Your 'torques' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Hello Prof. Smith,
Here is my Torques lab.
Tanya
Positions of the three points of application, lengths of systems B, A and C (left to right), the forces in Newtons exerted by those systems, description of the reference point:
1.7, 12.2, 13.5
7.8, 8.8, 11
-1.95, 2.2, -2.75
the left end of the rob
I divided the length of each rubber band by 4.
Net force and net force as a percent of the sum of the magnitudes of all forces:
-2.5
36.23%
I believe the middle rubber band was doubled, and the 2.2 N force should really be more like 4.4 N. This would make the sum of the forces about -.3 N, and the error would be small.
Moment arms for rubber band systems B and C
10.4
1.3
Lengths in cm of force vectors in 4 cm to 1 N scale drawing, distances from the fulcrum to points B and C.
12, 13.2, 15.6
10.4, 1.3
Torque produced by B, torque produced by C:
-20.28, -3.575
One of the torques should be positive, the other negative (one counterclockwise, the other clockwise). Otherwise these results follow from your data; however the torques should be much more nearly equal and opposite.
Net torque, net torque as percent of the sum of the magnitudes of the torques:
- 23.855, 100, because it is the same number, but with the opposite sign
Forces, distances from equilibrium and torques exerted by A, B, C, D:
2.625, -2.5, -2.3, 2.25
1, 4.5, 12.6, 13.2
2.625, -11.25, -28.98, 29.7
The sum of the vertical forces on the rod, and your discussion of the extent to which your picture fails to accurately describe the forces:
0.075
Because I obtained this small number, I can say that the left two forces are almost in equilibrium with the other two forces, which are on the right.
Net torque for given picture; your discussion of whether this figure could be accurate for a stationary rod:
6
Because I obtained this number I can make a conclusion about the torques. Since the number is positive, the net torque is acting in counterclockwise position.
For first setup: Sum of torques for your setup; magnitude of resultant and sum of magnitudes of forces; magnitude of resultant as percent of sum of magnitudes of forces; magnitude of resultant torque, sum of magnitudes of torques, magnitude of resultant torque as percent of the sum of the magnitudes:
-7.98
0.075, -2.55
2.94%
-10.155, -7.98, 127.26%
For second setup: Sum of torques for your setup; magnitude of resultant and sum of magnitudes of forces; magnitude of resultant as percent of sum of magnitudes of forces; magnitude of resultant torque, sum of magnitudes of torques, magnitude of resultant torque as percent of the sum of the magnitudes:
3.05
0.075, -2.175
3.44%
2.12, 3.05, 69.5%
In the second setup, were the forces all parallel to one another?
No, in the second setup, the forces were not all parallel to one another.
About 5 degrees
Estimated angles of the four forces; short discussion of accuracy of estimates.
85 degrees
I used a protractor to estimate the angles. I think the uncertainty could be about 2-3 degrees.
x and y coordinates of both ends of each rubber band, in cm
-4.5, -14.5, -4.5, -4.2
3.6, 1.7, 12.4, 0
1.8, -3.9, 6.7, -13.5
Lengths and forces exerted systems B, A and C:.
10.4
8.96
10.776
B: I obtained the result by subtracting the upper y coordinate from the lower one.
A: I obtained the result by finding the square root of the sum of squares of x and y length, which I found by subtracting the lower y coordinate from the upper one and the lower x coordinate form the upper one.
C: I obtained the result by finding the square root of the sum of squares of x and y length of the big triangle and than I found by using the same way of finding the unknown side of the smaller triangle. The length I found by using the same way as I used in “A” triangle. Then I just subtracted the length of the side in the small triangle from the length of the side in the big triangle.
Sines and cosines of systems B, A and C:
1, 0
0.98, 0.19
0.91, 0.44
Magnitude, angle with horizontal and angle in the plane for each force:
90, 10.93, -65.22
I followed the instructions and in the last case I obtained 24.78. I subtracted this number from 90 and used this angle as a negative equivalent for 24.78 angle.
x and y components of sketch, x and y components of force from sketch components, x and y components from magnitude, sine and cosine (lines in order B, A, C):
0, 10.4, 0, 10.4, 0, 1
1.7, 8.8, 1.7, 8.8, .98, 0.19
6.7, 13.5, 6.7, 13.5, 0.91, 0.44
Sum of x components, ideal sum, how close are you to the ideal; then the same for y components.
8.4, it is the same
22.3, it si the same
Distance of the point of action from that of the leftmost force, component perpendicular to the rod, and torque for each force:
1.1, 8.1, 12.3
-2.86, 18.144, -33.1362
Sum of torques, ideal sum, how close are you to the ideal.
-14.9922, -17.8422, -3 units
How long did it take you to complete this experiment?
two days because my PC crashed with all my data.
Optional additional comments and/or questions:
Your data are mostly reasonable, and most of your analysis is good. However some of the torque totals seem to be inconsistent with what is expected.