The coefficient of static friction between the m = 3.20 kg crate and the 35.0° incline of the figure below is 0.340. What minimum force must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

I assume a crate is on an incline.

In this case, fricion + force applied has to be greater than the component of weight down the plane.

friction: mg*mu*CosTheta
weight down plane: mgSinTheta

force+friction=weight down plane
solve for force.

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

First, let's define the forces involved:

1. Weight (W): This is the force acting vertically downward due to gravity, given by W = mg, where m is the mass of the crate (3.20 kg) and g is the acceleration due to gravity (9.8 m/s²).

2. Normal force (N): This is the force exerted by the incline perpendicular to the surface of the incline. Since the crate is on an incline, the normal force can be calculated using the formula N = mg * cosθ, where θ is the angle of the incline (35.0°).

3. Force of friction (Ff): This is the force that opposes the motion of the crate and acts parallel to the incline. The force of friction can be calculated using the formula Ff = μs * N, where μs is the coefficient of static friction (0.340).

Now, let's calculate the minimum force (F) required to prevent the crate from sliding down the incline:

Since the crate is in equilibrium, the net force acting along the incline must be zero. Therefore, we can set up the following equation:

F - Ff = 0

Substituting the values, we have:

F - (μs * N) = 0

Since the force (F) is perpendicular to the incline, it can be written as:

F = N * sinθ

Substituting the value of N from the formula above, we get:

F = (mg * cosθ) * sinθ

Now, substituting the values of m, g, and θ, we can calculate the minimum force (F):

F = (3.20 kg * 9.8 m/s² * cos35.0°) * sin35.0°

Solving this equation will give us the minimum force required to prevent the crate from sliding down the incline.