A person pulls with a force of 40 N on two boxes at rest. The top box is 3.0 kg and the bottom box is 6.0 kg. The coefficient of kinetic friction between the floor and the larger box is .20 and the angle is 30 degrees.
a) if the smaller box doesn't slide on top of the larger box, what is the acceleration of the larger box?
b) Given your answer to part a, what must be the minimum value of the coefficient of static friction between the smaller box and the larger box?
a) To find the acceleration of the larger box, we need to calculate the net force acting on it.
First, let's calculate the force of friction acting on the larger box. The force of kinetic friction (fk) can be found using the formula:
fk = μk * N
Where μk is the coefficient of kinetic friction and N is the normal force.
The normal force (N) can be calculated by multiplying the mass of the box (m) by the acceleration due to gravity (g).
m = 6.0 kg (mass of the larger box)
g = 9.8 m/s² (acceleration due to gravity)
N = m * g
N = 6.0 kg * 9.8 m/s²
N = 58.8 N
Using the given coefficient of kinetic friction, μk = 0.20, we can calculate the force of kinetic friction:
fk = μk * N
fk = 0.20 * 58.8 N
fk = 11.76 N
Next, we can calculate the net force (Fnet) acting on the larger box:
Fnet = Fapplied - fk
The applied force is given as 40 N:
Fapplied = 40 N
Fnet = 40 N - 11.76 N
Fnet = 28.24 N
Now, we can use Newton's second law of motion, F = ma, to find the acceleration (a) of the larger box.
Fnet = m * a
28.24 N = 6.0 kg * a
Solving for a:
a = 28.24 N / 6.0 kg
a ≈ 4.7 m/s²
Therefore, the acceleration of the larger box is approximately 4.7 m/s².
b) To find the minimum coefficient of static friction between the smaller box and the larger box, we need to consider the maximum force that can be exerted by static friction without causing the smaller box to slide. This occurs when the force of static friction is equal to the force applied on it.
Let's denote the force of static friction as fs and the minimum coefficient of static friction as μs.
The force applied on the smaller box, Fapplied, is the same as in part a, which is 40 N.
fs = μs * N
We can express N as the weight of the smaller box (m * g):
m = 3.0 kg (mass of the smaller box)
g = 9.8 m/s² (acceleration due to gravity)
N = m * g
N = 3.0 kg * 9.8 m/s²
N = 29.4 N
Therefore, the force of static friction is given by:
fs = μs * 29.4 N
Since the smaller box does not slide on top of the larger box, the force of static friction must be equal to the force applied:
fs = 40 N
Therefore, we can set up the equation:
40 N = μs * 29.4 N
Solving for μs:
μs = 40 N / 29.4 N
μs ≈ 1.36
Therefore, the minimum value of the coefficient of static friction between the smaller box and the larger box is approximately 1.36.
To find the acceleration of the larger box in part a, we need to determine the net force acting on it. We'll use Newton's second law, which states that the net force on an object is equal to the product of its mass and acceleration.
a) To begin, we need to calculate the force of friction between the larger box and the floor. The formula for kinetic friction is given by the product of the coefficient of kinetic friction (µ) and the normal force (N).
The normal force (N) can be found using the formula N = m * g, where m is the mass of the box and g is the acceleration due to gravity (approximately 9.8 m/s^2).
For the larger box:
m = 6.0 kg
g = 9.8 m/s^2
N = (6.0 kg) * (9.8 m/s^2) = 58.8 N
Now, we can calculate the force of kinetic friction (F_friction) using the formula F_friction = µ * N.
µ (coefficient of kinetic friction) = 0.20
F_friction = (0.20) * (58.8 N) = 11.76 N
Since the smaller box doesn't slide on top of the larger box, the frictional force between them will be equal to the force of kinetic friction (F_friction).
Now, we need to calculate the net force acting on the larger box. There are two forces involved: the applied force (40 N in this case) and the frictional force (11.76 N).
The net force (F_net) is given by the formula F_net = F_applied - F_friction.
F_applied = 40 N
F_friction = 11.76 N
F_net = (40 N) - (11.76 N) = 28.24 N
Finally, we can use Newton's second law to calculate the acceleration (a) of the larger box.
F_net = m * a
m = 6.0 kg
a = ?
28.24 N = (6.0 kg) * a
a = 28.24 N / 6.0 kg
a = 4.74 m/s^2
Therefore, the acceleration of the larger box is approximately 4.74 m/s^2.
b) To find the minimum value of the coefficient of static friction between the smaller box and the larger box, we need to consider the maximum force of static friction (F_friction_max) that can be exerted.
The formula for the maximum force of static friction is given by F_friction_max = µ * N. In this case, the normal force (N) acting on the smaller box is equal to its weight, which can be calculated as N = m * g.
For the smaller box:
m = 3.0 kg
g = 9.8 m/s^2
N = (3.0 kg) * (9.8 m/s^2) = 29.4 N
Substituting the values into the formula, we get:
F_friction_max = µ * N
F_friction_max = µ * 29.4 N
Since the smaller box is at rest on the larger box, the applied force (40 N) is balanced by the maximum force of static friction (F_friction_max). Therefore:
F_friction_max = 40 N
Substituting µ * 29.4 N = 40 N, we can solve for the minimum value of the coefficient of static friction (µ).
µ * 29.4 N = 40 N
µ = 40 N / 29.4 N
µ ≈ 1.36
Therefore, the minimum value of the coefficient of static friction between the smaller box and the larger box must be approximately 1.36.