Lab 7 Assignment-revv

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VPL Lab – Archimedes. Buoyancy – mod 1 Rev 12/19/18 Mod: 2/9/19 Physical Science Laboratory: PSC 151 Lab 7 Name Archimedes’ Principle Date Purpose To conne ct the words of Archimedes’ Prin ciple to the actual behavior of submerged objects. To use Archimedes’ Principle to determine the density of an unknown material. Explore the Apparatus and Theory 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. Figure 1 – The Buoyancy Apparatus 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. Three simple and three compound objects are available for study. They are composed of cork, aluminum, wood, and two unknown materials. These objects can be weighed in air and when partially or fully submerged. The hanging scale is raised and lowered by clicking and dragging 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. Virtual Buoyancy Apparatus PENCIL

VPL Lab – Archimedes. Buoyancy – mod 2 Rev 12/19/18 Mod: 2/9/19 Physical Science Laboratory: PSC 151 Lab 7 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. Now drag the cork and aluminum compound object and drop it somewhere below the hanger. It should become attached to the hanger. You should find that it weighs about 4.4 N. Drag and drop it on the digital scale and you’ll get 4.41 N. Now for the “(submerged) weight when fully or partially immersed in the water.” Drag the object back to the hanging scale. Slowly drag 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 weigh t of, 3.73 N. Since it’s in vertical equilibrium, we can say Σ F y = 0. We know the object has some actual weight pulling downward, W. When we hang the weight from the hook (and scale), it is countered by an upward force called W’. However, we just saw that when the object is somewhat submerged the scale exerts a lesser upward force. 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 forces represented by arrows in Figure 2 must add to zero. This is similar to how the force from the hangers on the left and right of the cart had to be equal if the cart was not accelerating. In scalar form, which we’ll use fro m here on, we add the magnitudes of the forces together with signs indicating their directions. Since only the weight is downward, W' + F b – W = 0 so F b = W – W' Getting back to our measured values, a. F b = W – W' = 4.41 N – 3.73 N = 0.68 N 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 farther the buoyant force becomes larger and the submerged weight, W' , is reduced by an equal amount. Try it. Archimedes’ Principle : A body wholly or partially immersed in a fluid will experience a buoyant force equal to the weight of the fluid displaced. Figure 2

VPL Lab – Archimedes. Buoyancy – mod 3 Rev 12/19/18 Mod: 2/9/19 Physical Science Laboratory: PSC 151 Lab 7 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, r aise 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 displac ed 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. Insert your value from 1a above in the following blank. b. Drag the graduated cylinder onto the digital scale. Scale reading with cylinder and water = 2.35 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. c. The displaced water’s weight = the buoyant force on the disk: F B = 2.35 N – 0.98 N = N d. Weight of empty graduated cylinder, W c = 0.98N N e. F b = weight of water displaced = W water + cylinder – W cylinder = 2.35 N – 0.98 N (Ex. 1.69N – .98N = .71N) 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 water g = ρ water g V water . [Note that 𝜌 is used to represent densities.] Since we know that the density of water is 1000 kg/m 3 , we could calculate the weight of the water if we knew its volume. The first thing you need to do is lower the masses back into the water to about where they were before. f. F b = W water displaced = ρ water g V water = 1000 kg/m 3 × 9.80 N/kg × m 3 = N (Ex. 1000 kg/m 3 × 9.80 N/kg × (72 × 10 -6 m 3 ) = .71 N) 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. NOTE: The graduated cylinder measures volume in mL. You’ll always need to convert it to m 3 . (1 mL = 1 cm 3 = 10 -6 m 3 .)

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VPL Lab – Archimedes. Buoyancy – mod 4 Rev 12/19/18 Mod: 2/9/19 Physical Science Laboratory: PSC 151 Lab 7 Procedure 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 a bout finding an equation that fits. Rather you try to find a way that your information fits into Archimedes’ Principle. You’ll find that in general, Archimedes’ Principle will boil down to verifying that “The sum of the upward forces = the sum of the downward forces. ” I. Confirm A rchimedes’ 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’l l 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. 7a 7b 7c 7d Figure 7 Direct measurement of the buoyant force: Weighing the aluminum disk in air (Fig. 7a) gives us W . Weighing it when submerged (Fig. 7c) gives us W' . Because the weight is not moving and in equilibrium, the upward forces must equal the downward forces. From Figure 7e we can see that F b + W' al = W al , So by subtracting W’ al from both sides, we get F b = W al – W' al (EQUATION 2) 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 cyl + water displaced . F b = W water displaced = ( W cyl + water displaced ) – W cyl (EQUATION 3) 1. Collect and record the data below. W al = 3.70 N W' al = 2.35 N W cyl = 0.98 N W cyl + water = 2.35 N 2. Calculate the buoyant force, F b . twice, using Equation 2 and Equation 3. F b = W al – W' al = 3.70 N – 2.35 N = 1.35 N (using equation 2 -- F b = W al – W' al ) F b = W cyl + water displaced ) – W cyl = 2.35 N – 0.98 N = 1.35 N (using equation 3 -- F b = ( W cyl + water displaced ) – W cyl ) Figure 7e

