## 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

1. anonymous

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

2. 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

3. 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

4. anonymous

Ah nice work @Nishant_Garg :)

5. anonymous

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

6. 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, $l=9m$ $b=4m$

7. anonymous

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

8. 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 $F=\int\limits_{0}^{F}dF=b \rho g \int\limits_{0}^{2}h.dh$

9. anonymous

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

10. anonymous

@jayzdd

11. anonymous

hi

12. anonymous

Hey

13. anonymous

would you like help with this

14. anonymous

Yasss

15. 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

16. anonymous

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

17. 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)

18. anonymous

784.8? for the force on the bottom?

19. anonymous

F = Pressure * area total force = integral|dw:1439786892973:dw| rho * g *y * (dy * 5)

20. anonymous

Webwork says it's not

21. anonymous

right

22. anonymous

are you given special values for gravity

23. anonymous

No, I just assumed it would be 9.8

24. 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

25. anonymous

see if that is correct

26. anonymous

Rho is 1? It still says it's wrong

27. anonymous

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

28. anonymous

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

29. anonymous

Thereee we go

30. anonymous

That one worked

31. anonymous

which one worked?

32. anonymous

705600

33. anonymous

ok great

34. anonymous

so for the side I would use the same thing?

35. anonymous

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

36. anonymous

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

37. 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

38. anonymous

1000*9.8*2*(2*4) ? does the depth stay in the equation?

39. anonymous

we have to integrate now

40. anonymous

Oh okay, here's the integration

41. anonymous

|dw:1439787788080:dw|

42. anonymous

In that graphic, where does the 5 come from?

43. anonymous

F_i = P*A_i $$\Large \int \rho g y \cdot A \\ \Large \int \rho g y ( 4 \cdot dy)$$

44. anonymous

|dw:1439787946825:dw|

45. anonymous

the force on a horizontal slab along the side end of the aquarium is F_i = P_i*A_i $\Large \int \rho g y \cdot A \\ \Large \int_{0}^{2} 1000 \cdot 9.8 y ( 4 \cdot dy)$

46. anonymous

Thanks @jayzdd !!! You were really helpful!

47. anonymous

did you get 78400

48. anonymous

yeah

49. anonymous

you can also write the integral as $\Large \int_{0}^{2} 1000 \cdot 9.8 (2-y) ( 4 \cdot dy)$

50. anonymous

Thanks man! I'm gonna head to bed now

51. anonymous

Cheers :)

52. anonymous

*Or woman since I don't actually know*