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@iambatman I'm becoming a fan just for your name

Ah nice work @Nishant_Garg :)

That seems a bit more complicated than I was expecting...

Realized how dumb that last question was. Is the force an integral with P varying with respect to h?

So should that look like \[F = 4 * 10^3 * 9.8 \int\limits_{0}^{2} 2 dh\]

hi

Hey

would you like help with this

Yasss

784.8? for the force on the bottom?

F = Pressure * area
total force = integral|dw:1439786892973:dw| rho * g *y * (dy * 5)

Webwork says it's not

right

are you given special values for gravity

No, I just assumed it would be 9.8

see if that is correct

Rho is 1? It still says it's wrong

No, I'm surprised this question would use any of this, this is from a calc class, not physics

oh woops, i thought it was 1 kg /m^3

Thereee we go

That one worked

which one worked?

705600

ok great

so for the side I would use the same thing?

they did
F = 1000 * 9.8 * 2 * (9 * 4) = 705600

1000*9.8*2*(2*4) ? does the depth stay in the equation?

we have to integrate now

Oh okay, here's the integration

|dw:1439787788080:dw|

In that graphic, where does the 5 come from?

F_i = P*A_i
$$
\Large \int \rho g y \cdot A
\\ \Large \int \rho g y ( 4 \cdot dy)
$$

|dw:1439787946825:dw|

did you get 78400

yeah

you can also write the integral as
\[
\Large \int_{0}^{2} 1000 \cdot 9.8 (2-y) ( 4 \cdot dy)
\]

Thanks man! I'm gonna head to bed now

Cheers :)

*Or woman since I don't actually know*