VPL Lab – Archimedes. Buoyancy – mod 5 Rev 12/19/18 Mod: 2/9/19 Physical Science Laboratory: PSC 151 Lab 7 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 fill in all the missing pieces. Use part I as a model. Collect and record the data required below. 1. In the questions that follow you’ll submit your responses for Figure 8a, 8c, and 8d by supplying the missing pieces – scale pointer or digital scale reading. Either draw in the missing pieces or paste copies of Screenshots over the partial images below. For screenshots of 8a and 8c just capture an image of the scale. For 8d, capture an image of the digital scale reading. Figure 8a 8b 8c 8d 2. Weighing the cork disk in air (Fig. 8a) gives us W cork . Weighing it when partially submerged (Fig. 8c) gives us W' cork . This time it floats. (Note the slack string.) So W' cork = 0.00 N 3. From Figure 8e, we can see that F b is balanced by W c , so we can say that F b = W c (EQUATION 4) Weighing the empty graduated cylinder (8b) gives us W cyl . Weighing the cylinder after receiving the overflow water (8d) gives us W cyl + water displaced . Just as in the previous example, F b = W water displaced = ( W cyl + water displaced ) – W cyl (EQUATION 5) 4. W cork = 0.71 N 5. W cyl = 0.98 N 6. W' cork = 0.00 N 7. W cyl + water = 1,69 N 8. Volume of cork: V c = 70 mL = 70 x 10 -6 m 3 Calculate the buoyant force, F b . twice, using Equation 4 and Equation 5. 9. F b = W cork – W’ cork = 0.71 N- 0.00 N = 0.71 (Equation 4 -- F b = W c ) 10. F b = W cyl + water displaced – W cyl = 1.69 N -0.98 N = 0.71 N (Equation 5 -- F b = W water displaced = ( W cyl + water displaced ) – W cyl ) Figure 8e

VPL Lab – Archimedes. Buoyancy – mod 6 Rev 12/19/18 Mod: 2/9/19 Physical Science Laboratory: PSC 151 Lab 7 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 straight-forward. 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 at the bottom of the right -hand column and referred to as unknown material #1. (The lower, black part is the actual unknown material.) The unknown material is denser than cork; otherwise the system would try to flip upside-down. Your task is to determine the density of the unknown material. In the process you’ll first have to determine the density of the co rk. To make things a little easier, we will give you the density of the cork: Density of the cork = 241 kg/m 3 Finding the Density of an Unknown Material 10. Empty the graduated cylinder. Then attach unknown #1 (bottom right) to the hanging scale. Gradually lower it as far down as it will go. Measurements of W ud+c , W’ ud , F b , and volume for combined unknown black disk and cork Weight of unknown black disk + cork in air: W ud+c = 3.86 N Weight of unknown black disk in air: W ud = W ud+c – W c = 3.86 N – 0.71 N = 3.15 N Weight of unknown black disk fully immersed in water: W’ ud = 2.50 N Volume of water displaced by unknown disk = Volume of unknown black disk: V ud = 130 mL = 130 x 10 -6 m 3 Weight of graduated cylinder + water displaced by unknown black disk: W cyl+water displaced = 2.35 N Weight of graduated cylinder: W cyl = 0.98 N Weight of water displaced by unknown black disk = F B = 2.35 N – 0.98 N = 1.37 N 11. We want to know the density of the unknown material, ρ ud . If we could find its weight we could use W ud = ρ ud g V u since we know g and we can measure V u using our overflow system. So to find ρ ud we just need W u . We know W c , so if we knew F b we could find ρ ud using Equation 6. From equations 4 and 5 you know two different ways of finding F b . So you’re ready to go. So we have two equations, F b = W c + W ud , and W ud = ρ ud g V ud Combining the two equations and solving for ρ u we get: 𝜌 𝑢? = 𝐹 ? − ? ? 𝑔? 𝑢? We will use: ρ ud = W ud /gV ud = (3.15 N)/[(9.81 m/s 2 )(130 x 10 -6 m 3 )] = Take the data required below to determine the density of the unknown material. You’ll need to determine the total buoyant force, F b by either of the methods suggested in Part I. Record F b and ρ ud in the table below. Table 1 Data for Calculating Density of Unknown Substance W c = 0. 71 N (from part II) F b(ud+c) = 4.86 N 14. V ud = 130.0 mL 15. V ud = 130 x 10 ^-6 m 3 16. W ud = 3.15 N 17. ρ ud = 2470kg/m^3 = 2.47 g/cm^3 18. The unknown material is concrete. Check online to see if your result is reasonable. Note that concrete varies in density depending on its manufacture. Check information below https://civiltoday.com/civil-engineering-materials/concrete/361-density-of-concrete

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VPL Lab – Archimedes. Buoyancy – mod 7 Rev 12/19/18 Mod: 2/9/19 Physical Science Laboratory: PSC 151 Lab 7 https://sciencenotes.org/table-of-density-of-common-materials/ For Students Not Present During Lab Experiment Discussion of Experimental Results (1) Summarize how Archimedes’ Principle for the case where ρ object > ρ liquid was confirmed – the measurements that showed the confirmation. Archimedes principle for the case where object greater then liquid was confirmed was that although liquid has force that comes from everywhere. If a object has great weight, it will begin to sink. (2) How Archimedes’ Principle for ρ object < ρ liquid was confirmed – the measurements that showed the confirmation. An object that contains greater air then weight will cause water to push the object to the sky. (3) Summarize how the density of the unknown solid was found – the process and measurements necessary for determination. How the density of the unknown object was found by dividing the mass by volume to calculate the density of the object.