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Deck of Cards
idbigdeck
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/s2. 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?

Solution

Card
labelPart A
Wiki Markup
Toggle Cloak
idsysa
System:
Cloak
idsysa

We will treat the box as a .

Toggle Cloak
idinta
Interactions:
Cloak
idinta

External influence from the earth (gravity) and the truck bed (normal force and friction force).

Toggle Cloak
idmoda
Model:
Cloak
idmoda

.

Toggle Cloak
idappa
Approach:

Cloak
idappa

Toggle Cloak
iddiaga
Diagrammatic Representation

Cloak
iddiaga

We 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!

Cloak
diaga
diaga

Toggle Cloak
idmatha
Mathematical Representation

Cloak
idmatha

Using the free-body diagram, we construct the equations of Newton's Second Law applied to the box:

Latex
{^}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

{toggle-cloak:id=sysa} *System:* {cloak:id=sysa}We will treat the box as a [point particle].{cloak}

{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}External influence from the earth (gravity) and the truck bed (normal force _and_ friction force).{cloak}

{toggle-cloak:id=moda} *Model:* {cloak:id=moda}[Point Particle Dynamics].{cloak}

{toggle-cloak:id=appa} *Approach:*
{cloak:id=appa}
{toggle-cloak:id=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~ =

ay = 0.

Thus,

we

know

that

the

normal

force

acting

on

the

box

will

equal

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}{latex}
{warning}
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, Ff = 53 N, is compatible with the static friction limit.

Cloak
matha
matha

_F{_}{~}f~ = 53 N, _is_ compatible with the static friction limit. {cloak:matha} {cloak:appa}
Cloak
appa
appa

Card
labelPart B

Part B

Consider the same basic situation as above, but now suppose that the truck accelerates at a rate of 4.0 m/s2 rather than 3.5 m/s2. If the center of mass of the box is located 2.0 m horizontally from the edge of the truck bed, how much time will elapse from the instant the truck begins to accelerate until the instant the box falls off the truck bed?

Solution

Toggle Cloak
idsysintb
System and Interactions:
Cloak
idsysintb

As defined in Part A. We will also have to model the truck as a separate in order to determine the time for the box to fall off.

Toggle Cloak
idmodb
Models:
Cloak
idmodb

and One-Dimensional Motion with Constant Acceleration.

Toggle Cloak
idappb
Approach:

Cloak
idappb