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Deck of Cards
idbigdeck
h3.

Part

A |!parachute_accelerate.jpg!| |Photo by Cpl. Sean Capogreco courtesy

A

Card
labelPart A
Wiki Markup

Image Added

Photo by Cpl. Sean Capogreco courtesy U.S.

Army

|

In

the

photo

above,

Sgt.

1st

Class

Cheryl

Stearns

of

the

U.S.

Army's

Golden

Knights

parachute

team

demonstrates

how

to

maximize

acceleration

during

a

skydive.

Suppose

that

a

75

kg

skydiver

was

using

this

technique.

If

the

skydiver's

current

acceleration

is

2.0

m/s

{^}

2

{^}

toward

the

ground,

what

is

the

force

of

air

resistance

acting

on

the

skydiver?

h4. Solution {

Solution

Toggle Cloak

:

id

=

sysa

} *

System:
Cloak
idsysa

The skydiver will be treated as a .

Toggle Cloak
idinta
Interactions:
Cloak
idinta

External influences from the earth (gravity) and the air.

Toggle Cloak
idmoda
Model:
Cloak
idmoda

.

Toggle Cloak
idappa
Approach:

Cloak
idappa

Toggle Cloak
iddiaga
Diagrammatic Representation

Cloak
iddiaga

The free body diagram for this situation is:

Image Added

Note

When drawing a free body diagram for an object moving vertically under the influence of gravity and some other force(s), it is important to try to decide whether gravity is "winning" over the other force(s) or not. In this case, we know that the skydiver is accelerating toward the ground, so we expect that the force of gravity is larger than the force of air resistance. If you can come to such a conclusion, it is a good idea to represent it in your free body diagram.

Cloak
diaga
diaga

Toggle Cloak
idmatha
Mathematical Representation

Cloak
idmatha

The x-direction is unimportant in this problem, so we write only the y-component equation of Newton's Second Law:

Latex
*  {cloak:id=sysa}The skydiver will be treated as a [point particle].{cloak}

{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}External influences from the earth (gravity) and the air.{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}
The free body diagram for this situation is:

!skydiver_accelerateFBD.jpg!

{note}When drawing a free body diagram for an object moving vertically under the influence of gravity and some other force(s), it is important to try to decide whether gravity is "winning" over the other force(s) or not.  In this case, we know that the skydiver is accelerating toward the ground, so we expect that the force of gravity is larger than the force of air resistance.  If you can come to such a conclusion, it is a good idea to represent it in your free body diagram.{note}

{cloak:diaga}

{toggle-cloak:id=matha} {color:red} *Mathematical Representation* {color}

{cloak:id=matha}

The x-direction is unimportant in this problem, so we write only the y-component equation of Newton's Second Law:

{latex}\begin{large}\[ \sum F_{y} = F_{\rm air} - mg = ma_{y} \] \end{large}{latex}

This

equation

has

only

one

unknown,

so

we

can

solve

for

_

Fair.

Latex
_~air~.  

{latex}\begin{large}\[ F_{\rm air} = mg+ma_{y} = 735\:{\rm N} + (75\:{\rm kg})(-2.0\:{\rm m/s}^{2}) = 590\:{\rm N}\] \end{large}{latex}

{warning}This problem illustrates the importance of choosing a coordinate system.  Only by explicitly choosing the y-direction (as we did in the free body diagram) will you remind yourself that the acceleration is downward, and so an appropriate sign must be applied.  In our case, because we chose the positive y-direction to be upward, the acceleration must be negative.{warning}

{tip}As we predicted in the note above, the force of air resistance is smaller than the force of gravity on the skydiver in this case.{tip}

{cloak:matha}
{cloak:appa}
Warning

This problem illustrates the importance of choosing a coordinate system. Only by explicitly choosing the y-direction (as we did in the free body diagram) will you remind yourself that the acceleration is downward, and so an appropriate sign must be applied. In our case, because we chose the positive y-direction to be upward, the acceleration must be negative.

Tip

As we predicted in the note above, the force of air resistance is smaller than the force of gravity on the skydiver in this case.

Cloak
matha
matha

Cloak
appa
appa

h3.

Part

B |!parachute_terminal.jpg|width=700!| |Photo by Shane Hollar courtesy

B

Card
labelPart B
Wiki Markup

Image Added

Photo by Shane Hollar courtesy U.S.

Navy.

|

If

they

dive

far

enough,

even

without

parachutes,

skydivers

will

reach

a

constant

velocity

known

as

terminal

velocity.

Suppose

that

a

75

kg

skydiver

is

falling

at

a

constant

velocity

of

55

m/s

toward

the

ground.

What

is

the

force

of

air

resistance

acting

on

the

skydiver?

h4. Solution {

Solution

Toggle Cloak

:

id

=

sysb

} *

System:

* {

Cloak

:

id

=

sysb

}

The

skydiver

will

be

treated

as

a

[point particle].{cloak} {

.

Toggle Cloak

:

id

=

intb

} *

Interactions:

* {

Cloak

:

id

=

intb

}

External

influences

from

the

earth

(gravity)

and

the

air.

{cloak} {

Toggle Cloak

:

id

=

modb

} *

Model:

* {cloak:id=modb}[Point Particle Dynamics].{cloak} {

Cloak
idmodb

.

Toggle Cloak

:

id

=

appb

} *

Approach:

* {cloak:id=appb} {

Cloak
idappb

Toggle Cloak

:

iddiagb
Diagrammatic Representation

Cloak
iddiagb

The free body diagram for this situation is:

Image Added

Note

Once again, we can use cues from the problem to decide whether gravity is "winning" over the other force(s) or not. In this case, we know that the skydiver is moving at constant speed, so we expect that the force of gravity is precisely balanced by the force of air resistance.

