Archimedes’ Principle & Buoyancy
Inquiry Lab Kit for AP® Physics 2
Materials Included In Kit
Sodium chloride, 300 g Clay, 225 g Paperclips, jumbo, 20 Sand, black, 1 kg
Spring scale, 2.5 N* Thread, spool Vial, snap-on cap, 50 mL, 12 *Additional spring scales are required to complete the experiment.
Additional Materials Required
Balance, 0.01-g precision† Beaker, 400-mL* Hot glue (optional) Hydrometer, universal (optional)† Knife† Magnetic stirrer or stirring rod† Metric ruler*
Powder funnel† Scissors* Spring scale, 2.5-N* Tape, masking or electrical† Thermometer* *for each lab group †for Prelab Preparation
Prelab Preparation
Prepare 5.0% w/w sodium chloride solution by dissolving 200 g of sodium chloride in 3800-mL of water. (Optional: Test the specific gravity of the resulting solution using a hydrometer. See the Lab Hints section for details.) Cut each clay block into roughly six equal pieces, for a total of 12 pieces.
Vials
- Place a small piece of tape in the center of the underside of the vial caps.
- Bend a paperclip into a hook shape (see Figure 3).
{14007_Preparation_Figure_3}
- Carefully push the larger hook end of the paperclip through the tape and the center of a cap from the underside (see Figure 4). If necessary, use a pushpin to create a hole.
{14007_Preparation_Figure_4}
- Repeat steps 2 and 3 for each cap. (Optional: Place a drop of hot glue around the paper clip to completely seal the hole in the cap.)
- Using a funnel, fill the vials nearly full with sand, approximately 80 grams.
- Securely fit the cap with the paperclip hook on each vial.
Safety Precautions
Wear chemical splash goggles or safety glasses. Instruct students to wipe up any spills immediately. Please follow all laboratory safety guidelines.
Disposal
Please consult your current Flinn Scientific Catalog/Reference Manual for specific procedures and general guidelines, and review all federal, state and local regulations that may apply, before proceeding. All salt solutions may be rinsed down the drain according to Flinn Suggested Disposal Method #26b. All other materials may be saved and stored for future use.
Lab Hints
- This laboratory activity can be completed in two 50-minute class periods. It is important to allow time between the Introductory Activity and the Guided-Inquiry Activity for students to discuss and design the guided-inquiry procedures. Also, all student-designed procedures must be approved for safety before students are allowed to implement them in the lab. Prelab Questions may be completed before lab begins the first day, and analysis of the results may be completed the day after the lab or as homework. An additional lab period would be needed for students to complete an optional inquiry investigation (see Opportunities for Inquiry in Further Extensions).
- Enough materials are provided in this kit for 24 students working in pairs, or 12 groups of students.
- The specific gravity of the sodium chloride solution may be measured using a hydrometer (Flinn Catalog No. AP4314). To use the hydrometer, follow these steps.
- Fill a tall cylinder about three-fourths full of the liquid to be tested.
- Slowly immerse the hydrometer in the liquid to a point just below where it naturally sinks.
- When the liquid and hydrometer are in equilibrium, take the reading at the liquid surface level.
- Specific gravity is the ratio of the liquid density to water density (at standard conditions). The density of water is 1 g/cm3; therefore, the specific gravity reading is the density of the sodium chloride solution.
- An additional inquiry opportunity for students is to design a vial to neither sink nor float in the water or other liquid. To accomplish this, students need to conclude that sinking occurs when the vial weight exceeds the buoyant force and floating occurs when the vial weight is less than the buoyant force. To neither sink nor float, the vial weight and buoyant force must be equal.
- Possible sources of error present in this investigation include the effect of the cap on the total volume, the limitations of the accuracy of the spring scale, etc.
Teacher Tips
- Some accounts describe Archimedes as having placed the crown in a full container of water and measuring the overflow, then comparing the overflow from the same mass of pure gold. This method has met with some criticism, as the difference in the displaced water would have been too small to measure accurately, especially with the surface tension of the water creating a source of error. This method also does not employ the law of buoyancy. Since Archimedes also worked with levers, he may have suspended the crown on one side of a balance and a nugget of gold of equal weight on the opposite side. Both could be submerged in water at the same time. If the two objects remained balanced, that would mean the buoyant force on each was the same, and their volumes would be equal. If the crown had a greater volume than the mass of pure gold, then the buoyant force on the crown would be greater, and the balance would tip down on the gold nugget side.
