Hovering kite height, using forces

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The discussion revolves around solving a physics problem involving a kite hovering at a specific height, with given forces acting on it. The original poster expresses concern about their solution being too easy for a problem rated with two stars, indicating difficulty. They calculate the height of the kite using the tension in the line and the wind force, arriving at a height of approximately 32.15 meters. Other participants suggest alternative methods and interpretations, highlighting the importance of considering the kite's weight and the angles involved in the forces. The conversation emphasizes the complexity of force equilibrium in such problems and the potential for different approaches leading to varied answers.
jehan4141
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This problem has 2 stars in my book, which means it is supposed to pretty difficult. I got the answer fairly easily and am worried that maybe I missed something important and thus got the incorrect answer too easily. No answer is provided to self-check my answer to see whether or not I am approaching the problem correctly, so would you mind telling me if my work is correct? I know this site isn't a place to have your work checked, but I just want to make sure I am doing this correctly because the problem seemed easier than 2 stars. I usually can't finish a 2 star problem. Thank you!

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A kite is hovering over the ground at the end of a 43-m line. The tension in the line has a magnitude of 16 N. Wind blowing on the kite exerts a force of 19 N, directed 56 degrees above the horizontal. Note that the line attached to the kites is NOT oriented at an angle of 56 degrees above the horizontal. find the height of the kite, relative to the person holding the line.

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MY APPROACH:


Givens:
Length, L = 43 m
Tension, T = 16 N
Force of wind, Fw = 19 N, 56 degrees above the horizontal


From the problem, I understand that the kite is hovering in one spot. Thus, all horizontal and vertical forces equal to zero.

By setting the horizontal forces to zero, I get:
ΣFx = (Fw)cos56 - TcosΘ = 0
Fwcos56 = TcosΘ
19cos56 = 16cosΘ
Θ = 48.39116057

Using the value of Θ we have just found, I can use
sinΘ = height / hypotenuse to find the height of the kite.
sin(48.39116057) = height / 43
height = 32.15091308
 
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Looks good to me. :smile:
 
jehan4141 said:
This problem has 2 stars in my book, which means it is supposed to pretty difficult. I got the answer fairly easily and am worried that maybe I missed something important and thus got the incorrect answer too easily. No answer is provided to self-check my answer to see whether or not I am approaching the problem correctly, so would you mind telling me if my work is correct? I know this site isn't a place to have your work checked, but I just want to make sure I am doing this correctly because the problem seemed easier than 2 stars. I usually can't finish a 2 star problem. Thank you!

------------------------------------------------------------------------------------------

A kite is hovering over the ground at the end of a 43-m line. The tension in the line has a magnitude of 16 N. Wind blowing on the kite exerts a force of 19 N, directed 56 degrees above the horizontal. Note that the line attached to the kites is NOT oriented at an angle of 56 degrees above the horizontal. find the height of the kite, relative to the person holding the line.

-----------------------------------------------------------------------------------------

MY APPROACH:


Givens:
Length, L = 43 m
Tension, T = 16 N
Force of wind, Fw = 19 N, 56 degrees above the horizontal


From the problem, I understand that the kite is hovering in one spot. Thus, all horizontal and vertical forces equal to zero.

By setting the horizontal forces to zero, I get:
ΣFx = (Fw)cos56 - TcosΘ = 0
Fwcos56 = TcosΘ
19cos56 = 16cosΘ
Θ = 48.39116057

Using the value of Θ we have just found, I can use
sinΘ = height / hypotenuse to find the height of the kite.
sin(48.39116057) = height / 43
height = 32.15091308

I got a different answer, but I used a slightly different method. I'm not seeing right now why we would get different answers, though.

Maybe try it my way to see if I just made a math error. Instead of summing the horizontal forces to zero, I figured out what the angle between the wind and the kite string would need to be, to lower the wind force of 19N down to 16N. That gave me the angle between the wind and the string, which gives the angle of the string to the ground.

Do you get your same answer with that method?
 
Thank you red belly! :D I will try your way and get back to you in a few minutes berkeman. Thank you all!
 
Last edited:
Berkeman: I do not see why I should try and make lower the wind force from 19 N to 16 N. What does it mean to have the forces equal?
 
jehan4141 said:
Berkeman: I do not see why I should try and make lower the wind force from 19 N to 16 N. What does it mean to have the forces equal?

I was thinking that the component of the 19N wind force in the direction of the tension force has to be 16N, since the tension is caused by the wind.
 
How I picture this problem is that the kite is hovering at one point above the ground. In my mental picture I see a person holding a kite, with the string directed (or flowing) towards the upper right corner of my mental image. I think this where our pictures differ: I see the force of wind coming from the far right side of my mental picture, say from, I don't know...100 yards to the person's right. Thus, in my mental picture, the tension in the string isn't caused by the force of wind. Rather, the force of wind is what causes the kite to remain in one place above the ground.

I think I see your mental picture, and yes, that does open up a whole 'nother can of worms. AHHHHH! I see your interpretation of the problem and see how you could get another answer :/
 
If you look at forces in any direction other than horizontal, then you need to consider the (unknown) weight of the kite.
 
Redbelly98 said:
If you look at forces in any direction other than horizontal, then you need to consider the (unknown) weight of the kite.

Ah, thanks for that RB!
 

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