Explore BrainMass
Share

# Newton Laws, conveyer belt

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

1. A package is at rest on a conveyer belt which is initially at rest. The belt is started and moves to the right for 1.5s with a constant acceleration of 3.2m/s^2. The belt then moves with a constant deceleration a_2 and comes to a stop after a total displacement of 4.6m. Knowing that the coefficients of friction between the package and the belt are μ = 0.35 and μ = 0.25, determine (a) the deceleration a_2 of the belt, (b) the displacement of the package relative to the belt as the belt comes to a stop.

2. To transport a series of bundles of shingles A to a roof, a contractor uses a motor-driven lift consisting of a horizontal platform BC which rides on rails attached to the sides of a ladder. The lift starts from rest and initially moves with a constant acceleration a_1 as shown. The lift then decelerates at a constant rate a_2 and comes to rest at D, near the top of the ladder. Knowing that the coefficient of static friction between the bundle of shingles and the horizontal platform is 0.30, determine the largest allowable acceleration a_1 an d the largest allowable deceleration a_2 if the bundle is not to slide on the platform.

https://brainmass.com/engineering/mechanical-engineering/newton-laws-conveyer-belt-67054

#### Solution Preview

(1). At the end of the acceleration, the displacement of the belt is:

d = ½ * a* t = ½ * 3.2 * 1.5* 1.5 = 3.6 m

So the displacement of the belt ...

#### Solution Summary

Solution provides detailed, clear explanation of how to calculate a package deceleration and displacement on a conveyer belt, as well as transport times for a bundle of shingles.

\$2.19