- #1

tenbee

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## Homework Statement

Half-Passage: In the packing industry, processing packages via conveyor belts is immensely practical and vital for operations. Maintaining packages in a neat and file line is no easy task, considering that many conveyor belts are non-linear and require the traversal of hills and valleys. As a general rule of thumb, boxes are kept no less than 5m apart so that collisions do not occur and skew the entire process. In the following diagram, box A is at the very top of the conveyor belt and box B is at the very bottom. The height of the conveyor belt system is h and the slops of the hill is r.

http://www.mcatquestionaday.com/pictures/080925.gif

Question 2: Assume that the conveyor belt of the incline is distinct and moves separately from the conveyor belt on which B rests. If the conveyor belt underneath A breaks, how long will it take for box A to hit box B assuming that box B remains stationary and the coefficient of kinetic friction of the conveyor belt is 0.4? Also assume that the mass of box A is m. For simplicity, assume that the distance down the ramp is d.

## Homework Equations

## The Attempt at a Solution

answer: {2d/[gsin(r) - 0.4gcos(r)]}1/2

Okay... I understand how to solve this equation, but I don't understand why it's '2d' instead of 'd'. A little help please : )

To find time...

a = v/t and v= d/t, so a = (d/t)/t

F = ma --> a = F/m

F

_{fr}= µ

_{k}*F

_{norm}

So for force --> (mgsinθ - mgcosθ)/m, then cancel the m --> a = (gsinθ - gcosθ)

(d/t)/t = (gsinθ - gcosθ) --> (d/t) = t(gsinθ - gcosθ) --> d = t*t(gsinθ - gcosθ --> d/(gsinθ - gcosθ) = t

^{2}--> [d/(gsinθ - gcosθ)]

^{1/2}= t

Where do they get 2d from?!?

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