Alexander & Kite: Force Needed to Hold Back

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Homework Help Overview

The discussion revolves around a physics problem involving forces acting on a kite being held by a person named Alexander. The scenario includes a kite with a specific weight, a string at a given angle, and wind forces affecting the kite's position. Participants are exploring the equilibrium of forces and the calculations necessary to determine the tension in the string.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to calculate the tension in the string, considering the kite's weight and the wind force. There are attempts to draw free body diagrams and set up equilibrium equations. Questions arise about the correct interpretation of forces acting on the kite, particularly regarding the wind's influence and whether all information from the problem statement is accurate.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have provided guidance on drawing force diagrams and using Newton's laws, while others express confusion about the problem's setup and the forces involved. There is no explicit consensus, as differing views on the forces acting on the kite are being debated.

Contextual Notes

Participants note potential issues with the problem's information, particularly regarding the wind's direction and its effect on the kite's equilibrium. There is a suggestion that the problem may not support a solution where the kite remains aloft, raising questions about the assumptions made in the analysis.

anthroxy
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Hi there i kinda got stuck on a couple of questions in my course book and don't understand how i am supposed to do i guess i'll focus on the first one now and post the second later

Alexander is running around with his kite, the string is 25m long and the kite weighs 450g

a) The string is completely stretched out between Alexander and the kite, the angle is 65 degrees from the horizontal plane. A force of wind with 10N is affecting the kite and has a direction of 25 degrees on top of the horizontal plane. How big is the force that Alexander is using to hold the kite back?
 
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you should focus on calculating the tension in the string which will be the force that Alexander will be applying to hold the kite back. We can assume the kite to be in equilibrium if Alexander is running with constant speed.The tension will act away from kite.Make free body diagram of kite and set up equilibrium equation to calculate tension.I hope so that this much will do.
 
Yeah that far i understood but i think my problem lies in drawing the full body diagram, which i guess would be kinda hard for you to explain =/
 
Is it possible to post the free body diagram that i made, thought i could do it through paint but the file became to big
 
Okey i guess i'll try to explain without showing my diagram.

Since we know the weight of the object we can use Newtons second law to find out the Fg
Fg = ma which in this case was Fg = 0,45 * 9,82 which gives us Fg= 4,419

From there i used my diagram to find out the tension on the string/rope and since we have the angle and the opposite we can use sin 65 = 4,419/c to find out the hypotenuse (and the hypotenuse being F-tension) which gave me F-tension = 4,87N

So Alexander is pulling with 4,87N is this correct?
 
I don't think so. For one thing, it appears that you are not accounting for the 10N, 25 degree force from the wind acting on the kite.

However, when I work through this problem myself, it appears impossible to solve. Are you sure all the information is copied correctly from the problem statement?
 
anthroxy said:
Is it possible to post the free body diagram that i made, thought i could do it through paint but the file became to big

You can always print it and then scan it as a .jpg.
 
Paint can save in .jpg format, unless you're using a really old version - or better yet, use .png, which generally gives smaller file sizes for pictures with only a few colors.
 
Last edited:
Redbelly98 said:
I don't think so. For one thing, it appears that you are not accounting for the 10N, 25 degree force from the wind acting on the kite.

However, when I work through this problem myself, it appears impossible to solve. Are you sure all the information is copied correctly from the problem statement?
yeah its really impossible to solve because there is always a net downward force acting on the kite which implies that it is falling down.
 
  • #10
for the kite to be in equilibrium there must be an upward force created by the wind to balance the weight of the kite
 
  • #11
Redbelly98 said:
I don't think so. For one thing, it appears that you are not accounting for the 10N, 25 degree force from the wind acting on the kite.

However, when I work through this problem myself, it appears impossible to solve. Are you sure all the information is copied correctly from the problem statement?

Yeah i completely agree with you, and that's what i though was so strange, since i didn't take account to the wind and don't really get how I'm supposed to do it either.

Everything is completely copied from the book, haven't missed anything as far as i see
 
  • #12
Is this a proper depiction of the original problem?
 

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  • #13
anthroxy said:
Yeah i completely agree with you, and that's what i though was so strange, since i didn't take account to the wind and don't really get how I'm supposed to do it either.
.

1. In general, you start by drawing a force diagram showing every force acting on the object (the kite, in this problem). A good way to display your diagram is to upload it to a site like www.photobucket.com, and then post a link to the uploaded image in your post at Physics Forums.
.

2. Next, think in terms of the horizontal and vertical components of those forces, using trig and the given angles. Set up 2 equations using Newton's 2nd Law, equating the net forces with ma as follows:

∑ [horizontal force components] = m ax

∑ [vertical force components] = m ay

(Be sure to pay attention to whether each force component is in the positive or negative direction.)
.

3. Try to determine whether ax and ay should be zero, or not. Does anything in the problem statement indicate that the kite is accelerating? Is Alexander running with a constant velocity? There should be something in the wording of the question that addresses this.
.

4. Finally, you work with those two force equations to solve the problem. If necessary, look for any other information in the problem statement that might be relevant.
.

Everything is completely copied from the book, haven't missed anything as far as i see
Okay.
 
  • #14
This is beginning to look like a pulley question to me. Maybe I'm wrong, but doesn't Alex have to exert a force equal to the two forces pushing down on the kite (the weight of the kite and SIN 25 * the 10N wind)?
 
  • #15
AC130Nav said:
This is beginning to look like a pulley question to me. Maybe I'm wrong, but doesn't Alex have to exert a force equal to the two forces pushing down on the kite (the weight of the kite and SIN 25 * the 10N wind)?
Why do you think the wind exerts a downward force?

The string can only pull down on the kite, not push up. So if the wind, weight, and string tension were all downward, there is no way for the kite to stay aloft. Something must have an upward force component; it isn't the string tension, and it sure isn't gravity.

Unfortunately, even if the string tension were zero here, the downward weight of the kite overcomes the upward force of the wind (10 N at 25°). So the information given in the problem statement does not support a solution where the kite stays aloft. (That's why I asked the OP if he had provided the information correctly.[/size])
 
  • #16
Redbelly98 said:
Why do you think the wind exerts a downward force?
I have reread the original post. I thought the wind described was the area prevailing wind, but it does appear to be the relative wind (which includes Alex pulling on the kite).

An observation: 65 + 25 = 90. If the kite is perpendicular to the string, that relative wind is down its axis (edge-on).
 

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