Further Extensions
Opportunities for Inquiry Using the same clay piece from the Guided-Inquiry Activity, design a vessel that will float on water and be able to carry additional mass. Optimize the design so that it has a large ratio of mass added to vessel mass. Include an explanation of your design and optimization procedures with respect to Archimedes’ principle.
Alignment to the Curriculum Framework for AP® Physics 2
Enduring Understandings and Essential Knowledge Materials have many macroscopic properties that result from the arrangement and interactions of the atoms and molecules that make up the material. (1E) 1E1: Matter has a property called density.
At the macroscopic level, forces can be categorized as either long-range (action-at-a-distance) forces or contact forces. (3C) 3C4: Contact forces result from the interaction of one object touching another object, and they arise from interatomic electric forces. These forces include tension, friction, normal, spring (Physics 1), and buoyant (Physics 2).
Learning Objectives 1E1.1: The student is able to predict the densities, differences in densities, or changes in densities under different conditions for natural phenomena and design an investigation to verify the prediction. 1E1.2: The student is able to select from experimental data the information necessary to determine the density of an object and/or compare densities of several objects. 3C4.1: The student is able to make claims about various contact forces between objects based on the microscopic cause of those forces. 3C4.2: The student is able to explain contact forces (tension, friction, normal, buoyant, spring) as arising from interatomic electric forces and that they therefore have certain directions.
Science Practices 2.2 The student can apply mathematical routines to quantities that describe natural phenomena. 2.3 The student can estimate numerically quantities that describe natural phenomena. 4.1 The student can justify the selection of the kind of data needed to answer a particular scientific question. 4.2 The student can design a plan for collecting data to answer a particular scientific question. 4.3 The student can collect data to answer a particular scientific question. 4.4 The student can evaluate sources of data to answer a particular scientific question. 6.1 The student can justify claims with evidence. 6.2 The student can construct explanations of phenomena based on evidence produced through scientific practices. 6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 7.2 The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understandings and/or big ideas.
Correlation to Next Generation Science Standards (NGSS)†
Science & Engineering Practices
Asking questions and defining problems Developing and using models Planning and carrying out investigations Analyzing and interpreting data Using mathematics and computational thinking Engaging in argument from evidence Obtaining, evaluation, and communicating information
Disciplinary Core Ideas
MS-PS1.A: Structure and Properties of Matter MS-PS2.A: Forces and Motion HS-PS1.A: Structure and Properties of Matter
Crosscutting Concepts
Patterns Cause and effect Scale, proportion, and quantity Systems and system models
Performance Expectations
MS-PS1-2. Analyze and interpret data on the properties of substances before and after the substances interact to determine if a chemical reaction has occurred. MS-PS2-2. Plan an investigation to provide evidence that the change in an object’s motion depends on the sum of the forces on the object and the mass of the object
Answers to Prelab Questions
- Predict the net force on an object in the following situations. Describe the motion of the object after release. Assume the object is held under water and then released.
- Fg > FB
The force due to gravity is greater than the buoyant force, therefore the force in the downward direction is greater than the force in the upward direction. The net force on the object is in the downward direction, therefore the object would sink when released underwater.
- Fg < FB
The buoyant force is greater than the force due to gravity, therefore the force in the upward direction is greater than the force in the downward direction. Because the net force is in the upward direction, the object would float to the surface of the water and remain on the surface.
- Fg = FB
The buoyant force and force due to gravity are equal, therefore the net force is neither in the upward nor downward direction. Because the net force is equal to zero, the object would remain stationary in the water, neither floating nor sinking.
- A 33.5 cm3 metal block is suspended from a scale by a massless string. The block is slowly lowered into water at 23 °C (dH2O = 0.998 g/cm3). The scale is allowed to reach equilibrium and reads 2.61 N.
- What volume of water is displaced by the block as it is lowered into the container?
The volume of water displaced will be equivalent to the volume of the metal block. Therefore, 33.5 cm3 of water is displaced by the block.
- Describe, in your own words, the source of the buoyant force. What causes the buoyant force?
