Question

A large aquarium of height 5.00 m is filled with fresh water to a depth of 2.00 m. One wall of the aquarium consists of thick plastic 8.00 m wide. By how much does the total force on that wall increase if the aquarium is next filled to a depth of 4.00 m

Answer 1

answer given is 4.69 × 10^5 N

Answer 2

the force acting on the wall will not be constant
the force acting on differential element of height say dh will be -\[dF=p \times (b \times dh)\]
b=base length
p=pressure at base
\[dF=\rho gh(b dh)\]\[dF=\rho gbhdh\]
rho=density of liquid
now we can integrate and to limits till where we want to find the force in both the cases.

Answer 3

Awesome ! let me setup the bounds :)

Answer 4

(:

Answer 5

Force when the aquarium is filled to a depth of 2m :

\[ F = 8\int\limits_0^2\rho g h \,dh = 8\rho g \dfrac{h^2}{2} \Bigg|_0^2 = 16\rho g\]

Answer 6

Force when the aquarium is filled to a depth of 4m :

\[ F = 8\int\limits_0^4\rho g h \,dh = 8\rho g \dfrac{h^2}{2} \Bigg|_0^4 = 64\rho g\]

Answer 7

do they look good ?

Answer 8

increase in force = \(64\rho g-16\rho g = 48\rho g \)

Answer 9

well \[F=\rho gb \int\limits_{0}^{H}hdh\]\[F=\frac{ 1 }{ 2 }\rho bgH^2\]

density is not given right?

Answer 10

what we did is correct tho

Answer 11

i can google for fresh water density..

Answer 12

lol i forgot we use water in aquariums

Answer 13

google says \(\rho\) = 983.854 kg/m^3

Answer 14

yeah putting them in we get this->462804.922=4.6 x10^5

Answer 15

wonderful ! thanks !

Answer 16

yw (:

Answer 17

guess thats a huge foce! more than 10% of the force due to atmospheric pressure :

force due to atmospheric pressure = pA = 10^5*(8*4) = 32*10^5

Answer 18

i wonder how much will be the blood pressure of fishes located deep in the ocean (:

Answer 19

lets caculate

\(p = p_0 + \rho g h \)

letting \(h=3000\)

\(p = p_0 + 10^3*10*3000 \)

Answer 20

lol thats too much

Answer 21

\(p = p_0 + 3*10^6 = 31p_0\)

the pressure is 31 times greater than the atmospheric pressure at a depth 3000 meters !

Answer 22

i have no idea how fish or anything could possibly survive at such huge pressures...

Answer 23

they might have strong cardiac muscles to pump blood at high pressure

Answer 24

wow did i miss counting a 0 ?

Answer 25

\(p = p_0 + 3*10^7 = 310p_0\)

the pressure is 310 times greater than the atmospheric pressure at a depth 3000 meters !

now that is really huge

Answer 26

yes
it will be 3x10^7+10^5

Answer 27

how much work will be done by the heart to pump blood at this high pressure

Answer 28

pressure=force/area
something over 10^7 =force/(surface area of fish)

surface area of fish lets say 10m^2
so it will be like
F=10^8N

W=FS
hmm now what

Answer 29

shuld we take S=length of arteries or veins?

Answer 30

I think they don't have to do any extra work if their internal pressure is also of that order

Answer 31

but what will create that internal pressure?

Answer 32

fishes have no air inside them nor their skin is so tough

Answer 33

These animals have evolved to survive the extreme pressure of the sub-photic zones. The pressure increases by about one bar every ten meters. To cope with the pressure, many fish are rather small, usually not exceeding 25 cm in length. Also, scientists have discovered that the deeper these creatures live, the more gelatinous their flesh and more minimal their skeletal structure. These creatures have also eliminated all excess cavities that would collapse under the pressure, such as swim bladders.[1]

Answer 34

https://en.wikipedia.org/wiki/Deep_sea_creature#Barometric_pressure

Answer 35

btw the calculation of pressure is wrong :)

Answer 36

the gravity at 3000m below sea level and at sea level varies greatly

Answer 37

thats a good point, but our calculations are still pessimistic... gravity increases till it reaches the dense core part of the earth center... so whatever number we got isn't an overestimate..

Answer 38

yes (:

Answer 39

once you go below the surface of the earth, gravity decreases linearly to zero at the centre. the water that is above you is pulling you back up.

given how far you are away from the centre, i would not be surprised to find that your calcs are v accurate.

my question is this....\(dF=(\rho g \, b \, h \color{red}{+ P_{atm}}) \, dh\) ?

my sense is that if you follow this through \(P_{atm}\) is small change.