1. ## hydrostatic

Here is another hydrostatic force problem, I hope this isn't double posting. It's not an identical thread, so it shoudn't be right? Well here it is:
A tank is 7 m long, 5 m wide, 3 m high, and contains kerosene with density 820 kg/m3 to a depth of 2.5 m. (Give your answers correct to three significant figures.)
It first asks to give the hydrostatic pressure at the bottom of the tank. Isn't that just 820*9.8*2.5?? Thats what I thought, so I put down for my answer 2.01*10^4 is that 3 sig figs? but it was wrong.
Then, it asks for the hydrostatic force at the bottom of the tank.
I think you pull out the 820*9.8 and you have the integral from 0 to 2.5 of (x* some area, I don't know how to find that)
Finally, it asks for the hydrostatic force on one end of the tank, which I'm not sure about. Whats the difference between this question and the previous one?
Thanks!

2. The side of the tank is a rectangle. That is lot better than some wacky shaped surface. For real, who ever seen a tank with a parabolic side. Come on.

The force on a side:

The tank is 3 feet deep and filled to 2.5 feet. It has width 5 and length 7

820(5)(7)=28700

Anyway, the rectangle maintains a constant width.

$28700\int_{0}^{\frac{5}{2}}(3-y)dy$

3. haha Yeah exactly!! Okay, well so the pressure would be 28700? And I integrated the integral for the end of the tank hydro. force and got 1.26*10^5 and it was wrong!! What did I do!! I just integrated adding an x to the 3 and a squared to the y(and dividing y by 2). I think I'm typing in the answers wrong.

4. Did you get 125562.5?.

Is the tank full or filled to 2.5 feet when they ask for the force on the side?.

5. Yes, I got that, but I typed it in as 2.56*10^5 because it wants it to 3 sigfigs!

6. I think the kerosene is filled to a depth of 2.5 So it's filled, to answer your above question (sorry I didn't see it!)