The buoyant force results from unbalanced forces on the top and bottom of the block. When the block is submerged in water, the water exerts a pressure on the surface of the block. Water pressure increases as depth increases. The side surfaces of the block experience balanced forces due to pressure. The top and bottom surfaces experience unbalanced forces due to pressure because the bottom of the block is at a greater depth than the top. The pressure on the bottom surface is greater than the pressure on the top surface, resulting in a net force in the upward direction. This is called the buoyant force.
- Determine the magnitude of the buoyant force.
FB = mH2O x g mH2O = VH2O x dH2O x (1 kg/1000 g) FB = VH2O x dH2O x (1 kg/1000 g) x g FB = 33.5 cm3 x 0.998 g/cm3 x (1 kg/1000 g) x 9.81 m /s2 FB = 0.328 N
- Determine the weight of the block outside of the water.
Fnet = Fg – FB Fg = Fnet + FB Fg = 2.61 N + 0.328 N Fg = 2.94 N
- What are the mass and density values of the block?
Fg = mblock x g mblock = Fg / g = 2.94 N / 9.81 m/s2 mblock = 0.300 kg = 300.0 g dblock = mblock /Vblock = 300.0 g/33.5 cm3 dblock = 8.96 g/cm3
Sample Data
Discussion for Introductory Activity
{14007_Data_Table_1}
*Volume = πr 2h = π(1.55 cm)2 x 7.18 cm = 54.2 cm3
Water temperature: 24.6 °C
{14007_Data_Table_2}
Analyze the Results Water density: 0.9971 g/cm 3
Mass of water = Vdisplaced x dwater Mwater = 54.2 cm3 x 0.9971 g/cm3 = 54.0 g = 0.0540 kg Weight of water = Mwater x g Wwater = 0.0540 kg x 9.81 m/s2 = 0.530 N
Buoyant force:
FB = WAir – WWater FB = 0.80 N – 0.21 N = 0.59 N
The buoyant force and weight of displaced water are nearly equal. The volume of the cap was not taken into consideration when calculating the weight of the displaced water. The near equality of the buoyant force and weight of displaced water agree with Archimedes’ principle. Procedure to Determine Density of Sodium Chloride Solution
- Fill a 400-mL beaker about three-quarters full with the unknown sodium chloride solution.
- Zero the spring scale.
- Using the vial from the Introductory Activity, suspend the vial from the spring scale.
- Record the weight of the vial in air.
- Slowly submerge the vial in the unknown solution until the cap is below the surface. Allow the vial and solution to settle. Record the weight of the vial while it is submerged.
- Remove the vial from the solution and completely dry the vial.
- Repeat the previous two steps for a total of three trials.
It is necessary to check that the spring scale is zeroed between each trial so that the measured weights are correct. In order to determine the density of an unknown liquid, the volume of the submerged object must be known. The vial from the Introductory Activity was previously measured and its volume calculated. Using the two weight measurements (in air and submerged), the buoyant force can be calculated. Using the buoyant force in conjunction with the volume of the vial, the density of the solution can determined. This procedure is similar to Question 6, parts a, b and c. Procedure to Determine Volume of Clay
- Tie a string around the clay so that the spring scale can hook onto the clay.
- Zero the spring scale.
- Attach the hook of the spring scale to the string around the clay. Record the weight of the clay.
- Slowly submerge the clay into the unknown solution. Allow the clay and solution to come to rest.
- Record the weight of the clay in the solution.
- Remove the clay from the solution. Pat dry the piece of clay.
- Repeat the previous steps for a total of four trials.
Analyze the Results
{14007_Data_Table_3}
Average density of sodium chloride solution:
FB = msoln x g msoln = dsoln x V FB = dsoln x V x g dsoln = FB /(V x g) = 0.61 N x (1000 g / 1 kg) x (1 / 54.2 cm3) x (1 / 9.81 m/s2) dsoln = 1.15 g /cm3
Percent Error:
% Error = |Experimental – Actual| / Actual x 100 % Error = |1.15 g/cm3 – 1.050 g/cm3|/1.050 g/cm3 x 100 % Error = 9.52%
{14007_Data_Table_4}
Average volume:
FB = dsoln x V x g V = FB / (dsoln x g) V = 0.10 N / (1.15 g/cm3 x 9.81 m/s2) x (1000 g/1 kg) V = 8.9 cm3
Average density:
dclay = mclay /Vclay mclay = (WIn Air / g) x (1000 g/1 kg) mclay = (0.15 N/9.81 m/s2) x (1000 g/1 kg) = 15 g dclay = 15 g /8.9 cm3 = 1.7 g/cm3
Procedure Analysis In general, the procedure provided repeatable data points. The accuracy of the measured values was hindered by the use of a spring scale. The spring scale did not provide small enough graduations to measure the very small changes in the weight of the vial and clay. The readability of the spring scale was only to 0.01-N. The materials used in the experiment allowed for two significant digits throughout the experiment. Greater masses could be used to increase the amount of significant digits. Additionally, a digital scale or similar device with a greater degree of precision would improve the accuracy of the measurements.
