---
title: Factory Conveyor Belt
topic: Force
author: James Ropotar
source: original
template_version: 1.4
attribution: standard
singleVariant: false
showCorrectAnswer: false
outcomes:
- 6.6.2.0
difficulty:
- undefined
randomization:
- undefined
taxonomy:
- undefined
span:
- chapter
length:
- long
tags:
- APSC181
- JR
assets:
- Conveyor1.png
- Conveyor2.png
part1:
type: number-input
pl-customizations:
rtol: 0.05
weight: 1
allow-blank: true
label: $\theta = $
suffix: $\circ$
part2:
type: number-input
pl-customizations:
rtol: 0.05
weight: 1
allow-blank: true
label: $x = $
suffix: $\rm{m}$
myst:
substitutions:
params:
vars:
title: Atwood Machine Difference
mu: 0.25
v: 4
M: 10
h: 5
t: 0.052
---
# {{ params.vars.title }}
To minimize space usage and for effective distributed controls, packages on a factory conveyor belt descend via multi-belt sections placed horizontally on different levels with the packages changing directions when dropping from one belt to the other.
## Part 1
If the belts are moving at a speed $v = {{ params.v }} \ \rm{m/s}$ and the package of mass $m = {{ params.M }} \ \rm{kg}$ does not slip relative to the belt prior to descent, what is the angle $\theta$ the resultant contact force vector makes with the belt for the time of impact.
The height between the belts is $h = {{ params.h }} \ \rm{m}$, and the coefficient of friction is $\mu = {{ params.mu }}$
(Hint: The resultant is the vector sum of the normal reaction and friction force)
### Answer Section
Please enter in a numeric value in $\rm{N}$.
## Part 2
If the impact time is $t = {{ params.t }} \ \rm{s}$, determine the minimum extra length $x$ of the conveyor belt for the package to not fall off the belt.
### Answer Section
Please enter in a numeric value in $\rm{N}$.
## Attribution
Problem is licensed under the [CC-BY-NC-SA 4.0 license](https://creativecommons.org/licenses/by-nc-sa/4.0/).
