h3. Part A
A 15 kg box is sitting in the bed of a pickup truck. The truck begins to accelerate at a constant rate of 3.5 m/s {^}2 {^}. Given that the friction between the box and the truck bed is characterized by a coefficient of kinetic friction of 0.25 and a coefficient of static friction equal to 0.40, what is the magnitude of the friction force acting on the box once the truck begins its acceleration?
h4. Solution
{Solution:=} *System:* {:=} [point particle].{cloak}
{:=} *Interactions:* {:=} __ {cloak}
{:=} *Model:* {cloak:id=moda}[Point Particle Dynamics].{cloak}
{:=} *Approach:*
{cloak:id=appa}
{: Diagrammatic RepresentationWe begin with a sketch that represents the situation, and then create the appropriate free body diagram.  Image Added
| Note |
|---|
It is important to note that friction works to prevent movement along the interface between the box and the truck bed. The truck bed is moving forward, so friction will attempt to pull the box forward as well. If the box moves at the same rate as bed, then the interface is static. For this reason, "static" friction will actually cause motion of the box in this case! |
Mathematical RepresentationUsing the free-body diagram, we construct the equations of Newton's Second Law applied to the box: | Latex |
|---|
=diaga} {color:red} *Diagrammatic Representation* {color}
{cloak:id=diaga}
We begin with a sketch that represents the situation, and then create the appropriate free body diagram.
!closethegate.jpg!
{note}It is important to note that friction works to prevent movement along the interface between the box and the truck bed. The truck bed is moving forward, so friction will attempt to pull the box forward as well. If the box moves at the same rate as bed, then the interface is static. For this reason, "static" friction will actually _cause_ motion of the box in this case!{note}
{cloak:diaga}
{toggle-cloak:id=matha} {color:red} *Mathematical Representation* {color}
{cloak:id=matha}
Using the free-body diagram, we construct the equations of Newton's Second Law applied to the box:
{latex}\begin{large}\[\sum F_{x} = F_{f} = ma_{x} \]\[\sum F_{y} = N - mg = ma_{y} \] \end{large}{latex}
|
Since the truck is moving only in the x-direction, we expect _a _~y~ y = 0. Thus, we know that the normal force acting on the box will equal it its weight. If friction is adequate, we expect that the box will accelerate in the x-direction at the same rate as the truck does. In that case, we expect:
{| Latex |
|---|
}\begin{large}\[ F_{f} = (15\:{\rm kg})(3.5\:{\rm m/s}^{2}) = 53\:{\rm N} \]\end{large} | | Warning |
|---|
| {latex}
{warning} We're not finished yet! {warning}
|
It is important now to check that this result does not conflict with the requirement that
{| Latex |
|---|
}\begin{large} \[ F_{f} \le \mu_{s} N\] \end{large}{latex}
| Since we have already used the y-direction equation of Newton's Second Law to conclude that the normal force on the box is equal to _mg _, we find:
{| Latex |
|---|
}\begin{large} \[ F_{f} = 53\:{\rm N} < \mu_{s} N = 0.40(15\:{\rm kg})(9.8\:{\rm m/s}^{2}) = 59\:{\rm N} \]\end{large}{latex}
| We therefore conclude that our answer, _F _~f~ f = 53 N, _is _ compatible with the static friction limit. | Cloak |
|---|
|
{cloak:matha}
{cloak:appa}
| |
|