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viola_85
Apr4-04, 11:26 PM
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

cookiemonster
Apr4-04, 11:40 PM
Don't forget that gravity is also acting on the helicopter, and the propellers must overcome the force of gravity as well.

cookiemonster

viola_85
Apr4-04, 11:46 PM
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?

cookiemonster
Apr4-04, 11:52 PM
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

viola_85
Apr4-04, 11:54 PM
But Newton's Second Law is Force = Mass times Acceleration. Which I used in the first place.

cookiemonster
Apr5-04, 12:02 AM
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

ShawnD
Apr5-04, 12:19 AM
I think cookie is saying this

F = (4500)(9.8 + 2)

Chen
Apr5-04, 05:30 AM
The law is:
\Sigma F = ma
But the lift force is not the only one acting on the helicopter:
\Sigma F = F_{lift} - mg

ShawnD
Apr5-04, 05:47 AM
The law is:
\Sigma F = ma
But the lift force is not the only one acting on the helicopter:
\Sigma F = F_{lift} - mg

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

Chen
Apr5-04, 05:50 AM
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.