Fluids Virtual Lab

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Arkansas Tech University *

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2000

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Mechanical Engineering

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Dec 6, 2023

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NAME BUOYANT FORCE S TRUCTURED Driving Question | Objective What are the relationships between the buoyant force on an object submerged in a fluid and a) the volume of the submerged object, and b) the weight of the fluid displaced by the submerged object? Perform an experiment whose data will help determine both relationships. Materials and Equipment  Data collection system  Thread, 60 cm  PASCO Smart Cart with hook 1  Beaker, 100-mL  PASCO Overflow Can 2  Beaker, 1-L  PASCO Smart Cart Rod Stand Adapter 3  Glass stir rod  Brass cylinder  Felt-tipped pen with permanent ink  Aluminum cylinder  Liquid dish soap, 3 mL  Balance, 0.01-g resolution (1 per class)  Water, 500 mL  Table clamp or large base  Paper towel, several sheets  Support rod, 45-cm (2)  Meter stick  Right angle clamp 1 www.pasco.com/ap37 2 www.pasco.com/ap2 5 3 www.pasco.com/ap40 PASCO Smart Cart PASCO Overflow Can PASCO Smart Cart Rod Stand Adapter Background Fluids are generally thought of as liquids; however, this is a common misconception. A fluid is anything that can flow, which includes gasses as well as liquids. When an object is submerged in a fluid, it experiences an upward buoyant force F b that opposes gravitational force F g . This is the reason ice floats on the top of water, and a balloon filled with helium rises in air. If we define F g in the negative direction, a submerged object will rise in the fluid if the net force is positive (the condition of ice rising in water, or a helium balloon rising in air), and sink if it is negative (the condition of a rock sinking in a pond). The magnitude of the gravitational force acting on an object is proportional to its mass, but it is easily observable that the buoyant force acting on a submerged object is not proportional to the object's mass: a small rock may have the same mass as a tennis ball, but a tennis ball floats in water and the rock does not. So, what is different between these two objects? Their masses may be the same but their volumes are different, and so is the volume of water displaced by each once submerged. In this activity you will explore the relationship between the buoyant force acting on an object and the volume of fluid displaced by the object, and draw conclusions that help establish the mathematical relationship between buoyant force and a) the volume of the submerged object, and b) the weight of the fluid displaced by the submerged object. Safety Follow this important safety precaution in addition to your regular classroom procedures: PASCO / PS-3815 1

BUOYANT FORCE / STRUCTURED  Make necessary arrangements to your workstation to avoid getting water on any electronic equipment. Procedure Part 1 – Brass Cylinder S ET U P 1. Fill the 1-L beaker with approximately 500 mL of water. 2. Slowly add approximately 3 mL of liquid dish soap to the water, and then use the stir rod to slowly mix the soap into the water being very careful not to make the soapy water foamy. Set the soapy water aside for a moment. 3. Use the meter stick to measure the length l and radius r of the brass and aluminum cylinders. Record the values for the brass cylinder in cm in the spaces above Table 1 in the Data Analysis section below. Record the values for the aluminum cylinder in cm above Table 2. 4. Measure and make small marks on the sides of both cylinders at 1 4 l , 1 2 l , and 3 4 l locations. 5. Assemble your equipment similar to the diagram at right:  Use the thread to hang the brass cylinder from the Smart Cart force sensor hook so that it hangs vertically with its top surface approximately 5 cm to 10 cm from the sensor.  Place the 100-mL beaker under the spout of the overflow can so it will catch water as it pours out. 6. Connect the Smart Cart force sensor to the data collection system, and then create a digits display showing Force in newtons. 7. Remove the brass cylinder from the force sensor hook, and then zero the Smart Cart force sensor in your data collection system. Rehang the cylinder after the sensor is zeroed. 8. Using the soapy water you just made, slowly fill the overflow can (being very careful not to make the soapy water foamy) until water starts to pour from its spout into the 100-mL beaker. The water will continue to drip into the beaker until it reaches the exact level of the spout inside the can. 9. Once the overflow can has finished dripping, empty the 100-mL beaker into the 1-L beaker, dry the inside of the 100-mL beaker, and place it on the balance. 10. Tare/zero the balance so that it reads zero with the dry 100-mL beaker on it, and then replace the beaker under the spout of the overflow can. C OLLECT D ATA 11. Start recording data and observe the force measured by the Smart Cart force sensor. This measured force value is equal to the tension F T in the string. Record the first tension value F T1 of the brass cylinder suspended above the water in the overflow can (corresponding to a Depth of 0 cm) into Table 1. 2 PASCO / PS-3815

