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h2. Part A

!normalbox1.png|width=40%!

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

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

*Interactions:* External influences from the earth (gravity) and the floor (normal force).

*Model:*  [Point Particle Dynamics].

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

!normalfbd1.png!

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}

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h2. Part B

!normalbox2.png|width=40%!

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

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

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

*Model:*  [Point Particle Dynamics].

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

!normalfbd2.png!

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

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

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h2. Part C

!normalbox3.png|width=40%!

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

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

*Interactions:* external influences from the earth (gravity), the floor (normal force) and the rope (tension).

*Model:*  [Point Particle Dynamics].

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

!normalfbd3.png!

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}

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h2. Part D

!normalbox4.png|width=40%!

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

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

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

*Model:*  [Point Particle Dynamics].

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

!normalfbd4.png!

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

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