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h2. Part A

!thatnormal1.png|width=40%!

A person holds a 10 kg box against a 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?

System:  Box as [point particle] subject to 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:

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

Because the box is held at rest, it has no acceleration at all (_a_~x~ = _a_~y~ = 0).  Setting _a_~x~ = 0 in the _x_ direction equation gives:

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

h2. Part A

!thatnormal1.png|width=40%!

A person holds up a 10 kg box by applying a perfectly horizontal force of 300 N to press it against a rough wall. What is the magnitude of the normal force exerted on the box by the wall?

System:  Box as [point particle] subject to 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:

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

Because the box is held at rest, it has no acceleration at all (_a_~x~ = _a_~y~ = 0).  Setting _a_~x~ = 0 in the _x_ direction equation gives:

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