Hydrostatic pressure on triangular plate

revolve
Messages
19
Reaction score
0

Homework Statement



A triangle with base 3 m and height 4 m is submerged vertically in water so that the tip is even with the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it.

Homework Equations



<br /> \int_{a}^{b}{\rho}g(x)w(x)dx<br />

The Attempt at a Solution



I keep getting the wrong answer. Does it look like I'm setting it up right?

Using similar triangles

I have a = \frac{3}{4\sqrt{2}}\left(4-xi^{*}\right)

wi^{*}=2\left(\frac{\sqrt{3}}{2}-a\right)

a\mbox{re}a=wi\; \Delta x=\left( \sqrt{3}\; -\; 3\frac{\sqrt{2}}{2}+\frac{3}{4\sqrt{2}}x \right)

Pressure = 1000gx

Force = P*A
 

Attachments

  • hydrostatic pressure.GIF
    hydrostatic pressure.GIF
    7 KB · Views: 2,091
Physics news on Phys.org
No. I'm not sure where you're getting the sqrt(3)/2 from in your formulas for a and w. Also, why not use similar triangles to find w directly as a function of x instead of going through the mess with a?
 
a=3/8(4-xi)

Sorry, I don't know where I got that either. I must need some rest.
 
Area=3/4xi(delta x)

Does this look right?
 
Looks good.
 
Got it. Thank you for your help.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top