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 packages and the belt is 0.75, determine the time at which the first package slips.

The given answer is 11.32 seconds.

The friction force on a package not slipping must be M*0.15 m/s^2, and the maximum static friction force is M*9.8 m/s^2. I don't see why slipping would ever occur, or what the radius of the "idler pully" bas to do with the problem.

To determine the time at which the first package slips, we can use the concept of friction and the equations of motion.

Let's start by finding the maximum static friction force that can act between the belt and the package. We can calculate this using the equation:

Maximum static friction force (Ff) = coefficient of static friction (μ) * normal force (N)

The normal force (N) acting on the package is equal to the weight of the package. Since there is no vertical acceleration, the weight of the package is balanced by the normal force:

N = mg

Next, we need to calculate the speed of the belt as a function of time. Since the belt starts from rest and has a constant speed increase, we can use the equation of motion for linear acceleration:

v = u + at

Where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time.

In this case, the initial velocity (u) is zero, the acceleration (a) is 150 mm/s^2, and the final velocity (v) is the speed of the belt at any given time t.

Now, we can find the time at which the package slips. The package will begin to slip when the maximum static friction force is exceeded by the force required to accelerate the package linearly with the increasing speed of the belt:

Ff = m * a (where m is the mass of the package)

Rearranging this equation, we can solve for the time (t):

t = Ff / (m * a)

Finally, substituting the given values into the equation, we can calculate the time at which the first package slips.

Given:
Radius of the idler pulley (r) = 300 mm
Coefficient of static friction (μ) = 0.75
Acceleration (a) = 150 mm/s^2

We can convert the radius to meters for consistency:

r = 300 mm = 0.3 meters

Now, let's calculate the time at which the first package slips:

First, let's assume the mass of the package is 1 kg for simplicity (you can adjust the mass value based on the specific situation):

N = mg
N = 1 kg * 9.8 m/s^2
N = 9.8 N

Maximum static friction force (Ff) = μ * N
Ff = 0.75 * 9.8 N
Ff = 7.35 N

The time at which the package slips can be calculated by:
t = Ff / (m * a)
t = 7.35 N / (1 kg * 0.15 m/s^2)
t = 49 s

Please note that the calculated time seems to be incorrect based on the given answer. Double-check the given values and the equations used to confirm the accuracy of the answer.