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Excerpt
hiddentrue

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

Composition Setup
 Deck of Cards
idbigdeck
 Wiki Markup
{excerpt:hidden=true}An introduction to determining the size of the static friction force.{excerpt}

{composition-setup}{composition-setup}

{deck:id=bigdeck}
{card:label=Part 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~ of 25 N directed due north and a second person pushes on the box with a force _F_~2~ = 25 N directed due south, what is the magnitude and direction of the force of static friction acting on the box?

h4. Solution

{toggle-cloak:id=sysa} *System:*  {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}
 Card
labelPart A
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 F1 of 25 N directed due north and a second person pushes on the box with a force F2 = 25 N directed due south, what is the magnitude and direction of the force of static friction acting on the box?
Solution
 Toggle Cloak
idsysa
 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
\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}
{cloak:appa}
{card}
{card:label=Part B}

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~ of 25 N directed due north and a second person pushes on the box with a force _F_~2~ = 25 N directed due east, what is the magnitude and direction of the force of static friction acting on the box?

h4. Solution

{toggle-cloak:id=sysb} *System:*  {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}
 Cloak
matha
matha
 Cloak
appa
appa
 Card
labelPart B
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 F1 of 25 N directed due north and a second person pushes on the box with a force F2 = 25 N directed due east, what is the magnitude and direction of the force of static friction acting on the box?
Solution
 Toggle Cloak
idsysb
 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
\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}
{latex}

{warning}
 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 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}
 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
\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&deg; S of W.

{cloak:mathb}
{cloak:appb}
{card}
{card:label=Part 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~ of 25 N directed due north and a second person pushes on the box with a force _F_~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

{toggle-cloak:id=sysc} *System:*  {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}
at 45° S of W.
 Cloak
mathb
mathb
 Cloak
appb
appb
 Card
labelPart C
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 F1 of 25 N directed due north and a second person pushes on the box with a force F2 = 25 N also directed due north, what is the magnitude and direction of the force of static friction acting on the box?
Solution
 Toggle Cloak
idsysc
 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
\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!

{cloak:mathc}
{cloak:appc}
{card}
{deck}

 Cloak
mathc
mathc
 Cloak
appc
appc