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Excerpt
hiddentrue

Several examples showing how to find the normal force in common situations.

Composition Setup
 Deck of Cards
idbigdeck
 Wiki Markup
{excerpt:hidden=true}Several examples showing how to find the normal force in common situations.{excerpt}

{table:align=right|cellspacing=0|cellpadding=1|border=1|frame=box|width=30%}
{tr}
{td:align=center|bgcolor=#F2F2F2}{*}[Examples from Dynamics]*
{td}
{tr}
{tr}
{td}
{pagetree:root=Examples from Dynamics}
{search-box}
{td}
{tr}
{table}


{composition-setup}{composition-setup}

{deck:id=bigdeck}
{card:label=Part A}

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

{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}
 Card
labelPart A
Part A
Image Added
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?
Solution
 Toggle Cloak
idsysa
 System: 
 Cloak
idsysa
Box as .
 Toggle Cloak
idinta
 Interactions: 
 Cloak
idinta
External influences from the earth (gravity) and the floor (normal force).
 Toggle Cloak
idmoda
 Model: 
 Cloak
idmoda
.
 Toggle Cloak
idappa
 Approach: 
 Cloak
idappa
 Toggle Cloak
iddiaga
  Diagrammatic Representation 
 Cloak
iddiaga
We begin with a free body diagram for the box:
Image Added
 Cloak
diaga
diaga
 Toggle Cloak
idmatha
  Mathematical Representation 
 Cloak
idmatha
From the free body diagram, we can write the equations of Newton's 2nd 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~ 
y = 
 
 0 
 
 in 
 
 the 
 
 above 
 
 equation 
 
 gives:


{
 Latex
}
\begin{large}\[ N = mg = \mbox{98 N}\]\end{large}
{latex}

{cloak:matha}
{cloak:appa}

{card}
{card:label=Part B}

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

{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}
 Cloak
matha
matha
 Cloak
appa
appa
 Card
labelPart B
Part B
Image Added
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/s2. What is the magnitude of the normal force exerted on the box by the floor?
Solution
 Toggle Cloak
idsysb
 System: 
 Cloak
idsysb
Box as .
 Toggle Cloak
idintb
 Interactions: 
 Cloak
idintb
External influences from the earth (gravity), the floor (normal force) and the person (applied force).
 Toggle Cloak
idmodb
 Model: 
 Cloak
idmodb
.
 Toggle Cloak
idappb
 Approach: 
 Cloak
idappb
 Toggle Cloak
iddiagb
  Diagrammatic Representation 
 Cloak
iddiagb
We begin with a free body diagram for the box:
Image Added
 Cloak
diagb
diagb
 Toggle Cloak
idmathb
  Mathematical Representation 
 Cloak
idmathb
From the free body diagram, we can write the equations of Newton's 2nd 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~ 
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.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

{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}
 Cloak
mathb
mathb
 Cloak
appb
appb
 Card
labelPart C
Part C
Image Added
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?
Solution
 Toggle Cloak
idsysc
 System: 
 Cloak
idsysc
Box as .
 Toggle Cloak
idintc
 Interactions: 
 Cloak
idintc
External influences from the earth (gravity), the floor (normal force) and the rope (tension).
 Toggle Cloak
idmodc
 Model: 
 Cloak
idmodc
.
 Toggle Cloak
idappc
 Approach: 
 Cloak
idappc
 Toggle Cloak
iddiagc
  Diagrammatic Representation 
 Cloak
iddiagc
We begin with a free body diagram for the box:
Image Added
 Cloak
diagc
diagc
 Toggle Cloak
idmathc
  Mathematical Representation 
 Cloak
idmathc
From the free body diagram, we can write the equations of Newton's 2nd 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~ 
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}

{card}
{card:label=Part D}

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

{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}
 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).
 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?
 Cloak
mathc
mathc
 Cloak
appc
appc
 Card
labelPart D
Part D
Image Added
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?
Solution
 Toggle Cloak
idsysd
 System: 
 Cloak
idsysd
Box as .
 Toggle Cloak
idintd
 Interactions: 
 Cloak
idintd
 External influences from the earth (gravity), the floor (normal force) and the person (applied force).
 Toggle Cloak
idmodd
 Model: 
 Cloak
idmodd
.
 Toggle Cloak
idappd
 Approach: 
 Cloak
idappd
 Toggle Cloak
iddiagd
  Diagrammatic Representation 
 Cloak
iddiagd
We begin with a free body diagram for the box:
Image Added
 Cloak
diagd
diagd
 Toggle Cloak
idmathd
  Mathematical Representation 
 Cloak
idmathd
From the free body diagram, we can write the equations of Newton's 2nd 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~ 
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}
 Tip
Again, 
 
 we 
 
 can 
 
 check 
 
 the 
 
 force 
 
 balance 
 
 in 
 
 the 
 _
y
_ 
 direction. 
  
 In 
 
 this 
 
 case 
 _
mg
_ 
 (98 
 
 N) 
 
 and 
 _
F
_~A
A,
y~ 
y (10 
 
 N) 
 
 act 
 
 to 
 
 balance 
 
 N 
 
 (108 
 
 N).
{tip}

{cloak:mathd}
{cloak:appd}
{card}
{deck}




 Cloak
mathd
mathd
 Cloak
appd
appd