{excerpt:hidden=true}An introduction to determining the size of the static friction force.{excerpt}

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{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}

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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!

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{toggle-cloak:id=matha} {color:red} *Mathematical Representation* {color}

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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.
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{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}\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}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}\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.

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{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}\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~ = 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!
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