Insect wing physics? Average force?

In summary, the flying insect applies an average force equal to twice its weight during each downward stroke while hovering. Assuming 100 downward strokes per second, estimate the average power output of the insect.
  • #1
algar32
89
1

Homework Statement



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!
Thanks.


Homework Equations





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
 
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  • #2
Hey
What exact problem are you having?

(anyways hint:You have calculated the force wrong.Units are not right.)
 
  • #3
I am sorry, I am a little confused. What units should it be?
 
  • #4
Mass in SI units is taken in kg not grams :-)
 
  • #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?
 
  • #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...
thanks.
 
  • #7
You have forgotten work done in lifting wing up :-)
 
  • #8
emailanmol said:
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?

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?
 
  • #9
algar32 said:
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?


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 .
 
  • #10
emailanmol said:
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 .
mg*d=
.01kg*9.81*.02m

Is that *100 the right answer?
What happened to the times 2 on the force?
 
Last edited:
  • #11
No no.

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

You can't 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
 
  • #12
emailanmol said:
No no.

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

You can't 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

2* .01kg *9.81*.02m = .005886
full cycle = .--5886 x 100 = .5886

this isn't right ^^
not sure what i did wrong :(
 
  • #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
 
  • #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 I am coming up is mgd.

Time for once cycle is 1/100 s.

Total power is (2mgd+mgd)*100
 
  • #15
emailanmol said:
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 I am coming up is mgd.

Time for once cycle is 1/100 s.

Total power is (2mgd+mgd)*100

so the answer is (2 x .01 x 9.81 x .01 + .01 x 9.81 x .01) x 100
=.2943
?
 
  • #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
 
  • #17
yah I thought it was strange that you couldn't find the upward force information.
Also 2.94 Isn't correct. What else should I try?
 
  • #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 textbook 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 let's just skip power delivered during upward motion)
 
  • #19
emailanmol said:
What is the correct answer.Let me know so that I can figure out what the question actually wants.

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

I am not sure what the answer is supposed to be or how to approach it :(
 
  • #20
emailanmol said:
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 textbook 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 let's just skip power delivered during upward motion)
so 200 x .01 x 9.81 x .01 = .1962 Is this that right?
 
  • #21
algar32 said:
I am not sure what the answer is supposed to be or how to approach it :(

Hey. So how did you come to know 2.94 isn't correct?


The approach to power is to realize 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
so

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 textbook is wrong according to me :-)
 
  • #22
Yes
0.1962 should be right :-)
 
  • #23
Thanks friend :)
 
  • #24
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:

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The main factors that determine the aerodynamics of insect wings are the shape and size of the wing, the speed at which the insect is flying, and the air density and viscosity.

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The shape of an insect wing is crucial for its flight performance. It affects the lift and drag forces acting on the wing, as well as the stability and maneuverability of the insect in flight.

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Insect wings generate lift through a combination of factors, including the angle of attack, the shape of the wing, and the flapping motion. This creates a difference in air pressure above and below the wing, resulting in lift.

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