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{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}

We begin with a free body diagram for the box:

!normalfbd1.jpg!

{cloak:diaga}

{toggle-cloak:id=

matha} 

{color:red} *Mathematical Representation* {color}

{cloak:id=

matha}

From 

the 

free body diagram, we can 

write the

 equations of [Newton's 2nd Law|Newton's Second Law].  We ignore the x-direction, since there are no forces acting.

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

Because the box is sliding over level ground, it is not moving at all in the _y_ direction.  Thus, it certainly has no y-acceleration.  Setting _a_~y~ = 0 in the above equation gives:

{latex}\begin{large}\[ 

N = mg = 

\mbox{98 N}\]\end{large}{latex}

{cloak:matha}
{cloak:appa}

appb} *

Approach:*  

{cloak:id=

appb}

{toggle-cloak:id=

diagb} {color:red} *

Diagrammatic Representation* {color}

{cloak:id=

diagb}

We begin with a free body diagram for the box:

!normalfbd2.jpg!

{cloak:diagb}

{toggle-cloak:id=

mathb} 

{color:red} *Mathematical Representation* {color}

{cloak:id=

mathb}

From the free body diagram, we can write the equations of [Newton's 2nd Law|Newton's Second Law]. 

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

Because the box is sliding over level ground, it is not moving at all in the _y_ direction.  Thus, it certainly has no y-acceleration.  Setting _a_~y~ = 0 in the _y_ direction equation gives:

{latex}\begin{large}\[ 

N = 

mg = 

\mbox{98 N}\]\end{large}{latex}

{cloak:mathb}
{cloak:appb}

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}

We begin with a free body diagram for the box:

!normalfbd3.jpg!

{cloak:diagc}
{toggle-cloak:id=

mathc} 

{color:red} *Mathematical Representation* {color}

{

cloak:id=

mathc}

From the free body diagram, we can write the equations of [Newton's 2nd Law|Newton's Second Law]. We ignore the x-direction, since there are no forces acting.

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

Because the box is sliding over level ground, it is not moving at all in the _y_ direction.  Thus, it certainly has no y-acceleration.  Setting _a_~y~ = 0 in the _y_ direction equation gives:

{latex}\begin{large}\[ T + N - mg = 

0 \]\end{large}{latex}

Solving for the 

normal 

force gives:

{latex}\begin{large}\[ N = mg - T = \mbox{78 N}\]\end{large}{latex}

{tip}When three or more forces act in a direction with zero acceleration, it is always a good idea to check your answer by putting the numbers on the free body diagram and making sure that they balance.  In this case, T (20 N) and N (78 N) act to balance mg (98 N).{tip}

{note}Follow up question:  The floor no longer supports the entire weight of the box (98 N) because the rope is carrying some of the weight (20 N).  How will the _person's_ normal force be affected in this situation?  If the floor is carrying so much less weight, what part of the building is now feeling an extra load?{note}

{cloak:mathc}
{cloak:appc}

 N - mg -F_{

A}\sin\theta = 

0 \]

\end{large}{latex}

Solving for the normal force gives:

{latex}\begin{large}\[ N = mg + F_{A}\sin\theta = \mbox{108 N}\]\end{large}{latex}

{tip}Again, we can check the force balance in the _y_ direction.  In this case _mg_ (98 N) and _F_~A,y~ (10 N) act to balance N (108 N).{tip}

{cloak:mathd}
{cloak:appd}