The coefficient of static friction between the m = 2.95-kg crate and the 35.0° incline is 0.345. What minimum force must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

The static friction force must equal the weight component dwn the incline.

(F + M*g*cos35)*Us = M g sin35

Solve for F

Post it.

Nice thank you! This worked perfectly!

what is g

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.

First, let's break down the weight of the crate into components. The component parallel to the incline can be found using the equation: weight_parallel = m * g * sin(θ), where m is the mass of the crate, g is the acceleration due to gravity (approximately 9.8 m/s^2), and θ is the angle of the incline (35.0°).

weight_parallel = 2.95 kg * 9.8 m/s^2 * sin(35.0°)
weight_parallel ≈ 16.07 N

Next, let's calculate the maximum force of static friction that can act on the crate. The maximum force of static friction can be determined using the equation: friction_max = μ * normal_force, where μ is the coefficient of static friction and normal_force is the force perpendicular to the incline.

To find the normal force, we need to break down the weight of the crate into its components. The component perpendicular to the incline is given by the equation: weight_perpendicular = m * g * cos(θ).

weight_perpendicular = 2.95 kg * 9.8 m/s^2 * cos(35.0°)
weight_perpendicular ≈ 24.85 N

Now, we can determine the maximum force of static friction:

friction_max = 0.345 * 24.85 N
friction_max ≈ 8.57 N

Since we want to prevent the crate from sliding down the incline, the applied force must be equal to or greater than the maximum force of static friction.

Therefore, the minimum force that must be applied perpendicular to the incline to prevent the crate from sliding down is approximately 8.57 N.