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PI Name
DA Name
R Name
~
Carlie ArlinghausDalton Shepler N'Diah Jones
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~
M1 (g)
δM1
F1
M2 (g)
δM2
F2
~
180
99.95
0.05
979.51
20
50
0.05
490
M3_min (g)
M3_max (g)
F3min
F3max
F3_measured
δF3
52.95
53.05
518.91
519.89
519.4
0.49
346.5
0.5
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~
~
X component
Y component
~
F1
-979.51
1.20004516307E-13
F3x
F3y
F3_calc
δF3
~
F2
460.4493841851 167.589870229578
519.0606 -167.58987023 545.44503618 -0.9559665
342.1062 0.047428776134412
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~
DA 1: Include the two scatter plots that compare the magnitudes and angles.
These charts should have titles, axes, labels, units, etc, as shown in Figure 2.
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~
F1x_max
F1x_min
F2x_max
F2x_min
F1y_max
F1y_min
F2y_max
F2y_min
F3x_max
F3x_min
F3y_max
F3y_min
~
-979.02
-980
460.90983356928 459.9889
1.2006455E-13 1.199445E-13
167.75746 167.4223
520.01106519909 518.11016643 -167.4223 -167.7575
~
F3x
F3y
F3_min
F3_max
F3
δF3
θ3max
θ3min
519.06061581491
-167.58987023
546.401018802702 544.4891 545.445052285 -0.955966518
-17.846434 -17.94129
-17.8938626344282 0.0474287761
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Researcher: Explain in detail how your group determined θ_max and θ_min.
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~
Large surface area
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~
θmax
θmin
μs_large max
μs_large mμs_large
δμs
~
17.5
16.5
1.51371554438863 1.510264
1.5119899257 0.0017256187
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~
~
Medium surface area
~
~
θmax
θmin
μs_Medium max
μs_Mediumμs_Medium
δμs
~
12.5
11.5
1.49096634108266 1.484058
1.4875121646 0.0034541765
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~
small surface area
~
~
θmax
θmin
μs_small max
μs_small mμs_small
δμs
~
11.5
10.5
1.48405798811891 1.475845 1.47995130429 0.0041066838
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~
~
~
~
~
~
~
~
Large surfa
1.51199
~
medium su
1.487512
~
small surfa
1.479951
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~
~
~
~
~
~
~
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Researcher: Create a careful scale drawing, similar to Figure 1, of the three forces, F_1,F_2, and F_3 that you have selected.
Then create a second figure
where these forces are rearranged without changing the magnitudes or directions so they follow the “tip-to-tail” method of addition of vectors.
Explicitly
explain the concept(s) which connect(s) your drawing to Newton’s Second Law.
Experimental Determination of F
3
and θ
3
θ
1
θ
2
θ
3_measured
δθ
3
Calculation of F
3
and θ
3
F
3
theory
θ
3
_calc
δθ
3
θ
3
δθ
3
We adjusted the inclined plane for the wooden block until it slid down. Then
proceeded repeat this procedure five times, we were able to see in our data, the
lowest and highest angles possible that would cause the wooden block to slide
down the plane.
DA2:
One striking feature of Figure 2 is the fact that the “Measured” points have error bars that are much larger than the “Calculated” points. Highlight
rows 31
through 37, right-click, and select “Unhide.”
Use these additional cells to determine
why
the “Measured” results have uncertainties that are so
much bigger than the “Calculated” results.
Ultimately you want to determine which cells in row 5 control the size of the error bars for each type of result.
The cells that include everything with the angle of F2. Since F1 Contributes nothing to the x direction, then F2 should be what is making the error bars for
figure 2 bigger than the calculated version.
DA: Create a scatter plot to compare μs based on the surfaces in contact. Label
all axes. In your caption, comment on whether the three μ measurements agree
or not.
PI: Using the charts provided by the DA, explain why the measured and calculated values of F_3 and θ_3 overlap or not. In your explanation, use terms
such as overlap and error bars etc.
The error bars do not overlap showing that there is a significant difference in values. The measured and calculated values do not overlap because they
represent separate things so they are not similar, hence them not overlapping.
PI: Using the chart provided by the DA, explain whether μs depends on the surface area of the
wooden block. In your explanation, use terms such as overlap and error bars etc.
The μs doesn't seem to depend on the surface area as all three values are around the same, ranging from
1.48-1.51, based on the data from the chart created by the DA. Because surface area doesn't play a role in
determining the force caused by friction, that static coefficient shouldn't depend on the surface area of the
block. Between the medium and small surface areas, the error bars overlap, showing that there is not a
significant difference in these values. This helps to give evidence to support that the μs doesn't depend on
surface area, since the differences aren't significant.
PI:
The formula μ_s≥tanθ relies on the fact that there are exactly three forces involved on the
incline plan, which form a triangle in exactly the same way that the Researcher described at the
end of part 1.
Build upon the Researchers work within the context of static friction to determine
WHY μ_s≥tanθ.
No the three measurements do not agree. The measurements are suppose to
be different due to the different amounts of surface area.
When something is on a ramp, "creating a triangle", due to the laws of static friction, the
coefficient of static friction is going to be the same as the tangent of an angle. Therefore, in order
for the formula μ_s≥tanθ to work, there needs to be exactly three forces, so the coefficient must be
greater than tanθ so that this triangle can be formed with the ramp.
0.8
1
1.2
1.4
1.6
1.8
2
339
340
341
342
343
344
345
346
347
342.106211535916
346.5
Angle difference between calc & measured
θ3_measured
θ3_calc
Degrees
0.8
1
1.2
1.4
1.6
1.8
2
505
510
515
520
525
530
535
540
545
550
545.445036180287
519.4
Difference of magnitude between calc & measured
F3_measured
F3_calc
Magnitude of orce
0.8
1
1.2
1.4
1.6
1.8
2
1.46
1.47
1.48
1.49
1.5
1.51
1.52
Static Coefficient
Large surface area
medium surface area
small surface area
µs
0.8
1
1.2
1.4
1.6
1.8
2
505
510
515
520
525
530
535
540
545
550
545.445036180287
519.4
Difference of magnitude between calc & measured
F3_measured
F3_calc
Magnitude of orce
0.8
1
1.2 1.4
1.6 1.8
2
339
340
341
342
343
344
345
346
347
342.106211535916
346.5
Angle difference between calc & measured
θ3_measured
θ3_calc
Degrees
I decided to move F2 and F3 because it was easier to make these connections “tip to tail” without disrupting the
direction and magnitudes of the vectors. Considering F2 and F3 are similar in size and direction
its best we
make those conections to F1.
This is the perfect representation of newtons second law, the F1=F2+F3 for both
mass and forces.
