Fluid Statics

In this experiment we measure the buoyant force on a cylinder experimentally.

The weight of a metal cylinder in air is 1.047 N. the weight of a metal cylinder in water is 0.68 N.

The buoyant force is equal to W-T=1.047 -.6700 = .377
Here is the propogated error for this calculation using MATLAB:

>> properror(B,[w t],[1.047 .670],[.001 .001])
ans =
    [      0.377]    '+/-'    [0.0014]
    'Percent Error'    '+/-'    [0.2111]


The second method for determining the buoyant force involves using archimedes principle. We measured the mass of a dry beaker to be .1909 kg and the mass with water to be .230 kg. the displaced water is .0391 and .384 N using this method.

error:

B = (981*w)/100 - (981*t)/100

>> properror(B,[w t],[.230 .1909],[.001 .0001])
ans =
    [       0.3836]    '+/-'    [0.0099]
    'Percent Error'    '+/-'    [2.5703]


The third method is to use volume of the cylinder. V=pi*r^2*h.
we measured the height to be .0765 m, the diameter to be .025m.
calculating volume in MATLAB:

>> properror(V,[r h],[.0125 .0765], [.001 .0001])
ans =
    [  3.7552e-005]    '+/-'    [6.0085e-006]
    'Percent Error'    '+/-'    [    16.0005]
Water displaced = V_cyl * density_water*9.81. using 1000kg/m^3 calculating the buoyant force in MATLAB:

>> properror(V,[r h],[.0125 .0765], [.001 .0001])
ans =
    [       0.3684]    '+/-'    [ 0.0589]
    'Percent Error'    '+/-'    [16.0005]

Summary:

The buoyant forces all agree to within +_.01. Calculations made with force analysis had the least error at .0014 of the calculated value as well as the least percent error. probably due to the least number of variables and most precise of measurements taken.

If the cylinder had touched our water container our values would have been too low as normal force would also have been contributing to our upward force.