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New York University Archimedes Principle Density Lab Questions

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“Archimedes’ Principle

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Borough of Manhattan Community College Science Department General Physics (PHY 109 and PHY 110) Laboratory Experiment #7 Archimedes’ Principle OBJECTIVES: To test experimentally the validity of Archimedes’ Principle. EQUIPMENT NEEDED: 250 ml beaker (1), 50 ml graduated cylinder (1), Cent-o-gram balance (1), Hooked metal cylinder (3) THEORY: When an object of mass m and volume Vo is fully, or partly, immersed inside a fluid (liquid or gas) of density f, a net upward force acts on it. This force is called buoyant force Fb. Archimedes’ principle provides a way of calculating the magnitude of Fb. According to this principle, Fb has magnitude of the weight of the displaced fluid mfg i.e. Fb  m f g   f V f g . (1) In (1), f is the density of fluid (for water f = 1,000 kg/m3), Vf is the volume of the displaced fluid, and g is the magnitude of the gravitational acceleration (g = 9.80 m/s2). If the object is fully submerged in the fluid, then Vf = Vo, so the magnitude of the buoyant force becomes Fb   f Vo g . (2) When an object is fully or partly immersed in a fluid (be it air, water etc.) its weight “appears” to be less than its actual value of mg. This is due to the presence of the buoyant force Fb. In general, the magnitude of the “apparent” weight (reading of the scale) Fscale of the immersed object is given by Fscale = mg – Fb (Figure 1). 47 Figure 1. The magnitude of the “apparent” weight (reading of the scale Fscale) of the immersed object is lower than its weight mg by the magnitude of the buoyant force Fb. Since the density of air is much smaller than the density of the liquid that we use, we can ignore the buoyant force due to the air in this experiment. For our purpose, let Fg be the magnitude of the “apparent” weight of the object in air, and Fg apparent be the magnitude of the “apparent” weight of the same object when fully submerged in the fluid. Then Fg – Fg apparent = Fb. (3) And, by substituting (2) into (3), we find Fg – Fg apparent = fVog. (4) Figure 2. Archimedes’ Principle Set. 48 PROCEDURE: Measurements: 1. Fill, almost halfway, the graduated cylinder with tap water. Record this volume of water in Data Table. 2. Fully immerse the hooked metal cylinder in the water. Record the new volume. This is the volume of water and the metal cylinder. 3. Find the difference between these readings. The difference is the volume of the metal cylinder Vo. Convert the result of this measurement to m3. 4. Repeat Steps 1 - 3 for two other hooked metal cylinders. 5. Empty the graduated cylinder. 6. Using the Cent-o-gram balance, weigh and record the mass of the hooked cylinder in the air. Convert the result of this measurement in kg. 7. Repeat Step 6 for two other hooked metal cylinders. 8. Fill, almost 250 ml, the beaker with tap water. 9. Place the beaker at the Cent-o-gram balance. Connect the hooked metal cylinder to a wire. Submerge completely the hooked cylinder into the water and hang the wire from the balance (Figure 2). 10. Record the “apparent” mass of the hooked cylinder when it is completely submerged in the water. Convert the result of this measurement in kg. 11. Repeat Steps 9 - 10 for two other hooked metal cylinders. Empty the beaker. Calculations: 12. Calculate the magnitude of the buoyant force Fb for all metal cylinders using equation (2). In that equation substitute Vo from Step 3 (in m3). 13. Multiply the mass in air of the hooked cylinder by 9.80 m/s2 to get the weight in the air. Record this result Fg. Multiply the “apparent” mass by 9.80 m/s2 to get the “apparent” weight. Record this result Fg apparent. 