A 1000-N crate is being pushed across a level floor at a constant speed by a force F of 300 N at an angle of 20.0 degrees below the horizontal, as shown in Figure P4.41a. (a) What is as the coefficient of kinetic friction between the crate and the floor? (b) If the 300-N force is instead pulling the block at an angle of 20.0 degrees above the horizontal, as shown in Figure P4.41b, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a).
component of push along floor = 300 cos 20
component of push normal to floor = 300 sin 20
friction force holding back = (1000 + 300 sin 20) * mu
300 cos20 = mu (1000 + 300 sin 20)
solve for mu
then I am sure you can do the rest.
To solve this problem, we'll use Newton's second law of motion and the concepts of friction.
(a) To find the coefficient of kinetic friction between the crate and the floor, we need to consider the forces acting on the crate. The crate is being pushed across the floor at a constant speed, so the net horizontal force on the crate must be zero.
The horizontal force acting on the crate is given as F = 300 N at an angle of 20.0 degrees below the horizontal. Let's resolve this force into its horizontal and vertical components.
Horizontal component of force (F_h) = F * cos(theta)
F_h = 300 N * cos(20 degrees)
F_h = 300 N * 0.9397
F_h = 281.91 N (approximately)
Vertical component of force (F_v) = F * sin(theta)
F_v = 300 N * sin(20 degrees)
F_v = 300 N * 0.3420
F_v = 102.60 N (approximately)
The weight of the crate acting vertically downward is given as 1000 N.
The normal force acting upward on the crate must be equal to the vertical component of force (F_v) plus the weight (1000 N).
Normal force = F_v + weight
Normal force = 102.60 N + 1000 N
Normal force = 1102.60 N (approximately)
Now, we can calculate the force of kinetic friction (f_k) acting horizontally against the crate.
Force of kinetic friction (f_k) = coefficient of kinetic friction * normal force
Since the crate is moving at a constant speed, the force of kinetic friction is equal in magnitude but opposite in direction to the horizontal component of force (F_h).
f_k = F_h
Also, f_k = coefficient of kinetic friction * normal force
Therefore, coefficient of kinetic friction = F_h / Normal force
Coefficient of kinetic friction = 281.91 N / 1102.60 N
Coefficient of kinetic friction ≈ 0.256
(b) If the 300-N force is pulling the crate at an angle of 20.0 degrees above the horizontal, we can calculate the acceleration of the crate using the following steps:
First, let's resolve the applied force into its horizontal and vertical components.
Horizontal component of force (F_h) = F * cos(theta)
F_h = 300 N * cos(20 degrees)
F_h = 300 N * 0.9397
F_h = 281.91 N (approximately)
Vertical component of force (F_v) = F * sin(theta)
F_v = 300 N * sin(20 degrees)
F_v = 300 N * 0.3420
F_v = 102.60 N (approximately)
Now, we can calculate the net force acting on the crate horizontally.
Net force (F_net) = F_h - Force of kinetic friction
From part (a), we know that the force of kinetic friction is approximately 281.91 N.
Therefore, F_net = 281.91 N - 281.91 N
F_net = 0 N
Since the net force acting on the crate horizontally is zero, the acceleration of the crate will also be zero. Hence, the crate will not experience any acceleration when pulled at an angle of 20.0 degrees above the horizontal.
To solve part (a), we need to determine the coefficient of kinetic friction between the crate and the floor. The fact that the crate is being pushed at a constant speed tells us that the net force on the crate is zero. We can break down the forces acting on the crate in the horizontal direction.
Let's calculate the net force in the horizontal direction:
1. Calculate the horizontal component of the pushing force (F) by using the angle given:
F_horizontal = F * cos(angle)
F_horizontal = 300 N * cos(20°)
F_horizontal = 274.52 N (rounded to two decimal places)
2. The only other horizontal force acting on the crate is the force of kinetic friction (f_k), which is in the opposite direction of the pushing force. So,
F_kinetic_friction = f_k
3. Since the crate is moving at a constant speed, the magnitude of the force of kinetic friction is equal to the magnitude of the pushing force.
F_kinetic_friction = 274.52 N
4. Lastly, we can use the equation for the force of kinetic friction:
F_kinetic_friction = μ_k * F_normal
where μ_k is the coefficient of kinetic friction and F_normal is the normal force.
Since the crate is on a level floor and not accelerating vertically, the normal force is equal to its weight. Thus,
F_normal = weight of the crate
F_normal = 1000 N
Substituting the values into the equation, we have:
274.52 N = μ_k * 1000 N
Now we can solve for the coefficient of kinetic friction (μ_k):
μ_k = 274.52 N / 1000 N
μ_k = 0.275 (rounded to three decimal places)
Therefore, the coefficient of kinetic friction between the crate and the floor is 0.275.
Moving on to part (b), where the 300 N force is pulling the crate at an angle of 20.0 degrees above the horizontal:
To find the acceleration of the crate, we need to find the net force acting on it. The forces acting on the crate in the horizontal direction are:
1. The pulling force (F_pull) has a horizontal component given by:
F_pull_horizontal = F_pull * cos(angle)
F_pull_horizontal = 300 N * cos(20°)
F_pull_horizontal = 282.84 N (rounded to two decimal places)
2. The force of kinetic friction (f_k) is still present and acts in the opposite direction. So,
F_kinetic_friction = f_k = 274.52 N (same as part a)
3. The net horizontal force is the vector sum of the horizontal components of the pulling force and the force of kinetic friction:
Net_force_horizontal = F_pull_horizontal - F_kinetic_friction
Net_force_horizontal = 282.84 N - 274.52 N
Net_force_horizontal = 8.32 N (rounded to two decimal places)
Since the crate is being pulled, the net force will cause acceleration. Let's assume the mass of the crate is m. The net force is related to the acceleration (a) by Newton's second law:
Net_force_horizontal = m * a
Using this equation, we can solve for the acceleration:
8.32 N = m * a
However, we don't know the value of m. What we do know is the weight of the crate, which is equal to the force of gravity acting on it. So,
Weight = m * g
Weight = 1000 N
To eliminate the unknown mass (m), we can express it in terms of weight and acceleration:
m = Weight / g
Substituting this into the equation for net force, we have:
8.32 N = (Weight / g) * a
8.32 N = (1000 N / g) * a
And we need to solve for a:
a = (8.32 N * g) / 1000 N
Now, we need to know the acceleration due to gravity (g), which is approximately 9.8 m/s^2.
a = (8.32 N * 9.8 m/s^2) / 1000 N
a ≈ 0.081 m/s^2 (rounded to three decimal places)
Therefore, the acceleration of the crate when pulled at an angle of 20.0 degrees above the horizontal is approximately 0.081 m/s^2.