Question: Helicopter over a scale

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  • Thread starter HawkHunt
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  • #1
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Hey all,

This is clearly a very simple question but for some reason my roommate and I just can't agree on this.

The scenario is as follows: There is a helicopter sitting on a scale. The helicopter weighs 700kg and thus has 700*g = 6867 N of force acting on it by gravity. In order to fly the helicopter must overcome its own weight and so it must produce >6867N of lift.

The question is, how much does the scale register on the millisecond the helicopter leaves the scale. My reasoning is that the scale must register 6868N atleast.

Please enlighten me.
 

Answers and Replies

  • #2
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It would help if you told us why you think that. I have a theory that might not be correct, but I'd like to hear your thoughts first, so as to not poison the water.
 
  • #3
.Scott
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Assuming the scale is fully supporting the downstream or air, it would depend on the vertical force applied to the helicopter during the period leading up to T+0.001 seconds and the time it takes for that force to be transfered through the air.

If the rotor generated 0 lift at T=-0.0001 and 2G's starting at T=0, then the scale would likely register 0 at T=0.001 - because no downdraft would have been generated in time to reach the scale that soon.
 
  • #4
berkeman
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The question is, how much does the scale register on the millisecond the helicopter leaves the scale. My reasoning is that the scale must register 6868N atleast.
F=ma. :smile:

(with what vertical acceleration does the helo leave the pad?)
 
  • #5
.Scott
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F=ma. :smile:

(with what vertical acceleration does the helo leave the pad?)
@HawkHunt did not pose this question in the classical way - that is, as a steady state problem. So secondary effects (like the mass of the air and how quickly the rotor has spun up) need to be considered as well.
 

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