Integration question:
An aquarium 9 m long, 4 m wide, and 2 m deep is full of water. Find the following: hydrostatic pressure on the bottom of the aquarium, hydrostatic force on the bottom of the aquarium, and hydrostatic force on one end of the aquarium

- anonymous

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- anonymous

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

- anonymous

Pressure varies with depth as
\[P(h)=h \rho g\]
Where g is acceleration due to gravity and rho is density of the liquid
Since the liquid is water, we know that it's density is 1g/cm^3 or
\[\rho=\frac{1g}{cm^3}=\frac{10^{-3}Kg}{(10^{-2}m)^3}=\frac{10^{-3}Kg}{10^{-6}m^3}=\frac{1000Kg}{m^3}\]
\[g=9.8ms^{-2}\]
and h is given as \[h=2m\]
Using this u can calculate the pressure
Force can be calculated as
\[P=\frac{F}{A}\]
Where A is the area of the bottom surface of the tank
\[\therefore F=P \times A=h \rho g \times A=\rho \times g \times (l \times b \times h)=\rho g V\]
V=volume of tank

- anonymous

Thank you @RunawayGalaxy this is a fun problem unfortunately I do not have time to go over this right now but I will help you out later when I do have time. :) You need to know that force = pressure x area

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## More answers

- anonymous

Ah nice work @Nishant_Garg :)

- anonymous

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

- anonymous

Really all you have to do is plug in
rho h and g
in the equation
\[P=h \rho g\]
\[h=2m\]
\[\rho=10^3Kgm^{-3}\]
\[g=9.8ms^{-2}\]
With this you can calculate the pressure
Now \[F=P \times A=P \times l \times b\]
The area of the bottom floor of the tank is given by length times breadth,
\[l=9m\]
\[b=4m\]

- anonymous

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

- anonymous

To find force on one end of the tank, notice that
\[F=P \times A\]
Now the area of the surface on the side of the tank is given by breadth times height and is constant
\[A=b \times h\]|dw:1439782096654:dw|
But from the bottom of the tank to the top, the pressure is varying with height
\[P(h)=h \rho g\]
Consider a small height element dh,
for which the small pressure is given by
\[dP=\rho g.dh\]
Then the small force is given by
\[dF=b \times h \times \rho \times g \times dh\]
\[dF=b \rho g(h.dh)\]
Now the height is varying from 0 to 2m in the tank
so integrating within the limits 0 to 2 we can find the force
\[F=\int\limits_{0}^{F}dF=b \rho g \int\limits_{0}^{2}h.dh\]

- anonymous

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

- anonymous

@jayzdd

- anonymous

hi

- anonymous

Hey

- anonymous

would you like help with this

- anonymous

Yasss

- anonymous

let me do a quick refresher
from physics we know
pressure = force / area
here the force is due to weight
force = mass * gravity constant
mass = density * volume
volume = Area * height
P = (m*g) / A = (rho * Volume* g) / A = (rho * A* height * g) / A = rho * h * g

- anonymous

Yup. Nishant helped me get the answer to the first one, I'm just trying to answer the force questions

- anonymous

since P = 1 kg/m^3 (9.81 ) * 2 m = 19.62 N/m^2 (also known as pascals)
P = F / A
Area* P = F
5*8* 19.62 = 784.8 Newtons force on bottom
for the force on the side, I see the next part we have to integrate (catching up here)

- anonymous

784.8? for the force on the bottom?

- anonymous

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

- anonymous

Webwork says it's not

- anonymous

right

- anonymous

are you given special values for gravity

- anonymous

No, I just assumed it would be 9.8

- anonymous

I see, I used different values for the dimensions of the aquarium
should be the dimensions are
9 m long, 4 m wide, and 2 m
Force on bottom = P*A
F = rho * g * height * Area of base
F = 1 * 9.81 * 2 * (9 * 4) = 706.32

- anonymous

see if that is correct

- anonymous

Rho is 1? It still says it's wrong

- anonymous

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

- anonymous

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

- anonymous

Thereee we go

- anonymous

That one worked

- anonymous

which one worked?

- anonymous

705600

- anonymous

ok great

- anonymous

so for the side I would use the same thing?

- anonymous

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

- anonymous

for the side, we are asking for the total force on the side of the window, so to speak, of the aquarium

- anonymous

since pressure acts the same in all directions, it is going to act on the side of the aquarium the same as it acts on the floor, but it changes by depth

- anonymous

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

- anonymous

we have to integrate now

- anonymous

Oh okay, here's the integration

- anonymous

|dw:1439787788080:dw|

- anonymous

In that graphic, where does the 5 come from?

- anonymous

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

- anonymous

|dw:1439787946825:dw|

- anonymous

the force on a horizontal slab along the side end of the aquarium is
F_i = P_i*A_i
\[
\Large \int \rho g y \cdot A
\\ \Large \int_{0}^{2} 1000 \cdot 9.8 y ( 4 \cdot dy)
\]

- anonymous

Thanks @jayzdd !!! You were really helpful!

- anonymous

did you get 78400

- anonymous

yeah

- anonymous

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

- anonymous

Thanks man! I'm gonna head to bed now

- anonymous

Cheers :)

- anonymous

*Or woman since I don't actually know*

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