A 20 kg crate is placed on a horizontal conveyor belt. The materials are such that us = 0.47 and uk = 0.21.
What is the maximum acceleration the belt can have without the crate slipping?
If acceleration of the belt exceeds the value determined in part A, what is the acceleration of the crate?
Well, well, well, looks like we're dealing with some physics here! Let's clown around with the numbers, shall we?
Part A: To prevent the crate from slipping, we need to find the maximum acceleration. Now, "us" is the coefficient of static friction and "uk" is the coefficient of kinetic friction. So, let's find what those mean first.
If the crate is at rest, the maximum static frictional force that can act on it is given by:
\( F_{\text{max-friction}} = \text{us} \times m \times g \),
where m is the mass of the crate (20 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Once the crate starts moving, the frictional force changes to the kinetic frictional force given by:
\( F_{\text{kinetic-friction}} = \text{uk} \times m \times g \).
For the crate to stay put without slipping, the maximum static frictional force should be equal to the applied force, which is the mass of the crate times the acceleration of the belt. So we have:
\( F_{\text{max-friction}} = m \times a \).
Where "a" is the acceleration of the belt that we want to find.
Setting the maximum static frictional force equal to the applied force, we get:
\( \text{us} \times m \times g = m \times a \).
Simplifying, we find:
\( a = \text{us} \times g \).
Plugging in the given values: us = 0.47 and g ≈ 9.8 m/s^2, we can calculate the maximum acceleration.
Part B: If the acceleration of the belt exceeds the value we found in part A, then the crate will start to slip. In that case, the frictional force acting on the crate will change to the kinetic frictional force. So, the acceleration of the crate would be equal to:
\( a_{\text{crate}} = \text{uk} \times g \).
Using the given value uk = 0.21 and g ≈ 9.8 m/s^2, we can calculate the acceleration of the crate in this case.
Voila! Physics with a touch of comedy. Hope that helps!
To determine the maximum acceleration the belt can have without the crate slipping, we need to consider the static friction between the crate and the belt. The maximum static friction force that can be applied to the crate without it slipping is given by:
f_s,max = μ_s * N
where f_s,max is the maximum static friction force, μ_s is the coefficient of static friction, and N is the normal force acting on the crate.
In this case, the normal force N is equal to the weight of the crate:
N = m * g
where m is the mass of the crate and g is the acceleration due to gravity.
Substituting the values into the formula:
N = 20 kg * 9.8 m/s^2 = 196 N
Now, we can find the maximum static friction force:
f_s,max = 0.47 * 196 N = 91.92 N
The maximum acceleration the belt can have without the crate slipping is equal to the acceleration due to this maximum static friction force:
a_max = f_s,max / m
Substituting the values:
a_max = 91.92 N / 20 kg ≈ 4.596 m/s^2
Therefore, the maximum acceleration the belt can have without the crate slipping is approximately 4.596 m/s^2.
If the acceleration of the belt exceeds this maximum value, the crate will start to slip. In this case, the friction force acting on the crate will be kinetic friction, which is given by:
f_k = μ_k * N
where f_k is the kinetic friction force and μ_k is the coefficient of kinetic friction.
The acceleration of the crate can be calculated by subtracting the force due to kinetic friction from the net force acting on the crate and dividing it by the mass of the crate:
a_crate = (F_net - f_k) / m
Given that the maximum acceleration of the crate without slipping is 4.596 m/s^2, any acceleration greater than this will cause slipping to occur.
Let's assume the acceleration of the belt exceeds the maximum value and is equal to 5.0 m/s^2. In this case, we can calculate the kinetic friction force:
f_k = μ_k * N = 0.21 * 196 N = 41.16 N
The net force acting on the crate can be calculated by multiplying the mass of the crate by the acceleration of the belt:
F_net = m * a_belt = 20 kg * 5.0 m/s^2 = 100 N
Now, we can calculate the acceleration of the crate:
a_crate = (F_net - f_k) / m
= (100 N - 41.16 N) / 20 kg
≈ 3.0 m/s^2
Therefore, if the acceleration of the belt exceeds the maximum value determined in part A, the acceleration of the crate will be approximately 3.0 m/s^2.
To find the maximum acceleration the belt can have without the crate slipping, we need to consider the forces acting on the crate. The two relevant forces are the normal force (Fn) and the force of friction (Ff).
1. Normal Force (Fn): The normal force is equal to the weight of the object. In this case, the weight of the crate is given by the formula: Fn = m * g, where m is the mass of the crate (20 kg) and g is the acceleration due to gravity (9.8 m/s^2). Therefore, Fn = 20 kg * 9.8 m/s^2 = 196 N.
2. Force of Friction (Ff): The force of friction can be calculated using the equation Ff = us * Fn, where us is the coefficient of static friction (0.47). Therefore, Ff = 0.47 * 196 N = 92.12 N.
Since we want to find the maximum acceleration without the crate slipping, the force of friction must be equal to the maximum force that can be exerted without slipping, which is the product of the coefficient of static friction and the normal force (us * Fn). Therefore, us * Fn = Ff.
Using this equation, we can find the acceleration (a) at the maximum frictional force. Rearranging the equation, we have a = Ff / m. Plugging in the numbers, a = 92.12 N / 20 kg = 4.61 m/s².
So, the maximum acceleration the belt can have without the crate slipping is 4.61 m/s².
Now, if the acceleration of the belt exceeds this value, the crate will start to slip. In this case, we need to consider the coefficient of kinetic friction (uk). For a slipping crate, the force of friction can be calculated using Ff = uk * Fn.
To find the acceleration of the crate when slipping, we can use the equation Ff = m * a, where a is the acceleration of the crate. Rearranging the equation, we have a = Ff / m.
Plugging in the values, a = (uk * Fn) / m = (0.21 * 196 N) / 20 kg = 2.058 m/s².
Therefore, if the acceleration of the belt exceeds the value determined in part A, the acceleration of the crate will be 2.058 m/s².
Normal force = 20 * 9.81 = 196 N
.47 * 196 = 92.2 = max force without slipping
20 a = 92.2
a = 4.61 m/s^2 max without slip (we could have just said .47 g)
196 (.21) = 20 a
a = 2.06 m/s^2 while slipping