Question
A rigid square sheet of size 2 m × 2 m is hinged at the middle of the vertical edges to serve as a door which can turn about the horizontal axis OO′. A fluid of density ρ fills the space to the left of the sheet up to its top. The horizontal force F required (to be applied at the lower edge) to hold the sheet vertical is :
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Solution
[Key Concept: Hydrostatic pressure grows with depth, so the lower half of a mid-hinged gate gets the larger torque and must be balanced by a force at the bottom edge.]
$$\begin{matrix} \tau_{\text{upper~}} & \ = \int_{0}^{1}\mspace{2mu}\mspace{2mu}\rho\text{ }g(1 - x)(2dx)(x) \\ & \ = \rho g2\left\lbrack \frac{1}{2} - \frac{1}{3} \right\rbrack = \frac{\rho g}{3}(cw) \end{matrix}$$
$$\begin{matrix} \tau_{\text{lower~}} & \ = \int_{0}^{1}\mspace{2mu}\mspace{2mu}\rho\text{ }g\left( 1 + x^{'} \right)\left( 2{dx}^{'} \right)\left( x^{'} \right) \\ & \ = \rho g2\left\lbrack \frac{1}{2} + \frac{1}{3} \right\rbrack = \rho g \times \frac{5}{3}(\text{~Acw.~}) \\ \tau_{\text{net~}} & \ = \rho g\left( \frac{5}{3} - \frac{1}{3} \right) \\ & \ = \frac{4}{3}\rho\text{ }g = F \times 1 \\ \text{ }F = & \frac{4}{3}\rho\text{ }g \end{matrix}$$
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