A 20 kg crate is placed on a horizontal conveyor belt. The materials are such that μs = 0.6 and μk = 0.3.
(c) What is the maximum acceleration the belt can have without the crate slipping?
(d) If acceleration of the belt exceeds the value determined in part (c), what is the acceleration of the crate?
Normal force = m g
friction force static = .6 mg
for no slip = m a
.6 m g = m a
a = .6 g or about .6*9.8 = 5.88 m/s^2
F sliding friction = .3 m g = m a
a = .3 g = 2.94 g
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2.34m/s^2
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To answer these questions, we need to use the concepts of friction and the coefficients of static and kinetic friction.
(c) To determine the maximum acceleration the belt can have without the crate slipping, we need to consider static friction. The maximum static friction force (Fmax_s) can be calculated using the formula:
Fmax_s = μs * N
where μs is the coefficient of static friction and N is the normal force. The normal force is equal to the weight of the crate, which is given by:
N = m * g
where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the given values, we have:
m = 20 kg
μs = 0.6
g = 9.8 m/s^2
N = m * g = 20 kg * 9.8 m/s^2 = 196 N
Fmax_s = μs * N = 0.6 * 196 N ≈ 117.6 N
The maximum acceleration without slipping is equal to the force of static friction divided by the mass of the crate:
a_max = Fmax_s / m = 117.6 N / 20 kg ≈ 5.88 m/s^2
Therefore, the maximum acceleration the belt can have without the crate slipping is approximately 5.88 m/s^2.
(d) If the acceleration of the belt exceeds the value determined in part (c), the crate will start sliding, and we need to consider the kinetic friction force. The kinetic friction force (F_k) can be calculated using the formula:
F_k = μk * N
where μk is the coefficient of kinetic friction and N is the normal force (equal to the weight of the crate).
Plugging in the given values, we have:
m = 20 kg
μk = 0.3
g = 9.8 m/s^2
N = m * g = 20 kg * 9.8 m/s^2 = 196 N
F_k = μk * N = 0.3 * 196 N ≈ 58.8 N
The acceleration of the crate when the belt exceeds the maximum acceleration without slipping is equal to the force of kinetic friction divided by the mass of the crate:
a = F_k / m = 58.8 N / 20 kg ≈ 2.94 m/s^2
Therefore, if the acceleration of the belt exceeds the value determined in part (c), the acceleration of the crate will be approximately 2.94 m/s^2.