A 17 kg crate is placed on a horizontal conveyor belt. The materials are such that μs = 0.44 and μk = 0.21. A) What is the maximum acceleration the belt can have without the crate slipping? B) If acceleration of the belt exceeds the value determined in part A, what is the acceleration of the crate?
A) .44 * 17 * 9.81 = 17 a
solve for a
B) .21 * 17 * 9.81 = 17 a
(less once you get it moving)
To find the maximum acceleration the belt can have without the crate slipping, we need to consider the friction between the crate and the belt.
A) The maximum acceleration without slipping occurs when the force of static friction (fs) reaches its maximum value. The maximum force of static friction can be calculated using the formula:
fs = μs * N
where μs is the coefficient of static friction and N is the normal force acting on the crate.
The normal force is equal to the weight of the crate, which can be calculated using the formula:
N = m * g
where m is the mass of the crate and g is the acceleration due to gravity.
In this case, the mass of the crate is given as 17 kg. The acceleration due to gravity is approximately 9.8 m/s^2.
N = 17 kg * 9.8 m/s^2
N = 166.6 N
Now we can calculate the maximum force of static friction:
fs = μs * N
fs = 0.44 * 166.6 N
fs = 73.3 N
The maximum acceleration without slipping is equal to the force of static friction divided by the mass of the crate:
a(max) = fs / m
a(max) = 73.3 N / 17 kg
a(max) ≈ 4.31 m/s^2
Therefore, the maximum acceleration the belt can have without the crate slipping is approximately 4.31 m/s^2.
B) If the acceleration of the belt exceeds the value determined in part A, the crate will start to slip. In this case, we need to consider the force of kinetic friction (fk).
The force of kinetic friction can be calculated using the formula:
fk = μk * N
where μk is the coefficient of kinetic friction and N is the normal force.
We already know the value of N (166.6 N), and the coefficient of kinetic friction (μk) is given as 0.21.
fk = μk * N
fk = 0.21 * 166.6 N
fk = 34.9 N
Since the acceleration of the belt exceeds the maximum acceleration without slipping, the force of kinetic friction is acting on the crate in the opposite direction of motion. Therefore, the acceleration of the crate can be calculated using Newton's second law of motion:
fk = m * a
34.9 N = 17 kg * a
a = 2.05 m/s^2
Therefore, if the acceleration of the belt exceeds the value determined in part A, the acceleration of the crate is approximately 2.05 m/s^2.