Archimedes' Principle

Purpose

• •
  To connect the words of Archimedes' Principle to the actual behavior of submerged objects.
• •
  To examine the cause of buoyancy; that is, the variation of pressure with depth in a fluid.
• •
  To use Archimedes' Principle to determine the density of an unknown material.

Equipment

Simulation and Tools

Open the Buoyancy Apparatus simulation to do this lab.

Explore the Apparatus

We'll use the Buoyancy Apparatus in this lab activity. You can get quick access to help by rolling your mouse over most objects on the screen. To investigate buoyant forces, we need to measure the weight and volume of objects as well as their submerged weight when fully or partially immersed in a fluid. We'll use water as our fluid in this lab. We also need to measure the weight and volume of the fluid displaced. A hanging scale and a digital scale are available for measuring the weights of our objects. The overflow of water from the tank spills into a graduated cylinder. The volume of this displaced water can be read from the graduated cylinder, and the weight of the water can be found by weighing the graduated cylinder before and after overflow. Two simple and two compound objects are available for study. They are composed of cork, aluminum, and an unknown material. These objects can be weighed in air and when partially or fully submerged. The hanging scale is raised and lowered by dragging the metal arm that supports it. Both the hanging scale and the graduated cylinder can be read more precisely by zooming in. The graduated cylinder is emptied by dragging and releasing it above the tank. You'll find that some actions that could be done with real apparatus are forbidden. For example, if you remove the cork and aluminum object from the hanging scale and then attach a new object to the scale, you'll find no further overflow is possible without first emptying the cylinder. Similarly, you can't weigh more than one object on the digital scale at a time, and you can't make a mess by lowering an object into the water while the cylinder is on the digital scale. Another thing that is largely forbidden is dragging the screen when you're zoomed in. When you try to move an onscreen object and it doesn't behave as expected, this means that something else needs to be done first. There aren't too many choices. This and many other phrases you'll hear when studying buoyancy are mostly meaningless without direct experience. So, our exploration of the apparatus will mainly focus on the various situations you'll encounter and the related terminology. While we're at it, our use of submerged in some cases and immersed in others is not meant to suggest two different phenomena. Both refer to an object fully or partially in a fluid.

1. What's a "Buoyant Force", and What Exerts it?

Let's try it with the system in Figure 1. Right (or Ctrl) click and zoom in on the hanging scale. It appears to read zero initially. This sort of scale is not very precise. Our simulation reflects that level of precision. (Zoom back out.) Now drag the cork and aluminum compound object and drop it somewhere below the hanger. It should become attached to the hanger. (If it refuses to attach, you probably have the scale already lowered somewhat, or there might be some water in the graduated cylinder.) You should find that it weighs about 4.4 N. Zooming in and estimating a digit we might have 4.41 N. Drag and drop it on the digital scale and you'll get a more precise answer of about 4.41 N. Which scale should you use? We'll always use the value from the hanging scale, 4.41 N in this case, unless we have to use the digital scale for a measurement. This will only happen when we need to weigh the graduated cylinder. Now for the "(submerged) weight when fully or partially immersed in the water." Drag the object back to the hanging scale. Slowly drag the support arm above the scale downward until the aluminum disk is about half submerged. The scale might read about 3.73 N. This is a judgment call since the location of the half-way point is uncertain. Let's use that value in the following discussion. That's the "submerged weight." We'll use the term

W'

("W prime") to represent the submerged weight. So, its actual weight is 4.41 N and it appears lighter when partially submerged, maybe 3.73 N. It didn't lose weight. The water is providing an upward force on the bottom of the disk. So, we would say that a 4.41-N object was buoyed up by a force F_b so that it appears to weigh, or has a submerged weight of, 3.73 N. Since it's in vertical equilibrium, we can say

	F

_y = 0.

