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Excerpt
hiddentrue

An introduction to determining the size of the static friction force.

Composition Setup
Section
Column
width70%
Deck of Cards
idbigdeck
unmigrated-wiki-markup
Card
labelPart A
h2.

Part

A

Suppose

a

10

kg

cubic

box

is

at

rest

on

a

horizontal

surface.

There

is

friction

between

the

box

and

the

surface

characterized

by

a

coefficient

of

static

friction

equal

to

0.50.

If

one

person

pushes

on

the

box

with

a

force

_

F

_~1~

1 of

25

N

directed

due

north

and

a

second

person

pushes

on

the

box

with

a

force

_

F

_~2~

2 =

25

N

directed

due

south,

what

is

the

magnitude

and

direction

of

the

force

of

static

friction

acting

on

the

box?

h4. Solution {

Solution

Toggle Cloak

:

id

=

sysa

} *

System:
Cloak
idsysa

Box as .

Toggle Cloak
idinta
Interactions:
Cloak
idinta

External influences from the earth (gravity) the surface (normal force and friction) and the two people (applied forces).

Toggle Cloak
idmoda
Model:
Cloak
idmoda

.

Toggle Cloak
idappa
Approach:

Cloak
idappa

Toggle Cloak
iddiaga
Diagrammatic Representation

Cloak
iddiaga

To determine the force of static friction, we first find the net force in the absence of friction. We first draw the situation. A top view (physical representation) and a side view (free body diagram) of the box ignoring any contribution from friction are shown here.

Image Added

Cloak
diaga
diaga

Toggle Cloak
idmatha
Mathematical Representation

Cloak
idmatha

The net force parallel to the surface in the absence of friction is then:

Latex
* {cloak:id=sysa}Box as [point particle].{cloak} {toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}External influences from the earth (gravity) the surface (normal force and friction) and the two people (applied forces).{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} To determine the force of static friction, we first find the net force _in the absence of friction_. We first draw the situation. A top view (physical representation) and a side view (free body diagram) of the box _ignoring any contribution from friction_ are shown here. !basicstatic1.jpg! {cloak:diaga} {toggle-cloak:id=matha} {color:red} *Mathematical Representation* {color} {cloak:id=matha} The net force parallel to the surface in the absence of friction is then: {latex}
\begin{large}\[ \sum_{F \ne F_{f}} F_{x} = 0 \]
\[\sum_{F \ne F_{f}} F_{y} = F_{1} - F_{2} = 0 \]\end{large}
{latex}

Thus,

the

net

force

along

the

surface

is

zero

_

without

_

the

influence

of

static

friction,

and

so

the

static

friction

force

will

also

be

0.

Cloak
matha
matha

{cloak:matha} {cloak:appa}
Cloak
appa
appa

h2.

Part

B

Suppose

a

10

kg

cubic

box

is

at

rest

on

a

horizontal

surface.

There

is

friction

between

the

box

and

the

surface

characterized

by

a

coefficient

of

static

friction

equal

to

0.50.

If

one

person

pushes

on

the

box

with

a

force

_

F

_~1~

1 of

25

N

directed

due

north

and

a

second

person

pushes

on

the

box

with

a

force

_

F

_~2~

2 =

25

N

directed

due

east,

what

is

the

magnitude

and

direction

of

the

force

of

static

friction

acting

on

the

box?

h4. Solution {

Solution

Card
labelPart B
Wiki Markup
Toggle Cloak

:

id

=

sysb

} *

System:
Cloak
idsysb

Box as .

Toggle Cloak
idintb
Interactions:
Cloak
idintb

External influences from the earth (gravity) the surface (normal force and friction) and the two people (applied forces).

Toggle Cloak
idmodb
Model:
Cloak
idmodb

.

Toggle Cloak
idappb
Approach:

Cloak
idappb

Toggle Cloak
iddiagb
Diagrammatic Representation

Cloak
iddiagb

To determine the force of static friction, we first find the net force in the absence of friction. We first draw the situation. A top view (physical representation) and a side view (partial free body diagram) of the box ignoring any contribution from friction are shown here.

Image Added

Cloak
diagb
diagb

Toggle Cloak
idmathb
Mathematical Representation

Cloak
idmathb

The net force parallel to the surface in the absence of friction is then:

Latex
* {cloak:id=sysb}Box as [point particle].{cloak} {toggle-cloak:id=intb} *Interactions:* {cloak:id=intb}External influences from the earth (gravity) the surface (normal force and friction) and the two people (applied forces).{cloak} {toggle-cloak:id=modb} *Model:* {cloak:id=modb}[Point Particle Dynamics].{cloak} {toggle-cloak:id=appb} *Approach:* {cloak:id=appb} {toggle-cloak:id=diagb} {color:red} *Diagrammatic Representation* {color} {cloak:id=diagb} To determine the force of static friction, we first find the net force _in the absence of friction_. We first draw the situation. A top view (physical representation) and a side view (partial free body diagram) of the box _ignoring any contribution from friction_ are shown here. !basicstatic2.jpg! {cloak:diagb} {toggle-cloak:id=mathb} {color:red} *Mathematical Representation* {color} {cloak:id=mathb} The net force parallel to the surface in the absence of friction is then: {latex}
\begin{large}\[ \sum_{F \ne F_{f}} F_{x} = F_{2} = \mbox{25 N} \]
\[\sum_{F \ne F_{f}} F_{y} = F_{1} = \mbox{25 N} \]\end{large}
{latex}

In

order

to

prevent

the

box

from

moving,

then,

static

friction

would

have

to

satisfy:

{
Latex
}
\begin{large}\[ F_{s} = - \mbox{25 N }\hat{x} - \mbox{25 N }\hat{y} = \mbox{35.4 N at 45}^{\circ}\mbox{ S of W}.\]\end{large}
Warning
{latex} {warning}

We're

not

finished

yet!

