Lab 8_ Fluid Statics

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Temple University *

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Mechanical Engineering

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Dec 6, 2023

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Lab 8: Fluid Statics Group Members : Kaira Torres and Oswaldo Anazgo Salas Goal The goal of this experiment is to better understand static equilibrium principles, buoyant force principles, and applications of Archimedes’ method. Additionally, we will learn how to use specific gravity to calculate density and how pressure is dependent on depth. Procedure For Part I, we will calculate the density of the metal slug by measuring and subtracting its weight in air from its weight when submerged in water. We will use a hanging force sensor to calculate the slug’s weight in the air and while submerged. Using our experimental density value, we will infer the material of the slug. Then, we will repeat the experiment using the density of a wooden cylinder, which will float in the water. Part II, we will use an absolute pressure sensor to measure the pressure exerted on a tube as depth increases. Error and Precautions The precautions would be to make sure to use the brass instead of the gray metal since different objects have different densities. Possible sources of error are calibration errors in our force sensor and Capstone. Additionally, if our cylinder was moving while we calculated force, the acceleration could have skewed our results for our total buoyant force. For Part II, there is a possible calibration error for the pressure sensor and human error in measuring depth on the meter stick we had to calculate the depth accurately using a pressure sensor and meter stick. Precautions would be to make sure that the brass and the wood block don’t touch the bottom of the beaker. Results Object Weight in air (N) Weight in water (N) Buoyant Force (N) Density (g/cm^3) Metal Cylinder 2.23 1.42 .81 2.75 Wooden Cylinder .36 -.09 .45 0.8 Metal Slug 1.66 1.47 .19 8.73 Wooden Cylinder and Metal Slug 2.02 1.38 .64 3.16

Data Table of Pressure (kPa-1)vs Depth (cm) Scatterplot of Pressure (kPa-1)vs Depth (cm) Slope = 127 kg/(ms)^2 Y-int: 102 =P0 Questions 1. Use the expression 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.

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: 3. Do you think the metal is gold? To answer this, use a reliable online source 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. According to Pearson.com, the density of gold is 19.3g/cm^3. Our experimental value of density for the metal slug was 8.73 g/cm^3. Due to the large difference of density, we do not believe the metal slug is gold. According to Pearson.com, the density of brass is 8.73 g/cm^3.We believe, due to the identical density values, that our metal slug is most likely brass. 4. The total buoyant force acting on both objects is simply the sum of the buoyant force on each individual object: Fb total = Fb wood + Fb slug. Explain why this is true using the fact that the buoyant force is due to the volume of the fluid displaced.

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5. Is the value of the density of the wood you found greater 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? Our experimental value of the wood’s density was smaller than the density of water, which you would expect since wood floats. When fully submerged, the buoyant force on wood is larger than the gravitational force, which causes the wood to float. This explains why, without being weighted down, wood will float on water. 6. How does the pressure vary with depth? Support your claims with your results. Pressure increases with depth since, as you go deeper into water, more water is above you pushing down on you. Our results support this claim since the slope of our scatterplot is positive, meaning that as depth increases so did pressure. 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? The hydrostatic pressure equation is P(fluid) = P0+ rho (fluid) x g x depth. According to this equation, the slope of a pressure vs depth graph is the product of the density and free- fall (g). The y-intercept is the measured atmospheric pressure or the pressure the object will experience on earth in the absence of depth or pressure from any fluids. The value of the slope and y-intercept should be Expected atmospheric pressure at sea level = 101.3 kPa Expected slope = 98 kg/ (m s)^2 Experimental Slope = 127 kg/(ms)^2 Experimental Y-int: 102 kPa =P0 Percent Error for Slope: 25.4% Percent Error for P0= .7%

Discussions In Part I, we recorded the weight of the cylinder in the air and when submerged in water. Subtracting these values, we calculated the buoyant force, which we then used to calculate the density of the cylinder. Using the experimental value for density, we were able to infer the material of our cylinder. We had a 0% percent error for the reported density value of brass; therefore, we are confident in our assumption that the cylinder is brass. Possible sources of error are calibration errors in our force sensor and Capstone. Additionally, if our cylinder was moving while we calculated force, the acceleration could have skewed our results. However, we cannot be confident in the reproducibility of our results since we only ran the experiment one time. Following, we repeated the experiment with just the metal slug and then with the wooden cylinder and metal slug. Subtracting these values, we were able to obtain experimental values for the buoyant force and density of only the wooden cylinder. We are confident in our results since our value for density was .8 g/cm^3, which is less than 1 g/cm^3, the density of water. Knowing that wood floats, we assume that our results are reasonable. In Part II, we used tubing and calculated the absolute pressure as a function of the depth. We observed the pressure increase as depth increased, which was expected due to the hydrostatic pressure equation that states pressure and depth are proportional. Due to our high percent error for rho x g (25.4%), we cannot be confident in our results.