Emergency Homework Help: Forces

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Homework Help Overview

The discussion revolves around calculating the lift force exerted on a helicopter that is accelerating upward. The subject area includes forces, specifically Newton's Second Law of Motion and the effects of gravity on the helicopter's motion.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the application of Newton's Second Law, questioning how to account for both the lift force and the gravitational force acting on the helicopter. There are attempts to clarify the relationship between weight, mass, and acceleration.

Discussion Status

The discussion is active, with participants providing insights and corrections regarding the calculations. There is a focus on ensuring that all forces acting on the helicopter are considered, particularly the gravitational force, while exploring different interpretations of the equations involved.

Contextual Notes

Participants are navigating the complexities of the problem, including the need to consider both the lift force required for upward acceleration and the weight of the helicopter. There is an emphasis on understanding the correct application of Newton's laws without reaching a definitive conclusion.

viola_85
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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
 
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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
 
  • #10
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.
 

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