Easy problem, but cant figure it out

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The discussion revolves around calculating the lift force and resultant vector for a helicopter moving horizontally. Participants emphasize the need to resolve the lift force into its vertical and horizontal components, noting that the vertical component must equal the helicopter's weight of 52,100 N for equilibrium. Confusion arises regarding the correct trigonometric functions to use, with participants correcting each other's calculations. Ultimately, the correct approach involves using the angle of lift to find the magnitude of the lift force and its components. The conversation highlights the importance of understanding force equilibrium in physics problems.
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I've been trying out this problem for hours and can't figure it out.

A helicopter is moving horizontally to the right at a constant velocity. The weight of the helicopter is W = 52100 N. The lift force vector L generated by the rotating blade makes an angle of 21.0° with respect to the vertical.


Find the magnitude of the Lift force. And find the resultant vector R.

http://img237.imageshack.us/img237/9747/0455vd2.gif



At first i tried doing 52100sin69 for the lift..but that is not the correct answer...STUCK!
 
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What must the sum of the vertical forces be? What's the vertical component of the lift force? (Call the lift force "L".)
 
Try resolving the lift into horizontal and vertical components. If the helicopter is moving horizontally, with constant velocity, what does that tell you about the net vertical force? What about the net horizontal force?

Edit: beaten by a matter of seconds there :biggrin:
 
Vertical component of Lift must be equivalent to the magnitude of the weight.
 
cristo said:
Try resolving the lift into horizontal and vertical components. If the helicopter is moving horizontally, with constant velocity, what does that tell you about the net vertical force? What about the net horizontal force?

Edit: beaten by a matter of seconds there :biggrin:


Thats what i tried doing...


the net vertical force is 0, the horizontal force must then be 52100Ncos(69)..however i inputed this answer and it was marked wrong.
 
By the way I'm so confused now..
 
What forces act on the helicopter? (List them.)

What is the vertical component of each force?

Add up those vertical components and solve for L.
 
spidey12 said:
Thats what i tried doing...


the net vertical force is 0, the horizontal force must then be 52100Ncos(69)..however i inputed this answer and it was marked wrong.

52100Ncos(69) is incorrect...
 
Doc Al said:
What forces act on the helicopter? (List them.)

What is the vertical component of each force?

Add up those vertical components and solve for L.

I'm kind of at a loss but I'll try:


FyNet=W-fn

In the x direction: Fx=R-v

Right?
 
  • #10
Velocity is not a force. R is related to the horizontal component of the Lift.

Your Trig. was also incorrect.
 
  • #11
Vikingjl11 said:
Velocity is not a force. R is related to the horizontal component of the Lift.

Your Trig. was also incorrect.

I don't understand at all. How should I arrange my equation then?
 
  • #12
Vikingjl11 said:
Velocity is not a force. R is related to the horizontal component of the Lift.

Your Trig. was also incorrect.

Is the lift force going to be 52100N since the helicopter is not moving up or down?
 
  • #13
spidey12 said:
I'm kind of at a loss but I'll try:


FyNet=W-fn
What is fn?

In the x direction: Fx=R-v
v is a velocity, and thus should not be in a force equation. You are missing a term including the lift.


The easier way to do this, is to forget FyNet and Fx, since everything is in equilibrium. You should resolve the lift (L) into a horizontal and vertical component. Then, since the forces are in equilibrium, you should be able to write two equations:

Vertical: forces pointing up = forces pointing down
Horizontal: forces pointing right = forces pointing left.
 
  • #14
There is no acceleration in either the x or y direction. This means the net force must be zero in both the x and y directions. Therefore, the upwards force must be equivalent to the magnitude(absolute value) of the downforce or weight(weight = M*g). Also, R must be equal to the magnitude of the x component of the lift.

Use 51200/cos(21) = L to find magnitude of lift

51200N is y component of Lift.
 
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  • #15
cristo said:
What is fn?


v is a velocity, and thus should not be in a force equation. You are missing a term including the lift.


The easier way to do this, is to forget FyNet and Fx, since everything is in equilibrium. You should resolve the lift (L) into a horizontal and vertical component. Then, since the forces are in equilibrium, you should be able to write two equations:

Vertical: forces pointing up = forces pointing down
Horizontal: forces pointing right = forces pointing left.


this problem has been eating away at me for four damn hours now..ive tried all of this and I've read more than need to know about helicopters and i still can't seem to find that missing piece of info.


forces point up=forced pointing down: L=W

Horizontal: R=52100cos69

is that right?
 
  • #16
spidey12 said:
this problem has been eating away at me for four damn hours now..ive tried all of this and I've read more than need to know about helicopters and i still can't seem to find that missing piece of info.


forces point up=forced pointing down: L=W
No. There is only a component of L pointing vertically upwards, not the entire force, L. Try drawing a triangle with L as the hypotenuse, and the angle between the veritcal and the hypotenuse is 21 degrees. What is the length of the veritcal side?

Horizontal: R=52100cos69

is that right?
This is wrong since the above equation is wrong.
 
  • #17
Doc Al said:
What forces act on the helicopter? (List them.)

What is the vertical component of each force?

Add up those vertical components and solve for L.

spidey12 said:
I'm kind of at a loss but I'll try:


FyNet=W-fn

In the x direction: Fx=R-v

Right?
No. Try answering each question that I asked exactly as I asked it. Hint: Three forces act on the helicopter, all of which are labeled on your diagram. (The velocity is not one of them--that's why its arrow is drawn in a different color than the force arrows.)
 
  • #18
cristo said:
No. There is only a component of L pointing vertically upwards, not the entire force, L. Try drawing a triangle with L as the hypotenuse, and the angle between the veritcal and the hypotenuse is 21 degrees. What is the length of the veritcal side?


This is wrong since the above equation is wrong.

Ok that helped alot. Cos(x)=adjacent/hypotneuse so cos 21=52100/L or L=52100/cos21

is that correct?



For the second part, how is R related to this triangle?
 
  • #19
Doc Al said:


No. Try answering each question that I asked exactly as I asked it. Hint: Three forces act on the helicopter, all of which are labeled on your diagram. (The velocity is not one of them--that's why its arrow is drawn in a different color than the force arrows.)

Assuming my above post is correct...

would the Resultant be: 52100tan21=R
 
  • #20
spidey12 said:
Assuming my above post is correct...

would the Resultant be: 52100tan21=R

Yea, that would be equal to the x-component of L(lift) and -R.
 
  • #21
Vikingjl11 said:
Yea, that would be equal to the x-component of L(lift) and -R.


Ok thank you for helping me with this. Its a super easy problem but I am having major brain cramps today.

Thanks all!
 
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