# Hydrostatic forces

1. Dec 20, 2009

I need to find the hydrostatic force exerted on a plane submerged vertically in water. I attached a diagram of the problem.

Here are the basic definitions:
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d=distance from surface, p=density, P=pressure

$$p=\frac{m}{V}$$

$$P=pgd=\delta d$$

$$F=mg=pgAd$$

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The area of the ith strip is $$A_i=6\Delta y$$ so the pressure exerted on the ith strip is $$\delta d_i=pgd_i=pg(6-y_i^*)$$

The hydrostatic force on the ith strip is $$F_i=\delta_iA_i=6pg(6-y_i)\Deltay$$

The approximate force along the entire surface is therefore:

$$F_{net}=\lim_{n-\infty}\Sigma_{i=1}^n6pg(6-y_i)\Delta y$$

$$=6pg\int_0^4(6-y)dy$$

Am I setting this up correctly?

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2. Dec 20, 2009

### rock.freak667

It looks like you are doing it correctly from first principles, but I think this line should be (not sure on notation but this is how I saw a similar summation in a math book)

$$F_{net}=\lim_{\Delta y \rightarrow 0} \sum_{y=0} ^{y=4} 6pg(6-y_i)\Delta y$$

3. Dec 20, 2009

I was writing out the limit of the Riemann sum. There are $$n$$ subdivisions and $$\Delta y=\frac{4-0}{n}$$. So I think what you wrote was equivalent to the Riemann sum.

4. Dec 20, 2009

### rock.freak667

It probably is, I was never taught the Riemann Sum, but you are correct though.

5. Dec 20, 2009

### ideasrule

You don't need to write out the Riemann sum; that just makes things unnecessarily complicated. I find that going directly to Fnet$=6pg\int_0^4(6-y)dy$ is much easier and more intuitive. (BTW, that integral gives the exact force on the plate, not the approximate force.)