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The given figure shows, water stands at depth D=35.0 m behind the vertical upstream face of a dam of width W=314 m. Find the net horizontal force on the dam from the gauge pressure of the water.

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Determining the total force of the water on the dam, specifically the net horizontal force using the equations of gauge pressure and force on an irregular area namely;

pgauge=ρgydF=p dAF=ρgyW dy=12ρgWD2\begin{align*}p_{gauge}&=\rho gy\\dF&=p\ dA\\F&=\rho gyW\ dy\\&=\dfrac{1}{2} \rho gWD^{2}\end{align*}

Solving for the total force using the above-derived equation yields;

F=12ρgWD2=12(1000 kgm3)(9.8 ms2)(314 m)(352 m2)=1.88 ×109 N\begin{align*}F&=\dfrac{1}{2} \rho gWD^{2}\\&=\dfrac{1}{2}\left(1000\ \dfrac{\text{kg}}{\text{m}^{3}}\right)\left(9.8\ \dfrac{\text{m}}{\text{s}^{2}}\right)\left(314\ \text{m}\right)\left(35^{2}\ \text{m}^{2}\right)\\&=\boxed{1.88\ \times 10^{9}\ \text{N}}\end{align*}

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