Fluid Statics (1)

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Jan 9, 2024

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Title: Fluid Statics Group 33: Van Tran, Elyse Gallagher, Shrenik Patel Experiment Date: October 31st, 2023 Goals: In today’s lab, we will understand how to apply static equilibrium concepts to the buoyant force and hoe pressure changes with depth in a fluid. We will also be able to find the density of an unknown object by measuring its specific gravity. Procedure: Part I: Suspend the brass cylinder from the force sensor and record the value as the weight of the brass in air Suspend the brass in the water and record its weight in water Use your equation and the measured values to calculate the buoyant force on the brass Calculate the specific gravity and density of brass from your measurements Hang the brass below the wood so the wood will sink in water Measure and record the weight in the air and in water, then calculate the total buoyant force acting on both objects. Then use the known buoyant force on the brass to find the buoyant force on the wood Use equation 3 to calculate the specific gravity of the wood Tabulate and report the buoyant force, specific gravity in water, and density for all the materials you studied in this experiment Part II: Connect the absolute pressure sensor to the computer. Check the model of your pressure sensor, open the hardware setup menu, then locate and select it from the menu Fill the 1000 mL graduated cylinder with water, then slowly lower the tubing and record the absolute pressure as a function of the depth at 5 cm intervals. Record your data in an excel table. Make a scatterplot of the pressure vs. depth with pressure on the y-axis Precautions and Sources of Error : Sources of error can be human error. There was a lot of submerging and taking our best guest when it was submerged enough. We also had to interpret our graph as best as possible.
Data: Brass W air = 1.88 N Brass W water = 1.67 N
W air wood = 2.25 N W water wood = 0.65 N W air wood = 0.33 N
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BF brass = 1.88 -1.67 = 0.21N BF wood = 1.60 - 0.21 = 1.39 N p object =p F (W object /F B ) Density of Brass: 1*1.88 N/0.21 N = 8.95 g/cm^3 Density of Wood: 0.33 N/ 0.39 = 0.85 g/cm^3
Questions: Question 1. Use the expressions for W in air and W in water from the figure above to show that the buoyant force is equal to the difference between the weight in air and the weight in water. Show your work. F T, air = mg vs F T, water = mg - F B F B = W in air - W in water Question 2. Show that one can find the density of the unknown object by dividing its weight by the buoyant force: W object /F B =p object /p fluid To do this, use the following substitutions W object = p object Vobject g F B = W fluid = p fluid V fluid g V object =V fluid . (if the object is fully submerged) Show your work. p object =p F (W object /F B ) p object =W object (p Ft *V object ) Question 3. Do you think the metal is gold? To answer this, use a reliable online resource to look up the density of gold and other common metals. If you don’t think your metal is gold, what is it likely to be? Use your data to support your answer. Density of Brass: 1*1.88 N/0.21 N = 8.95 g/cm^3 No, I do not think the metal is gold. The metal is not gold as the density of gold is 19.3 g/cm^3 and the density of our metal is not close to that value. The metal is likely to be brass. Since the density of brass is 8.73 g/cm³, we got 8.95 g/cm³. Our value was closer to the value of brass. Question 4. The total buoyant force acting on both objects is simply the sum of the buoyant force on each individual object: F B , total = F B , wood + F B , slug . Explain why this is true using the fact that the buoyant force is due to the volume of the fluid displaced. This is true because of the Archimedes Principle. The total buoyant force (FB, total) acting on both objects, the wood and the slug, is the sum of the buoyant forces acting on each individual object, FB, wood and FB, slug. This is true because the buoyant force is due to the volume of fluid displaced by each object. Now, when we consider two objects immersed in the same fluid, such as the wood and the slug, the total buoyant force on both objects is the combined effect of each object displacing its own volume of fluid. So, we can write the total buoyant force as:
FB, total = FB, wood + FB, slug This is because each object contributes its own buoyant force based on the volume of fluid it displaces. When you sum these two buoyant forces together, you get the total buoyant force acting on both objects in the fluid. The principle behind this is that the buoyant force depends on the volume of fluid displaced by each object. So, when multiple objects are in the same fluid, you can calculate the total buoyant force by adding up the individual buoyant forces, which are determined by the volumes of fluid displaced by each object. Question 5. Is the value of the density of the wood you found greater than or less than that of water? When fully submerged, how does the buoyant force acting on the wood compare to the gravitational force acting on the wood? Does this explain why it floats to the surface when not weighted down? Density of Wood: 0.33 N/ 0.39 = 0.85 g/cm^3 The density of the wood is less than the density of water from our calculations. When an object is fully submerged in a fluid, the buoyant force acting on the object is equal in magnitude to the gravitational force acting on the object. The buoyant force counteracts the gravitational force, and the object doesn't experience a net force causing it to sink or rise; it remains in a state of equilibrium within the fluid. This upward buoyant force will cause the object to rise to the surface of the fluid until it reaches a point where the forces are balanced, and the object will then float. So, when the piece of wood is not weighted down, its average density is less than that of the surrounding water, and this difference in density results in the buoyant force pushing the wood upward and causing it to float to the surface. Question 6. How does the pressure vary with depth? Support your claims with your results. The deeper you are in a liquid the more pressure is above you because of the weight of the liquid. After looking at our data from part 2 it is proven the deeper you are the more pressure there is. Question 7. Look up the hydrostatic pressure equation. According to the equation, what physical quantities are represented by the slope and y-intercept of the plot? What should the value of the slope and y-intercept be? P=P 0 +pgh, The y intercept is atmospheric pressure, the slope is the density. The value of the slope is 0.0863 g/cm^3 and the y-intercept is 102.3 kPa. Discussion:
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In this lab we observed buoyant force and the relation it has with objects. We found that the buoyant force is equal to the difference of W in air and W in water . We were able to use the different equations to find the density of brass and wood. Our density of wood was found to be 8.95 g/cm 3 and wood to be 0.85 g/cm 3 . Our density of the slug showed that it was made of brass and not gold since the density was not close to the gold. We found that pressure and depth are directly related to one another. According to our graph as the depth increases, the pressure increases as well. Some sources of error can include human error where we had to use our best judgment when submerging the objects in water. Overall the lab allowed us to better understand the concept of fluid dynamics.