Phy 231
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: **
1.2, 8.2, 14.0
7.3, 7.4, 7.5
.3, .4, .5
The reference point is the left end of the rod.
The rubber bands are 7cm when they are not stretched, each cm the rubber band is stretched, is equal to 1N in force. So if the rubber band is stretched .3cm, the rubber band is exerting .3N of force.
** Net force and net force as a percent of the sum of the magnitudes of all forces: **
-.4
33
The first line is the net force of the system; I used .4 as the positive upward force, and .5 and .3 as the negative downward force. The second number is the net force as a percent of the sum of all forces: .4/1.2 * 100%.
** Moment arms for rubber band systems B and C **
7.0, 5.8
The first number is the distance in cm from the fulcrum to B. This number was found by subtracting 1.2 from 8.2cm. The second number is the distance in cm from the fulcrum to C. This number was found by subtracting 8.2 from 14.0cm.
** Lengths in cm of force vectors in 4 cm to 1 N scale drawing, distances from the fulcrum to points B and C. **
1, 1.5, 2
7.0, 5.8
The first line is number of cm representing each Newton. The second line is the distances from the fulcrum to the two downward forces found in the previous box.
** Torque produced by B, torque produced by C: **
+2.1, -2.9
These numbers are the distance from the fulcrum to either B or C times the force in Newtons of the associated rubber bands. These two numbers are in Ncm.
** Net torque, net torque as percent of the sum of the magnitudes of the torques: **
-.8
16
I divided .8 by 5 and then multiplied this by 100%.
This is the percent error in my experiment.
** Forces, distances from equilibrium and torques exerted by A, B, C, D: **
(-1.5N, 10cm, -15Ncm)
(1.5N, 8cm, 12Ncm)
(-.7N, 5cm, -3.5Ncm)
(1.9N, 10cm, 19Ncm)
The first number in each line are the Newton force when there are 8 dominos, 8 dominos, 4 dominos, and 10 dominos. The second number in each line is the distance in cm between the fulcrum and the end of the rubber band. The third number in each line is the torque found by: torque = distance * force.
** 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: **
1.2N
My first two forces (A, B) are medium sized. C is the smallest arrow. D is the largest arrow. The larger the arrow, the greater the force. The picture doesn't take into account where the forces are being applied. The farther away from the fulcrum the rubber band is, the less force the rubber band must act to keep the rod at equilibrium.
** Net torque for given picture; your discussion of whether this figure could be accurate for a stationary rod: **
14Ncm
I would think that this result would be wrong because instead of finding the torque around the fulcrum, we are calculating the torque around A. So we don't calculate A's torque at all because the distance from A is 0cm.
** 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: **
+24.5
1.2N, 5.6N
21%
24.5Ncm, 51.5Ncm, 48%
Sum of the torques: B: 2cm*-1.5N=-3Ncm. C: 15cm*-.7N=-10.5Ncm. D: 20cm*1.9N=38Ncm. -3Ncm-10.5Ncm+38Ncm=+24.5Ncm. I believe the percents are the percent error.
** 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: **
-8Ncm
-1N, 5N
20%
-8Ncm, 70Ncm, 11%
I found the percent of the sum of magnitudes by taking the resultant torque / sum of the magnitudes * 100%.
** In the second setup, were the forces all parallel to one another? **
The rod slanted towards the third quadrant. The ramp slants around 30 degrees. I wrote a horizontal line that the rod should be and a line the actual rod is and estimated how many degrees the rod slanted.
** Estimated angles of the four forces; short discussion of accuracy of estimates. **
60, 120, 60, 120
These numbers are the angle in degrees of the rubber bands that are exerting a force on the rod. I drew a horizontal where 90deg would be then estimated the difference in the actual angle.
** x and y coordinates of both ends of each rubber band, in cm **
(3.5, 3.2), (3.9, 11.4)
(10.7, 18.5), (10.4, 26.6)
(20.6, 4.0), (16.8, 13.8)
I drew an x y coordinate system with the origin to the left and bottom of the rubber band points. These are the x and y coordinates in cm of rubber bands B, A, and C.
** Lengths and forces exerted systems B, A and C:. **
8.2, .3
8.1, .3
10.5, .9
The first number in each line is the distance the rubber band is stretched found by the eqn: c^2 = a^2 + b^2. The second number in each line is the force in Newtons made by stretched the rubber band. 1N = 4cm. The rubber band starts at length 7cm.
** Sines and cosines of systems B, A and C: **
1, .05
1, .04
.9, .4
These are the magnitude of the sines and cosines of each rubber band. I divided y component/hypotenuse to find the magnitude of sine, and x component/hypotenuse to find the magnitude of cosine. The three rubber bands are slightly tilting towards the 2nd quadrant with rubber band C having a larger slant.
** Magnitude, angle with horizontal and angle in the plane for each force: **
.3, 87, 93
.3, 88, 272
.9, 69, 291
The first number in each line is force in Newtons that I calculated earlier. The second number in each line is the angle with the x axis for each rubber band found by arctan(y/x). The third number in each line is the angle in the plane for each of the force vectors found by starting at the positive x axis and counting how many degrees the rubber band is away from it. I used the angle with the x axis to make it more exact.
** 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): **
.3, 8.1, .08, 2, -.4, 8.1
.4, 8.2, .1, 2, .3, -8.2
3.8, 9.8, 1, 2.5, 3.8, -9.8
The first two numbers in each line are the x and y components of my graph. The next two numbers are the same numbers but divided by four to represent the force components. The next two numbers in each line were found by the equation: x or y comp = magnitude*sin or cos(theta). I took the theta from the second angles I found in the question above.
** Sum of x components, ideal sum, how close are you to the ideal; then the same for y components. **
3.7, 0, 3.7
-9.9, 0, -9.9
The sum of the x components should be zero, and the same goes for the y components. Since the rod is at equilibrium, the bands are producing just as much force in one direction as they are in the other. To find the sum of the x and y components, I added them from each rubber band.
** 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? **
3 hours
** Optional additional comments and/or questions: **
I performed this experiment last semester and copied it to a word document. I know I left the last two questions blank but I cant really remember what was going on and if it doesnt hurt my grade alot, I'd rather leave them blank. Sorry.
No problem; very unlikely to hurt your grade and you understand what you're doing. Your work looks good.