The coefficient of static friction between the 3.0 kg crate and the 31° incline shown below is 0.300. What is the magnitude of the minimum force, F, that must be applied to the crate perpendicularly to the incline to prevent the crate from sliding down the incline?

Let F be the applied force

[(M*g cos31) + F]*0.300 > M*g*sin31

to prevent sliding. That ensure that the friction force exceeds the weight force component down the slid.

Solve for F, in Newtons

169.2

To find the magnitude of the minimum force (F) that must be applied to the crate perpendicularly to the incline to prevent it from sliding down, we need to consider the forces acting on the crate along the incline.

Let's break down the forces involved:

1. Weight force (W):
The weight force acts vertically downwards and is given by:
W = m * g
Where,
m = mass of the crate = 3.0 kg
g = acceleration due to gravity = 9.8 m/s²

W = 3.0 kg * 9.8 m/s²
W = 29.4 N

2. Normal force (N):
The normal force acts perpendicular to the incline and counters the weight force.
N = W * cos(θ)
Where,
θ = angle of the incline = 31°

N = 29.4 N * cos(31°)
N = 25.4 N

3. Frictional force (Ff):
The frictional force opposes the motion of the crate and acts along the incline.
Ff = coefficient of static friction * N
The coefficient of static friction is given as 0.300.

Ff = 0.300 * 25.4 N
Ff = 7.62 N

To prevent the crate from sliding down, the minimum force (F) must counterbalance the frictional force (Ff).

Therefore,
F = Ff
F = 7.62 N

Hence, the magnitude of the minimum force (F) that must be applied to the crate perpendicularly to the incline to prevent the crate from sliding down the incline is 7.62 N.

To find the magnitude of the minimum force, F, required to prevent the crate from sliding down the incline, we need to use the concept of static friction.

The formula for static friction is:

F_static = µ_s * N

where F_static is the magnitude of the static friction force, µ_s is the coefficient of static friction, and N is the normal force acting on the crate.

To find the normal force, we need to consider the forces acting on the crate in the vertical direction. These forces include the weight of the crate (mg) and the normal force (N) exerted by the incline.

The weight of the crate can be calculated using the formula:

weight = m * g

where m is the mass of the crate and g is the acceleration due to gravity (-9.8 m/s^2).

weight = 3.0 kg * (-9.8 m/s^2) = -29.4 N

Since the incline is at an angle of 31° with respect to the horizontal, we can decompose the weight force into its components. The normal force (N) will act perpendicular to the incline, and the weight component (mg * sinθ) will act parallel to the incline.

N = weight * cosθ
N = -29.4 N * cos(31°)
N ≈ -25.4 N

Note that the negative sign indicates that the forces are acting in the opposite direction of the reference frame (downward).

Now, we can calculate the magnitude of the static friction force:

F_static = µ_s * N
F_static = 0.300 * (-25.4 N)
F_static ≈ -7.62 N

The negative sign indicates that the static friction force acts in the opposite direction to prevent sliding.

Finally, the magnitude of the minimum force, F, required to prevent the crate from sliding down the incline perpendicularly can be determined by taking the absolute value of the static friction force:

|F| = |F_static|
|F| = |(-7.62 N)|
|F| ≈ 7.62 N

Therefore, the magnitude of the minimum force, F, that must be applied to the crate perpendicularly to the incline to prevent sliding is approximately 7.62 N.