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EcKoh
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Homework Statement
A fire helicopter carries a 564 kg bucket of water at the end of a 19.2 m long cable. Flying back from a fire at a constant speed of 41.6 m/s, the cable makes an angle of 46.0° with respect to the vertical. Determine the force of air resistance on the bucket.
Knowns:
m = 564 kg
d1 = 19.2 m
v = 41.6 m/s
[tex]\vartheta[/tex] = 46.0
Homework Equations
For Drag:
Fd =(1/2)CdpAv2
The Attempt at a Solution
So first I started by making a triangle out of the vertical and findin the values of the sides, the first being d1:
d2 = 19.2 sin 46.0 = 13.81
d3 = 19.2 cos 46.0 = 13.38
Then I used the pythagorean theorem (a2 + b2 = c2) to check the values. I also subtracted to find the last angle of the triangle (180-90-46 = 44), but I don't think I needed that.
So now I used the drag formula (Fd =(1/2)CdpAv2). I used p =1.29 because that's what it equals at sea level, but since its a helicopter I don't know if I used the right value. How to I calculate this?
Then I made Cd 1.15 because that is the drag coefficient of a small cylinder according to my textbook. Is this right?
Next, I used the formula A=1/2bh to get A = 92.39
Plugging this in I get Fd = (1/2)(1.15)(1.29)(92.39)(41.62) = 124366.0155
And I know this is wrong. The answer should be 5.72×103 N. Where did I go wrong? And how would I go about fixing this?