Answers to Questions
Guided-Inquiry Activity
- In the Introductory Activity, the vial was submerged so that the cap was just below the surface of the water. If the vial were submerged to a greater depth, would the buoyant force increase, decrease, or stay the same? Explain how you made your determination and include a free-body diagram(s).
The depth of the vial, so long as the entire vial is submerged, does not affect the buoyant force. The buoyant force only depends on the volume of liquid displaced. Although the pressure of the water on the vial increases as depth increases, the difference between the pressure on the top and bottom of the vial remains unchanged. See the free-body diagram.
{14007_Answers_Figure_5}
- Calculate the apparent density of the vial from the Introductory Activity.
d = mass/volume mass = weight/g = 0.80 N / 9.81 m/s2 mass = 0.082 kg = 82 g d = 82 g / 54.2 cm3 = 1.5 g / cm3
- Define density in your own words.
Density is the ratio of the mass of a material to its volume. The greater the mass in a smaller volume, the more dense the material. Density also makes something feel “heavier” or “lighter” when comparing two different materials of the same size.
- Given the following densities, explain what happens to the mass of a material when its volume is increased.
- d = 1 g/cm3
The mass of the material will increase by the same amount as the volume increases. If the material doubles in volume, then the mass also doubles.
- d > 1 g/cm3
The mass of the material will increase by a greater amount as the volume increases. If the material doubles in volume, then the mass will increase by more than double.
- d < 1 g/cm3
The mass of the material will increase by a smaller amount as the volume increases. If the material doubles in volume, then the mass will increase by less than double.
- When a solid dissolves in water, it separates into its constituent parts and spreads throughout the liquid. The volume of the liquid increases by a small factor after the solid fully dissolves.
- Explain what happens to the mass of the solid as it dissolves.
According to the law of conservation of matter, the mass of the solid is still present in the solution. The individual constituent parts in the solid disperse into the water. The parts carry a small fraction of the total mass of the solid.
- Did the density of the solution stay the same, increase or decrease? Justify your prediction.
The mass of the water–solid system increased. Although the volume of the system increased as well, it changed by a small amount. It can be reasoned that there was a greater change in mass compared to volume. If the mass of the system is increased and its volume is only slightly increased, then the density of the system most likely increased.
- Block A is attached by a string to a spring scale and has a weight of 3.55 N in air. When the block is completely submerged in a liquid, it has an apparent weight of 3.00 N. The block has a volume of 22.4 cm3.
- Draw a free-body diagram of the forces on Block A when it is suspended in air and submerged in the liquid.
{14007_Answers_Figure_6}
- What is the buoyant force on the block?
Fnet = Fg – FB FB = Fg – Fnet = 3.55 N – 3.00 N FB = 0.55 N
- What is the density of the liquid in g/cm3?
FB = mliquid x g mliquid = dliquid x Vliquid FB = dliquid x Vliquid x g dliquid= FB /(Vliquid x g) = 0.55 N/(22.4 cm3 x 9.81 m/s2) dliquid = 0.0025 kg/cm3 = 2.5 g/cm3
Block B has the same weight as Block A. When B is submerged in the liquid, its apparent weight is 2.15 N.
- Predict the volume of Block B relative to the volume of Block A. Justify your determination with respect to Archimedes’ principle.
The weights of blocks A and B are equal when measured in air. When submerged in the liquid, the apparent weight of Block B is less than the apparent weight of Block A. In order for Block B to have a smaller weight in the liquid, the buoyant force acting on it is greater than the buoyant force acting on Block A. To increase the buoyant force, the volume of liquid displaced must be increased. Therefore, the volume of Block B is greater than that of Block A.
- Calculate the volume of Block B. Does this value match your prediction?
