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Excerpt

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

Composition Setup

Deck of Cards

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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]. *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}

Card

label	Part 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

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id	sysa

System:

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id	sysa

Box as .

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id	inta

Interactions:

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id	inta

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

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id	moda

Model:

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id	moda

.

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id	appa

Approach:

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id	appa

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id	diaga

Diagrammatic Representation

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id	diaga

We begin with a free body diagram for the box:

Image Added

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	diaga
	diaga

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id	matha

Mathematical Representation

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id	matha

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} ---- 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]. *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}

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	matha
	matha

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	appa

Card

label	Part 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

id	sysb

System:

Cloak

id	sysb

Box as .

Toggle Cloak

id	intb

Interactions:

Cloak

id	intb

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

Toggle Cloak

id	modb

Model:

Cloak

id	modb

.

Toggle Cloak

id	appb

Approach:

Cloak

id	appb

Toggle Cloak

id	diagb

Diagrammatic Representation

Cloak

id	diagb

We begin with a free body diagram for the box:

Image Added

Cloak

	diagb
	diagb

Toggle Cloak

id	mathb

Mathematical Representation

Cloak

id	mathb

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} ---- 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]. *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}

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	mathb

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	appb

Card

label	Part 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

id	sysc

System:

Cloak

id	sysc

Box as .

Toggle Cloak

id	intc

Interactions:

Cloak

id	intc

External influences from the earth (gravity), the floor (normal force) and the rope (tension).

Toggle Cloak

id	modc

Model:

Cloak

id	modc

.

Toggle Cloak

id	appc

Approach:

Cloak

id	appc

Toggle Cloak

id	diagc

Diagrammatic Representation

Cloak

id	diagc

We begin with a free body diagram for the box:

Image Added

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. 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} ---- 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]. *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}

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?

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	mathc

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	appc
	appc

Card

label	Part 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

id	sysd

System:

Cloak

id	sysd

Box as .

Toggle Cloak

id	intd

Interactions:

Cloak

id	intd

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

Toggle Cloak

id	modd

Model:

Cloak

id	modd

.

Toggle Cloak

id	appd

Approach:

Cloak

id	appd

Toggle Cloak

id	diagd

Diagrammatic Representation

Cloak

id	diagd

We begin with a free body diagram for the box:

Image Added

Cloak

	diagd
	diagd

Toggle Cloak

id	mathd

Mathematical Representation

Cloak

id	mathd

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}

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Key

Part A

Solution

Part B

Solution

Part C

Solution

Part D

Solution