Cloak
diagb
diagb

Toggle Cloak
idmathb
Mathematical Representation

Cloak
idmathb

The x-direction is unimportant in this problem, so we write only the y-component equation of Newton's Second Law:

Latex
=diagb} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diagb}

The free body diagram for this situation is:

!skydiver_terminalFBD.jpg!

{note}Once again, we can use cues from the problem to decide whether gravity is "winning" over the other force(s) or not.  In this case, we know that the skydiver is moving at constant speed, so we expect that the force of gravity is precisely balanced by the force of air resistance.{note}

{cloak:diagb}

{toggle-cloak:id=mathb} {color:red} *Mathematical Representation* {color}

{cloak:id=mathb}

The x-direction is unimportant in this problem, so we write only the y-component equation of Newton's Second Law:

{latex}\begin{large}\[ \sum F_{y} = F_{\rm air} - mg = ma_{y} \] \end{large}{latex}

Now,

since

the

problem

tells

us

the

skydiver

is

falling

with

a

constant

velocity,

we

know

that

the

y-accleration

is

zero.

Cancelling

the

_

ma

_~_

y

_~

term,

we

can

solve

for

_

Fair:

Latex
_~air~:

{latex}\begin{large}\[ F_{\rm air} = mg = 740\:{\rm N}\] \end{large}{latex}

{note}The speed of the skydiver is irrelevant to this problem.  As long as the acceleration is zero, the forces balance.{note}

{cloak:mathb}
{cloak:appb}
Note

The speed of the skydiver is irrelevant to this problem. As long as the acceleration is zero, the forces balance.

Cloak
mathb
mathb

Cloak
appb
appb

unmigrated-wiki-markup
Card
labelPart C
h3.

Part

C |!parachute_decel.jpg|width=700!| |Photo by Mass Communication Specialist 2nd Class Christopher Stephens courtesy

C

Image Added

Photo by Mass Communication Specialist 2nd Class Christopher Stephens courtesy U.S.

Navy.

|

Immediately

after

opening

the

parachute,

the

skydiver

will

begin

to

decelerate.

The

parachute

has

effectively

increased

the

skydiver's

air

resistance,

which

lowers

the

terminal

velocity.

Suppose

that

a

75

kg

skydiver

has

recently

pulled

the

rip

cord

and

is

currently

decelerating

at

2.0

m/s

{^}

2

{^}

.

What

is

the

force

of

air

resistance

acting

on

the

skydiver

(including

parachute)?

h4. Solution {

Solution

Toggle Cloak

:

id

=

sysc

} *

System:
Cloak
idsysc

The skydiver and parachute will be treated as a single .

Toggle Cloak
idintc
Interactions:
Cloak
idintc

External influences from the earth (gravity) and the air.

Toggle Cloak
idmodc
Model:
Cloak
idmodc

.

Toggle Cloak
idappc
Approach:

Cloak
idappc

Toggle Cloak
iddiagc
Diagrammatic Representation

Cloak
iddiagc

The free body diagram for this situation is:

Image Added

Note

Again, we try to decide whether gravity is "winning" over the other force(s) or not. In this case, we know that the skydiver's deceleration is equivalent to an upward acceleration. Thus, we expect that the (downward) force of gravity is less than the (upward) force of air resistance in this case.

Cloak
diagc
diagc

Toggle Cloak
idmathc
Mathematical Representation

Cloak
idmathc

The x-direction is unimportant in this problem, so we write only the y-component equation of Newton's Second Law:

Latex
*  {cloak:id=sysc}The skydiver and parachute will be treated as a _single_ [point particle].{cloak}

{toggle-cloak:id=intc} *Interactions:* {cloak:id=intc}External influences from the earth (gravity) and the air.{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}

The free body diagram for this situation is:

!skydiver_decelFBD.jpg!

{note}Again, we try to decide whether gravity is "winning" over the other force(s) or not.  In this case, we know that the skydiver's deceleration is equivalent to an _upward_ acceleration.  Thus, we expect that the (downward) force of gravity is less than the (upward) force of air resistance in this case.{note}

{cloak:diagc}

{toggle-cloak:id=mathc} {color:red} *Mathematical Representation* {color}

{cloak:id=mathc}
The x-direction is unimportant in this problem, so we write only the y-component equation of Newton's Second Law:

{latex}\begin{large}\[ \sum F_{y} = F_{\rm air} - mg = ma_{y} \] \end{large}{latex}

This

equation

has

only

one

unknown,

so

we

can

solve

for

_

Fair.

Latex
_~air~.  

{latex}\begin{large}\[ F_{\rm air} = mg+ma_{y} = 735\:{\rm N} + (75\:{\rm kg})(+2.0\:{\rm m/s}^{2}) = 890\:{\rm N}\] \end{large}{latex}

{warning}
Warning

Again,

it

is

important

to

consider

your

coordinate

system.

It

is

common

to

consider

a

deceleration

to

correspond

to

a

negative

acceleration.

In

this

case,

however,

because

the

skydiver

is

_

moving

in

the

negative

direction

_

,

the

deceleration

corresponds

to

a

positive

(upward)

acceleration

in

our

coordinate

system.

{warning} {tip}As we predicted in the note above, the force of air resistance is larger than the force of gravity on the skydiver in this case.{tip} {cloak:mathc} {cloak:appc}

Tip

As we predicted in the note above, the force of air resistance is larger than the force of gravity on the skydiver in this case.

Cloak
mathc
mathc

Cloak
appc
appc