www.webassign.net/sercp9/4-p-047-alt.gifThe coefficient of static friction between the m = 3.60−kg
crate and the 35.0° incline of the figure below is 0.330. What minimum force F must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?
normal force = 3.6 * 9.81 * cos 35 + F
friction force = .33 ( 3.6 * 9.81 * cos 35 + F) = 9.55+.33 F
tat has to balance component of weight down slope
9.55 + .33 F = 3.6 * 9.81 * sin 35 = 20.3
.33 F = 10.7
F = 32.4 N
To find the minimum force F required to prevent the crate from sliding down the incline, we need to consider the forces acting on the crate.
1. Gravity force (mg): This force acts vertically downwards and can be calculated as the product of the mass (m = 3.60 kg) and the acceleration due to gravity (g = 9.8 m/s²) using the formula mg.
2. Normal force (N): This force acts perpendicular to the incline and can be calculated using the formula N = mg * cos(θ), where θ is the angle of the incline (35.0°).
3. Friction force (Ff): This force opposes the motion of the crate and can be calculated using the formula Ff = μs * N, where μs is the coefficient of static friction (0.330). Since we want the minimum force to prevent sliding, we will use the maximum possible static friction force.
4. Force applied perpendicular to the incline (F): This is the force we are trying to find.
In order to prevent the crate from sliding, the force of static friction (Ff) must be equal to the force applied perpendicular to the incline (F).
Now, let's calculate the forces and find the minimum force F needed to prevent sliding.
1. Gravity force (mg):
mg = (3.60 kg) * (9.8 m/s²) = 35.28 N
2. Normal force (N):
N = mg * cos(θ)
N = (35.28 N) * cos(35.0°) = 28.93 N
3. Friction force (Ff):
Ff = μs * N
Ff = (0.330) * (28.93 N) = 9.54 N
4. Minimum force to prevent sliding (F):
F = Ff = 9.54 N
Therefore, the minimum force F that must be applied perpendicular to the incline to prevent the crate from sliding down the incline is 9.54 N.
To find the minimum force F required to prevent the crate from sliding down the incline, we will use the concept of static friction and the force equations involved.
First, let's understand the forces acting on the crate on the inclined plane:
1. The gravitational force (mg): This force pulls the crate downward vertically and can be calculated as the product of the mass of the crate (m) and the acceleration due to gravity (g) which is approximately 9.8 m/s².
mg = 3.60 kg * 9.8 m/s² = 35.28 N
2. The normal force (N): This force acts perpendicular to the incline and counterbalances the vertical component of the gravitational force. It can be calculated using the following equation:
N = mg * cos(θ)
Here, θ is the angle of the incline (35.0°).
N = 35.28 N * cos(35.0°) = 28.939 N
3. The force of static friction (fs): This force acts parallel to the incline and opposes the impending motion of the crate. According to the given information, the coefficient of static friction (μs) between the crate and the incline is 0.330.
The equation for static friction is:
fs ≤ μs * N
Since we want to find the minimum force F, we want to consider the maximum static friction that can be achieved. So, we use the inequality sign (≤).
fs ≤ 0.330 * 28.939 N
fs ≤ 9.538 N
Now, we know that the force F perpendicular to the incline counteracts the force of static friction. Therefore, the minimum force F required to prevent the crate from sliding down the incline is equal to the maximum force of static friction:
F = fs = 9.538 N
Hence, a minimum force of 9.538 N must be applied perpendicular to the incline to prevent the crate from sliding down.