Let's look at the forces and draw vector arrows to represent each force. We know the object has some actual weight, a downward force, W. We measure this force with our scale by finding the upward force that balances it. But the weight, W, is a downward force. When the object is somewhat submerged, the scale exerts a lesser upward force, W′. Since the pair of disks is still in equilibrium, some additional upward force must make up the difference. This is the buoyant force, F_b. Since the object is still in equilibrium, the three vector arrows shown in Figure 2a must vectorially add to equal zero. This is shown in two ways in Figures 2b and 2c. Figure 2c is a free-body diagram, FBD, where vectors radiate away from a point representing the object that the forces act on. We'll make use of FBDs throughout this lab to help us picture the forces acting on an object and to help us create equations to relate the force involved. Our FBD guides us to add the three force vectors together to equal a net force of zero. In scalar form, which we'll use from here on, we add the magnitudes of the forces together with signs indicating their directions. Since only the weight is downward, so Getting back to our measured values, This value, 0.68 N, is the buoyant force. The buoyant force is the part of the weight that the scale no longer has to support because a new upward force is present. It's an upward force caused by the pressure on the bottom of the disk. If you lower the disk further, the buoyant force becomes larger and the submerged weight, W ′, is reduced by an equal amount. Try it.

2. What is the "Fluid Displaced" and How is its Weight Related to the Buoyant Force?

In the preceding section, you focused on the buoyant force as an apparent reduction in the weight of the submerged object. What's the source of this force? With the cork and aluminum disk pair attached to the scale, lower the scale as far as it will go. A lot of water will overflow into the cylinder. When that's done, raise the scale back up. Now gradually lower the disks into the water. Notice the behavior of the water. Its level rises because the disks are "displacing" some of the water. That is, they're taking the water's place. So it rises to make room. When the water occupied that space, it was supported by the water below it. When the disks occupy the same space, the water below them provides exactly the same amount of upward force. That's the buoyant force. It was there all the time, but it was just holding up the water until the disks arrived. So "the buoyant force is equal to the weight of the fluid displaced!" We need to find out how to determine that weight. Refill the tank. To do this just drag the graduated cylinder somewhere above the tank and release it. It will empty itself and return to its post. Now gradually lower the disks into the water and notice what happens to the "displaced water." Since our tank was initially full this time, the displaced water overflowed into the graduated cylinder. We did that so that we could weigh it or find its volume. Let's use our previous results to see the relationship between the buoyant force and the displaced fluid's weight. If your apparatus is not right where you left it at the end of Part 1, repeat Part 1 to restore it. Be sure to zoom in once for better precision. Insert your value from 1a in the following blank. That's the weight of the fluid plus the weight of the graduated cylinder. So we need to find the weight of the graduated cylinder and subtract it from the total to find the weight of the water. Empty the cylinder as before. Place the cylinder on the digital scale. Got it? We'll see why your values for (b) and (e) were about the same shortly. But for now, you should understand what we mean by the statement that the buoyant force equals the weight of the fluid displaced.

3. Finding and Using the Volume of the Fluid Displaced

Now suppose we didn't have either of our scales. Could you still determine F_b? We know that the weight of the water can be found from W_water = m_waterg = ρ_watergV_water. Since we know that the density of water is 1000 kg/m³, we could calculate the weight of the water if we knew its volume. You can zoom in on the cylinder to read the volume reading, but there's a "Cylinder Zoom" feature that does that for you. You can now read the volume in mL. How about the aluminum disk? If you weighed it and then completely submerged it, you could measure its volume and then compute its density and convince the king that his disk is not silver. Incidentally, we went to extra expense in providing a non-wetting polypropylene graduated cylinder in this apparatus to avoid the need to account for a meniscus. You're welcome.

Theory

Archimedes' Principle

Archimedes' Principle is very easy to state in words, but learning how to use it takes practice in a variety of situations. In this lab, we'll first explore the literal meaning of this principle. Then we'll put it to use in a couple of situations. You'll use the wording of the principle to draw free-body diagrams (FBDs) and create equations that can be used to find unknown quantities such as the volume, mass, or density of an object or of the fluid it's immersed in. Solving problems using Archimedes' Principle can be really puzzling. The most important step is to use the words of the principle to create an equation. One important tool to help you create your equation is a free-body diagram. Archimedes' Principle usually involves objects in vertical equilibrium. And the terms that go into

	F

_y = 0

will likely come from the words of Archimedes' Principle—a body (mass, weight) buoyed up by (force), etc. In case you've forgotten... Without a bit more information, this is little more than magical incantation! We need to dig a little deeper and learn what causes the buoyant force and why the force varies as objects are gradually immersed in a fluid. We'll start with a couple of interactive animations built into the lab apparatus.

