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The Flying Leap of the Flea

  • Thread starter pepov
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Homework Statement



High-speed motion pictures (3500 frames/second) of a jumping 240μg flea yielded the data to plot the flea's acceleration as a function of time as shown in the figure (Figure 1) . (See "The Flying Leap of the Flea," by M. Rothschild et al. in the November 1973 Scientific American.) This flea was about 2 mm long and jumped at a nearly vertical takeoff angle. Use the measurements shown on the graph to answer the questions.

1) Find the initial net external force on the flea

2) How does it compare to the flea's weight

3) Find the maximum net external force on this jumping flea

4) When does this maximum force occur?

5) Use the graph to find the flea's maximum speed.

physics_zpseae922ed.jpg


Homework Equations



F = ma
a = Δv/Δt

The Attempt at a Solution



1)

m = .24 mg

I began by looking up the article because I did not understand what a/g on the y-axis means. I found that this is "Accerleration (Gravities)". It looks to me like the graph begins at ~60 a/g so the force of the jump is:

aj = 60*9.81 = 588.6 m/s2
Fj = m*aj = 588.6 m/s2*.24mg = 141.26 N

This is the force in the vertical direction. So I added the force of the flea's weight:

Fw = .24mg * -9.81 m/s2
Fw = -2.35 N

Fw + Fj = Fnet
-2.35 N + 141.26 N = 138.91 N

This answer is incorrect. Perhaps the initial force should only include the weight of the flea? Although this wouldn't make sense to me because then the graph would be irrelevant.

2)

Given the answer to (1) I should divide by Fw:

[Answer to (1)] / 2.35 = __ w

3)

The maximum net external force would be mass times the flea's highest a/g which looks to be around 140 a/g.

140 a/g * 9.81 = 1373.4 m/s2
Fmax = 1373.4 m/s2*.24mg
Fmax = 329.616 N

Following the same procedure as before (which will change when I figure out what I did wrong):

Fmax + Fw = FnetMAX
329.616 - 2.35 = 327.266 N

This is also incorrect.

4)

I will figure this out when I figure out 3 by looking at the graph.

5)

This was easy and I just looked at the graph for the answer.


Thanks very much for all your help,

Pepov
 
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Answers and Replies

  • #2
haruspex
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I'd say the graph starts closer to 63 than 60.
Does gravity count as an external force? I thought it would, but you may be right. If it doesn't, then extra thrust from the ground is needed to overcome it and still produce the required acceleration, so you should be increasing the force, not reducing it.
For 2), since a is given in gs, isn't this just the number read off the graph, plus 1!
For 5), I'm curious as to how you calculated the speed. It's not hard, but I wouldn't have said it was competely trivial.
 
  • #3
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The acceleration is given by the net force. So what you get my multiplying the acceleration from the graph by mass IS the net force. Why do you think you need to add or subtract to it?
The question is also about the net force, isn't it?
 
  • #4
haruspex
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The acceleration is given by the net force. So what you get my multiplying the acceleration from the graph by mass IS the net force. Why do you think you need to add or subtract to it?
The question is also about the net force, isn't it?
It says "net external force". Does gravity count as an external force? I would have said yes, but maybe the questioner thinks otherwise.
 
  • #5
collinsmark
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Fw = .24mg * -9.81 m/s2
Fw = -2.35 N
In addition to others' comments, you are treating milligrams like kilograms. The quote above is one example, but there are others. How many milligrams are in a kilogram? :wink:

[Edit: Welcome to Physics Forums, btw!]
 
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  • #6
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If the "net external force" represents the contact force (rather than mg, which is a body force), then you should be adding mg to ma, not subtracting it. However, I tend to agree with Nasu's interpretation of the net external force being ma. The ratio of the net external force to the weight is just the ordinate on the graph, a/g.
 
  • #7
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The opposite to "external" forces should be "internal" forces. These (external) do not change the momentum of the system. The external ones do.
So the change in momentum (and then the acceleration) is due to external forces.
At least this is what I thought is the meaning of external and internal.

However reading some sites about internal versus internal it seems that there is some confusions in textbooks. Here for example they use internal and external as other names for conservative and non-conservative forces:
http://www.physicsclassroom.com/class/energy/u5l2a.cfm
According to this, if I look at a falling body, the gravity is "internal". :smile:
 
  • #8
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As far as my understanding of my class thus far is concerned, this external force should be a sum of the forces acting on the flea. In fact the question asks for force on the flea. In a free-body diagram, I drew a force of gravity acting downward on the flea. I drew a pushing force of the ground against the flea which is equal to the force of the flea's jump. These should be the only two forces acting on the flea if air resistance is negligible.

In addition to others' comments, you are treating milligrams like kilograms. The quote above is one example, but there are others. How many milligrams are in a kilogram? :wink:
Thanks for pointing this out; what a stupid mistake -__-. I fixed it and moved the decimal place and the answer is still incorrect so there is definitely an issue with my free body diagram.

The new mass I used is:

m = 2.4e-7

[Edit: Welcome to Physics Forums, btw!]
Thanks!

[EDIT:]
For 5), I'm curious as to how you calculated the speed. It's not hard, but I wouldn't have said it was competely trivial.
When I looked up the article (which is at http://www-bsac.eecs.berkeley.edu/~sbergbre/research/publications/FlyingFlea1973.pdf [Broken]) I found a position v time and a velocity v time graph for this same study. I just used the velocity v time graph. That's why it was so easy :)
 
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  • #9
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Sorry to double-post, but what I wrote regarding the free-body diagram got me thinking...there should be one additional force acting on the flea, which would be the normal force counteracting gravity. It turns out that the flea's weight does not need to be involved in this equation because the normal force acting on the flea and the force of gravity cancel out leaving only the push force against the ground. So Fj is the correct answer (after you move the decimal over).
 
  • #10
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Thank you all for all your help I figured out all the answers after that realization.

1) 1.47e-4 N

2) 62.5*w

3) 3.29e-4 N

4) 1.2 ms

5) 1.2 m/s

I really appreciate you all for taking the time to respond!
 
  • #11
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Sorry to double-post, but what I wrote regarding the free-body diagram got me thinking...there should be one additional force acting on the flea, which would be the normal force counteracting gravity. It turns out that the flea's weight does not need to be involved in this equation because the normal force acting on the flea and the force of gravity cancel out leaving only the push force against the ground. So Fj is the correct answer (after you move the decimal over).
The normal force IS the force pushing the flea up. It is larger than gravity so that the net force is upward and accelerates the flea.
There is no other external force, as I understand. The flea jumps by its own.
 

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