A community for students.
Here's the question you clicked on:
 0 viewing
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
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

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@iambatman I'm becoming a fan just for your name

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Pressure 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
 one year ago
Best ResponseYou've already chosen the best response.0Thank 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah nice work @Nishant_Garg :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That seems a bit more complicated than I was expecting...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Really 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
 one year ago
Best ResponseYou've already chosen the best response.0Realized how dumb that last question was. Is the force an integral with P varying with respect to h?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0To 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
 one year ago
Best ResponseYou've already chosen the best response.0So should that look like \[F = 4 * 10^3 * 9.8 \int\limits_{0}^{2} 2 dh\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would you like help with this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let 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
 one year ago
Best ResponseYou've already chosen the best response.0Yup. Nishant helped me get the answer to the first one, I'm just trying to answer the force questions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since 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
 one year ago
Best ResponseYou've already chosen the best response.0784.8? for the force on the bottom?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0F = Pressure * area total force = integraldw:1439786892973:dw rho * g *y * (dy * 5)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Webwork says it's not

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0are you given special values for gravity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, I just assumed it would be 9.8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0see if that is correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Rho is 1? It still says it's wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, I'm surprised this question would use any of this, this is from a calc class, not physics

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh woops, i thought it was 1 kg /m^3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for the side I would use the same thing?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0they did F = 1000 * 9.8 * 2 * (9 * 4) = 705600

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for the side, we are asking for the total force on the side of the window, so to speak, of the aquarium

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since 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
 one year ago
Best ResponseYou've already chosen the best response.01000*9.8*2*(2*4) ? does the depth stay in the equation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we have to integrate now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay, here's the integration

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439787788080:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In that graphic, where does the 5 come from?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0F_i = P*A_i $$ \Large \int \rho g y \cdot A \\ \Large \int \rho g y ( 4 \cdot dy) $$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439787946825:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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
 one year ago
Best ResponseYou've already chosen the best response.0Thanks @jayzdd !!! You were really helpful!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can also write the integral as \[ \Large \int_{0}^{2} 1000 \cdot 9.8 (2y) ( 4 \cdot dy) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks man! I'm gonna head to bed now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0*Or woman since I don't actually know*
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.