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 cm, 11 cm, 16 cm
7 cm, 8 cm, 8.5 cm
1.3 N, 1.5 N, 1.6 N
The reference point was the point A where the rubber band crossed with the threaded rod.
** Net force and net force as a percent of the sum of the magnitudes of all forces: **
+4.4 net force
&#Some of the forces are upward and some are downward; also the force at A should be that of the two rubber bands acting at that point.
You need to choose a positive direction and make each force positive or negative, depending on whether it acts in this direction or opposite to this direction.
This conclusion will need to be modified. The net force is the sum of all the forces; when finding the sum you have to take signs into account. &#
10.3%
I obtained the first value by adding together the forces. I obtained the second number by taking the percent of the sum of the magnitude of the forces of all three of the rubber band systems.
** Moment arms for rubber band systems B and C **
12 cm, 13 cm
I obtained these numbers by measuring the position of the central fulcrum to the force from rubber band B and rubber band C.
The fulcrum is at A, and the moment arms are from B to A and from A to C. What are those distances?
** Lengths in cm of force vectors in 4 cm to 1 N scale drawing, distances from the fulcrum to points B and C. **
10 cm, 13 cm, 12 cm
12 cm, 13 cm
The first set of numbers are the lengths in cm of the vectors representing the forces exerted by the B, A, and C systems. The second set of numbers are the distances from the fulcrum to the point of application of the two 'downward' forces.
** Torque produced by B, torque produced by C: **
-15.6 ,+20.8
These results are the product of the moment-arm and the force, which is the torque.
Once moment-arms are corrected, this can be easily corrected.
** Net torque, net torque as percent of the sum of the magnitudes of the torques: **
5.2
1.25 %
I obtained the first result by adding together the two torques. I obtained the second number by find the percent of the sum of the percent of the magnitudes.
** Forces, distances from equilibrium and torques exerted by A, B, C, D: **
2.7 N, 0 cm, 2.7 cm * N
-4.7 N , 4 cm, -18.8 cm * N
-3.7 N, 12 cm, -44.4 cm * N
2.3 N, 17 cm, 39 cm * N
I obtained the first number in each line by using the 1 N to 4 cm conversion and measured the lengths. I obtained the second number by measuring the distances from equilibrium. I obtained the third number by measuring the torque. which was the meters * Newtons.
** 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: **
-3.4 N
The picture shows the upward forces and downward forces acted up on the threaded rod. I believe that the picture depicts the forces acting upon the rod to an accurate extent.
** Net torque for given picture; your discussion of whether this figure could be accurate for a stationary rod: **
2.7 cm * N, -18.8 cm * N, -44.4 cm * N, 39 cm * N
There is nothing in any picture or on the rod you used that permits a 44.4 cm moment arm. Nothing is that long.
Please explain in detail how you are measuring the moment arms.
The figure could be an accurate depiction of the torques actually acting on a stationary rod. Two of the torques are negative and two are positive, standing for the four hooks and rubber bands acting on the 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: **
-21,5 cm * N
-4 cm/N, 1.9 cm/N
1.8%
** 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: **
-2 cm * N
-1.35 cm/N
1.97%
** In the second setup, were the forces all parallel to one another? **
No, all the forces were not parallel to one another. I think they vary by about 10 degrees. I made my estimates by looking at the setup and estimating by how many degrees that the forces varied from being parallel to one another.
** Estimated angles of the four forces; short discussion of accuracy of estimates. **
about 90 degrees, 85 degrees, 85 degrees, 90 degrees
I made my estimates by examining the setup and seeing how far away or close they were to the 90 degrees. I believe that i was accurate in making the estimates.
** x and y coordinates of both ends of each rubber band, in cm **
( 3 cm, 1.5 cm) (3 cm, 9.5 cm)
(7.5 cm, 12 cm) ( 7.5 cm, 18 cm)
(14 cm, 2 cm) (11.5 cm, 9.5 cm)
These numbers are the x and y coordinates on the graph of the positions of the rubber bands. I obtained these numbers by making an x and y axis just below and to the left of all my points, then evaluating the coordinates for each system of rubber bands.
** Lengths and forces exerted systems B, A and C:. **
6 cm, 58.8 N
8 cm, 78.4 N
7.9 cm, 77.4 N
None of these rubber bands will exert a force of over 50 N without breaking, or at least being severaly stretched.
The second rubber band you report appears to have a length of 6 cm; be sure this is correct, since most of the rubber bands have unstretched lengths of over 7 cm.
The first number in each line are the lengths of the rubber band systems. The second number in each line is the force exerted by the rubber band systems. The first number was found by using the Pythagorean Theorem and coordinates to find the length. The second number was found by using the force equation to find the force.
** Sines and cosines of systems B, A and C: **
I do not understand how to find the sine and cosine for the systems. Could you please help me understand how to do this part of the lab? Thank you!
The forces are directed parallel to the rubber bands, so each force is parallel to a vectors between the ends of the corresonding rubber band.
If for example a rubber band extends between the points (5 cm, 6 cm) and ( 12 cm, 14 cm) then the line segment corresponding to the rubber band has x displacement 12 cm - 5 cm = 7 cm, and y displacement 14 cm - 6 cm = 8 cm. This line therefore has length sqrt( (7 cm)^2 + (8 cm)^2 ) = 10.6 cm, approx., and makes angle arcTan(8 cm / (7 cm) ) = 49 deg, approx., with the direction of the positive x axis.
The magnitude of the force exerted by this rubber band is found from the force vs. length calibration graph for the band. A typical force corresponding to a 10.6 cm length would be around 3 Newtons; that will be used in this example.
The force exerted by the rubber band on the first end will either be directed at the 49 deg angle. So the force vector is about 3 Newtons at about 49 degrees.
The force exerted on the other end will be equal and opposite to this--about 3 Newtons at about 180 deg + 49 deg = 229 deg.
The components of these forces would be found using the most basic methods of analysis of vectors, as for example in Introductory Problem Set 5.
** Magnitude, angle with horizontal and angle in the plane for each force: **
58.8 N, 0 deg
78.4 N, 0 deg
77.4 N, 18.43 deg
I determined the first numbers in each line by using the force equation to find the force in Newtons. I determined the second number in each line by using the arcTan button on my calculator to determine the degrees, by dividing the y component by the x.
See my preceding note. If the x axis is parallel to the rod, then your first two angles would each be 90 deg..
** 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: **
I do no understand how to do this section
** Sum of torques, ideal sum, how close are you to the ideal. **
** How long did it take you to complete this experiment? **
2 hours
** Optional additional comments and/or questions: **
I had a few problems with this lab, could you please help me by giving me some feedback so i can do those sections of the lab correctly? thank you!
Please see my notes and send a revision, which may of course include additional questions.
&#Please respond with a copy of this document, including my comments. Insert your revisions and/or questions and mark them with &#&#. &#