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

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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

Mathematics
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@iambatman I'm becoming a fan just for your name
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
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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Ah nice work @Nishant_Garg :)
That seems a bit more complicated than I was expecting...
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, \[l=9m\] \[b=4m\]
Realized how dumb that last question was. Is the force an integral with P varying with respect to h?
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 \[F=\int\limits_{0}^{F}dF=b \rho g \int\limits_{0}^{2}h.dh\]
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
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
Yup. Nishant helped me get the answer to the first one, I'm just trying to answer the force questions
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)
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
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
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
for the side, we are asking for the total force on the side of the window, so to speak, of the aquarium
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
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|
the force on a horizontal slab along the side end of the aquarium is F_i = P_i*A_i \[ \Large \int \rho g y \cdot A \\ \Large \int_{0}^{2} 1000 \cdot 9.8 y ( 4 \cdot dy) \]
Thanks @jayzdd !!! You were really helpful!
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*

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