# Hydrostatics pressure forces

I can deduce the Forces for F1 and F2:

\begin{align} F_1 &= \dfrac{1}{2}\rho g d_1^2 \\ &= \dfrac{1}{2} \cdot (1000 \cdot 9.81) \cdot d_1^2 \cdot 8 \\ \dfrac{F_1}{d_1^2} &= 39240\text{ N} \end{align}

I've moved the $$d_1^2$$ over to left and worked out the rest.

\begin{align} F_2 &= \dfrac{1}{2}\rho g \dfrac{d_2^2}{2} \cdot 8 \\ &= \dfrac{1}{2} \cdot (1000 \cdot 9.81) \cdot \dfrac{d_2^2}{2} \cdot 8 \\ 2\dfrac{F_2}{d_2^2} &= 39240\text{ N} \end{align}

Any ideas on solving b and c?

• Note that the 39240 isn't in Newtons, but in $\text{N/m}^2$ (since it's a force $F_1$ divided by the square of a distance $d_1$). Only after multiplying 39240 by that squared distance will you get the result in Newtons. Also, shouldn't the second fraction for $F_2$ be with $d_1$ instead of $d_2$, since $d_2$ is stated as equal to half of $d_1$?
– Wasabi
Jan 31, 2019 at 2:11

$$F_1= 4*F_2$$ because the thrust is related to h squared.
So they create a resultant acting at the height of (D1/3 - D1/(6*4) = 7/24D1. The resultant is $$F_{final}= 8*(D_1^2/2-D_{1}^2/8)= 4*D_1^2(3/4)*9.8$$