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 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.

- 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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