The coefficient of static friction between the m = 3.50 kg crate and the 35.0° incline of Figure P4.41 is 0.340. What minimum force must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?
Your response differs from the correct answer by more than 100%. N
Figure P4.41
Hint: Active Figure 4.21
Please show what you have attempted.
26.28
.08
To find the minimum force necessary to prevent the crate from sliding down the incline, we need to consider the forces acting on the crate.
First, let's break down the gravitational force into its components. The weight of the crate can be calculated using the formula:
Weight = mass * gravity
Given that the mass of the crate (m) is 3.50 kg and the acceleration due to gravity (g) is approximately 9.8 m/s², we can calculate the weight:
Weight = 3.50 kg * 9.8 m/s²
Next, we need to determine the normal force acting on the crate. The normal force is perpendicular to the incline and cancels out the perpendicular component of the weight. In this case, it is equal to the weight times the cosine of the angle of inclination:
Normal force = Weight * cos(angle of inclination)
Given that the angle of inclination is 35.0°, we can calculate the normal force:
Normal force = Weight * cos(35.0°)
Now, we can determine the maximum force of static friction. The coefficient of static friction (μ) is given as 0.340. The maximum force of static friction can be calculated by multiplying the coefficient of static friction by the normal force:
Maximum force of static friction = μ * Normal force
Finally, the minimum force required to prevent the crate from sliding down the incline is equal to the force of static friction, so it is given by:
Minimum force = Maximum force of static friction
By calculating these values, you should be able to find the correct answer.