- #1

wahaj

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## Homework Statement

Find F, the force acting at the top of the gate.

The gate is 2m high and 2m into the page. It is hinged at the bottom (red dot in the diagram)

Density of fluid varies linearly. 1000 kg/m

^{3}at the top and 1600 kg/m

^{3}at the bottom (depth of 4m)

## Homework Equations

[tex] M_s=-\int \int \vec{r} \times \hat{n} \ P dA [/tex]

## The Attempt at a Solution

My origin is at the red dot. z is positive upwards and x is positive towards the left into the fluids

[tex] \vec{r} \times \hat{n} = z \hat{k} \times \hat{i} = z \hat{j} [/tex]

I'm going to ignore the vector part and also the negative sign of this since I only need the magnitude.

[tex] P= -\rho g z + C \ (negative\ because\ g\ is\ negative) \\

P = 0\ when\ z = 4 \\

0= -4 \rho g + C \\ C = 4 \rho g \\

P = \rho g (4 - z) \\

\rho = 1000 \ when\ z = 4 \ \ \ \rho = 1600 \ when\ z = 0 \\

\rho = mz +b \\

1000 = 4m +b \ \ \ \ \ \ \ \ \ \ \ 1600 = 0m+b \\

\rho = 1600 - 150z \\

dA = dz dy , \ limits: \ [0,2],[0,2] \\

M = g \int_0^2 \int_0^2 z(1600 - 150z)(4 - z) \ dzdy \\

M = 147.804 kN \\

147.804kN = 2F \\

F = 73.902 kN [/tex]

The actual answer is 63.1 kN. What am I doing wrong?

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