crate from sliding down the incline, we need to consider the forces acting on the crate.

1. Identify the relevant forces:
- Weight (mg): The force due to gravity acting vertically downward, where m is the mass of the crate and g is the acceleration due to gravity.
- Normal force (N): The force exerted by the incline surface perpendicular to it.
- Friction force (f): The force of static friction between the crate and the incline, which acts parallel to the incline.
- Force applied (F): The force that needs to be exerted perpendicular to the incline to prevent the crate from sliding down.

2. Determine the components of the weight and normal force:
- Weight (mg): The weight can be divided into two components: one parallel to the incline (mg*sinθ) and one perpendicular to the incline (mg*cosθ), where θ is the angle of the incline.
- Normal force (N): The normal force acts perpendicular to the incline and is equal to mg*cosθ.

3. Analyze the forces in the horizontal direction:
- The force applied (F) and the friction force (f) are the only forces acting horizontally.
- The friction force (f) is equal to the coefficient of static friction (µ) multiplied by the normal force (N), which can be written as f = µN.
- The force applied (F) can be determined as F = f to prevent the crate from sliding.

4. Solve for the minimum force (F):
- In this case, the force applied (F) is equal to the friction force (f) required to prevent the crate from sliding.
- So, F = f = µN.

Now let's substitute the known values into the equations.

Given:
- Mass of the crate (m) = 3.78 kg
- Coefficient of static friction (µ) = 0.306
- Incline angle (θ) = 38°
- Acceleration due to gravity (g) = 9.8 m/s^2

First, calculate the normal force (N):
N = mg*cosθ
= (3.78 kg * 9.8 m/s^2) * cos(38°)

Then, substitute the value of N into the equation for the force applied (F):
F = µN
= 0.306 * N

Finally, plug in the value of N and solve for F.

Therefore, the minimum force required to prevent the crate from sliding down the incline is F = 0.306 * (3.78 kg * 9.8 m/s^2) * cos(38°). This will give you the answer in units of Newtons (N).