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Homework Help: Insect wing physics? Average force?

  1. Mar 19, 2012 #1
    1. The problem statement, all variables and given/known data

    A typical flying insect applies an average force equal to twice its weight during each downward stroke while hovering. Take the mass of the insect to be 10 g, and assume the wings move an average downward distance of 1.0 cm during each stroke
    Assuming 100 downward strokes per second, estimate the average power output of the insect. Answer must be in watts!

    2. Relevant equations

    3. The attempt at a solution
    watts = work joules in a second
    100 downward strokes in a second.
    work= force x distance
    force = (2*10*100)
    d= .01m * 100
    work= 2000j in a second
    answer= 2000 watts
  2. jcsd
  3. Mar 19, 2012 #2
    What exact problem are you having?

    (anyways hint:You have calculated the force wrong.Units are not right.)
  4. Mar 19, 2012 #3
    I am sorry, I am a little confused. What units should it be?
  5. Mar 19, 2012 #4
    Mass in SI units is taken in kg not grams :-)
  6. Mar 19, 2012 #5
    You will write force in one cycle while going downwards as 2mg.Displacement as 0.01 m.
    Time for one cycle as 1/100 seconds

    What is force while going up?
    What is work done in bringing thw wing up?
  7. Mar 19, 2012 #6
    I changed it to kg and it still came out wrong :(
    force = (2*.01kg*100)
    d= .01m * 100
    work= 2 j in a second
    answer= 2 watts

    Not sure what I am doing wrong...
  8. Mar 19, 2012 #7
    You have forgotten work done in lifting wing up :-)
  9. Mar 19, 2012 #8
    the force going up... I am not sure.. .But do I need that to solve this? Could you write out the basic equation so I can understand what to plug where?

    Thanks so much.

    Also how Would i find the upward force? Would it just be double? Also if we counted everything as double wouldn't the displacement be double as well? 1cm on the way down one on teh way up?
  10. Mar 19, 2012 #9

    Force acting on way up can be taken as mg (assuming no acceleration takes place)
    While coming up displacement is again 1cm.
    So work is mg*d.
    Time is again 1/100 seconds .
  11. Mar 19, 2012 #10

    Is that *100 the right answer?
    What happened to the times 2 on the force?
    Last edited: Mar 19, 2012
  12. Mar 19, 2012 #11
    No no.

    You will have to calculate work done seperately for upward and downward motion.

    You cant take net because

    1) net displacement for a cycle is 0

    2) force changes value and direction midway.

    For part one work done is 2mg*d* cos(0) i.e. 2mgd
    Cos 0 as displacement and for e in same direction

    For way up work is mg*dcos (0)
    Cos 0 as displacement and firce in same direction.

    So net work done in one cycle is 3mg*d .time is 1/100 s
  13. Mar 19, 2012 #12
    2* .01kg *9.81*.02m = .005886
    full cycle = .--5886 x 100 = .5886

    this isn't right ^^
    not sure what i did wrong :(
  14. Mar 19, 2012 #13
    What is the answer you are expecting(the textbook answer ).Let me see if it matches my value.

    As i said net power is 100*3*mg*d where d is 1cm and not 2
  15. Mar 19, 2012 #14
    You are using coeffiecientas 2.
    Its not 2.Its 3.

    You are using d as 2 cm its 1cm.

    Thats where you went wromg.

    Work done in going down is 2mgd (d is 0.01 cm)
    Work done im coming up is mgd.

    Time for once cycle is 1/100 s.

    Total power is (2mgd+mgd)*100
  16. Mar 19, 2012 #15
    so the answer is (2 x .01 x 9.81 x .01 + .01 x 9.81 x .01) x 100
  17. Mar 19, 2012 #16
    Yes Its 0.294 Watts (Units are imp).

    You follow the answer right?

    [to point out this question has not been correctly framed as one cannot determine forve acting during upward motion
  18. Mar 19, 2012 #17
    yah I thought it was strange that you couldnt find the upward force information.
    Also 2.94 Isn't correct. What else should I try?
  19. Mar 19, 2012 #18
    What is the correct answer.Let me know so that I can figure out what the question actually wants to ask.

    The other answer should be power=2*mg*d*100

    (Also does your text book have frequent mistakes? Like the kinetic friction question , this could also have a wrong answer.)

    I think the answer should 200mgd.

    (Question speaks only ablut downward movement of wings so lets just skip power delivered during upward motion)
  20. Mar 19, 2012 #19
    I am not sure what the answer is supposed to be or how to approach it :(
  21. Mar 19, 2012 #20
    so 200 x .01 x 9.81 x .01 = .1962 Is this that right?
  22. Mar 19, 2012 #21
    Hey. So how did you come to know 2.94 isn't correct?

    The approach to power is to realise the basic concept that power is work done in unit time.

    Work is force vector dot product displacement vector.

    In general dW=F*ds*cos(theta) where theta is angle between force and displacement vector

    W=integration of [F*ds*cos(theta)]

    Power is P=dW/dt.

    here, for downward F is constant and is given as
    -2mg j
    Displacement is - d j

    So work for one downward motion is 2mgd
    Power is 2mgd/t.

    Plug in d as 0.01m and t as 1/100 s.

    If the answer doesn't match, the text book is wrong according to me :-)
  23. Mar 19, 2012 #22
    0.1962 should be right :-)
  24. Mar 19, 2012 #23
    Thanks friend :)
  25. Mar 19, 2012 #24


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    Homework Helper

    It would be easier to consider the power = work done on the air, and assume no work done when the wings flap upwards.

    average force = 2 x 9.81 m / s2 x .010 kg = 0.1962 N.
    average distance = 1.0 cm = .01 m
    average frequency = 100hz = 100 / s

    Normally average force would mean average force versus time, but with this problem, apparently it means average force vesus distance on downstroke.

    power = work done per cycle x frequency = force x average downstroke distance per cycle x frequency
    power = .1962N x .01 m x 100/s = .1962 watts

    (It doesn't matter how long each downstroke takes, just the frequency).
    Last edited: Mar 19, 2012
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