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h2. Part A

!thatnormal1.png|width=40%!

A person holds a 10 kg box against a smooth wall (as it slides down) by applying a perfectly horizontal force of 300 N. What is the magnitude of the normal force exerted on the box by the wall?

h4. Solution

*System:*  Box as [point particle] subject to external .

*Interactions:* External influences from the earth (gravity), the wall (normal force) and the person (applied force).

*Model:*  [Point Particle Dynamics].

*Approach:*  We begin with a free body diagram for the box:

!thatfbd1.png!

{note}It is important to note that any surface has the potential to exert a normal force and that the normal is always perpendicular to the plane of the surface.  If the wall did not exert a normal force, the box would simply pass through it.{note}

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

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

Because the box is held against the wall, it has no movement (and no acceleration) in the _x_ direction (_a_~x~ = 0).  Setting _a_~x~ = 0 in the _x_ direction equation gives:

{latex}\begin{large}\[ N = F_{A} = \mbox{300 N} \]\end{large}{latex}

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h2. Part B

!thatnormal2.png|width=40%!

A person moves a 10 kg box up a smooth wall by applying a force of 300 N. The force is applied at an angle of 60° above the horizontal.  What is the magnitude of the normal force exerted on the box by the wall?

h4. Solution

*System:*  Box as [point particle] subject to external .

*Interactions:* External influences from the earth (gravity), the wall (normal force and friction) and the person (applied force).

*Model:*  [Point Particle Dynamics].

*Approach:*  We begin with a free body diagram for the box:

!thatfbd2.png!

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

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

Because Because the box is held against the wall, it has no movement (and no acceleration) in the _x_ direction (_a_~x~ = 0).  Setting _a_~x~ = 0 in the _x_ direction equation gives:

{latex}\begin{large}\[ N = F_{A}\cos\theta = \mbox{150 N} \]\end{large}{latex}

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h2. Part C

!thatnormal3.png|width=40%!

A person scrapes a 10 kg box along a low, smooth ceiling by applying a force of 300 N at an angle of 30° above the horizontal.  What is the magnitude of the normal force exerted on the box by the ceiling?

h4. Solution

*System:*  Box as [point particle] subject to external.

*Interactions:* External influences from the earth (gravity), the wall (normal force and friction) and the person (applied force).

*Model:*  [Point Particle Dynamics].

*Approach:*  We begin with a free body diagram for the box:

!thatfbd3.png!

{note}The ceiling must push down to prevent objects from moving up through it.{note}

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

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

Because Because the box is held against the ceiling, it has no movement (and no acceleration) in the _y_ direction (_a_~y~ = 0).  Setting _a_~y~ = 0 in the _y_ direction equation gives:

{latex}\begin{large}\[ F_{A}\sin\theta  - mg - N = 0 \]\end{large}{latex}

which we solve to find:

{latex}\begin{large}\[ N = F_{A}\sin\theta - mg = \mbox{52 N}\]\end{large}{latex}

{tip}We can check that the _y_ direction is in balance.  We have N (52 N) and mg (98 N) on one side, and _F_~A,y~ on the other (150 N).{tip}