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h2. Part A

!Pushing a Box Some More^pushbox2_1.png|width=40%!

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 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_?

System:  Box as [point particle] subject to external influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).

Model: [Point Particle Dynamics].

Approach:  We begin with a free body diagram:

 FBD

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}

h2. Part B

!pushblock2_2.png|width=40%!

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_?

System:  Box as [point particle] subject to external influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).

Model: [Point Particle Dynamics].

Approach:  Before writingWe again begin with a free body diagram:

 FBD

Which implies the form of [Newton's 2nd Law|Newton's Second Law] for the _x_ direction, we break the applied force _F_ into x- and y-components:

This implies:

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

so

{latex}\begin{large}\[ F_{xf} = \mu_{k}\left(mg - F\cossin\theta = ma_{x}\right)\] \end{large}{latex}

Solving for _F_which is substituted into the x-component equation and solved to give:

{latex}\begin{large}\[ F = \frac{ma_{x} +\mu_{k}mg}{\cos\theta + \mu_{k}\sin\theta} = \mbox{34.688 N}\]\end{large}{latex}