Calculating Lift Force and Air Resistance in a Moving Helicopter

AI Thread Summary
The discussion focuses on calculating the lift force and air resistance for a helicopter moving horizontally at a constant velocity. The weight of the helicopter is given as 57,600 N, and the lift force makes a 21° angle with the vertical. Participants clarify that since the helicopter is not accelerating, the sum of forces in both the X and Y directions must equal zero. The user initially struggles with the calculations but ultimately realizes they can solve for the lift force and air resistance by applying the correct equations. The conversation emphasizes the importance of understanding force components in static equilibrium scenarios.
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A helicopter is moving horizontally to the right at a constant velocity. The weight of the helicopter is W = 57600 N. The lift force L generated by the rotating blade makes an angle of 21.0° with respect to the vertical.

(a) What is the magnitude of the lift force in N?

(b) Determine the magnitude of the air resistance R that opposes the motion.


I used \SigmaFx= max--> 0 (because constant velocity means 0 acceleration)= Lsin(21)

and \SigmaFy= may-->Lcos21-57600


now I don't know what to do. What did i do wrong?
 
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Hi PhysicsFailure!

I am a new member and somewhat new to physics compared to most here but I think I can help you out on this one.

First, we know that the helicopter is not accelerating in either the X OR Y directions (it is moving strictly horizontally). For this reason, the sum of all forces in each direction is 0, not just X.

Have you tried drawing a simple diagram, with the helicopter represented as a point at the origin, the weight acting downward, and the force of lift acting upward at an angle of 21 degrees to the Y axis?

I think if you set the Y equation you already have to 0, the force of lift will become apparent. From here, figure what component of that lift on the X axis must be matched by the air resistance for the copter to not have any acceleration (sum of X forces=0)

Let me know if you get it!
 
So far so good! It looks as though you've done everything right.

"I used \SigmaFx= max--> 0 (because constant velocity means 0 acceleration)= Lsin(21)"

So then what is the force due to air resistance?

"and \SigmaFy= may-->Lcos21-57600"

Part a is asking for L. How would you solve for L in this case?

Keep in mind that because there is no vertical acceleration, the net force in the y direction will also be zero.

PhysicFailure said:
A helicopter is moving horizontally to the right at a constant velocity. The weight of the helicopter is W = 57600 N. The lift force L generated by the rotating blade makes an angle of 21.0° with respect to the vertical.

(a) What is the magnitude of the lift force in N?

(b) Determine the magnitude of the air resistance R that opposes the motion.I used \SigmaFx= max--> 0 (because constant velocity means 0 acceleration)= Lsin(21)

and \SigmaFy= may-->Lcos21-57600now I don't know what to do. What did i do wrong?

Homework Statement


Homework Equations


The Attempt at a Solution

 
haha thanks so much bchandler and mattowander. I actually figured out what I was not doing. I could just solve it! and for some reason i was just not thinking...THANK YOU!
 
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