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anonymous

  • one year ago

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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  1. anonymous
    • one year ago
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    @iambatman I'm becoming a fan just for your name

  2. anonymous
    • one year ago
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    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

  3. anonymous
    • one year ago
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    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

  4. anonymous
    • one year ago
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    Ah nice work @Nishant_Garg :)

  5. anonymous
    • one year ago
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    That seems a bit more complicated than I was expecting...

  6. anonymous
    • one year ago
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    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, \[l=9m\] \[b=4m\]

  7. anonymous
    • one year ago
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    Realized how dumb that last question was. Is the force an integral with P varying with respect to h?

  8. anonymous
    • one year ago
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    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 \[F=\int\limits_{0}^{F}dF=b \rho g \int\limits_{0}^{2}h.dh\]

  9. anonymous
    • one year ago
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    So should that look like \[F = 4 * 10^3 * 9.8 \int\limits_{0}^{2} 2 dh\]

  10. anonymous
    • one year ago
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    @jayzdd

  11. anonymous
    • one year ago
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    hi

  12. anonymous
    • one year ago
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    Hey

  13. anonymous
    • one year ago
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    would you like help with this

  14. anonymous
    • one year ago
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    Yasss

  15. anonymous
    • one year ago
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    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

  16. anonymous
    • one year ago
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    Yup. Nishant helped me get the answer to the first one, I'm just trying to answer the force questions

  17. anonymous
    • one year ago
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    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)

  18. anonymous
    • one year ago
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    784.8? for the force on the bottom?

  19. anonymous
    • one year ago
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    F = Pressure * area total force = integral|dw:1439786892973:dw| rho * g *y * (dy * 5)

  20. anonymous
    • one year ago
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    Webwork says it's not

  21. anonymous
    • one year ago
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    right

  22. anonymous
    • one year ago
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    are you given special values for gravity

  23. anonymous
    • one year ago
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    No, I just assumed it would be 9.8

  24. anonymous
    • one year ago
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    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

  25. anonymous
    • one year ago
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    see if that is correct

  26. anonymous
    • one year ago
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    Rho is 1? It still says it's wrong

  27. anonymous
    • one year ago
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    No, I'm surprised this question would use any of this, this is from a calc class, not physics

  28. anonymous
    • one year ago
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    oh woops, i thought it was 1 kg /m^3

  29. anonymous
    • one year ago
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    Thereee we go

  30. anonymous
    • one year ago
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    That one worked

  31. anonymous
    • one year ago
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    which one worked?

  32. anonymous
    • one year ago
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    705600

  33. anonymous
    • one year ago
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    ok great

  34. anonymous
    • one year ago
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    so for the side I would use the same thing?

  35. anonymous
    • one year ago
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    they did F = 1000 * 9.8 * 2 * (9 * 4) = 705600

  36. anonymous
    • one year ago
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    for the side, we are asking for the total force on the side of the window, so to speak, of the aquarium

  37. anonymous
    • one year ago
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    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

  38. anonymous
    • one year ago
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    1000*9.8*2*(2*4) ? does the depth stay in the equation?

  39. anonymous
    • one year ago
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    we have to integrate now

  40. anonymous
    • one year ago
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    Oh okay, here's the integration

  41. anonymous
    • one year ago
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    |dw:1439787788080:dw|

  42. anonymous
    • one year ago
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    In that graphic, where does the 5 come from?

  43. anonymous
    • one year ago
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    F_i = P*A_i $$ \Large \int \rho g y \cdot A \\ \Large \int \rho g y ( 4 \cdot dy) $$

  44. anonymous
    • one year ago
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    |dw:1439787946825:dw|

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

  46. anonymous
    • one year ago
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    Thanks @jayzdd !!! You were really helpful!

  47. anonymous
    • one year ago
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    did you get 78400

  48. anonymous
    • one year ago
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    yeah

  49. anonymous
    • one year ago
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    you can also write the integral as \[ \Large \int_{0}^{2} 1000 \cdot 9.8 (2-y) ( 4 \cdot dy) \]

  50. anonymous
    • one year ago
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    Thanks man! I'm gonna head to bed now

  51. anonymous
    • one year ago
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    Cheers :)

  52. anonymous
    • one year ago
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    *Or woman since I don't actually know*

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