The coefficient of static friction between 3.00 kg crate and the 35.0 incline is 0.300.what minimum force F must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline

force of gravity along the incline is

... m g sin(35º)

force of gravity perpendicular to the incline
... m g cos(35º)

.300 {x + [m g cos(35º)]} = m g sin(35º)

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

Let's break down the forces acting on the crate:

1. The force of gravity (weight): This force is acting vertically downward and can be calculated using the formula F_gravity = m * g, where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s^2).

F_gravity = 3.00 kg * 9.8 m/s^2 = 29.4 N

2. The normal force (N): This force is perpendicular to the incline and counteracts the component of the weight force that is acting perpendicular to the incline.

N = F_gravity * cos(theta), where theta is the angle of the incline (35.0 degrees).

N = 29.4 N * cos(35.0 degrees) = 24.06 N

3. The force of static friction (F_friction): This force acts parallel to the incline and opposes the motion of the crate.

F_friction = coefficient of static friction * N

F_friction = 0.300 * 24.06 N = 7.2 N

To prevent the crate from sliding down the incline, the minimum force (F) must be equal to or greater than the force of static friction.

Therefore, F >= F_friction = 7.2 N

Thus, a minimum force of 7.2 N must be applied perpendicular to the incline to prevent the crate from sliding down.