Step 1

We are given the discharge rate of flow at each stream as $Q = 0.25\;{{{{\rm{m}}^3}} \mathord{\left/ {\vphantom {{{{\rm{m}}^3}} {\rm{s}}}} \right. } {\rm{s}}}$, the nozzle velocity of each stream as $v = 50\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$ and the density of sea water as ${\rho _{sw}} = 1020\;{{{\rm{kg}}} \mathord{\left/ {\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right. } {{{\rm{m}}^3}}}$.

We are asked to determine the tension developed in the anchor chain.

 
Step 2

The free-body diagram of the fire boat can be drawn as:

Here, W is the weight, N is the normal force and T is the tension in the anchor chain.

 
Step 3

To calculate the mass flow rate of each stream, we have:

\[\frac{{dm}}{{dt}} = Q{\rho _{sw}}\]

Substitute the values in the above expression, we get:

\[\begin{array}{c} \frac{{dm}}{{dt}} = \left( {0.25\;{{{{\rm{m}}^3}} \mathord{\left/ {\vphantom {{{{\rm{m}}^3}} {\rm{s}}}} \right. } {\rm{s}}}} \right) \times \left( {1020\;{{{\rm{kg}}} \mathord{\left/ {\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right. } {{{\rm{m}}^3}}}} \right)\\ \frac{{dm}}{{dt}} = 255\;{{{\rm{kg}}} \mathord{\left/ {\vphantom {{{\rm{kg}}} {\rm{s}}}} \right. } {\rm{s}}} \end{array}\]
 
Step 4

To calculate the resultant of the forces in the x-direction, we have:

\[\sum {F_x} = \frac{{dm}}{{dt}}\left[ {{{\left( {{v_A}} \right)}_x} + {{\left( {{v_B}} \right)}_x}} \right]\]

Here, ${\left( {{v_A}} \right)_x}$ is the nozzle velocity of stream at point A in x-direction and ${\left( {{v_B}} \right)_x}$ is the nozzle velocity of stream at point B in x-direction.

Substitute the values in the above expression, we get:

\[\begin{array}{c} T\cos 60^\circ = \left( {255\;{{{\rm{kg}}} \mathord{\left/ {\vphantom {{{\rm{kg}}} {\rm{s}}}} \right. } {\rm{s}}}} \right) \times \left[ {{v_A}\cos 30^\circ + {v_B}\cos 45^\circ } \right]\\ T = \left( {510\;{{{\rm{kg}}} \mathord{\left/ {\vphantom {{{\rm{kg}}} {\rm{s}}}} \right. } {\rm{s}}}} \right) \times \left[ {{v_A}\cos 30^\circ + {v_B}\cos 45^\circ } \right] \end{array}\] … (1)
 
Step 5

As the nozzle velocity of stream at each point is same then, ${v_A} = {v_B} = v$. Hence, from equation (1), we get:

\[\begin{array}{c} T = \left( {510\;{{{\rm{kg}}} \mathord{\left/ {\vphantom {{{\rm{kg}}} {\rm{s}}}} \right. } {\rm{s}}}} \right) \times \left[ {v\cos 30^\circ + v\cos 45^\circ } \right]\\ T = \left( {510\;{{{\rm{kg}}} \mathord{\left/ {\vphantom {{{\rm{kg}}} {\rm{s}}}} \right. } {\rm{s}}}} \right) \times v \times \left( {\cos 30^\circ + \cos 45^\circ } \right)\\ T = \left( {510\;{{{\rm{kg}}} \mathord{\left/ {\vphantom {{{\rm{kg}}} {\rm{s}}}} \right. } {\rm{s}}}} \right) \times \left( {50\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}} \right) \times \left( {1.573} \right) \times \left( {\frac{{{\rm{1}}\;{\rm{N}}}}{{{\rm{1}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right)\\ T = 40.11 \times {10^3}\;{\rm{N}} \times \left( {\frac{{{{10}^{ - 3}}\;{\rm{kN}}}}{{1\;{\rm{N}}}}} \right)\\ T = 40.11\;{\rm{kN}} \end{array}\]