Experiment 1 : Fluid Statics

In this experiment, Fluid Statics, the buoyant force (B) is going to be obtained by three different ways. Then the final results obtained from each way will be compared to see how different the buoyant force can vary by applying different calculation or measurement. The primary devices used in this experiment were metal cylinder graduated cylinder, and beaker. For part A of this experiment, the metal cylinder was weighed in air and also weighed after it was submerged in the water. Then the concept B=T-W will be applied to calculate the buoyant force. The part B of the experiment, the weight of the beaker was measured first, then it was again measured after the metal cylinder was put in the cylinder and the water was spilled out from the graduated cylinder. After applying some calculations, the displaced water, which should be equal to the buoyant force can also be obtained. Lastly, for the way of getting buoyant force in part C, the formula, ρgV will be used. After the width and the length of the metal cylinder were measured, the volume was calculated and the mentioned formula was applied to get the final result for the buoyant force.

pictures taken from part1 of this experiment :

 the metal cylinder was hung so that its weight could be measured

picture taken from part two of the experiment:

submerged the metal cylinder into the graduated cylinder and then weighted the beaker with the spilled water

Part A data and calculation 

Wair=1.13N	Wwater=0.67N
B=T-W
B=1.13-0.67=0.46N

Part B data and calculation 

Beaker m :	0.15086 (kg)
Beaker + water :	0.19177(kg)
0.19177(kg)-0.15086(kg)=0.04091(kg)
Water m:	0.04091(kg)
0.04091x9.8=0.400918(N)
Wf=B:	0.400918(N)

Part C data and calculation

Equation for volume:	V=πr2h
Height(h):	3.6(m)
Diameter(d):	0.025(m)
V=π(0.0125)2(3.6)=3.73x10-5(m3)
Equation for Wf:	ρgV
Wf =B=(1000 kg/ m3)(9.8m/s2)( 3.73x10-5m3)=0.366(N)

D.Summary

Results for Buoyant Force

	Part A	Part B	Part C
Buoyant force (B)	0.46(N)	0.400918(N)	0.366(N)

  

Part A uncertainty:

Air=1.13

u{Air}=0.015

Water=0.67

u{Water}=0.035

radical(((1-Air)*u{Air})^2+((Water-1)*(0.035)*u{Water})^2)=0.007

---> 0.45 +/- 0.007

Part B uncertainty:

let W_beaker+water=A

let W_beaker=B

displaced water= (A-B)

A=0.19177

u{A}=0.000035

B=0.15086

u{B}=0.00003

radical(((1-B)*9.8*u{A})^2+((A-1)*9.8*u{B})^2)=3.8*10^(-4)

--->0.400918 +/- 3.8*10^(-4)

Part C uncertainty:

h:3.6

u{h}:0.3

D:0.025

r=0.0125

u{r}:0.00025

radical((2*pi*rho*g*r*h*u{r})^2+(pi*rho*g*r^2*u{h})^2)=1.6

--->0.366 +/- 1.6

Questions
1.

	Part A	Part B	Part C
Results	0.45 +/- 0.007	0.400918 +/- 3.8*10^(-4)	0.366 +/- 1.6

2.

I think the result obtained from Part B is the most accurate because based on the calculated uncertainties, Part B has the smallest value for its uncertainty.

3.
The buoyant force should be smaller since there is an extra force points upward (same direction as the buoyant force) by the bottom of the water container. Therefor, now B=T-W-Forceupward