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: **
This is part 1. I lost my fourth push pin but will obtain one by Friday.
** 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: **
.5,6,12
8,7.5,7.5
2,1.5,1.6
left end of the horizontal line
I used the lengths for the corresponding graphs of the rubber bands to calculate the force in Newton.
These are where the vertical lines intersect the horizontal line, from left to right (where the paper clips were.) Next are the lengths of the rubber bands. Then are the forces corresponding to the rubber band lengths. I used the leftmost corner of the horizontal line as my reference point. I used the callibration graphs for the rubber bands I used to get the forces.
** Net force and net force as a percent of the sum of the magnitudes of all forces: **
-2N.
40%
A is going up by 1.5N
The B and C(2N +1.6N) are directed downward by 3.6N.
** Moment arms for rubber band systems B and C **
5.5, 6
These are the moment-arm for the force of the B then C rubber bands. I measured them from where the rod was held with the paper clips attached to A band.
** Lengths in cm of force vectors in 4 cm to 1 N scale drawing, distances from the fulcrum to points B and C. **
8,6,6.24
5.5,6
The first line is the lengths in cm of the vectors representing the forces exerted by systems B,A,and C. The next line is the distance of B and C from the fulcrum. I used the 4cm=1N to get the first row. I measured from the fulcrum with a ruler to get the distance of B and C from the fulcrum.
** Torque produced by B, torque produced by C: **
11Ncm
-9.6Ncm
For B, 2N*5.5cm=11Ncm
For C, 6cm *1.6N=9.6Ncm
C is negative becuase it would make the rod go in a clockwise direction.
** Net torque, net torque as percent of the sum of the magnitudes of the torques: **
1.4Ncm
6.7%
11Ncm-9.6Ncm=1.4Ncm
1.4Ncm/20.6Ncm=.067=6.7%
1.4Ncm is the calculated net torque and 6.7% is the percent of the sum of the magnitudes in the second line.
** Forces, distances from equilibrium and torques exerted by A, B, C, D: **
** 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: **
** Net torque for given picture; your discussion of whether this figure could be accurate for a stationary rod: **
** 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: **
** 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: **
** In the second setup, were the forces all parallel to one another? **
** Estimated angles of the four forces; short discussion of accuracy of estimates. **
** x and y coordinates of both ends of each rubber band, in cm **
** Lengths and forces exerted systems B, A and C:. **
** Sines and cosines of systems B, A and C: **
** Magnitude, angle with horizontal and angle in the plane for each force: **
** 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): **
** Sum of x components, ideal sum, how close are you to the ideal; then the same for y components. **
** Distance of the point of action from that of the leftmost force, component perpendicular to the rod, and torque for each force: **
** Sum of torques, ideal sum, how close are you to the ideal. **
** How long did it take you to complete this experiment? **
** Optional additional comments and/or questions: **
Very good. Note that the upward force at A is exerted by two rubber bands, each probably exerting a force of about 2 N, giving you a total upward force is 4 N, total downward force 3.6 N. These are nearly equal and opposite, which is of course the case here.