How to Calculate Drag Coefficient for a Fire Helicopter's Bucket?

AI Thread Summary
To calculate the drag coefficient for a fire helicopter's bucket, the resistive force is proportional to the square of the bucket's speed. The equation provided, C_d = 2mg / (V_T)^2Aρ, requires terminal velocity, but it's noted that a more general expression for drag can be used instead. The drag force can be expressed as F_drag = (C_D / A) * (ρ * v^2 / 2). Participants clarify that the coefficient of friction is not applicable in this context, emphasizing the need to focus on drag calculations. Understanding these principles allows for the determination of the drag coefficient without needing terminal velocity.
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A fire helicopter carries a 560-kg bucket at the end of a cable 20.6 m long as in the figure below. As the helicopter flies to a fire at a constant speed of 39.2 m/s ,the cable makes an angle of 39.6 with respect to the vertical. The bucket presents a cross-sectional area of 3.96m^2 in a plane perpendicular to the air moving past it. Determine the drag coefficient, assuming that the resistive force is proportional to the square of the bucket's speed.

I've got that C_d = 2mg / (V_T)^2A\rho, but I don't know how to find the terminal velocity of the bucket to find the coefficient of drag. Can anyone help?
 

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You don't need to use the terminal verlocity at all, but a more general expression for the force of friction. You have enough information to determine the force of friction, and you know the airspeed.
 
But the equation for the force of friction is f_f = \mu N. I don't know the coefficent of friction... unless it's 1 since it's flying through the air?
 
Remember Newton's 1st Law!

\sum F_{x} = 0

\sum F_{y} =0
 
Sorry, what I meant is that you need a more general expression for drag (rather than friction). You should be able to calculate the force that the drag exerts and work from there.
 
What do you mean a more general equation for drag? That's all my book gives me...
 
Try

F_{drag} = \frac{C_{D}}{A} \frac{\rho v^2}{2}
 
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