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hydrostatics image

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?

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  • $\begingroup$ 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 $d_1$? $\endgroup$
    – Wasabi
    Jan 31 '19 at 2:11
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$ F_1= 4*F_2 $ because the thrust is related to h squared.

The forces acting on the gate are -8*F1 at the height of D1/3 and (+8/4)*F1 = 2*F1 acting at 1/6*D1.

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 $

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Magnitude of resultant is the algebraic sum.There are expressions to find the depth of the points F1 and F2 (centre of pressures).You have the magnitude of the three forces (F1,F2 and resultant), assume right direction for resultant.Then apply the concept that "the sum of moments due to individual forces equal to the moment due to the resultant force".

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  • $\begingroup$ @hazzey I agree with you.But dont delete the post (I know that its not your decision, its based on voting) because I might never know my mistake.Instead downvote and if possible comment on it. $\endgroup$
    – user17332
    Jan 31 '19 at 18:05

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