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

[finlejb]

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?

 

[NateTG]

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.

 

[finlejb]

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?

 

[Pyrrhus]

Remember Newton's 1st Law!

$$\sum F_{x} = 0$$

$$\sum F_{y} =0$$

 

[NateTG]

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.

 

[finlejb]

What do you mean a more general equation for drag? That's all my book gives me...

 

[Pyrrhus]

Try

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