Simple Free Body Diagram question

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SUMMARY

The discussion centers on calculating the lift force exerted by air on a 6500-kg helicopter that accelerates upward at 0.6 m/s² while lifting a 1200-kg car. The correct approach involves recognizing that both the helicopter's weight (6500 kg * 9.8 m/s²) and the car's weight (1200 kg * 9.8 m/s²) act downward, while the lift force acts upward. The net force is determined by the equation: Lift Force - (Weight of Helicopter + Weight of Car) = Net Force, where the net force equals the total mass times the acceleration (ma).

PREREQUISITES
  • Understanding of Newton's Second Law of Motion (F = ma)
  • Basic knowledge of free body diagrams
  • Familiarity with gravitational force calculations (Weight = mass * gravity)
  • Concept of net force and its direction in physics
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  • Study free body diagram examples in physics textbooks
  • Learn about Newton's laws of motion in detail
  • Explore the concept of lift and its applications in aerodynamics
  • Practice problems involving forces and acceleration in various contexts
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to forces and motion.

psycovic23
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I have a question in my textbook that I can't answer for the life of me.

"A 6500-kg helicopter accelerates upward at .6 m/s^2 while lifting a 1200-kg car. (a) what is the lift force exerted by the air on the rotors?"

Now, my diagram has a vector pointing up with a force of (6500*9.8)N. There is also a vector pointing down at (1200*9.8)N. Wouldn't the net force then be equal to the force of the pull minus the force of the weight? Fn-(6500*9.8)-(1200*9.8) = net force

Or am I completely wrong? At the moment, having read this section multiple times, I'm still having trouble grasping this whole natural force and the summation of the forces = ma. My teacher doesn't really help much as she isn't the greatest teacher either...
 
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weight acts down

psycovic23 said:
Now, my diagram has a vector pointing up with a force of (6500*9.8)N.
Why would the weight of the helicopter point up?
There is also a vector pointing down at (1200*9.8)N.
Right.
Wouldn't the net force then be equal to the force of the pull minus the force of the weight? Fn-(6500*9.8)-(1200*9.8) = net force
Yes, if by Fn you mean the force of the air on the rotors. (It looks like you changed your mind about the direction of the helicopter's weight!)


Your force diagram should show the weight of both acting down, and the force of the air acting up. The net force acts up, since we know the acceleration is up.
 
Oooohhhhhhhh...I see what I did wrong! For some reason, I was thinking the weight of the helicopter was...um...tension up or something. I don't know, but thank you very much!
 

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