anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@iambatman I'm becoming a fan just for your name
anonymous
  • 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
  • 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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anonymous
  • anonymous
Ah nice work @Nishant_Garg :)
anonymous
  • anonymous
That seems a bit more complicated than I was expecting...
anonymous
  • 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
  • anonymous
Realized how dumb that last question was. Is the force an integral with P varying with respect to h?
anonymous
  • 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
  • anonymous
So should that look like \[F = 4 * 10^3 * 9.8 \int\limits_{0}^{2} 2 dh\]
anonymous
  • anonymous
@jayzdd
anonymous
  • anonymous
hi
anonymous
  • anonymous
Hey
anonymous
  • anonymous
would you like help with this
anonymous
  • anonymous
Yasss
anonymous
  • 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
  • anonymous
Yup. Nishant helped me get the answer to the first one, I'm just trying to answer the force questions
anonymous
  • 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
  • anonymous
784.8? for the force on the bottom?
anonymous
  • anonymous
F = Pressure * area total force = integral|dw:1439786892973:dw| rho * g *y * (dy * 5)
anonymous
  • anonymous
Webwork says it's not
anonymous
  • anonymous
right
anonymous
  • anonymous
are you given special values for gravity
anonymous
  • anonymous
No, I just assumed it would be 9.8
anonymous
  • 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
  • anonymous
see if that is correct
anonymous
  • anonymous
Rho is 1? It still says it's wrong
anonymous
  • anonymous
No, I'm surprised this question would use any of this, this is from a calc class, not physics
anonymous
  • anonymous
oh woops, i thought it was 1 kg /m^3
anonymous
  • anonymous
Thereee we go
anonymous
  • anonymous
That one worked
anonymous
  • anonymous
which one worked?
anonymous
  • anonymous
705600
anonymous
  • anonymous
ok great
anonymous
  • anonymous
so for the side I would use the same thing?
anonymous
  • anonymous
they did F = 1000 * 9.8 * 2 * (9 * 4) = 705600
anonymous
  • anonymous
for the side, we are asking for the total force on the side of the window, so to speak, of the aquarium
anonymous
  • 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
  • anonymous
1000*9.8*2*(2*4) ? does the depth stay in the equation?
anonymous
  • anonymous
we have to integrate now
anonymous
  • anonymous
Oh okay, here's the integration
anonymous
  • anonymous
|dw:1439787788080:dw|
anonymous
  • anonymous
In that graphic, where does the 5 come from?
anonymous
  • anonymous
F_i = P*A_i $$ \Large \int \rho g y \cdot A \\ \Large \int \rho g y ( 4 \cdot dy) $$
anonymous
  • anonymous
|dw:1439787946825:dw|
anonymous
  • 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
  • anonymous
Thanks @jayzdd !!! You were really helpful!
anonymous
  • anonymous
did you get 78400
anonymous
  • anonymous
yeah
anonymous
  • anonymous
you can also write the integral as \[ \Large \int_{0}^{2} 1000 \cdot 9.8 (2-y) ( 4 \cdot dy) \]
anonymous
  • anonymous
Thanks man! I'm gonna head to bed now
anonymous
  • anonymous
Cheers :)
anonymous
  • anonymous
*Or woman since I don't actually know*

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