Posted by Hal on Monday, April 7, 2008 at 12:47am.
A series of small packages are being moved by a thin conveyor belt that passes over a 300mm 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.

Dynamics  Hal, Monday, April 7, 2008 at 12:49am
I should add, the answer is given as 11.32s, I just don't know how to get there. Thanks again.

Dynamics  Damon, Monday, April 7, 2008 at 5:31am
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

Dynamics  Hal, Monday, April 7, 2008 at 1:13pm
The acceleration increases at a constant rate, it does not remain constant. Thanks for replying, though. I'm still having trouble with it.

Dynamics  Hal, Monday, April 7, 2008 at 1:32pm
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.
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