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{composition-setup}{composition-setup}

{deck:id=partdeck}
{card:label=Part A}


h3. Part A

!Pushing a Box Some More^pushbox2_1.png|width=400!

{excerpt}A person pushes a box of mass 15 kg along a floor by applying a force _F_ at an angle of 30° below the horizontal.  There is friction between the box and the floor{excerpt} characterized by a coefficient of kinetic friction of 0.45.  The box accelerates horizontally at a rate of 2.0 m/s{color:black}^2^{color}.  What is the magnitude of _F_?

h4. Solution

{toggle-cloak:id=sysA} *System:*  {cloak:id=sysA} Box as [point particle].{cloak:sysA}

{toggle-cloak:id=intA} *Interactions:* {cloak:id=intA} External influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).{cloak}

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

{toggle-cloak:id=appA} *Approach:* 

{cloak:id=appA}

{toggle-cloak:id=diagA} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diagA}

 We begin with a free body diagram:

!pushfrictionmore1.png!

{cloak:diagA}

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

{cloak:id=mathA}


With the free body diagram as a guide, we write the equations of [Newton's 2nd Law|Newton's Second Law]:

{latex}\begin{large}\[ \sum F_{x} = F\cos\theta - F_{f} = ma_{x}\] 
\[ \sum F_{y} = N - F\sin\theta - mg = ma_{y}\] \end{large}{latex}

We can now use the fact that the box is sliding over level ground to tell us that _a_~y~ = 0 (the box is not moving at all in the y-direction).  Thus:

{latex}\begin{large}\[ N = F\sin\theta + mg \]\end{large}{latex}

Now, we can write the friction force in terms of _F_ and known quantities:

{latex}\begin{large}\[ F_{f} = \mu_{k}N = \mu_{k}\left(F\sin\theta + mg\right)\]\end{large}{latex}

Substituting into the x-component equation yields:

{latex}\begin{large}\[ F\cos\theta - \mu_{k}\left(F\sin\theta + mg\right) = ma_{x}\]\end{large}{latex}

which is solved to obtain:

{latex}\begin{large}\[ F = \frac{ma_{x} +\mu_{k}mg}{\cos\theta - \mu_{k}\sin\theta} = \mbox{150 N}\]\end{large}{latex}

{cloak:mathA}
{cloak:appA}
{card:Part A}
{card:label=Part B}

h3. Part B

!pushblock2_2.png|width=400!

A person pulls a box of mass 15 kg along a floor by applying a force _F_ at an angle of 30° above the horizontal.  There is friction between the box and the floor characterized by a coefficient of kinetic friction of 0.45.  The box accelerates horizontally at a rate of 2.0 m/s{color:black}^2^{color}.  What is the magnitude of _F_?

h4. Solution

{toggle-cloak:id=sysB} *System:*  {cloak:id=sysB} Box as [point particle].{cloak:sysB}

{toggle-cloak:id=intB} *Interactions:* {cloak:id=intB}External influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).{cloak:intB}

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

{toggle-cloak:id=appB} *Approach:* 

{cloak:id=appB}

{toggle-cloak:id=diagB} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diagB}

 We again begin with a free body diagram:

!pushfrictionmore2.png!

{cloak:diagB}

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

{cloak:id=mathB}

The diagrammatic representation suggests the form of [Newton's 2nd Law|Newton's Second Law]:

{latex}\begin{large}\[ \sum F_{x} = F\cos\theta - F_{f} = ma_{x}\] 
\[ \sum F_{y} = N + F\sin\theta - mg = ma_{y}\] \end{large}{latex}

Again using the fact that _a_~y~ is zero if the box is moving along the level floor gives us:

{latex}\begin{large}\[ N = mg - F\sin\theta\]\end{large}{latex}

so

{latex}\begin{large}\[ F_{f} = \mu_{k}\left(mg - F\sin\theta\right)\]\end{large}{latex}

which is substituted into the x-component equation and solved to give:

{latex}\begin{large}\[ F = \frac{ma_{x} +\mu_{k}mg}{\cos\theta + \mu_{k}\sin\theta} = \mbox{88 N}\]\end{large}{latex}

{cloak:mathB}
{cloak:appB}
{card:Part B}
{deck:partdeck}