Calculate Hydrostatic Force Exerted on Plane Submerged in Water

[AdkinsJr]

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:
---------------------------
d=distance from surface, p=density, P=pressure

p=\frac{m}{V}

P=pgd=\delta d

F=mg=pgAd

---------------------------

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?

 

[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

 

[AdkinsJr]

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.

 

[rock.freak667]

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.

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

 

[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.)