The coefficient of static friction between the m = 3.20 kg crate and the 35.0° incline of Figure P4.41 is 0.260. What minimum force must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?
To solve this problem, we can break down the forces acting on the crate on the incline. We'll calculate the force parallel to the incline, which needs to be less than or equal to the maximum static friction, in order to prevent the crate from sliding down.
First, let's draw a free body diagram for the crate on the incline:
35°
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─────┘
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__|__ ← Normal force (N)
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│ → Force of gravity (mg)
v
Now, let's analyze the forces:
1. Force of gravity (mg):
The force of gravity acting vertically downward can be resolved into two components: one perpendicular to the incline, and one parallel to the incline. The component perpendicular to the incline (mg * cos(35°)) is balanced by the normal force (N) exerted by the incline. The component parallel to the incline (mg * sin(35°)) will cause the crate to slide down the incline.
2. Friction force (F_friction):
The friction force between the crate and the incline opposes the motion of the crate. The maximum value of static friction force is given by the equation F_friction = coefficient of static friction * N.
3. Force applied perpendicular to the incline (F_applied):
This is the force that we need to calculate. It acts in the direction of the normal force (N), preventing the crate from sliding down the incline.
Now, let's equate the forces parallel to the incline:
mg * sin(35°) = F_friction
And since F_friction = coefficient of static friction * N, we can substitute it in:
mg * sin(35°) = μ * N
The normal force (N) can be calculated by equating the forces perpendicular to the incline:
mg * cos(35°) = N
Rearranging this equation, we get:
N = mg * cos(35°)
Substitute this expression for N in the equation mg * sin(35°) = μ * N:
mg * sin(35°) = μ * (mg * cos(35°))
Now, let's solve for the force applied perpendicular to the incline (F_applied):
F_applied = mg * sin(35°) / cos(35°)
Substituting the given values:
m = 3.20 kg
μ = 0.260
θ = 35°
F_applied = (3.20 kg) * (9.8 m/s^2) * sin(35°) / cos(35°)
Calculating this expression will give you the minimum force required to prevent the crate from sliding down the incline.
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.
Let's break down the forces:
1. Weight of the crate (mg): This force acts vertically downward and can be calculated using the mass of the crate (m = 3.20 kg) and the acceleration due to gravity (g = 9.8 m/s²).
Weight (mg) = m * g
2. Normal force (N): This force acts perpendicular to the incline, pushing the crate against it. It can be calculated using the weight of the crate and the angle of incline.
Normal force (N) = mg * cos(θ), where θ is the angle of incline (35.0°).
3. Force of static friction (fs): This force acts parallel to the incline and opposes the component of the applied force that tries to slide the crate downward. The magnitude of fs can be found using the coefficient of static friction (μs) and the normal force.
fs = μs * N, where μs is the coefficient of static friction (0.260).
The minimum force required to prevent the crate from sliding down the incline is equal to the force of static friction:
Minimum force = fs
Now, let's calculate each force:
Weight (mg) = 3.20 kg * 9.8 m/s² = 31.36 N
Normal force (N) = 31.36 N * cos(35.0°) ≈ 25.61 N
Force of static friction (fs) = 0.260 * 25.61 N ≈ 6.66 N
Therefore, the minimum force that must be applied perpendicular to the incline to prevent the crate from sliding down is approximately 6.66 N.