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.
** Your optional message or comment: **
** 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: **
2.7 cm, 9 cm, 13.6 cm
8.2 cm, 7.5 cm, 10.4 cm
1.38 N, .80 N, 1.9 N
I just drew a line that I called origin that is on the left side and measured form there. I looked at the length on the x axis and found, approximately, N on the y. The results are merely the correlation between rubber band length and force. I set up the experiment, measured the bands, compared the lengths to the calibration graph to obtain the Newtons.I also multiplied the force on A by 2 since it has two bands.
** Net force and net force as a percent of the sum of the magnitudes of all forces: **
-2.43
60 percent
For the first, I made the forces negative for the bands B and C, then added them to positive A. For the percentage, I took the absolute value of the first number and dividedi tby the aboslute value of the sum of the forces.
** Moment arms for rubber band systems B and C **
63 cm
4.6 cm
These are the dsitances from A to B and from A to C respectively in terms of where the lines cross the x axis or the rod-line. They should be more even, but I apparently didn't do a good job of eyeballing the system when I set it up.
** Lengths in cm of force vectors in 4 cm to 1 N scale drawing, distances from the fulcrum to points B and C. **
5.52 cm, 3.2 cm, 7.6 cm
6.3 cm, 4.6 cm
The first line numbers are the Newtons x 4 cm. The second line is composed of the same numbers reported above, because the rod length didn't change.
** Torque produced by B, torque produced by C: **
-8.691
8.74
The results are derived by taking the moment-arm distance and multiplying it by the force that the one particular rubber band exerted. I am not quite sure about the signs. I know C is negative because of the clockwise direction, but does that make B positive?
** Net torque, net torque as percent of the sum of the magnitudes of the torques: **
.049
28 percent
I divided .049 by the sum of the torques (all positive this time). I think how close the results were indicates fairly decent accuracy.
** Forces, distances from equilibrium and torques exerted by A, B, C, D: **
.19 N, -1.50 N, -2.2 N, .23 N
0 cm, 3.6 cm, 19.4 cm, 25.4 cm
1.406, -12.45, -1.65, 1.725
I obtained the first two lines' results from the drawing, the last lines by multiplying the force by the distance.
** 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: **
-1.3
Although the picture makes it appear as though the positive and negative forces are equal, they aren't. Also, I don't think the drawing shows all positive forces.
** Net torque for given picture; your discussion of whether this figure could be accurate for a stationary rod: **
-.4
It could be fairly accurate since the torque sum is so close to 0.
** 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: **
-10.969
1.3, 2.14
61 percent
10.969, 17.231, 64 percent
These are the resultant forces/torques, divided by the sum of the forces/torques to get a percentage.
** 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: **
-1.396
.09, 4.11
2 percent
1.396, 33.904, 4 percent
These are the resultant forces/torques divided by the sum of the forces/torques to get the percentage.
** In the second setup, were the forces all parallel to one another? **
I would guess they were off by about 20 degrees. The B and C bands pulled the A and D bands inward at what I just guess to be about 20 degrees away from 90.
** Estimated angles of the four forces; short discussion of accuracy of estimates. **
90, 93, 94
I literally just guessed based on how close the angles were to 90 degrees. I think my estimates are accurate within 5 degrees.
** x and y coordinates of both ends of each rubber band, in cm **
(4.4, -12.2) (4.4, -4.2)
(7.0, 1) (5.0, 7.3)
(18.6, -10.6) (13.4, -.7)
These are the coordinates for the rubber bands used during the latest experiments.
** Lengths and forces exerted systems B, A and C:. **
8 cm, 1.14 N
6.6 cm, .1 N
11.2 cm, 3 N
Pythagorean Theorem for lengths and old calibrations information corresponding to these lengths for the force.
** Sines and cosines of systems B, A and C: **
-.017, +1
+.018, +.99
-.016, +.99
These are the sine and cosine of each angle found by using x and y components and the hypotenuse.
** Magnitude, angle with horizontal and angle in the plane for each force: **
1.14 N, 72.38, 82.5 degrees
.1 N, UND, 270 degrees
3 N, -62.28, 173 degrees
The force I determined by comparing the length of the hypotenuse to my calibrated graph data. The angles fro teh plane I found by taking arctan (y/x), and the final value I got by multiplying the first 2 together.
** 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, -8.4, 0, -8.4, .35, -1.09
.08, 4, .14, 4, 0, .1
.28, -10.8, .28, -10.80, 1.40, -2.66
For the first two numbers, I measured the original vectors, saw to what forces these lenghts would correspond, multiplied that number by four, and reported. FOr the second two numbers, I used the first two to find these x and y components. For the last two, I used the information in the preceeding blank to calculate the x and y components.
** Sum of x components, ideal sum, how close are you to the ideal; then the same for y components. **
2.17, 0, 2.17
-18.85, 0, 18.85
My results clearly are not very accurate. I obtained them by first marking above the directions of gthe x and y components using positive and negatives; then, I added them together.
** Distance of the point of action from that of the leftmost force, component perpendicular to the rod, and torque for each force: **
.3 cm, -2.1 N, -.63 N*cm
7.0 cm, 1 N, 7 N*cm
13.4 cm, -2.7 N, -36.18 N*cm
** Sum of torques, ideal sum, how close are you to the ideal. **
-29.81 N*cm, 0, 29.81 N*cm
Basically, my results are far off. This is the sum of the torques acting on the rod.
** How long did it take you to complete this experiment? **
3 hours
** Optional additional comments and/or questions: **
&#Good responses. Let me know if you have questions. &#