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{excerpt:hidden=true}Several examples showing how to find the normal force in common situations.{excerpt}
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h2. Part A

!normalbox1.png|width=500!

A 10 kg box slides at 

a constant speed of 2 m/s along a smooth floor.  What is the magnitude of the normal force exerted on the box by the floor?

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) and the floor (normal force).{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}

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}
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{card}
{card:label=Part B}

h2. Part B

!normalbox2.jpg|width=500!

A person pushes a 10 kg box along a smooth floor by applying a perfectly horizontal force of 20 N.  The box accelerates horizontally at 2 m/s{color:black}^2^{color}.  What is the magnitude of the normal force exerted on the box by the floor?

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 floor (normal force) and the person (applied force).{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}

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}

{card}

{card

:label

=Part

 C}

h2. Part C

!normalbox3.jpg|width=500!

A person is trying to lift a 10 kg box by applying a perfectly vertical force of 20 N with the help of a pulley. What is the magnitude of the normal force exerted on the box by the floor?

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 floor (normal force) and the rope (tension).{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}

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}

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{card

:label

=Part

 D}

h2. Part D

!normalbox4.jpg|width=500!

A person pushes a 10 kg box along a smooth floor by applying force of 20 N.  The force is applied at 30° below the horizontal.  What is the magnitude of the normal force exerted on the box by the floor?

h4. Solution

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

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

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

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

{cloak:id=appd}

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

{cloak:id=diagd}

We begin with a free body diagram for the box:

!normalfbd4.jpg!

{cloak:diagd}

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

{cloak:id=mathd}
From the free body diagram, we can write the equations of [Newton's 2nd Law|Newton's Second Law]. 

{latex}\begin{large}\[\sum F_{x} = F_{A}\cos\theta = ma_{x}\]
\[ \sum F_{y} = N - mg - F_{A}\sin\theta = ma_{y}\]\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}\[ N - mg -F_{A}\sin\theta = 0 \]\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}

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