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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 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 A !thatnormal2.png|width=40%! A person moves a 10 kg box up a 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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Box as .

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id	intb

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

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	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 = 0 in the x direction equation gives:

Latex
\begin{large}\[ N

Setting

= F_{A}\cos\theta = \mbox{150 N} \]\end{large}

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Card

label	Part C

Part C

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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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System:

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Box as .

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Interactions:

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External influences from the earth (gravity), the ceiling () 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

_~x~ = 0 in the _x_ direction equation gives: {latex}

free body diagram for the box:

Image Added

Note
The ceiling must push down to prevent objects from moving up through it.

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	diagc
	diagc

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

Because Because the box is held against the ceiling, it has no movement (and no acceleration) in the y direction (a_y = 0). Setting a_y = 0 in the y direction equation gives:

Latex
\begin{large}\[ F_{A}\sin\theta - mg - N = 0 \]\end{large}

which we solve to find:

Latex
\begin{large}\[ N = F_{A}\

cos

sin\theta - mg = \mbox{

150

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).

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Key

Part A

Solution

Part B

Solution

Part C

Solution