The coefficient of static friction between the 2.90 kg crate and the 35.0° incline of Figure P4.41 is 0.260. What minimum force F must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?
x: mgsinα = F(fr)
y: N=mgcosα + F
mgsinα = F(fr) =μN =μ(mgcosα + F),
F =mgsinα/μ - mgcosα =
=mg(sinα/μ – cosα)
To find the minimum force F that must be applied to the crate perpendicular to the incline, we can follow these steps:
Step 1: Determine the weight of the crate.
The weight can be calculated using the formula:
Weight = mass * gravity acceleration
Given:
Mass of the crate (m) = 2.90 kg
Gravity acceleration (g) = 9.8 m/s^2 (approximate value)
Weight = 2.90 kg * 9.8 m/s^2
Weight = 28.42 N
Step 2: Determine the component of the weight acting parallel to the incline.
Since the incline is at an angle of 35.0°, the component of the weight acting parallel to the incline can be calculated using the formula:
Weight(parallel) = Weight * sin(angle)
Weight(parallel) = 28.42 N * sin(35.0°)
Weight(parallel) = 16.22 N
Step 3: Determine the maximum force of static friction.
The maximum force of static friction can be found using the coefficient of static friction and the component of the weight acting parallel to the incline:
Maximum force of static friction = coefficient of static friction * Weight(parallel)
Coefficient of static friction (μ) = 0.260
Maximum force of static friction = 0.260 * 16.22 N
Maximum force of static friction = 4.22 N
Step 4: Determine the minimum force F required to prevent the crate from sliding down the incline.
The minimum force F can be found by summing up the forces acting parallel to the incline:
Minimum force F = Weight(parallel) + Maximum force of static friction
Minimum force F = 16.22 N + 4.22 N
Minimum force F = 20.44 N
Therefore, the minimum force F that must be applied to the crate perpendicular to the incline to prevent sliding down is 20.44 N.
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. There are two main forces at play: the force of gravity and the force of friction.
1. First, let's calculate the force of gravity acting on the crate. To do this, we use the equation:
F_gravity = m * g
where m is the mass of the crate (2.90 kg) and g is the acceleration due to gravity (9.8 m/s²).
F_gravity = 2.90 kg * 9.8 m/s² = 28.42 N
So, the force of gravity is 28.42 N acting vertically downward.
2. Next, we calculate the normal force (N) acting perpendicular to the incline. The normal force is equal in magnitude to the force of gravity, but acts perpendicular to the incline. Since the incline is at an angle of 35.0°, the normal force can be calculated using:
N = F_gravity * cos(angle)
N = 28.42 N * cos(35.0°) = 23.17 N
So, the normal force acting perpendicular to the incline is 23.17 N.
3. Now, we can calculate the maximum force of static friction (F_friction) that can act between the crate and the incline. The coefficient of static friction (μ) is given as 0.260. The formula for static friction is:
F_friction = μ * N
F_friction = 0.260 * 23.17 N = 6.02 N
Therefore, the maximum force of static friction is 6.02 N.
4. Lastly, to prevent the crate from sliding down the incline, the force applied perpendicular to the incline (F) should counteract the force of static friction. Since the crate is on the verge of sliding, the minimum force needed is equal to the force of static friction:
F = F_friction = 6.02 N
Therefore, the minimum force required to prevent the crate from sliding down the incline is 6.02 N.
Note: The direction of the force is perpendicular to the incline, so it is not along the incline as mentioned in the problem.