The shaft of a 6 m wide gate in the figure will fail at a moment of 3924 kN.m about the hinge P. The maximum value of water depth \( h \) (in m) that the gate can hold is ......... (round off to the nearest integer).
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When solving for the maximum water depth in gate problems, ensure to use the moment equilibrium method and account for the force acting on the gate, as well as the geometry of the gate and its position.
Let the force \( F \) act on a gate at \( B \).
From the figure, we know that:
\[
F = \rho g h A
\]
Where:
- \( \rho = 1000 \, {kg/m}^3 \) (density of water)
- \( g = 9.81 \, {m/s}^2 \) (acceleration due to gravity)
- \( h \) is the depth of the water
- \( A = 6 \times h \) is the area of the gate
Substituting the values:
\[
F = 1000 \times 9.81 \times (h - 2) \times 30 = 294.3 \times (h - 2) \, {kN}
\]
From the figure:
\[
\sin \theta = \frac{h - h_{cp}}{PB}
\]
where \( h_{cp} = h - 4 \). Substituting this into the equation:
\[
\sin \theta = \frac{h - (h - 4)}{PB} = \frac{4}{5}
\]
Next, using the moment equilibrium about \( P \), we take the moment about \( P \):
\[
F \times PB = 3924
\]
Substituting the value of \( F \):
\[
294.3 \times (h - 2) \times \frac{5}{4} \times \left[ 2 - \frac{4}{3(h - 2)} \right] = 3924
\]
Simplifying and solving for \( h \):
\[
(h - 2) \times \left( 2 - \frac{4}{3(h - 2)} \right) = 32/3
\]
\[
2h - 4 = 32/3
\]
\[
h = 8 \, {m}
\]
Thus, the maximum depth \( h \) that the gate can hold is \( 8 \, {m} \).
