Archimedes Lab

 

I. Data Analysis—Part A

 

a. Water

diameter = .019 m

A = pi*r² = 2.8*10^-4 m²

slope = .0291 N/cm = 2.91 N/m = p_fgA

p_f = slope/gA = (2.91 N/m)/(9.8 m/s²* 2.8*10^-4 m²) = 1050 kg/m³

% error = (1050 – 1000) kg/m³ / 1000 kg/m³ = 5.0 %

 

b. Oil

diameter = .019 m

A = 2.8*10^-4 m²

Slope = .0250 N/cm²= 2.50 N/m²

P_f = slope/gA = (2.50 N/m²) / (9.8 m/s² * 2.8*10^-4m²) = 900. kg/m³

% error = (910 – 900) kg/m³ / (910 kg/m³) = 1.1%

 

II. Data Analysis—Part B

 

a. small block

i. definition of density

p = m/V = .1298 kg / (.043 m * .043 m * .070 m) = 1.0*10³ kg /m³

ii. Archimedes’ principle

fraction submerged = (.070 m - .002 m)/(.070 m) = .97

p_wood / p_water = fraction submerged

p_wood= .97(1000 kg/m³) = 970 kg/m³

 

b. large block

i. definition of density

p = m/V = .1632 kg / (.038 m * .075 m * .075 m) = 764 kg/m³

ii. Archimedes’ principle

fraction submerged = (.038 m - .0095 m) / (.038 m) = .75

p_wood /p_water = fraction submerged

p_wood = .75(1000 kg/m³) = 750 kg/m³

 

III. Discussion questions

 

1. A submarine can submerge or surface by adjusting its density—i.e if the submarine lowers its density so that it is less than water’s, it will rise to the surface, and if its density is greater than water's, it will sink. So, all the submarine needs to do is find a way to adjust its weight (say, take on or remove water) to adjust its depth.

 

2. No; the buoyant force must always equal the weight of the block, otherwise the wood block would not float.

 

3. The measured weight would be equal to the sum of the individual weights. Because the water exerts an upward buoyant force on the block equal to the weight of the block, the block exerts a downward force equal to the weight of the block on the water. So there are two forces acting on the water (the weight of the water and the weight of the block) that the normal force must balance, making the measured weight equal to the sum of the weights.

 

4b) W – W’ = p_fgV

p_f = (W – W’)/(gV)

p_f = (m – m’)/(V)

p_f = p_o(m – m’)/(m)

p_f = p_o(1 – m’/m)

where p_o is the density of the solid, m is the solid’s mass, and m’ is the solid’s apparent mass

4a)

p_f  = (2700 kg/m³)(1 – 1.90 kg/2.80 kg) = 868 kg/m³

 

5. Yes; the buoyant force depends only on the density of the fluid, the acceleration due to gravity, and the volume displaced by the solid. So, since these quantities are equal for a solid and a hollow steel ball of the same size, the buoyant forces on the two are equal.