The coefficient of static friction between the 3.00 kg crate and the 35.0 degree incline is .300. What minimum force must be applied to the crate from sliding down the incline?

Ok I tried an didn't even come close to the right answer.

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

1. Weight (mg): The weight of the crate can be calculated using the mass of the crate (m) and the acceleration due to gravity (g). In this case, g is approximately 9.8 m/s^2.

Weight (mg) = m * g
= 3.00 kg * 9.8 m/s^2
= 29.4 N

2. Normal force (Fn): The normal force is the force exerted by the inclined surface perpendicular to the surface. It is equal in magnitude and opposite in direction to the component of the weight perpendicular to the surface.

Normal force (Fn) = Weight (mg) * cos(θ)
= 29.4 N * cos(35 degrees)
= 24.1 N

3. Force of friction (Ff): The force of friction acts parallel to the surface and opposes the motion of the crate. It can be calculated using the coefficient of static friction (μs) and the normal force.

Force of friction (Ff) = Coefficient of static friction (μs) * Normal force (Fn)
= 0.300 * 24.1 N
= 7.23 N

4. Force applied (Fapplied): The minimum force required to slide the crate down the incline is equal to the force of friction.

Force applied (Fapplied) = Force of friction (Ff)
= 7.23 N

Therefore, the minimum force that must be applied to the crate from sliding down the incline is 7.23 N.

To find the minimum force required to slide the crate down the incline, we need to consider the gravitational force acting on the crate and the friction force opposing its motion.

Here's how you can calculate it:

1. Start by determining the gravitational force acting on the crate. The gravitational force can be calculated using the formula:

F_gravity = mass * gravitational acceleration

Given that the mass of the crate is 3.00 kg and the gravitational acceleration is approximately 9.8 m/s², we can calculate the gravitational force:

F_gravity = 3.00 kg * 9.8 m/s² = 29.4 N

2. Next, determine the component of the gravitational force acting down the incline. To do this, find the force's parallel component:

F_parallel = F_gravity * sin(angle)

In this case, the angle of the incline is 35.0 degrees, so we can find the parallel component of the gravitational force:

F_parallel = 29.4 N * sin(35.0°) = 16.8 N

3. The friction force opposing the crate's motion can be calculated using the equation:

F_friction = coefficient of static friction * F_normal

The normal force (F_normal) is equal to the perpendicular component of the gravitational force:

F_normal = F_gravity * cos(angle)

F_normal = 29.4 N * cos(35.0°) = 24.0 N

Substituting the value for the coefficient of static friction given as 0.300:

F_friction = 0.300 * 24.0 N = 7.20 N

4. Finally, find the minimum force required to overcome the static friction and initiate motion. The minimum force required is equal to the sum of the friction force and the parallel component of the gravitational force:

F_min = F_friction + F_parallel

F_min = 7.20 N + 16.8 N = 24.0 N

Therefore, the minimum force that must be applied to the crate to slide it down the incline is approximately 24.0 N.

force gravity down plane: 3g*sin35

force friction up plane=mu*3g*cos35
forces up plane=forces down plane
so F+mu*3g*cos35=3g*sin35

so figure and solve for F