Page History

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*  {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}

*  {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}

}\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}{latex}

{warning}

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

{note}

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}

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

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

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

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

*  {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:

}\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:

}\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 =

0.

Thus,

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

With

this

information,

we

can

evaluate

the

limit:

}\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!