Calculating Work for Pumping Water from a Hemispherical Tank

In summary, the conversation discusses the work needed to pump water out of a hemispherical tank of radius 5 feet, filled with water weighing 62.5 pounds per cubic foot. The individual has attempted to solve this problem using calculus but has encountered difficulty. The question has been addressed in a forum dedicated to calculus and analysis, with a link provided for further discussion.
  • #1
drklrdbill
9
0
A hemispherical tank of radius 5 feet is situated so that its flat face is on top. It is full of water. Water weighs 62.5 pounds per cubic foot. The work needed to pump the water out of the lip of the tank is ? foot-pounds.


I tried evaluating the integral of pi(125x/3)(5-x)^2 from 0 to 5.

What am I doign wrong? I cannot figure this out for the life of me, spent about the last hour on a seemingly simple problem that continues to stump me.
 
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  • #3


Calculating the work for pumping water from a hemispherical tank can be a bit tricky, but it looks like you're on the right track with your integral. However, there are a few things you may be doing wrong that are causing you to get stuck.

Firstly, it's important to note that the integral you are evaluating is for the volume of the tank, not the work needed to pump the water out. In order to calculate the work, we need to consider the weight of the water being pumped out of the tank.

To do this, we can use the formula W = mgh, where W is the work, m is the mass of the water being pumped, g is the acceleration due to gravity (32.2 ft/s^2), and h is the height the water is being pumped to. In this case, h would be the radius of the tank, or 5 feet.

Next, we need to calculate the mass of the water being pumped. Since we know the weight of water per cubic foot (62.5 pounds), we can use the formula m = V * p, where m is the mass, V is the volume, and p is the density. In this case, the volume of water being pumped would be the volume of the hemispherical tank, or 2/3 * pi * r^3, where r is the radius of the tank (5 feet). So the mass of the water being pumped would be (2/3 * pi * 5^3) * 62.5 = 1304.16 pounds.

Now, we can plug in these values into the formula W = mgh to get the work needed to pump the water out of the tank. This would be (1304.16 pounds) * (32.2 ft/s^2) * (5 feet) = 210,309.12 foot-pounds.

So it looks like the work needed to pump the water out of the lip of the tank would be 210,309.12 foot-pounds. I hope this helps clarify the process for you. Keep practicing and don't get discouraged, math can be tricky at times but with practice and patience, you'll get it!
 

1. What is a work integral problem?

A work integral problem is a type of mathematical problem that involves finding the work done by a force on an object. It is typically solved using calculus, specifically integration.

2. How do you calculate the work integral?

The work integral is calculated by taking the integral of the dot product of the force and displacement vectors. This can be represented as W = ∫F · ds, where F is the force and ds is the infinitesimal displacement.

3. What is the difference between a conservative and non-conservative work integral?

A conservative work integral is one where the work done by a force is independent of the path taken, only depending on the initial and final positions. A non-conservative work integral is one where the work done can vary depending on the path taken.

4. What are some real-life applications of work integral problems?

Work integral problems are commonly used in physics and engineering to calculate the energy required to move an object, such as in designing machines and structures. They are also used in fields such as economics to calculate the amount of work done in financial transactions.

5. Are there any limitations to using work integral problems?

One limitation of work integral problems is that they assume a constant force, which may not always be the case in real-world scenarios. Additionally, they may not be useful for systems with changing or non-conservative forces. Other methods, such as using the work-energy theorem, may be more appropriate in these situations.

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