A series of small packages are being moved by a thin conveyor belt that passes over a 300-mm radius idler pulley. The belt starts from rest at time t=0 and its speed increases at a constant rate of 150mm/s^2. Knowing that the coefficient of static friction between the pacakges and the belt is 0.75, determine the time at which the first package slips.
Thanks in advance.
I should add, the answer is given as 11.32s, I just don't know how to get there. Thanks again.
I am sorry, the acceleration is constant and given. The force required between package and belt is therefore constant (m a) which is m g * coef. As the problem is stated, there is no particular time.
m a = . 75 m g
a = .75 (9.8)
a = 7.35 m/s^2 = 7350 mm/s^2 before it slips
The acceleration increases at a constant rate, it does not remain constant. Thanks for replying, though. I'm still having trouble with it.
Err, I typed that wrong. The acceleration is in fact constant. The velocity on the other hand is increasing at a constant rate so at some point the velocity will be greater than the friction force can hold. I just can't put it all together.
To determine the time at which the first package slips, we need to find the maximum acceleration the belt can have without exceeding the static friction force.
We can start by calculating the maximum static friction force using the equation:
F_s = µ_s * N
where F_s is the maximum static friction force, µ_s is the coefficient of static friction, and N is the normal force.
In this case, the normal force is equal to the weight of the package, which can be calculated as:
N = m * g
where m is the mass of the package and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Next, we need to determine the maximum acceleration the belt can have without exceeding the static friction force. This is given by:
a_max = F_s / m
where a_max is the maximum acceleration and m is the mass of the package.
Since the acceleration of the belt is increasing at a constant rate, we can use the equation of motion:
v = u + a*t
where v is the final velocity of the belt, u is the initial velocity of the belt (which is 0 in this case), a is the acceleration of the belt, and t is the time.
We can rearrange this equation to solve for t:
t = (v - u) / a
Substituting the given values, we have:
v = 150 mm/s^2 * t (since the speed increases at a constant rate)
u = 0 mm/s^2
a = 150 mm/s^2
t = (150 mm/s^2 * t - 0 mm/s^2) / 150 mm/s^2
Simplifying the equation and converting mm/s^2 to m/s^2:
t = t
Now, we need to find the time when the maximum acceleration of the belt equals the maximum acceleration without exceeding the static friction force. Therefore:
t = a_max / a
Substituting the given values:
t = (F_s / m) / a
t = (0.75 * m * g) / (m * 150 mm/s^2)
t = (0.75 * 9.8 m/s^2) / (150 mm/s^2)
t = 0.0493 s
Therefore, the first package slips at approximately t = 0.0493 seconds.