14. Calculate and record the buoyant force as the difference Fg –Fg apparent for all three metal cylinders. 15. Find the % Difference between Fb and Fg –Fg apparent for all three metal cylinders. 49 Student Name __________________________ Section _______________ Date ___________ Report on Laboratory Experiment “Archimedes’ Principle” DATA TABLES Unit conversion reminder: 1 ml = 1 cm3 = 110-6 m3. Symbol of hooked cylinder Material of hooked cylinder Volume of water ml Cu Copper T Tin Z Zinc Symbol of hooked cylinder In air Mass kg Volume of Volume of the water and cylinder, Vo the cylinder ml ml m3 Buoyant “Apparent” “Apparent” Force, Fg-Fg apparent mass weight, Fg apparent In water Weight Fg N kg N N Buoyant Force, Fb N % Difference of Buoyant Force Fb and Fg-Fg apparent % Cu T Z QUESTION: How do you argue that your results support the validity of Archimedes’ Principle? 50 Archimedes’ Principle. Background knowledge Pressure: Force divided by area (force per unit area). 𝑃= 𝐹 F = P*A 𝐴 Units: Pascal(Pa)= 1 N/m2 is the SI unit. Also in use are pounds per square inch (psi) and bars or millibars. Pascal(Pa) pound per square inch (psi) = 6,895 Pa bar = 14.5038 psi = 100,000 Pa  Atmospheric pressure at sea level is about 101,325 Pa and at 20 km high it is about 5,000 Pa.  Your car tire pressure should be between 32 psi and 35 psi.  Average atmospheric pressure at sea level is about 1 bar (1,000 millibar). Pressure calculation example: Calculate the pressure that you apply standing on the floor. Assume that your mass is 70 kg and that your shoes together have a contact area of 300 cm2 with the floor. Pressure in Pa. First, calculate your weight in Newton and the contact area in m2. F = W = m*g = (70 kg)*(9.8 m/s2) = 686 N A = 300 cm2 = 300*(0.01 m)2 = 0.03 m2 𝑃= 𝐹 686 𝑁 = = 𝟐𝟐, 𝟖𝟔𝟕 𝑷𝒂 𝐴 0.03 𝑚2 Pressure in psi: First, calculate the weight in lbf and the area in square inches. 70 kg = 70*(2.2 lb) = 154 lb; then F = weight = 154 lbf A = 300 cm2 = 300*(0.3937 in)2 = 300*(0.155 in2) = 46.5 in2 𝑃= 𝐹 154 𝑙𝑏𝑓 = = 𝟑. 𝟑 𝒑𝒔𝒊 𝐴 46.5 𝑖𝑛2 Mass density: Mass density (𝜌) is the mass per unit volume of a volume of material. 𝜌= 𝑀 𝑉 densities in g/cm3 M=𝜌*V M, V water steel lead mercury gold 1 7.6 11.3 13.7 19.3 Hydrostatic pressure: Pressure applied by a static fluid (liquid or gas) on the objects in contact with it. The hydrostatic force is perpendicular to the surface of the objects. The pressure increases as the depth in the fluid increases. For a fluid of constant mass density ρ, the hydrostatic pressures P1 and P2 at two points separated by a depth h is. P1 P2 = P1 + ρ*g* h h Pressure increases with depth. P2 > P1 P2 Hydrostatic pressure example: A one meter high tank is filled with water and its top is open to the atmosphere. Calculate the hydrostatic pressure inside the tank at its bottom. Note, the standard atmospheric pressure is 101,325 Pa and the density of water is 1,000 kg/m3. The surface of water is at a pressure P1=101,325 Pa equal to the atmospheric pressure. P2(at 1 m from the top) = P1 + ρ*g* h = 101,325 Pa + (1,000 kg/m3)*(9.8 m/s2)*(1 m) = 101,325 Pa + 9,800 Pa = 111,125 Pa Buoyancy force: This is the upward force that a fluid exerts on an object partially or fully submerged. This force is a result of the fact that hydrostatic pressure increases with depth in a fluid. Look at the fluid forces acting on a cube fully submerged. The lateral forces are equal in magnitude and compensate each other. The force at the bottom face is stronger than the one at the top face, therefore, there is a net upward force acting on the cube. F1 F = P*A Fhorizontal = F3 - F4 = 0 Fvertical = F2 – F1 ≠ 0 F4 F3 Fvertical = Fbuoyancy F2 Archimedes’ Principle: The buoyant force is equal to the weight of the fluid that the object displaces. Fb = (density of the fluid )(volume of fluid displaced by the submerged object)(gravitational acceleration) Fb = ρf Vdis g Example # buoyancy-1: Calculate