Helicopter lift force with upward acceleration

  • Thread starter Thread starter viola_85
  • Start date Start date
  • Tags Tags
    Forces Homework
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
9 replies · 3K views
viola_85
Messages
3
Reaction score
0
A 4500-kg helicopter accelerates upward at 2m/s^2. What lift force is exerted by the air on the propellers?

Is the answer just Newtons Second Law of Motion F=ma, so F=(4500kg)*(2)
, so F= 9000N
 
Physics news on Phys.org
Don't forget that gravity is also acting on the helicopter, and the propellers must overcome the force of gravity as well.

cookiemonster
 
so would I have to convert the initial mass into weight. i.e. W=MG

(4500)*(9.8)=44100
44100*2=88200N

Is that what I would have to do?
 
The lift must be greater than the weight enough to accelerate the helicopter 2 m/s^2.

F = aF is an equation you just made up and has no meaning. Try using Newton's 2nd Law.

cookiemonster
 
But Newton's Second Law is Force = Mass times Acceleration. Which I used in the first place.
 
Yes, you did a good job of finding the net force on the helicopter. But that's not the lift because you're not taking into consideration gravity.

cookiemonster
 
I think cookie is saying this

F = (4500)(9.8 + 2)
 
The law is:
[tex]\Sigma F = ma[/tex]
But the lift force is not the only one acting on the helicopter:
[tex]\Sigma F = F_{lift} - mg[/tex]
 
Chen said:
The law is:
[tex]\Sigma F = ma[/tex]
But the lift force is not the only one acting on the helicopter:
[tex]\Sigma F = F_{lift} - mg[/tex]

Although technically correct, your post can cause a great deal of confusion for somebody who doesn't fully understand the concept.

If I'm understanding correctly, you are saying this.
ma = F - mg
F = ma + mg
F = m(a+g) <----- like my post said
 
Yes, and why would my post cause confusion? Your post confuses me. :wink: I just thought it would be better to show how to get to the solution you posted, so that in future problems he can do the same.