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: **
.4,7.9,12
7.4,7.8,7.9
.58,2.8,1.5
The fixed reference point was .4cm left of the intersection of the bar and the paper clip attached to rubber band B.
The forces were obtained from the individual graphs for each rubber band. The force for the each of the two rubber bands in A for that stretch were added together. I suspect there is something wrong here because the net force should be 0 since the entire system is at rest but -.58 + -1.5 + 2.8 != 0
** Net force and net force as a percent of the sum of the magnitudes of all forces: **
.72
14%
The net force was obtained by adding the forces with negative signs on the ones on the downward forces. The percentage was obtained by taking the magnitude of all of the forces (without signs) and dividing the net force by that magnitude (.72/(2.8+2.08)) and multiplying by 100.
** Moment arms for rubber band systems B and C **
7.5,4.1
I measured the distance from the hook on rubber band B to the hook connecting the bar to the rubber bands at A and did the same between rubber band C and the rubber bands at A.
** Lengths in cm of force vectors in 4 cm to 1 N scale drawing, distances from the fulcrum to points B and C. **
1.16,5.6,3
7.5,4.1
The first line is the length of the force vectors using a scale of 2cm to 1N. The second line is the distance from the fulcrum to the application of the forces B and C, respectively.
** Torque produced by B, torque produced by C: **
+4.3,-6.2
These are the torques for B and C respectively. They were obtained by multiplying the moment-arm for each rubber band system by the force of the rubber band.
** Net torque, net torque as percent of the sum of the magnitudes of the torques: **
-1.9,
-18
I added the positive and negative torques for B and C to obtain the first line and then divided that by the sum of the magnitudes of the torques (6.2+4.3) and multiplied by 100 to obtain the second line.
** Forces, distances from equilibrium and torques exerted by A, B, C, D: **
+1.4,0
-1.4,2.8
-.69,9.1
+.63,14.9
These are the forces and the distances from the point where force A meets the rod.
** 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: **
-.06
The picture is much as the example with the forces and distances listed previously. It is fairly accurate. There is a -0.06 difference in the force. I would have expected the net force to be 0.
** Net torque for given picture; your discussion of whether this figure could be accurate for a stationary rod: **
6Ncm
This figure seems to be off by about 6Ncm. The net torque should come out to be 0 if the rod is stationary.
** 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: **
-.8
.06,4.1
1.5
.8,19.6,4.1
I added up all the torques with their appropriate signs to arrive at -.8Ncm. The second line is the net force in N and the sum of the magnitudes of all forces. The percentage of the net torque vs. total torque is -.8/(3.9+6.3+9.4) = .041 * 100 = 4.1%
** 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
0,5.8
The magnitude of the net force is 0.
1,36.8,2.7
The first line is the net torque from summing the torques with directions, the second line is the net force and the sum of all magnitudes of forces acting. The percentage of the net force to total of all magnitudes of force is therefore 0. The last line is the the magnitude of the net torque and the sum of all torques acting about point A.
** In the second setup, were the forces all parallel to one another? **
I would estimate the angles changed for C, and D by less about .7 degrees. I took the arctangent of the length of the second length and the length of the difference from 90 degrees (.1cm for those that changed) and used that for rise/run (.1/length of rubber band).
** Estimated angles of the four forces; short discussion of accuracy of estimates. **
90,90,89.3,89,3
I stated how the estimates were made in the previous entry.
** x and y coordinates of both ends of each rubber band, in cm **
1,.9,1,8.7
6.4,13.8,5.7,21.5
15,1.3,13.3,8.8
** Lengths and forces exerted systems B, A and C:. **
7.8,-1.3
7.7,2.4
7.7,-1.8
These are the lengths of the rubber bands obtained from using the coordinates of their ends and utilizing the Pythagorean Theorem rather than measuring the lengths.
** Sines and cosines of systems B, A and C: **
0,1
-.91,1
.22,-.97
** Magnitude, angle with horizontal and angle in the plane for each force: **
90
95
282
These values were obtained from the arctangent of y/x distances calculated between the points of each vector.
** 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,-5.2,0,-1.3,0,-1.3
-1,9.6,-.25,2.4,-.22,2.4
1.2,-4.7,.3,-1.05,.26,-1.2
These values are the length of x and y components from the drawing followed by the calculated length/4 to obtain a value in N, and then the x and y components as calculated using the magnitude and the cosine and sine.
** Sum of x components, ideal sum, how close are you to the ideal; then the same for y components. **
.4,0,+.4
-.1,0,-.1
The first number in each line is the sum of the x and y components respectively. The ideal sum would be 0. The x-component is off by +.4 and the y-component is off by -.1
** Distance of the point of action from that of the leftmost force, component perpendicular to the rod, and torque for each force: **
0,1.3,0
5.8,2.4,13.92
11.7,-1.2,-14.04
The above are the distance from the point at which force B came into contact with the rod in cm; the y-components of each vector representing B,A, and C; and the torque of each.
** Sum of torques, ideal sum, how close are you to the ideal. **
-.12,0,-.12
The first is the sum of the torques acting around the point where B meets the rod. The ideal torque would be 0. The calculations are off by -.12Ncm
** How long did it take you to complete this experiment? **
About 4 hours.
** Optional additional comments and/or questions: **
&#Very good work. Let me know if you have questions. &#