Archimedes' Principle Experiment: Unit IV - Fluid Statics and Dyamics: AP Physics B Web Portfolio

Archimedes' Principle Experiment

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Part A - Data Analysis (Water)

Graph 2.png

1. Slope = 0.05052 N/cm = 5.052 N/m

2. $Slope\:=\frac{F}{h}=\rho_FgA$

$\rho_{Water}=\frac{F}{hgA}=\frac{5.052\:\frac{N}{m}}{9.8\:\frac{m}{s^2}\cdot\pi\left(\frac{d}{2}\right)^2}=\frac{5.052\:\frac{N}{m}\:}{9.8\:\frac{m}{s^2}\cdot\pi\left(\frac{0.025m}{2}\right)^2}=1050\:\frac{kg}{m^3}$

3. $\%\:Error\:=\:\frac{\left|Experimental\:Value\:-\:Accepted\:Value\right|}{Accepted\:Value}\cdot100\%$

$\%\:Error\:=\frac{\left|1050\:\frac{kg}{m^3}-1000\:\frac{kg}{m^3}\right|}{1000\:\frac{kg}{m^3}}\cdot100\%\:=5.00\%\:error\:for\:water$

Part A - Data Analysis (Oil)

1. Slope = 0.04598 N/cm = 4.598 N/m

2. $Slope\:=\frac{F}{h}=\rho_FgA$

$\rho_{Oil}=\frac{F}{hgA}=\frac{4.598\:\frac{N}{m}}{9.8\:\frac{m}{s^2}\cdot\pi\left(\frac{d}{2}\right)^2}=\frac{4.598\:\frac{N}{m}\:}{9.8\:\frac{m}{s^2}\cdot\pi\left(\frac{0.025m}{2}\right)^2}=955.8\:\frac{kg}{m^3}$

3. $m=V\rho=\left(0.5L\right)\left(\frac{1\:m^3}{1000\:L}\right)\left(955.8\:\frac{kg}{m^3}\right)\left(\frac{1000\:g}{1\:kg}\right)=477.9\:g$ using Archimedes' Principle

$m=V\rho\:=\left(0.5\:L\right)\left(\frac{1\:m^3}{1000\:L}\right)\left(910\:\frac{kg}{m^3}\right)\left(\frac{1000\:g}{1\:kg}\right)=455\:g$ using mass of known 500 mL of oil

$\frac{\left|477.9\:g\:-\:455\:g\right|}{455\:g}\cdot100\%\:=\:5.04\%\:difference$

Part B - Data Analysis

$V_{small\:block}=lwh\:=\left(0.043\:m\right)\left(0.043\:m\right)\left(0.070\:m\right)\:=\:1.29\:\cdot10^{-4}\:m^3$

$V_{large\:block}=lwh=\left(0.075\:m\right)\left(0.075\:m\right)\left(0.038\:m\right)=2.14\:\cdot10^{-4}\:m^3$

$Fraction\:submerged_{small\:block}\:=\frac{h_{submerged}}{h_{total}}=\frac{6.8\:cm}{7.0\:cm}=97.14\%\:submerged$

$Fraction\:submerged_{large\:block}=\frac{h_{submerged}}{h_{total}}=\frac{2.85\:cm}{3.8\:cm}=75\%\:submerged$

$\rho_{small\:block}=\frac{m}{V}=\frac{0.1298\:kg}{1.29\:\cdot10^{-4}\:m^3}=1006.2\:\frac{kg}{m^3}$ according to definition of density

$\rho_{large\:block}=\frac{m}{V}=\frac{0.1632\:kg}{2.14\:\cdot10^{-4}\:m^3}=762.6\:\frac{kg}{m^3}$ according to definition of density

$\rho_{obj.}V_{obj.}=\rho_FV_{disp.\:F}$

$\rho_{small\:block}=\frac{\rho_FV_{disp.\:F}}{V_{small\:block}}=\frac{\left(1000\:\frac{kg}{m^3}\right)\left(0.9714\:V_{small\:block}\right)}{V_{small\:block}}=971.4\:\frac{kg}{m^3}$ according to Archimedes' Principle

$\rho_{large\:block}=\frac{\rho_FV_{disp.\:F}}{V_{large\:block}}=\frac{\left(1000\:\frac{kg}{m^3}\right)\left(0.750\:V_{large\:block}\right)}{V_{large\:block}}=750\:\frac{kg}{m^3}$ according to Archimedes' Principle

Discussion Questions

1. Submarines submerge and surface through use of ballast tanks which fill with water. To submerge, submarines open their ballast tanks, filling them with water, and increasing the total mass of the submarine. The volume of the submarine does not change throughout this process since the outside shell retains its shape. Therefore, the density of the submarine rapidly increases. When the density of the submarine is greater than that of water it starts to submerge. In order to surface, submarines use pumps to pump the water out of the ballast tanks, thus decreasing the total mass and density of the submarine. When the density decreases to that below water's, the submarine begins to rise to the surface.

2. Yes because the wood weighs the same in both configurations. In either configuration, the blocks of wood are in equilibrium, meaning that the buoyant force, the only other force acting upon the blocks, must equal the weight of the wood blocks. Since the weights are the same, the buoyant forces must be the same as well.

3. No because the weight of water and the weight of the block still have to be accounted for, a conservation of mass, per se. Even though the apparent mass of the block is less than it would be out of water, this means that the apparent mass of the water would be more due to Newton's Third Law, meaning the mass of the system would remain constant, the mass of the block + the mass of the water.

4a. $\rho=\frac{m}{V}$

$V=\frac{m}{\rho}=\frac{2.8\:kg}{2700\:\frac{kg}{m^3}}=0.00104\:m^3$

$\rho=\frac{m}{V}=\frac{m-m'}{V}=\frac{2.8\:kg-1.9\:kg}{0.00104\:m^3}=868\:\frac{kg}{m^3}$

4b. $\rho_F=\frac{m_F}{V}=\frac{m-m'}{\frac{m}{\rho_{solid}}}$

5. Yes, they experience the same buoyant force because they displace the same liquid. Due to Archimedes' Principle, because they displace the same liquid in the same quantities, they would have to have the same buoyant force.

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