Versions Compared

Old Version 22

changes.mady.by.user Andrew E Pawl

Saved on

compared with

New Version Current

changes.mady.by.user Andrew E Pawl

Saved on

Key

  • This line was added.
  • This line was removed.
  • Formatting was changed.
Comment: Migration of unmigrated content due to installation of a new plugin

Excerpt
hiddentrue

Explore the force from air resistance acting on a skydiver at various stages of the dive.

Composition Setup
30%
 Section
 Column
width70%
 Deck of Cards
idbigdeck
 Card
labelPart A
Part A
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/s2 toward the ground, what is the force of air resistance acting on the skydiver? 
Solution
 Toggle Cloak
idsysa
 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:


h3. Part A

{table:align=right}{tr}{td}!parachute_accelerate.jpg!{td}{tr}{table}
In the photo at right, Sgt. 1st Class Cheryl Stearns of the U.S. Army's Golden Knights parachute team demonstrates how to maximize acceleration during a skydive (photo by Cpl. Sean Capogreco courtesy U.S. Army).  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

{toggle-cloak:id=sysa} *System:*  {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.png!

{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}


 Latex
 Wiki Markup
This 
 
 equation 
 
 has 
 
 only 
 
 one 
 
 unknown, 
 
 so 
 
 we 
 
 can 
 
 solve 
 
 for 
 _
 Fair. 


h3. Part B

!parachute_terminal.jpg|width=700!
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?  (Photo by Shane Hollar courtesy U.S. Navy.) 

h4. 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}

{toggle-cloak:id=appb} *Approach:*  

{cloak:id=appb}

{toggle-cloak:id=diagb} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diagb}

The free body diagram for this situation is:

!skydiver_terminalFBD.png!

{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 
 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}


 Card
labelPart B
 Wiki Markup
 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
 Card
labelPart B
Part B
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? 
Solution
 Toggle Cloak
idsysb
 System: 
 Cloak
idsysb
The skydiver will be treated as a .
 Toggle Cloak
idintb
 Interactions: 
 Cloak
idintb
External influences from the earth (gravity) and the air.
 Toggle Cloak
idmodb
 Model: 
 Cloak
idmodb
.
 Toggle Cloak
idappb
 Approach: 
 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
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


h3. 
Part 
 C

|!parachute_decel.jpg|width=50%!|
|Photo by Mass Communication Specialist 2nd Class Christopher Stephens courtesy 
C
 Card
labelPart C
 Wiki Markup
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^
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.png!

{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}

 Column
 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
width