Variation of Pressure with Depth

Click the Pressure button to load an interactive simulation. Adjust the small cylinder up and down as needed to recreate the discussion that follows. In Figure 3a, a solid brass cylinder is shown above a tank completely filled with water. Atmospheric pressure acts essentially equally on all sides of the cylinder and on the water's surface as indicated by the five P_atm arrows in Figure 3a. At all times the horizontal pressures will cancel. So we need only examine the vertical pressures. Gradually lower the cylinder until about 1/4 of it is submerged, as shown in Figure 3b. (The water overflowing the sides is not shown. It also magically returns as needed.) Notice the vectors that appear and change as you move the cylinder up and down. A black vector, P_up, the upward pressure exerted by the water, is shown in two places along with a blue ΔP vector. At a given depth, h, in a container of water exposed to the atmosphere, there is a pressure due to the weight of the column of water above that point plus the weight of the atmosphere above the surface. This pressure is omnidirectional and given by We're only interested in the pressures at the top, P_down, and bottom, P_up, of the cylinder. We'll refer to these more generally as P₁ and P₂, respectively. With the cylinder partially submerged, only the upward pressure on the bottom of the cylinder appears. (See Figure 4a.) What about ΔP? ΔP is the difference in pressure between the top and bottom of the cylinder. Currently, the top of the cylinder has a pressure P_down = P_atm acting on it. The pressure acting on the bottom is P_up, which equals P_atm + ρ_wgh₂. ΔP, the difference in these two pressures, is just P_up – P_down = ρ_wgh₂. This explains why ΔP always (so far) equals P_up. Let's change that. As you continue to drag the cylinder downward, there will continue to be just the three vectors and each will continue to increase in length but remain equal to each other until a specific point is reached. Continue to lower the cylinder until it's about halfway to the bottom of the tank as shown in Figure 4b. You can now clearly see the downward vector P_down. Drag the cylinder up and down in this middle region of the tank. As you do this, the vector ΔP behaves differently than the way it behaved before. In summary, the difference in pressure from top to bottom is just

P₂ − P₁ = ρg(h₂ − h₁).

In the case of a partially submerged object,

h₁ = 0,

so this equation will work for either a partially or fully submerged object.

The Effect of the Pressure Difference, ΔP, on the Buoyant Force

We've found a simple equation that gives the pressure difference between the top and bottom of our cylinder when it's somewhat or totally submerged. So how does this relate to buoyancy? For a cylinder of top and bottom surface area, A_c, the total force exerted by the water pressure would be Consider a cylinder of height h_c. In Figure 5a, with

h₂ =

1

4

h_c,

ρ_wg(h₂ − h₁)

reduces to

ρ_wg

1

4

h_c

. The total force, F_{net water}, exerted by the water pressure on the cylinder would be Now consider the totally submerged cylinder in Figure 5b. In this case,

h₂ − h₁ = h_c

, so

ρ_wg(h₂ − h₁)

reduces to

ρ_wgh_c

. The total force, F_{net water}, exerted by the water pressure on the cylinder would be What about different shapes? How about a cone or a sphere? The result is exactly the same but the proof is not so simple. We won't pursue that any farther. Consider three cases: a cylinder of aluminum, a cylinder of liquid water, and a cylinder of cork. A cylinder of liquid water would be impossible to create, but we can imagine it to be enclosed in a thin, weightless, cylindrical shell. Click the Buoyancy button. Select the aluminum cylinder. Use this to test your reasoning and correct it if necessary. Select the water cylinder. Use this to test your reasoning and correct it if necessary. Select the cork cylinder. Use this to test your reasoning and correct it if necessary. One final suggestion: reopen the Pressure simulation. Lower the cylinder all the way to the bottom. Any thoughts?