Fnet , B = Fg – FB, B FB, B = dliquid x Vliquid, B x g Fnet, B = Fg – (dliquid x Vliquid, B x g) dliquid x Vliquid, B x g = Fg – Fnet, B Vliquid, B = (Fg – Fnet, B)/(dliquid x g) Vliquid, B = (3.55 N – 2.15 N)/(0.0025 kg/cm3 x 9.81 m/s2) Vliquid, B = 57.1 cm3 Yes, the volume of Block B is greater than the volume of Block A.
Review Questions for AP® Physics 2 Two balloons of negligible mass are filled with gases to equivalent volumes. Balloon A contains helium (d He = 0.163 g/L). Balloon B contains sulfur hexafluoride (d SF6 = 6.17 g/L). The density of air (at sea-level) is approximately 1.20 g/L.
- Predict what will occur when the balloons are released from a height of 1.0 m.
Balloon A will float, and Balloon B will sink.
- With respect to Archimedes’ principle, explain how you made your predictions.
When the balloons are filled, a certain amount of air is displaced. A buoyant force equal to the weight of the displaced air will be exerted on the balloons. The weight of Balloon A will be much less than the weight of the displaced air because the density of helium is significantly less than the density of air. The weight of Balloon B will be much greater than the weight of the displaced air because the density of sulfur hexafluoride is so large.
A new balloon, Balloon C, is filled with 1.5 L of sulfur hexafluoride.
- Draw a free-body diagram of Balloon C.
{14007_Answers_Figure_7}
- Predict some changes that could be made to Balloon C to cause it to no longer sink in air.
In order to make Balloon C no longer sink, the weight of the gas inside needs to be decreased while maintaining the same overall volume. This could be accomplished by replacing an equal volume of sulfur hexafluoride with helium. With less mass of gas in the same volume, the buoyant force will remain the same but will be greater than the weight of the gas in the balloon. Another method to cause Balloon C to no longer sink would be to increase the volume of the balloon by adding another gas. The other gas would have to have a density less than both sulfur hexafluoride and air. A less dense gas will increase the volume of the balloon while increasing the mass by a smaller amount. The additional volume will increase the buoyant force because the weight of the displaced air is greater than the weight of the added gas in the balloon.
- A student proposes the following change to Balloon C: “By adding air to the balloon, the increased volume of the balloon will increase the buoyant force.” Do you agree or disagree with the student’s proposal? Explain your reasoning.
The student is not correct. If air is introduced to the balloon, the volume of displaced air will be equal in weight to the gas doing the displacing. As an example, if 1 L of air is added, then 1 L of air is displaced. Those two volumes have the same density, and therefore, the same mass and weight.
- Another student suggests the following: “Adding helium to the balloon will increase the volume of the balloon and decrease the overall density of the contained gases. This will allow the balloon to float.”
- Do you agree or disagree with the student’s proposal? Explain your reasoning.
This student is correct. The density of helium is significantly less than that of air (0.163 g/L << 1.20 g/L). When helium displaces air, the weight of the helium is less than the weight of the air. Therefore, the buoyant force on the helium is much greater than its weight. A certain amount of helium will increase the volume of the balloon to the point that the buoyant force is great enough to lift the weight of the sulfur hexafluoride.
- Verify the student’s assertion by calculating the volume of helium needed so that Balloon C neither sinks nor floats.
{14007_Answers_Figure_8}
Fbuoy = Fg Fbuoy = Fbuoy, SF6 + Fbuoy, He Fg = Fg, SF6 + Fg, He Fbuoy = Fg Fbuoy, SF6 + Fbuoy, He = Fg, SF6 + Fg, He dairVSF6 x (g/1000) + dairVHe x (g/1000)= dSF6 VSF6 x (g/1000) + dHeVHe x (g/1000) dairVSF6 + dairVHe = dSF6 VSF6 + dHeVHe dairVSF6 – dSF6 VSF6 = dHeVHe – dairVHe (dairVSF6 – dSF6 VSF6)/ (dHe – dair ) = VHe VHe = (1.2g/L x 1.5 L – 6.17 g/L x 1.5 L)/ (0.163 g/L – 1.2 g/L) VHe = 7.2 L
References
AP® Physics 1: Algebra-Based and Physics 2: Algebra-Based Curriculum Framework; The College Board: New York, NY, 2014.
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