Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

Ah nice work @Nishant_Garg :)

Answer 5

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

Answer 6

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,
\[l=9m\]
\[b=4m\]

Answer 7

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

Answer 8

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
\[A=b \times h\]
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
\[F=\int\limits_{0}^{F}dF=b \rho g \int\limits_{0}^{2}h.dh\]

Answer 9

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

Answer 10

@jayzdd

Answer 11

hi

Answer 12

Hey

Answer 13

would you like help with this

Answer 14

Yasss

Answer 15

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

Answer 16

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

Answer 17

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)

Answer 18

784.8? for the force on the bottom?

Answer 19

F = Pressure * area
total force = integral rho * g *y * (dy * 5)

Answer 20

Webwork says it's not

Answer 21

right

Answer 22

are you given special values for gravity

Answer 23

No, I just assumed it would be 9.8

Answer 24

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

Answer 25

see if that is correct

Answer 26

Rho is 1? It still says it's wrong

Answer 27

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

Answer 28

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

Answer 29

Thereee we go

Answer 30

That one worked

Answer 31

which one worked?

Answer 32

705600

Answer 33

ok great

Answer 34

so for the side I would use the same thing?

Answer 35

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

Answer 36

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

Answer 37

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