Fbuoyancy on a cube with 1 m side fully submerged in water. The cube is fully submerged. The volume of water displaced is the volume of the cube ( 1 m3 ). The mass of 1 m3 of water is: Mass = density*volume = (1,000 kg/m3)*(1 m3) = 1,000 kg The weight of 1,000 kg of water is: W = m*g = (1,000 kg)*(9.8 m/s2) = 9,800 N Buoyancy force = weight of fluid displaced by the object = 9,800 N Example # buoyancy-2: A solid plastic cylinder having a radius of 2 cm and a length of 40 cm is floating with its symmetry axis perpendicular to the surface of water and with 10 cm above this surface. Calculate the buoyancy force. The cylinder is partially submerged with 30 cm of its length inside water. The volume of water displaced is: V = (𝜋 ∗ 𝑅 2 ) ∗ 𝐻 = (3.14 ∗ (0.02 𝑚)2 ) ∗ (0.3 𝑚) = 3.77 ∗ 10−4 ∗ 𝑚3 The mass of displaced water is: 𝑀 = 𝜌 * V = (1,000 kg/m3)*( 3.77 ∗ 10−4 ∗ 𝑚3) = 0.377 kg The weight of 0.377 kg of water is: W = m*g = (0.377 kg)*(9.8 m/s2) = 3.69 N Buoyancy force = weight of fluid displaced by the object = 3.69 N Predicting if an object will sink in a fluid: The maximum possible buoyancy force is compared to the weight of the object as follows: If F buo max > Wobject : The object floats (part of it stays above the fluid surface) If F buo max = Wobject : The object stays fully submerged at any depth but won’t sink to the bottom. If F buo max < Wobject : The object sinks to the bottom. Note that the maximum possible buoyancy force is obtained calculating the maximum possible volume of fluid displaced by the object. When the maximum possible volume of fluid displaced by the object is occupied only by the material of the object and not by other materials (like air spaces, for example), then the prediction can be done without calculations by comparing the density of the object with the density of the fluid as follows: If ρobject < ρfluid : The object floats. If ρobject = ρfluid : The object stays fully submerged at any depth but won’t sink to the bottom. If ρobject > ρfluid : The object sinks to the bottom. For example, a steel ball (ρobject = 7.6 g/cm3) sinks in water (ρfluid = 1 g/cm3) but floats in mercury (ρfluid = 13.7 g/cm3). However, a gold ball (ρobject = 19.3 g/cm3) sinks in water and also in mercury. Questions: 1. What is the weight of liquid mercury displaced by a 2 cm radius solid gold sphere fully submerged? Show your work. Note: the volume of a sphere of radius R is V=(4/3) π R3. Show your work. 2. A solid sphere of 2 cm radius floats on liquid mercury showing ½ of its shape above the surface of the liquid. Calculate the buoyancy force acting on the sphere. 3. Someone wants to use as a boat a hollow steel box with its top removed. The dimensions are length=2 m, width = 2m and height = 0.5 m. Calculate the maximum buoyancy force that water can apply on this box. Show your work. 4. The person in problem #3 above wants to load 1,500 kg of rocks in this “boat”. The empty boat has a mass of 200 kg. Predict if the loaded boat is able to float. Show your work. 5. Can a solid sphere of a material with a density of 0.9 g/cm3 float in water? Justify your answer. 6. Can a solid lead sphere float in liquid mercury? Justify your answer. 7. As you know, ice floats in water. What can you say about the density of ice compared to the density of liquid water? 8. A person wants to use a spherical helium balloon of radius 3 m3 to fly in air at normal pressure. The mass of the man is 70 kg. To simplify the analysis neglect the mass of the empty balloon. The densities of air and helium are: 𝜌air = 1.3 kg/m3 and 𝜌helium = 0.18 kg/m3. Determine if the balloon filled with helium is able to lift the person. This is an extra credit question. Interesting link: https://www.theguardian.com/world/video/2015/jul/07/calgary-balloon-man-helium-daniel-boriavideo ...
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Final Answer

