# Logic behind the Lifting Water problem

• Vlykarye
In summary, the conversation discusses the logic behind calculating the work required to move all water out of a tank using a water pump. The user must take into consideration the volume of water at sub-intervals of depth and the height the water is moving. There are two methods presented, one where the pump is constantly moving to adjust to the decreasing water level and one where the pump stays in a fixed position. The latter method is more practical in the real world and takes into account the siphon effect.
Vlykarye

## Homework Statement

I am trying to understand the logic behind the water pump that requires a user to take into consideration the volume of water at sub-intervals of depth of a tank. There are two types of problems, one where the volume changes at different depths, and one where it doesn't. In both cases, the user also has to consider the height the volume of water is moving. This height also seems to change at different depths of the tank. The end result is the calculation of work required to move all water out of the tank.

Ex: There is cylindrical tank with a height of 10 meters and radius of 5 meters. The tank is standing on its base, so the volume of water does not change at different depths. The volume is described by the equation: $A(y) = \pi r^2 \delta h$ where the height is approaching 0 and is dealt with through integration.

I do not need help solving this word problem. Rather, I need help understanding the logic behind it. We are using a water pump to move the water inside the cylindrical tank. Now, the most logical way one could visualize this problem is to imagine a pump that has a hose reaching to the bottom of the tank. In this case, the height the water must travel is constant (the length of the hose). However, the example uses a different method. So far, I have started a theory that the pump is moving along the height of the tank, in order to constantly be touching the uppermost sub-interval of water in the tank.

I don't understand the logic behind this method, because such an application seems impractical in the real world. I have come to a conclusion that either these examples are using this impractical form of pumping water, or that I'm confused on how the pump is actually being used.

## Homework Equations

$A(y) = \pi r^2 \delta h$

$W = \int_a^b ρ g A(y) D(y) dy$

## The Attempt at a Solution

Ex, Method 1:
$A(y) = 25 \pi$ | $D(y) = y$ | $ρ = 1000 kg / m^3$ | $g = 9.8 m / s^2$

Therefore, $W = 25 \pi (1000) (9.8) \int y dy$ from 0 to 10

$W = 12,250,000\pi J ≈ 38,484,510 J$ for the book's method.

Ex, Method 2:
Now, let's consider the method we come up with using logic. Using this method, our pump is pumping water through a hose or tube that reaches the bottom of the tank. The height the water is being lifted stays constant, the height of the tube/tank.
$A(y) = 25 \pi$ | $D(y) = 10$ | $ρ = 1000 kg / m^3$ | $g = 9.8 m / s^2$

Therefore, $W = 25 \pi (1000) (9.8) \int 10 dy$ from 0 to 10

$W = 24,500,000\pi J ≈ 76,969,020 J$ for this method.

From a quick glance, it's obvious that our second method is using an average height based formula. My question is not whether one way is better than the other (that's easy) nor whether one way is right or wrong ([strike]both ways must be right[/strike]), but rather: why would the first method ever be used to pump water from a tank, and how? Wouldn't the first method require that we are constantly moving the pump or a hose to adjust to the fact that the water level is decreasing? And if so, how do we calculate the extra work being done to move the pump or hose? And lastly, why is this fact never mentioned in any of the examples that I have found?

Side note: in the second method, I do not consider that the change in volume might affect the work required/done. In fact, I cannot consider this due to my lack of physics knowledge. If someone could clarify this, I'd appreciate it.

Last edited by a moderator:
To understand why you take into account the height of the water in the tank, imagine what happens when you first put the hose into the tank.

Even before you've done any work using your pump, the water level in the hose is equal to the water level in the tank. The additional work you put in is only to lift the water beyond the current water level (and over the side of the tank).

I see. I didn't take that into consideration. I assumed the tube was empty at the start of the procedure.

Let's assume that there is a device to keep the water out of the tube until the pump starts. Would the second method still be incorrect, and why? How does this affect the first method?

I ask this because I can't visually grasp the procedure of method 1.

I was at first puzzled by your even asking this question but then I saw the statement "The height the water is being lifted stays constant, the height of the tube/tank". You also say " such an application seems impractical in the real world. I have come to a conclusion that either these examples are using this impractical form of pumping water, or that I'm confused on how the pump is actually being used."

Perhaps you think that they are imagining starting with the hose at the top of the container and then lowering the hose so it always is taking water from the very top of the water in the container rather than starting with the hose at the bottom of the tank so they are always "lifting" the full height of the container.

It doesn't matter because in the scenario in which the hose is always at the bottom of the water, the height of water above the bottom pushes down on the water going into the hose, decreasing the force necessary to lift it (a "siphon" effect). So the force required is always that necessary to lift just the top "layer" of water.

So does that mean there is more work being done upon the water that we don't want to calculate? For example, work being done by gravity to push the water up the tube?

Yes, but that can be done most easily by recognizing that it will be just enough to lift the water in the pipe to the level of the water in the tank- that why we don't need to calculate it.

Alright. I think I understand now. Thank you for the explanation!

## What is the "Lifting Water problem"?

The Lifting Water problem is a classic physics problem that involves calculating the force required to lift a certain volume of water to a certain height using a lever.

## What is the logic behind solving the Lifting Water problem?

The logic behind solving the Lifting Water problem involves using the principles of equilibrium and mechanical advantage to determine the required force and length of the lever needed to lift the water. This includes understanding the weight and density of the water, as well as the distance between the fulcrum and the point where the force is applied.

## What are the key equations used to solve the Lifting Water problem?

The key equations used to solve the Lifting Water problem are the equation of equilibrium (F1 x d1 = F2 x d2) and the equation of mechanical advantage (MA = d2/d1). These equations allow us to calculate the required force and length of the lever based on the known variables.

## What are some common mistakes when solving the Lifting Water problem?

Some common mistakes when solving the Lifting Water problem include forgetting to take into account the weight of the lever or the density of the water, using incorrect units in calculations, or using the wrong formula. It is also important to carefully label all known and unknown variables to avoid confusion.

## How is the Lifting Water problem used in real life?

The Lifting Water problem is used in many real-life scenarios, such as designing cranes, pulleys, and other lifting mechanisms. It is also applicable in engineering and construction projects that involve moving heavy objects. Additionally, understanding the principles behind this problem can help in everyday tasks like lifting furniture or carrying heavy objects.

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