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Several examples illustrating how to find the normal force in not-so-common situations.

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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? System: Box as [point particle] subject to 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}

Card

label	Part A

Part A

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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?

Solution

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System:

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Box as .

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Interactions:

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External influences from the earth (gravity), the wall () and the person ().

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Model:

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.

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Approach:

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Diagrammatic Representation

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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.

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	diaga
	diaga

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Mathematical Representation

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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} 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? 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: !thatfbd2.png! From the free body diagram, we can write the equations of [Newton's 2nd Law|Newton's Second Law]. {latex}

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Card

label	Part 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

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id	sysb

System:

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id	sysb

Box as .

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Interactions:

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id	intb

External influences from the earth (gravity), the wall () and the person ().

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Model:

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.

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Approach:

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Diagrammatic Representation

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We begin with a free body diagram for the box:

Image Added

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	diagb
	diagb

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Mathematical Representation

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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

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} 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? 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: !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}

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Card

label	Part 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

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id	sysc

System:

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id	sysc

Box as .

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Interactions:

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id	intc

External influences from the earth (gravity), the ceiling () and the person ().

Toggle Cloak

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Model:

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.

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Approach:

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Toggle Cloak

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Diagrammatic Representation

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id	diagc

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

id	mathc

Mathematical Representation

Cloak

id	mathc

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

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}

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).

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Key

Part A

Solution

Part B

Solution

Part C

Solution