BUOYANT FORCE / STRUCTURED 12. Gently loosen the thumbscrew on the right angle clamp and slowly lower the cylinder into the water in the overflow can until the cylinder is submerged up to the first mark, l /4. Tighten the thumbscrew to hold the cylinder in place as water drips from the overflow can into the 100-mL beaker. 13. Once the overflow can has stopped dripping: in Table 1, record the new tension value F T2 from the force sensor, and then place the 100-mL beaker onto the balance and record the mass of the water that was displaced into the beaker. 14. Repeat the previous steps three additional times, lowering the cylinder to depths of l /2, 3 l /4, and l (cylinder completely submerged). Record the tension F Tn measured by the Smart Cart force sensor and the total mass of the displaced water each time. NOTE: Each time the cylinder is lowered, be sure the overflow can has stopped dripping before recording any measurements. 15. Stop recording data. Part 2 – Aluminum Cylinder S ET U P 16. Raise the brass cylinder out of the overflow can and remove it. Use the thread to hang the aluminum cylinder from the Smart Cart force sensor hook so that it hangs vertically with its top surface approximately 5 cm to 10 cm from the sensor. 17. Pour the soapy water from the 100-mL beaker into the 1-L beaker, place the 100-mL beaker back under the pour spout, and then slowly refill the overflow can from the 1-L beaker until water starts to pour from its spout into the 100-mL beaker. 18. Once the overflow can has finished dripping, empty the 100-mL beaker into the 1-L beaker, dry the inside of the 100- mL beaker, place it on the balance, and then tare/zero the balance so that it reads zero with the dry 100-mL beaker on it. Replace the dry beaker under the spout of the overflow can. C OLLECT D ATA 19. Follow the same Part 1 data collection steps using the aluminum cylinder. Record tension F Tn measured by the Smart Cart force sensor and the total mass of the displaced water at the same five depths: 0, l /4, l /2, 3 l /4, and l (cylinder completely submerged). Record all values using the aluminum cylinder into Table 2. PASCO / PS-3815 3

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BUOYANT FORCE / STRUCTURED Data Analysis Part 1 – Brass Cylinder Brass cylinder length l (cm): 6.5 Brass cylinder radius r (cm): 1.17 Brass cylinder area A cyl (cm 2 ): 4.30 Table 1: Buoyant force and displacement values for a brass cylinder submerged in a fluid Depth (cm) V subm (cm 3 ) Tension (N) m disp (g) F b (N) w disp (N) 0 0.00 0.00 F T1 2.01 0.00 0.00 0.00 1 4 l 1.6 22.167 F T2 1.96 5.26 0.05 0.05 1 2 l 3.15 43.641 F T3 1.9 11.24 0.11 0.11 3 4 l 4.7 65.116 F T4 1.84 16.76 0.17 0.16 l 6.5 87.283 F T5 1.77 23.95 0.24 0.23 Part 2 – Aluminum Cylinder Aluminum cylinder length l (cm): 6.5 Aluminum cylinder radius r (cm): 1.17 Aluminum cylinder area A cyl (cm 2 ): 4.30 Table 2: Buoyant force and displacement values for an aluminum cylinder submerged in a fluid Depth (cm) V subm (cm 3 ) Tension (N) m disp (g) F b (N) w disp (N) 0 0.00 0.00 F T1 0.66 0.00 0.00 0.00 1 4 l 1.6 22.167 F T2 0.61 5.27 0.05 0.05 1 2 l 3.15 43.641 F T3 0.55 11.38 0.11 0.11 3 4 l 4.7 65.116 F T4 0.49 17.72 0.17 0.17 l 6.5 87.283 F T5 0.42 24.55 0.24 0.24 4 PASCO / PS-3815

