Helicopter Problem: Calculating Lift Force on Rotors

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The problem involves calculating the lift force exerted by the air on the rotors of a helicopter that is accelerating upward while lifting an additional weight. The context includes a helicopter with a mass of 13,000 kg and a car weighing 900 kg, with an upward acceleration of 0.41 m/s².

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  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law and the forces acting on the helicopter, including gravitational force and lift force. There is confusion about how to incorporate the upward acceleration into the calculations.

Discussion Status

Some participants are exploring different interpretations of the problem and attempting to clarify the role of acceleration in the calculations. Hints have been offered regarding the relationship between the forces and the need to consider the direction of each force.

Contextual Notes

There are concerns about the precision of the data provided in the problem, particularly regarding significant figures and the expectation of rounding to the nearest Newton.

ebola_virus
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There were two posts reltaed to this question but both of them are so contradictory; the question goes

a 13,000kg helicopter accelerates upward at 0.41m/s^2 while lifting a 900kg car. to the nearest Newton what is the lift force exerted by the air on the rotors?

now one post would be

Fnet = (13000)(9.8) + (900)(9.8)

which in a sense is what we are looking for (but unfortunately, is wrong for some reason)

and so i don't know what piece of hte puzzple I am missing; I am sure the 0.41m/s^2 comes in at some point. Is it the force required to accerleate the helicopter? If so what would that be.. I am lost.
 
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You're right, the acceleration does come in at some point! :)

HINT: The sum of the forces is mass times acceleration.
 
i cn't tell what I am getting wrong because its an online question where i plug in the values and get answers.. and thing is i keep getting values off by like. twenty or something. what i did is, fully:

(13000)(9.8) + (900)(9.8) ------- (1)

then i obtained

(13909)(0.41) ------- (2)

added the two together. am i right?
 
Ask yourself what is the direction of each of the forces?
 
each of the forces as in?
 
As in the force of gravity and the force of lift. You were on the right track:

[tex]F_{lift} - M g = Ma[/tex]

where M is the combined mass so [itex]F_{lift} = M(g + a)[/itex]. I am not totally surprised that the numbers are off since it makes no sense to ask for the lift to the nearest Newton given data accurate to only two significant digits (shame on the authors for doing that!)

Using g = 9.8 m/s^2 I get 141,919 N (blindly retaining all the digits after the 4).
 

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