A person is using a massless string to fly a kite. The kite has a mass of 0.620kg and the string is attatched to the kite at an angle of 58 degrees below the horizontal. The tension in the string is 10 N. In what direction does the force from the wind have to be on the kite in order to hold the kite motionless? Choose the best answer.
The answer is 70 degrees above the horizontal but I would've thought that it would be 58 degrees above the horizontal, in order to balance out the tension force of the kite..? In the explanation the book said that For the kite to be stationary, the wind force must exactly oppose this resultant force, so it must be at an angle greater than 58 degrees above the horizontal.Why would the angle have to be greater?? I don't get it :S
The answer is 70 degrees above the horizontal but I would've thought that it would be 58 degrees above the horizontal, in order to balance out the tension force of the kite..? In the explanation the book said that For the kite to be stationary, the wind force must exactly oppose this resultant force, so it must be at an angle greater than 58 degrees above the horizontal.Why would the angle have to be greater?? I don't get it :S
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The kite also has weight. The wind must balance the combined effect of string tension and weight;
Fx = TCos(58)
Fy = mg + TSin(58)
Tan() = [mg + TSin(58)]/TCos(58)
T = string tension, Fx & Fy are components of the wind force.
Plug in the numbers to get Tan(), then take arctan to get your answer.
Fx = TCos(58)
Fy = mg + TSin(58)
Tan() = [mg + TSin(58)]/TCos(58)
T = string tension, Fx & Fy are components of the wind force.
Plug in the numbers to get Tan(), then take arctan to get your answer.