BUOYANT FORCE / STRUCTURED 1. Calculate the following using the measured values for each cylinder. Record the results into or above each cylinder’s respective table. a. The depth in centimeters in Tables 1 and 2 using cylinder length l . b. The cross-sectional area A , in cm 2 , of each cylinder using radius r .   2 cyl A r c. The volume of the cylinder submerged V subm , in cm 3 , at each depth.   subm cyl depth V A 2. If the tension F Tn measured when each cylinder was submerged is equal to the difference between the gravitational force and the buoyant force:   Tn g b F F F and the tension F T1 measured when each cylinder was suspended above the water is equal to the gravitational force:  T1 g F F then the buoyant force on each cylinder is equal to:   b T1 Tn F F F (1) For both cylinders, use Equation 1 to calculate the buoyant force F b at each depth. Record the results into each cylinder’s respective table. 3. In the blank Graph 1 axes, plot a graph of buoyant force versus submerged volume with two curves: one for the brass cylinder and one for the aluminum cylinder. Be sure to label both curves and both axes with the correct scale and units. [OR copy and paste graph from Excel or other graphing software]s Graph 1: Buoyant force on a cylinder versus volume of cylinder submerged y = 361.32x + 2.4508 PASCO / PS-3815 5

BUOYANT FORCE / STRUCTURED 4. Are the curves for the brass and aluminum cylinders in Graph 1 similar? o Yes 5. Based on your data, is it reasonable to assume that the relationship between buoyant force and submerged volume would be similar if you had used a third object with greater mass (greater density)? Explain your reasoning. o No, if the object is more dense than the fluid, it would have a downward acceleration, commonly referred to as sinking. 6. Calculate the weight w disp (in newtons) of the displaced fluid at each depth for both cylinders. Record your results into their respective columns. disp disp 1 kg 1,000 g w m g   7. In the blank Graph 2 axes, plot a graph of buoyant force versus weight of displaced fluid with two curves: one for the brass cylinder and one for the aluminum cylinder. Be sure to label both curves and both axes with the correct scale and units. Graph 2: Buoyant force on a cylinder versus weight (in newtons) of fluid displaced by the cylinder Brass : y = 98.966x + 0.1599 Aluminum : y = 102.51x + 0.097 8. Are the curves for the brass and aluminum cylinders in Graph 2 similar? o Yes 6 PASCO / PS-3815

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BUOYANT FORCE / STRUCTURED 9. Based on your data, is it reasonable to assume that the relationship between buoyant force and the weight of the displaced fluid would be similar for a third object with greater mass (greater density)? Explain your reasoning. o No, if the object is more dense than the fluid, it would have a downward acceleration, commonly referred to as sinking. Analysis Questions 1. What type of mathematical relationship (proportional, squared, inverse, inverse squared, et cetera) between buoyant force and submerged volume is implied by your data? o Proportional, as volume increases, so does the buoyant force. 2. Based on your data, express the relationship between buoyant force F b and submerged volume V subm by completing this proportionality statement:  b F ρ fluid , V subm 3. Convert the proportionality statement from the previous question into an equation by introducing a proportionality constant k :  b F V subm * k 4. The buoyant force F b acting on an object that is partially or completely submerged in a fluid is described by the equation: b F Vg   (2) where V is the submerged volume of the object and ρ is the density of the fluid in which the object is submerged. Which terms from this equation would be represented in your equation's proportionality constant k ? k = ρ*g 5. Use your data to determine the density of the soapy water. Hint: Use the equation of best fit from graph 1 and Eq (1). Given that the density of plain water is about 1 g/cm 3 , does your value make sense? Defend why or why not. If it does not make sense you should check your work and/or check with your instructor. slope = ρ g → ρ = slope/g = 361.32/9.81 = 36.86 N/m 3 → 3.68 N/cm 3 No, my value does not realy make sense because it should be less than 1, as the density decreases as you add more soap. 6. Archimedes’ principle states that an object completely or partially submerged in a fluid experiences an upward buoyant force equal in magnitude to the weight of the fluid displaced by the object. Does your data support this statement? If yes, explain how it supports it; if no, identify which data do not support it, and what may have caused this disagreement. The amount of water displaced when the cylinders are submerged depends upon the volume of the material, thus its yield is the same amount of water in both of the objects in which both have the same volume. So, yes, my data does support the statement. PASCO / PS-3815 7

