What is the required work to pump water out of a spherical tank?

In summary, to pump water out of a spout, you would need to do work equal to 1000* pi*(9-y^{2})*(7-y)*y. This would be negative if you tried to integrate from 0 to 6. If you chose y from the bottom of the tank, the distance to the top would be 7-y.
  • #1
RedBarchetta
50
1
[SOLVED] Work: Spherical Problem

Homework Statement


A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 for g and 3.14 for [PLAIN]http://www.webassign.net/images/pi.gif.[/URL] Round your answer to three significant digits.) W=_________

r=3
h=1
6-4-022alt.gif


The Attempt at a Solution


To set up this problem, I started by taking the area of one slice of water to be pi r[tex]^{2}[/tex] multiplied by an infitesimally small height [tex]\Delta y[/tex] to get volume. Then multiply this by the density of water; 1000 kg/m[tex]^{3}[/tex] to get the volume of one slice.

Now I want R as a function of, let's say y. We know the area of a circle is r[tex]^{2}[/tex]2+y[tex]^{2}[/tex]=3[tex]^{2}[/tex]. So r=sqrt(9-y[tex]^{2}[/tex]). Also, the distance for any slice from the top is 7-y.

After this I tried integrating from zero to six of the function 1000*pi*(9-y[tex]^{2}[/tex])*(7-y)*y. This came out to be negative...in fact -36000*pi..

Apparently, that is wrong. Where did I go wrong?

Thanks.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Your expression for R as a function of y is wrong.
 
  • #3
The above equation was really meant to be r[tex]^{2}[/tex]+y[tex]^{2}[/tex]=3[tex]^{2}[/tex]. I just noticed the typo. Or, are you saying that's wrong?
 
  • #4
Be careful about what y represents. [itex]r= \sqrt{9- y^2}[/itex] is correct if is the height above the center of the sphere. But in that case, integrating from 0 to 6 is wrong. Integrating from the bottom of the tank to the top takes y from -3 to 3.
 
Last edited by a moderator:
  • #5
I integrated and got 2469600*pi as my answer. My homework checker told me I was wrong. I even rounded to the proper amount of sig figs(3).

Heres what I have:
dA=(9-y[tex]^{2}[/tex])pi
(Multiplied by infitesimally thickness dy)
dV=(9-y[tex]^{2}[/tex])pi dy
(Since d=mv, I multiplied density of water (1000kg/m[tex]^{3}[/tex])
dM=1000(9-y[tex]^{2}[/tex])pi dy
(multiplied by gravity 9.8 m/s[tex]^{2}[/tex])
dF=9800(9-y[tex]^{2}[/tex])pi dy
(Now multiplied by distance (7-y))
dW=9800(9-y[tex]^{2}[/tex])(7-y) dyNow I integrated dW with the limits from -3 to 3 and got the above result. I'm lost. :confused: Thanks for the help Ivy.
 
  • #6
The distance to the top would be 7 - y if you chose y from the bottom. It appears that you have chosen y as the distance from the center.
 
  • #7
Great! 4-y worked.

I can finally put this problem to rest. Thanks for the help everyone!
 
Last edited:

1. What is "Work: Spherical Problem"?

"Work: Spherical Problem" refers to a mathematical problem that involves calculating the amount of work needed to move an object from one point on the surface of a sphere to another point on the surface of the same sphere.

2. What is the formula for solving the "Work: Spherical Problem"?

The formula for solving the "Work: Spherical Problem" is W = F * d * cos(theta), where W is the work done, F is the force applied, d is the displacement, and theta is the angle between the force and the displacement vectors.

3. What are some real-life applications of the "Work: Spherical Problem"?

The "Work: Spherical Problem" is commonly used in physics and engineering to calculate the work done in moving objects on curved surfaces such as spheres. It also has applications in astronomy and navigation, where the movement of objects on the surface of planets or stars is involved.

4. How does the "Work: Spherical Problem" differ from the "Work: Linear Problem"?

The "Work: Spherical Problem" involves calculating the work done on a curved surface, while the "Work: Linear Problem" involves calculating the work done on a straight line. The formula for the "Work: Spherical Problem" takes into account the angle between the force and displacement vectors, while the formula for the "Work: Linear Problem" does not.

5. Are there any limitations to using the "Work: Spherical Problem" formula?

Yes, the "Work: Spherical Problem" formula assumes that the force and displacement are constant and that there is no friction or other external factors affecting the movement of the object on the surface of the sphere. In real-life scenarios, these assumptions may not hold true and can affect the accuracy of the calculated work.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
1K
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
547
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
675
  • Calculus and Beyond Homework Help
Replies
1
Views
6K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
57K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Back
Top