{warning}

We

must

now

check

that

this

needed

friction

force

is

compatible

with

the

static

friction

limit.

Newton's

2nd

Law

for

the

_

z

_

direction

tells

us:

{
Latex
}
\begin{large}\[ \sum F_{z} = N - mg = ma_{z}\]\end{large}
{latex} {note}
Note

Note that friction from an xy surface cannot act in the z direction.

We know that the box will remain on the surface, so az = 0. Thus,

Latex
Note that friction from an _xy_ surface cannot act in the _z_ direction.{note} We know that the box will remain on the surface, so _a_~z~ = 0. Thus, {latex}
\begin{large}\[ N = mg = \mbox{98 N}\]\end{large}
{latex}

With

this

information,

we

can

evaluate

the

limit:

{
Latex
}
\begin{large}\[ F_{s} \le \mu_{s} N = \mbox{49 N} \]\end{large}
{latex}

Since

35.4

N

<

49

N,

we

conclude

that

the

friction

force

is

indeed

35.4

N at 45° S of W.

Cloak
mathb
mathb

at 45&deg; S of W. {cloak:mathb} {cloak:appb}
Cloak
appb
appb

unmigrated-wiki-markup
Card
labelPart C
h2.

Part

C

Suppose

a

10

kg

cubic

box

is

at

rest

on

a

horizontal

surface.

There

is

friction

between

the

box

and

the

surface

characterized

by

a

coefficient

of

static

friction

equal

to

0.50.

If

one

person

pushes

on

the

box

with

a

force

_

F

_~1~

1 of

25

N

directed

due

north

and

a

second

person

pushes

on

the

box

with

a

force

_

F

_~2~

2 =

25

N

also

directed

due

north,

what

is

the

magnitude

and

direction

of

the

force

of

static

friction

acting

on

the

box?

h4. Solution {

Solution

Toggle Cloak

:

id

=

sysc

} *

System:
Cloak
idsysc

Box as .

Toggle Cloak
idintc
Interactions:
Cloak
idintc

External influences from the earth (gravity) the surface (normal force and friction) and the two people (applied forces).

Toggle Cloak
idmodc
Model:
Cloak
idmodc

.

Toggle Cloak
idappc
Approach:

Cloak
idappc

Toggle Cloak
iddiagc
Diagrammatic Representation

Cloak
iddiagc

To determine the force of static friction, we first find the net force in the absence of friction. We first draw the situation. A top view (physical representation) and a side view (free body diagram) of the box ignoring any contribution from friction are shown here.

Image Added

Cloak
diagc
diagc

Toggle Cloak
idmathc
Mathematical Represenatation

Cloak
idmathc

The net force parallel to the surface in the absence of friction is then:

Latex
* {cloak:id=sysc}Box as [point particle].{cloak} {toggle-cloak:id=intc} *Interactions:* {cloak:id=intc} External influences from the earth (gravity) the surface (normal force and friction) and the two people (applied forces).{cloak} {toggle-cloak:id=modc} *Model:* {cloak:id=modc}[Point Particle Dynamics].{cloak} {toggle-cloak:id=appc} *Approach:* {cloak:id=appc} {toggle-cloak:id=diagc} {color:red} *Diagrammatic Representation* {color} {cloak:id=diagc} To determine the force of static friction, we first find the net force _in the absence of friction_. We first draw the situation. A top view (physical representation) and a side view (free body diagram) of the box _ignoring any contribution from friction_ are shown here. !basicstatic3.jpg! {cloak:diagc} {toggle-cloak:id=mathc} {color:red} *Mathematical Represenatation* {color} {cloak:id=mathc} The net force parallel to the surface in the absence of friction is then: {latex}
\begin{large}\[ \sum_{F \ne F_{f}} F_{x} = 0 \]
\[\sum_{F \ne F_{f}} F_{y} = F_{1}+F_{2} = \mbox{50 N} \]\end{large}
{latex}

In

order

to

prevent

the

box

from

moving,

then,

static

friction

would

have

to

satisfy:

{
Latex
}
\begin{large}\[ F_{s} = \mbox{0 N }\hat{x} - \mbox{50 N }\hat{y} = \mbox{50 N due south}.\]\end{large}
{latex}

Again,

we

must

check

that

this

needed

friction

force

is

compatible

with

the

static

friction

limit.

Again,

Newton's

2nd

Law

for

the

_

z

_

direction

tells

us:

{
Latex
}
\begin{large}\[ \sum F_{z} = N - mg = ma_{z}\]\end{large}
{latex}

and

know

that

the

box

will

remain

on

the

surface,

so

_

a

_~z~

z =

0.

Thus,

{
Latex
}
\begin{large}\[ N = mg = \mbox{98 N}\]\end{large}
{latex}

With

this

information,

we

can

evaluate

the

limit:

{
Latex
}
\begin{large}\[ F_{s} \le \mu_{s} N = \mbox{49 N} \]\end{large}
{latex}

Since

50

N

>

49

N,

we

conclude

that

the

static

friction

limit

is

violated.

The

box

will

move

and

kinetic

friction

will

apply

instead!

{:mathc} {cloak:appc}
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mathc

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