BUOYANT FORCE / STRUCTURED Synthesis Questions 1. A wood salvage company is hoisting an old tree trunk off the bottom and out of a lake. The cable from the hoist is tied around the log above its center of mass. The hoist applies a force of 9,800 N to the cable to suspend the log in the lake water ( F T water ), and a force of 29,000 N to suspend the log above the lake surface ( F T air ). What are the volume and density of the log? Assume the lake water has a density of 1,007 kg/m 3 . m = 29000 / 9.8 = 2959.19 kg Applied force = ρ Vg 9800 = 1007 * V * 9.8 → V = 9800 / 1007*9.8 = 0.99 kg/m 3 ρ = m / V ρ = 2958.19 / 0.99 = 2989.074 kg/m 3 2. A cylinder with radius 5.00 cm and length 20.0 cm is lowered into a tank of glucose, which has a density of 1,385 kg/m 3 . The cylinder is lowered in four stages: A) Zero submersion B) Half-submerged to a depth of 10.0 cm C) Fully submerged to a depth of 20.0 cm D) Fully submerged to a depth of 30.0 cm a. What is the buoyant force on the cylinder at each stage? A) body not submerged. F b A = 0 B) when volume L = 10.0 cm V = π * r ² * L → V = π (5.00 2 ) (10.0) → V ₂ = 785.4 cm 3 F bB = ρ *g*V² → F bB = (1,385) (9.8) (7.854x10 -6 ) → F bB = 1.07x10 2 N C) submerged body L = 20.0 V = π* r² *L → V = π (5.00 2 ) (20.0) → V ₃ = 1,570 cm3 F bC = ρ *g*V → F bC = (1,385) (9.8) (15.70x10 ⁻⁶) → F bC = 2.13x10 2 N D) The cylinder is still fully submerged, so the thrust does not change F bD = 2.13x10 2 N b. After being lowered to a depth of 30.0 cm, the string holding the cylinder is cut. If the net force on the cylinder after the string is cut is 1.00 N downward, what is the density of the cylinder material? F net = (mg – F b ) m = ρ Vg ρ = (1 + 2.13x10 2 ) / (7.85x10 -3 )(0.3)(9.8) = 1449.9 kg/m 3 3. A crab fisherman has built a crab trap out of plastic pipe and wire mesh. The overall mass and volume of the trap are 5.59 kg and 6,213 cm 3 , respectively. To catch crab, the trap must sink to the ocean floor. The fisherman has several lead weights to add to the trap to ensure it sinks. If sea water has a density of 1,021 kg/m 3 , and each lead weight has mass of 113.4 g and volume of 10.0 cm 3 , what is the minimum number of weights the fisherman must add so that the trap sinks to the ocean floor? 8 PASCO / PS-3815

BUOYANT FORCE / STRUCTURED mg=Vg 5.5g + N*113.4 / 1000 → ((6213+N*10) /10 6 )*1021 → 5.59 + 0.1134N = 6.343 + 0.01021N → N*0.1031g = 0.75 → N = 7.30 The minimum is 8 weights. PASCO / PS-3815 9

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