A 1100 N crate is being pushed across a level floor at a constant speed by a force of 500 N at an angle of 20.0° below the horizontal. (a) What is the coefficient of kinetic friction between the crate and the floor?
(b) If the 500 N force is instead pulling the block at an angle of 20.0° above the horizontal as shown in Figure P4.37b, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in (a).
a. M*g = 1100 N. = Wt. of crate.
M = 1100/g = 1100/9.8 = 112 kg = Mass of crate.
Fn = M*g + Fap*sin20 = Normal force.
Fn = 1100 + 500*sin20 = 1271 N.
Fk = u*Fn = u*1271 = 1271u. = Force of kinetic friction.
Fap*Cos20-Fk = M*a.
500*cos20-1271u = M*0
469.8-1271u = 0,
u = 0.370.
b. Fn = M*g - Fap*sin20.
Fn = 1100-500*sin20 = 929 N.
Fk = u*Fn = 0.370 * 929 = 343.7 N.
500*cos20-343.7 = 112a.
a = 1.13 m/s^2.
To find the coefficient of kinetic friction between the crate and the floor, we need to start by calculating the normal force acting on the crate.
(a) The normal force can be determined by decomposing the vertical forces acting on the crate. Since the crate is on a level floor, the normal force cancels out the weight of the crate.
The weight of an object is given by the formula:
Weight = mass x gravity
Given that the weight (W) of the crate is 1100 N and the acceleration due to gravity (g) is approximately 9.8 m/s^2, we can calculate the mass of the crate:
1100 N = mass x 9.8 m/s^2
mass = 1100 N / 9.8 m/s^2
mass ≈ 112.24 kg
Now, the normal force (N) can be calculated as:
N = mass x gravity
N = 112.24 kg x 9.8 m/s^2
N ≈ 1100 N
Next, we can determine the force of friction (Ff) between the crate and the floor. The force of friction can be calculated using the formula:
Ff = coefficient of friction x N
Since the crate is being pushed at a constant speed, the force of friction opposes the applied force. Thus, the force of friction is equal to the applied force (500 N) in this case.
Substituting the known values into the formula, we can solve for the coefficient of kinetic friction (μk):
500 N = μk x 1100 N
μk = 500 N / 1100 N
μk ≈ 0.45
Therefore, the coefficient of kinetic friction between the crate and the floor is approximately 0.45.
(b) If the 500 N force is instead pulling the crate at an angle of 20.0° above the horizontal, we can find the acceleration of the crate.
We need to analyze the forces acting along the horizontal axis. The horizontal component of the applied force is given by:
Force_horizontal = Force_applied x cos(angle)
Substituting the known values, we get:
Force_horizontal = 500 N x cos(20.0°)
Next, we need to consider the forces acting vertically. The net vertical force can be calculated as follows:
Force_vertical = Force_gravity - Force_applied x sin(angle)
Substituting the known values, we get:
Force_vertical = (mass x gravity) - (500 N x sin(20.0°))
Since the crate is on a level floor and the force of friction opposes the applied force, the net vertical force is zero. Therefore, we can set it equal to zero and solve for the acceleration.
0 = (mass x gravity) - (500 N x sin(20.0°))
Solving for the acceleration (a), we get:
a = (500 N x sin(20.0°)) / (mass x gravity)
Substituting the values, we have:
a ≈ (500 N x sin(20.0°)) / (112.24 kg x 9.8 m/s^2)
a ≈ 1.59 m/s^2
Therefore, the acceleration of the crate when it is pulled with a force of 500 N at an angle of 20.0° above the horizontal is approximately 1.59 m/s^2.
To find the coefficient of kinetic friction between the crate and the floor, you need to use the given information and apply Newton's second law of motion.
(a) The force pushing the crate across the floor is 500 N at an angle of 20.0° below the horizontal. First, find the vertical component of this force (Fv) and the horizontal component (Fh).
Fv = 500 N * sin(20.0°) = 170.2 N
Fh = 500 N * cos(20.0°) = 469.8 N
Since the crate is moving at a constant speed, the net force on the crate is zero. The force of kinetic friction (Fk) is equal in magnitude and opposite in direction to the horizontal component of the applied force.
Fk = -Fh = -469.8 N
The formula for the force of kinetic friction is Fk = μk * F_n, where μk is the coefficient of kinetic friction and F_n is the normal force exerted on the crate.
The normal force is equal to the weight of the crate since the crate is on a level floor. So, F_n = weight of the crate = mass * gravitational acceleration.
Given that the weight of the crate is 1100 N and the acceleration due to gravity is 9.8 m/s^2:
F_n = 1100 N
μk * 1100 N = -469.8 N
Solving for μk:
μk = -469.8 N / 1100 N = -0.427
However, the coefficient of kinetic friction cannot be negative. It represents the frictional interaction between two surfaces. So, the correct answer is μk = 0.427.
(b) If the 500 N force is pulling the crate at an angle of 20.0° above the horizontal, you can use the same approach and Newton's second law to find the acceleration of the crate.
The vertical component of the force (Fv') is:
Fv' = 500 N * sin(20.0°) = 170.2 N
Since the crate is still on a level floor, the normal force exerted on the crate remains the same.
F_n = 1100 N
The net horizontal force acting on the crate is the force pulling the crate horizontally minus the force of kinetic friction:
F_net = Fh - Fk
F_net = 500 N * cos(20.0°) - μk * F_n
Substituting the values:
F_net = 469.8 N - (0.427 * 1100 N) = 23 N
Finally, you can use Newton's second law, F_net = mass * acceleration, to find the acceleration (a) of the crate:
23 N = mass * a
Given that the mass of the crate is not provided, the provided information is insufficient to determine the exact value of the acceleration of the crate.