---
title: Truck Up Hill
topic: Force
author: John Hopkinson
source: PHYS 112 (Hopkinson) 2018W1 Midterm 2 Q3
template_version: 1.4
attribution: standard
partialCredit: true
singleVariant: true
outcomes:
- 6.1.1.4
- 6.7.1.0
- 6.9.1.3
difficulty:
- undefined
randomization:
- undefined
taxonomy:
- undefined
span:
- undefined
length:
- undefined
tags:
- SJ
assets:
- truckuphill1.png
part1:
type: multiple-choice
pl-customizations:
weight: 1
fixed-order: true
myst:
substitutions:
params_vars_title: Truck Up Hill
params_part1_ans1_value: ' a= $\mu_s g - g\sin$(10$^\circ$)'
params_part1_ans1_feedback: 'At the point of slipping, $f_s$ = $f_{s\ max}$=$\mu_s$
n . Also, $f_s$ - $w\sin$(10$^\circ$)$= $ma and $n - w\cos$(10$^\circ$) = 0
. Therefore, $\mu_s$ $w\cos$(10$^\circ$)$ - w\sin$(10$^\circ$) = $ma$ . Since
$w= mg$, $m$ is canceled out and we will have , a=$\mu_s g\cos$(10$^\circ$)
- $g\sin$(10$^\circ$). '
params_part1_ans2_value: ' a < $\mu_s g\cos$(10$^\circ$) - $g\sin$(10$^\circ$)'
params_part1_ans2_feedback: 'At the point of slipping, $f_s$ = $f_{s\ max}$=$\mu_s$
n . Also, $f_s$ - $w\sin$(10$^\circ$)$= $ma and $n - w\cos$(10$^\circ$) = 0
. Therefore, $\mu_s$ $w\cos$(10$^\circ$)$ - w\sin$(10$^\circ$) = $ma$ . Since
$w= mg$, $m$ is canceled out and we will have , a=$\mu_s g\cos$(10$^\circ$)
- $g\sin$(10$^\circ$). '
params_part1_ans3_value: ' a=$\mu_s g\cos$(10$^\circ$) - $g\sin$(10$^\circ$)'
params_part1_ans3_feedback: 'At the point of slipping, $f_s$ = $f_{s\ max}$=$\mu_s$
n . Also, $f_s$ - $w\sin$(10$^\circ$)$= $ma and $n - w\cos$(10$^\circ$) = 0
. Therefore, $\mu_s$ $w\cos$(10$^\circ$)$ - w\sin$(10$^\circ$) = $ma$ . Since
$w= mg$, $m$ is canceled out and we will have , a=$\mu_s g\cos$(10$^\circ$)
- $g\sin$(10$^\circ$). '
params_part1_ans4_value: ' a=$\mu_s g\sin$(10$^\circ$)- $g\cos$(10$^\circ$) '
params_part1_ans4_feedback: 'At the point of slipping, $f_s$ = $f_{s\ max}$=$\mu_s$
n . Also, $f_s$ - $w\sin$(10$^\circ$)$= $ma and $n - w\cos$(10$^\circ$) = 0
. Therefore, $\mu_s$ $w\cos$(10$^\circ$)$ - w\sin$(10$^\circ$) = $ma$ . Since
$w= mg$, $m$ is canceled out and we will have , a=$\mu_s g\cos$(10$^\circ$)
- $g\sin$(10$^\circ$). '
params_part1_ans5_value: ' This cannot be found from the information given.'
params_part1_ans5_feedback: ' At the point of slipping, $f_s$ = $f_{s\ max}$=$\mu_s$
n . Also, $f_s$ - $w\sin$(10$^\circ$)$= $ma and $n - w\cos$(10$^\circ$) = 0
. Therefore, $\mu_s$ $w\cos$(10$^\circ$)$ - w\sin$(10$^\circ$) = $ma$ . Since
$w= mg$, $m$ is canceled out and we will have , a=$\mu_s g\cos$(10$^\circ$)
- $g\sin$(10$^\circ$).'
---
# {{ params_vars_title }}
A truck accelerates up a 10 $^\circ$ incline. For a box in the back of the truck as shown in Fig. (i), the free body diagram is shown in Fig. (ii). The coefficient of friction between the box and the truck is $\mu_s$ . At the point of slipping, the acceleration of the truck is:
## Part 1
### Answer Section
- {{ params_part1_ans1_value }} {{ params.vars.units}}
- {{ params_part1_ans2_value }} {{ params.vars.units}}
- {{ params_part1_ans3_value }} {{ params.vars.units}}
- {{ params_part1_ans4_value }} {{ params.vars.units}}
- {{ params_part1_ans5_value }} {{ params.vars.units}}
## Attribution
Problem is licensed under the [CC-BY-NC-SA 4.0 license](https://creativecommons.org/licenses/by-nc-sa/4.0/).
![The Creative Commons 4.0 license requiring attribution-BY, non-commercial-NC, and share-alike-SA license.](https://raw.githubusercontent.com/firasm/bits/master/by-nc-sa.png)