= Solution
Use the ideal steady incompressible-flow model, neglecting height change across the smooth nozzle. <Flow conservation> gives inlet and outlet speeds $u_0=Q/A_0$ and $u_1=Q/A_1$. Set atmospheric pressure to zero. The <Bernoulli equation> between inlet and outlet gives
$$
\boxed{p_0=\frac\rho2(u_1^2-u_0^2)=\frac{\rho Q^2}{2}\left(\frac1{A_1^2}-\frac1{A_0^2}\right).}
$$
Choose a control volume containing the fluid in the nozzle, with the positive axis pointing toward the outlet. The <integral momentum equation> gives
$$
p_0A_0+F_{\mathrm{wall\ on\ fluid}}=\rho Q(u_1-u_0).
$$
Thus the fluid's force on the nozzle is $p_0A_0-\rho Q(u_1-u_0)$, in the outlet direction. Substitute the pressure and velocities:
$$
\begin{aligned}
F_{\mathrm{fluid\ on\ nozzle}}
&=\rho Q^2\left[\frac{A_0}{2A_1^2}-\frac1{A_1}+\frac1{2A_0}\right]\\
&=\frac{\rho Q^2}{2A_0}\left(\frac{A_0}{A_1}-1\right)^2.
\end{aligned}
$$
The fireman's holding force is equal and opposite. Therefore \b[the required force points upstream and has magnitude]
$$
\boxed{F=\frac{\rho Q^2}{2A_0}\left(\frac{A_0}{A_1}-1\right)^2.}
$$
This is the <holding force on a contracting nozzle>; retaining the inlet pressure force is essential to its sign and value.
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