The question is probably much more difficult than your teacher anticipated. Lift is in the vertical direction, and lift is equal to the sum of weight and the downward component of tension.
Assuming the tension is perpendicular to the kits, then the force in the direction of the wind is 100sin20. But This drag force is equal to the horizontal component of tension, so finally, the vertical component of tension is horizontal force*tan70 (draw a picture of the kite and string to ground) or
lift=100sin20*tan70
I am not certain what your teacher had in mind with this question.
A kite has a lifting surface area of 630 square meters. Suppose the wind is blowing against the kite with a force of 100 newtons at an angle of 20 degrees above the horizontal. How much force is lifting the kite?
I tried to draw a diagram, but I don't really understand what the question means. I got an answer of 88.889 newtons, but I'm pretty sure that my method for getting this answer is incorrect.
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Please check my analysis. I am worried that I missed something.
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