How Does Air Resistance Affect a Fire Helicopter's Bucket?

[MarkFL]

Here is this week's POTW:

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A fire helicopter carries a bucket of mass $m$ at the end of a cable. Flying back from a fire with an empty bucket at a constant speed, the cable makes an angle $\alpha$ with respect to the vertical.

(a) Determine the force of air resistance on the bucket.

(b) After filling the bucket with sea water, the helicopter returns to the fire at the same speed with the bucket now making an angle $\beta<\alpha$ with the vertical. What is the mass of the water in the bucket?

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[anemone]

Congratulations to Ackbach for his correct solution, which you can find below:

Spoiler

(a) We draw the Free Body Diagram (FBD):

\begin{tikzpicture}
\draw (0,0) node[above=0.5cm,right=0.7cm] {$m$} rectangle (2,1);
\draw [->] (1,1) node[above=0.9cm,right=0.35cm] {$T$} node[above=0.4cm,right=-0.05cm] {$\alpha$} -- (2,2);
\draw [->] (1,1) -- (1,2);
\draw [->] (1,0) node[below=0.5cm,left=0.01cm] {$mg$} -- (1,-1);
%\draw [->] (2,0.5) node[below=0.4cm,right=0.2cm] {$F$} -- (3,0.5);
\draw [->] (0,0.5) node[below=0.4cm,left=0.2cm] {$F_{r}$} -- (-1,0.5);
\end{tikzpicture}
Here $F_r$ is air resistance, and $T$ is the tension in the cable. As the velocity is constant, there is no acceleration, and Newton's Second Law becomes
\begin{align*}
T\cos(\alpha)-mg&=0 \\
T\sin(\alpha)-F_r&=0.
\end{align*}
Rearranging yields
\begin{align*}
T\cos(\alpha)&=mg \\
T\sin(\alpha)&=F_r.
\end{align*}
Dividing the second equation by the first gives
$$\tan(\alpha)=\frac{F_{r}}{mg},\quad\text{or}\quad F_r=mg\tan(\alpha).$$
The direction of $F_r$ is opposite the helicopter's direction of travel.

(b) The FBD is identical, except that instead of mass $m$, we have mass $m+M$, where $M$ is the mass of the water in the bucket, and we have angle $\beta$. The equations will therefore be
\begin{align*}
T\cos(\beta)-(m+M)g&=0 \\
T\sin(\beta)-F_r&=0.
\end{align*}
The same re-arranging yields
$$\tan(\beta)=\frac{F_r}{(m+M)g}\implies m+M=\frac{F_r\cot(\beta)}{g}\implies M=\frac{F_r\cot(\beta)}{g}-m=\frac{mg\tan(\alpha)\cot(\beta)}{g}-m=m\left(\frac{\tan(\alpha)}{\tan(\beta)}-1\right).$$