Archimedes' Principle with Additional Applied Forces

Most of our work with Archimedes' Principle includes additional forces such as the ones suggested at the end of each of the three sets of questions above. Typical examples include:

• •
  A lifeguard pulling a person up onto a dock will find that the person seems to get heavier as she's lifted further out of the water. This leads to the strategy of starting out with a large upward pull to give the swimmer an initial upward speed, making the rest of the pull easier.
• •
  A buoy in a river will be held in place by a heavy weight sitting on the river bottom. The "one final question" above will come into play in this case.
• •
  The buoyant force might be used to partially or fully support another object. For example, a life jacket has to supply enough buoyant force to support its own weight as well as the person wearing it. A waterlogged life jacket might continue to supply the same lift, but the added water will provide an additional undesirable downward force.

Analyzing such situations usually involves the use of translational equilibrium equations. That is, if a body is in equilibrium, the sum of the forces in any direction will equal zero. A buoyant force is just another force. It's just a bit trickier because it varies as an object is submerged to varying depths. Just as important, a careful free-body diagram, FBD, is essential in working out what objects and forces are involved. Example 1: A steel anchor weighs 2000 N in air. How much force would be required to support it when fully submerged in water? That is, what is its apparent weight, W ′, when submerged? W ′ is the unknown upward support force in this case. In addition to the upward support force, W ′, there is the upward buoyant force, F_b, and the downward gravitational force, that is, its weight, W. The sum of these forces equals zero. So, (using magnitudes)

F_b + W ′ = W.

Other considerations: since the anchor displaces an equal volume of water. And, since we know the weight and mass of the anchor, we can find its volume by looking up its density and using ρ_steel = m_anchorV_anchor. What will change as the anchor breaks the surface of the water?

Procedure

Please print the worksheet for this lab. You will need this sheet to record your data.

In the following sections, you'll collect and analyze data to investigate Archimedes' Principle in some typical situations, such as the one suggested in Example 1. You won't be provided step-by-step directions, but you'll have blanks provided for the data you need to collect. This method is being used because buoyancy problems are always sort of free-form. It's not about finding an equation that fits; rather, you'll try to find a way that your information fits into Archimedes' Principle. You'll also get guidance in developing FBDs as well as equations describing your FBDs. Note that the equations are in scalar form. That is, we'll say "The sum of the upward forces = the sum of the downward forces" (all positive magnitudes) rather than "The sum of the upward forces + the sum of the downward forces = 0" (all vectors).

I. Confirm Archimedes' Principle for the Case Where ρ_object > ρ_liquid Using Overflow

In this section, we'll directly measure the buoyant force and the weight of the water displaced for an object that sinks. We'll then confirm that they are equal. Our object will be an aluminum disk. If either the Buoyancy or Pressure simulation is still open, click the Close button. Direct measurement of the buoyant force: weighing the aluminum disk in air (Figure 7a) gives us W. Weighing it when submerged (7c) gives us W ′. From our FBD (7e), we can see that so Measurement of the weight of the water displaced: weighing the empty graduated cylinder (7b) gives us W_cyl. Weighing the cylinder after receiving the overflow water (7d) gives us W_cylw displaced.

II. Confirm Archimedes' Principle for ρ_object < ρ_liquid Using Overflow

In this section, we'll directly measure the buoyant force and the weight of the water displaced for a floating object, and we'll confirm that they are again equal. Our object will be a cork disk. This time, you'll fill in all the missing pieces. Use Part I as a model. Collect and record the data required below. Use equations 2

F_b = W_al − W_al'.

and 3

F_b = W_{water displaced} = (W_{cylw displaced}) − W_cyl

as a guide in producing equations 4 and 5.

III. Determine the Density of an Unknown Material

We've confirmed that Archimedes' Principle correctly describes the forces acting on a body wholly or partially submerged. That gives us a starting point for more complex situations that are not so straightforward. One of your object choices is a compound object with a cork on top and a rod of negligible volume connecting it to a large black unknown object below it. It's in the right-hand column. The unknown object is more dense than cork; otherwise, the system would try to flip upside down. Your task is to determine the density of the unknown object. In the process, you'll first have to determine the density of the cork.

Density of the Cork

You already know the weight of the cork from Part II. From the weight, you can calculate the mass. What you need is the cork's volume. You can find that using another compound object–the cork with the attached aluminum disk. It's in the middle of the left column of objects. Show the calculations for V_cork (mL), V_cork (m³) and ρ_cork.

Finding the Density of the Unknown Material

Return the cork/aluminum object to its home. Empty the graduated cylinder.

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