Hey,Find the below-attached solutions of the 8 Lab. Questions in PDF and DOCX formars.Good Luck

Archimedes’ Principle. Background knowledge
Questions’ Solutions

1.
The sphere is fully submerged in Mercury, so the volume of displaced Mercury equals the volume of
a sphere of radius 𝑅 = 2 𝑐𝑚 = 0.02 𝑚, so;
4
4
𝑉𝐷𝑖𝑠𝑝 = 𝜋𝑅 3 = 𝜋(0.02 𝑚)3 = 3.351 × 10−5 𝑚3
3
3
The mass of displaced Mercury is:
𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = 𝜌𝑀𝑒𝑟𝑐𝑢𝑟𝑦 × 𝑉𝑀𝑒𝑟𝑐𝑢𝑟𝑦
𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = 𝜌𝑀𝑒𝑟𝑐𝑢𝑟𝑦 × 𝑉𝐷𝑖𝑠𝑝
𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = (13.7 𝑔/𝑐𝑚3 ) × (3.351 × 10−5 𝑚3 )
𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = (13.7 𝑔⁄𝑐𝑚3 ) × (33.51 𝑐𝑚3 ) = 459.09 𝑔 = 0.45909 𝑘𝑔
Then;
𝑊𝑒𝑖𝑔ℎ𝑡𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = 𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 × 𝑔 = (0.45909 𝑘𝑔) × (9.80 𝑚⁄𝑠 2 ) = 4.4991 𝑁

2.
Half of the Gold sphere is submerged in Mercury, so the volume of displaced Mercury is;
1 4
4
𝑉𝐷𝑖𝑠𝑝 = ( 𝜋𝑅 3 ) = 𝜋(0.02 𝑚)3 = 1.6755 × 10−5 𝑚3
2 3
6
The mass of displaced Mercury is:
𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = 𝜌𝑀𝑒𝑟𝑐𝑢𝑟𝑦 × 𝑉𝑀𝑒𝑟𝑐𝑢𝑟𝑦
𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = 𝜌𝑀𝑒𝑟𝑐𝑢𝑟𝑦 × 𝑉𝐷𝑖𝑠𝑝
𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = (13.7 𝑔/𝑐𝑚3 ) × (1.6755 × 10−5 𝑚3 )
𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = (13.7 𝑔⁄𝑐𝑚3 ) × (16.755 𝑐𝑚3 ) = 229.54 𝑔 = 0.22954 𝑘𝑔
Then;
𝑊𝑒𝑖𝑔ℎ𝑡𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = 𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 × 𝑔 = (0.22954 𝑘𝑔) × (9.80 𝑚⁄𝑠 2 ) = 2.2495 𝑁
𝐵𝑢𝑜𝑦𝑎𝑛𝑐𝑦 𝑓𝑜𝑟𝑐𝑒 = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 = 2.2495 𝑁

3.
The maximum possible water volume to be displaced by the box is;
𝑉𝐷𝑖𝑠𝑝 = 𝑉𝐵𝑜𝑥 = 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑤𝑖𝑑𝑡ℎ × ℎ𝑒𝑖𝑔ℎ𝑡
𝑉𝐷𝑖𝑠𝑝 = (2 𝑚) × (2 𝑚) × (0.5 𝑚) = 2.0 𝑚3
The mass of displaced Water is:
𝑀𝑊𝑎𝑡𝑒𝑟 = 𝜌𝑊𝑎𝑡𝑒𝑟 × 𝑉𝑊𝑎𝑡𝑒𝑟
𝑀𝑊𝑎𝑡𝑒𝑟 = 𝜌𝑊𝑎𝑡𝑒𝑟 × 𝑉𝐷𝑖𝑠𝑝
𝑀𝑊𝑎𝑡𝑒𝑟 = (1000 𝑘𝑔⁄𝑚3 ) × (2.0 𝑚3 ) = 2000 𝑘𝑔
Then;
𝑊𝑒𝑖𝑔ℎ...

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