Determining Air Resistance Force on Fire Helicopter's Water Bucket

[EcKoh]

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
\vartheta = 46.0

Homework Equations

For Drag:
F_d =(1/2)C_dpA_v²

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 (F_d =(1/2)C_dpA_v²). 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 C_d 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 F_d = (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?

 

[jamesrc]

You're over-complicating things and incorrectly using the formula for force due to air resistance. Drag in that formula is a function of the cross-sectional area of the bucket as seen by the wind - that is information you do not have according to your problem statement - the area of the triangle you have calculated is not relevant.

Instead, draw a free body diagram for the bucket. What are the forces acting on it? The weight of the bucket, the force due to drag, and the tension from the cable. You should be able to draw that triangle of forces - use the magnitude of the force that you know and the angle of the triangle to find the magnitude of the other force you need.

 

[EcKoh]

Okay, I am drawing the free body diagram right now. Would I be correct to say that normal force isn't acting on the bucket?

 

[RoyalCat]

Okay, I am drawing the free body diagram right now. Would I be correct to say that normal force isn't acting on the bucket?

A normal force as a result of its contact with what surface, exactly..?

 

[EcKoh]

Okay so I redrew it, and finally figured it out hours later. Here is my solution, please let me know if I did this right or just happened to fall upon the correct answer:

First I redrew the free body diagram just as I had before, labeling tension (T) in the rope, force of drag (F_d) pointing away from the bucket and the direction of movement, and W=mg pointing down.

Then I used the x-axis and y-axis to get two formulas:

T sin \theta = F
T cos \theta = mg

This allowed for me to change it to tan \theta = F/mg

Then I plugged in 564 for mass 46 for \theta, and 9.8 for gravity, giving me an answer of 5723.58 N.

Is this the correct method?

 

[RoyalCat]

That's spot on, EcKoh. Well done. :)