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Part A
|
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?
Solution
System:
The skydiver will be treated as a point particle.
Interactions:
External influences from the earth (gravity) and the air.
Model:
Point Particle Dynamics].
Approach:
Diagrammatic Representation
The free body diagram for this situation is:
Mathematical Representation
The x-direction is unimportant in this problem, so we write only the y-component equation of Newton's Second Law:
Unknown macro: {latex} \begin
Unknown macro: {large} [ \sum F_
Unknown macro: {y} = F_
Unknown macro: {rm air}
- mg = ma_
] \end
This equation has only one unknown, so we can solve for Fair.
Unknown macro: {latex} \begin
Unknown macro: {large} [ F_
Unknown macro: {rm air}
= mg+ma_
Unknown macro: {y}
= 735\:
Unknown macro: {rm N} + (75\:
Unknown macro: {rm kg}
)(-2.0\:
Unknown macro: {rm m/s}
^
Unknown macro: {2}
) = 590\:
] \end
Part B
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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
System:
The skydiver will be treated as a point particle.
Interactions:
External influences from the earth (gravity) and the air.
Model:
Point Particle Dynamics.
Approach:
Diagrammatic Representation
The free body diagram for this situation is:
Mathematical Representation
The x-direction is unimportant in this problem, so we write only the y-component equation of Newton's Second Law:
Unknown macro: {latex} \begin
Unknown macro: {large} [ \sum F_
Unknown macro: {y} = F_
Unknown macro: {rm air}
- mg = ma_
] \end
Now, since the problem tells us the skydiver is falling with a constant velocity, we know that the y-accleration is zero. Cancelling the may term, we can solve for Fair:
Unknown macro: {latex} \begin
Unknown macro: {large} [ F_
Unknown macro: {rm air}
= mg = 740\:
Unknown macro: {rm N}
] \end
Part C
|
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)?
Solution
System:
The skydiver and parachute will be treated as a single point particle.
Interactions:
External influences from the earth (gravity) and the air.
Model:
Point Particle Dynamics.
Approach:
Diagrammatic Representation
The free body diagram for this situation is:
Mathematical Representation
The x-direction is unimportant in this problem, so we write only the y-component equation of Newton's Second Law:
Unknown macro: {latex} \begin
Unknown macro: {large} [ \sum F_
Unknown macro: {y} = F_
Unknown macro: {rm air}
- mg = ma_
] \end
This equation has only one unknown, so we can solve for Fair.
Unknown macro: {latex} \begin
Unknown macro: {large} [ F_
Unknown macro: {rm air}
= mg+ma_
Unknown macro: {y}
= 735\:
Unknown macro: {rm N} + (75\:
Unknown macro: {rm kg}
)(+2.0\:
Unknown macro: {rm m/s}
^
Unknown macro: {2}
) = 890\:
] \end
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