Buoyancy Lab

Data

Object Masses

Object	Symbol	Mass (kg)
Metal Cylinder	$M_C$	0.2245
Wooden Block	$M_B$	0.0322

Experimental Tension Values

Object	Scenario	Tension (g)	Symbol	Tension (N)
Metal Cylinder	Submerged	211.6	$T_C$	$0.211(9.81)=2.07$
Wooden Block	Sinker Submerged	85.5	$T_S$	$0.085(9.81)=0.833$
Wooden Block	Sinker & Block Submerged	34.75	$T_B$	$0.035(9.81)=0.343$

Free Body Diagram

Calculations

Weight of Metal Cylinder

$$W=mg$$ $$W=M_{C}g$$ $$W=2.20N$$

Tension in string with Metal Cylinder submerged

Multiply the reading of the balance in kg by gravity to determine tension in N

$$T_C(kg)\times 9.81=T_C(N)$$ $$T_C=2.07N$$

Experimental Buoyant Force on Metal Cylinder

Newton’s first law:

$$\Sigma F=ma$$

There is no acceleration:

$$\Sigma F=0$$

General Scenario:

$$0=T+B-mg$$

Specific Scenario with proper symbols:

$$0=T_C+B_C-M_{C}g$$

Solve for B:

$$B_C=M_{C}g-T_C$$

Substitute:

$$B_C=0.2245(9.81)-2.07=0.132N$$

Theoretical Volume and Buoyant Force for Metal Cylinder

Apparently the thing is tungsten

$$M=V\times \rho$$ $$M_C=V_C\times {\rho }_C$$ $$V_C=\frac{M_C}{{\rho }_C}$$

Substitute:

$$V_C=\frac{0.2245}{19.25\times 10^3}=0.00001166m^3$$

Equation for Buoyant Force

$$B={\rho }_{fluid}\times V_{displaced}\times g$$ $$B={\rho }_W\times V_C\times g$$

Substitute:

$$B_C=1000\times 0.00001166\times 9.81=0.114N$$

Error % In Buoyancy

$${\%}_E=\frac{E-T}{0.5(T+E)}$$ $${\%}_E=100\frac{0.132-0.114}{0.5(0.132+0.114)}=14.6\%$$

Block Density

Find buoyant force, which can be combined with mass to find density later:

$$\Sigma F=ma=0$$

Just sinker submerged:

$$0=T_S-M_{B}g-T_W$$ $$M_{B}g+T_W=T_S$$

Block and Sinker submerged:

$$0=T_B+B-M_{B}g-T_W$$ $$T_B+B=M_{B}g+T_W$$

Combine equations:

$$T_B+B=T_S$$ $$B=T_S-T_B$$

Relate buoyant force, volume:

$$B={\rho }_{fluid}\times V_{displaced}\times g$$

Substitute and solve for volume:

$${\rho }_{fluid}\times V_{displaced}\times g=T_S-T_B$$ $$V_{displaced}=\frac{T_S-T_B}{g{\rho }_{fluid}}$$

Relate density mass, volume:

$$\rho =\frac{m}{V}$$ $$V=\frac{m}{\rho }$$

Substitute:

$$\frac{M_B}{\rho }=\frac{T_S-T_B}{g{\rho }_{fluid}}$$

Solve for $\rho$:

$$\frac{M_B}{\frac{T_S-T_B}{g{\rho }_{fluid}}}=\rho$$ $$\frac{g{M}_B{\rho }_{fluid}}{T_S-T_B}=\rho$$

Substitute real values:

$${\rho }_B=\frac{9.81\times 0.0322\times 1000}{0.833-0.343}=644.7kg/m^3$$