Find the work required to pump all the water

In summary: It's from the height of the triangle- the base of the triangle is at 1, so \sqrt{3} is the height of the triangle multiplied by 3.
  • #1
nameVoid
241
0
the ends of an 8 foot long water trough are equilateral triangles having sides of length 2 feet. If the trough is full of water, find the work required to pump all the water over the top.
 
Physics news on Phys.org
  • #2


possible trapazoid section? area?
 
  • #3


I presume the trough has the vertex of the triangle down! Don't just try to remember formulas- think about what work means. The work done to lift a "piece of water" out of the trough is the weight of the piece times the height it must be lifted: the distance from the top base of the triangle to the piece. We can handle all "pieces of water" that have that same distance at once by thinking of a "layer of water" at a given distance from the top base. That will be a rectanglar layer having length 8 feet, depth dx, and width w, depending on what x is. That is, its volume will be 8w(x)dx and so its weight will be [itex]\gamma (8w(x)dx)[/itex] where [itex]\gamma[/itex] is the density of water (look it up) in pounds per cubic foot. That layer has to be lifted x feet to the top of the trough so the work done is [itex]8\gamma w(x)x dx[/itex]. Integrate that from 0 to the height of the triangle.

Now, determining w(x). Draw a picture: an equilateral triangle with vertex at the bottom, base at the height and a horizontal line x units below the base. The length of that line is w(x). "Similar triangles" is a good way to find its length. Another way would be to find the equations of the two lines, both through (0, 0) and one through [itex](1, \sqrt{3})[/itex], the other through [itex](-1, \sqrt{3})[/itex]. Be careful what you call "x" and what "w". Do you see where I got the [itex]\sqrt{3}[/itex]?
 

Related to Find the work required to pump all the water

1. How do you calculate the work required to pump all the water?

The work required to pump all the water can be calculated by multiplying the force required to move the water by the distance it needs to be pumped.

2. What is the force required to move the water?

The force required to move the water is equal to the weight of the water being pumped. This can be calculated by multiplying the density of water (1000 kg/m3) by the volume of water being pumped.

3. What is the distance that needs to be pumped?

The distance that needs to be pumped is the vertical height that the water needs to be lifted, which is also known as the "head". This can be measured in meters or feet.

4. What is the unit of measurement for work required to pump all the water?

The unit of measurement for work required to pump all the water is joules (J). However, in practical applications, it is often measured in kilojoules (kJ) or megajoules (MJ).

5. Can the efficiency of the pump affect the work required to pump all the water?

Yes, the efficiency of the pump can affect the work required to pump all the water. A more efficient pump will require less work to pump the same amount of water, while a less efficient pump will require more work.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Replies
1
Views
2K
Replies
40
Views
4K
Replies
2
Views
456
Replies
32
Views
4K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
3K
  • Other Physics Topics
Replies
30
Views
996
Back
Top