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Excerpt
hiddentrue

Several examples illustrating how to find the normal force in

...

not-so-common

...

situations.

{excerpt} {
Composition Setup
 Deck of Cards
idbigdeck
 Card
labelPart A
Part A
Image Added
A person holds a 10 kg box against a smooth 
}{composition-setup}
{deck:id=bigdeck}
{card:label=Part A}


h2. Part A

!thatnormal1.png|width=500!

A person holds a 10 kg box against a smooth 
 (i.e. 
 
 frictionless) 
 
 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

{
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|gravity (near-earth)]), the wall ([normal force]) and the person ([applied 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:

!thatfbd1.jpg!

{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}

{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]. 

{latex}
 Cloak
idsysa
Box as .
 Toggle Cloak
idinta
 Interactions: 
 Cloak
idinta
External influences from the earth (gravity), the wall () and the person ().
 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
 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.
 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. 
 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~ 
x = 
 
 0). 
  
 Setting 
 _
a
_~x~ 
x = 
 
 0 
 
 in 
 
 the 
 _
x
_ 
 direction 
 
 equation 
 
 gives:


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

{cloak:matha}
{cloak:appa}
{card}

{card:label=Part B}

h2. Part B

!thatnormal2.jpg|width=500!

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

{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|gravity (near-earth)]), the wall ([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:

!thatfbd2.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 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?
Solution
 Toggle Cloak
idsysb
 System: 
 Cloak
idsysb
Box as .
 Toggle Cloak
idintb
 Interactions: 
 Cloak
idintb
External influences from the earth (gravity), the wall () and the person ().
 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}\[\sum F_{x} = F_{A}\cos\theta - N = ma_{x}\]
\[ \sum F_{y} = F_{A}\sin\theta - mg = ma_{y}\]\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~ 
x = 
 
 0). 
  
 Setting 
 _
a
_~x~ 
x = 
 
 0 
 
 in 
 
 the 
 _
x
_ 
 direction 
 
 equation 
 
 gives:


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

{cloak:mathb}

{cloak:appb}

{card}
{card:label=Part C}

h2. Part C

!thatnormal3.jpg|width=500!

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

{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|gravity (near-earth)]), the ceiling ([normal force]) and the person ([applied force]).{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:

!thatfbd3.jpg!

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

{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]. 

{latex}
 Cloak
mathb
mathb
 Cloak
appb
appb
 Card
labelPart C
Part C
Image Added
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?
Solution
 Toggle Cloak
idsysc
 System: 
 Cloak
idsysc
Box as .
 Toggle Cloak
idintc
 Interactions: 
 Cloak
idintc
External influences from the earth (gravity), the ceiling () and the person ().
 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
 Note
The ceiling must push down to prevent objects from moving up through it.
 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. 
 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~ 
y = 
 
 0). 
  
 Setting 
 _
a
_~y~ 
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}

{cloak:mathc}
{cloak:appc}
{card}
{deck}

 Tip
We can check that the y direction is in balance. We have N (52 N) and mg (98 N) on one side, and FA,y on the other (150 N).
 Cloak
mathc
mathc
 Cloak
appc
appc