torques

Your work on torques 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:

1.5, 8, 11.5

10.1, 12, 14.2

0.74, 1.32, 1.9

I used the left end of the horizontal line as my reference point. The line for band B intersects the horizontal line 1.5cm from the left end of the line. The line for band A intersects the horizontal line 8cm from the left end. The line for band C intersects the horizontal line at 11.5cm from the left end.

I obtained the force in Newtons for each rubber band by finding the length on its own calibration graph and then estimating the corresponding force along the curve. For each rubber band I used its individual calibration graph.

The system A was to consist of two rubber bands in parallel. If this is the case, then I suspect your force from system A is more than that reported here--about twice as much, depending on the calibration of the other rubber band in A.

Net force and net force as a percent of the sum of the magnitudes of all forces:

-1.32

The sum of my forces was -0.74N + -1.9N + 1.32N = -1.32N. Therefore the sum of the magnitudes of my forces was equal to my net force. (-1.32/-1.32)100 = 100%

If the force at A was in fact double the reported force, your net force would be 0.

Moment arms for rubber band systems B and C

6.5, 3.5

Lengths in cm of force vectors in 4 cm to 1 N scale drawing, distances from the fulcrum to points B and C.

2.5, 3.0, 3.5

6.5cm, 3.5cm

Torque produced by B, torque produced by C:

4.81, -6.65

Net torque, net torque as percent of the sum of the magnitudes of the torques:

-1.84

16%

I found the net torque by adding the torques of the bands together: -4.81+6.65 = -1.84. Then I found the net torque as a percent of the magnitudes by adding the magnitudes of the torques, 4.81+6.65 = 11.46, and then dividing the magnitude of the net torque by this number and multiplying by 100 to get a percentage: (1.84/11.46)100 = 16%

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):