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: **
resubmit w/&&&&&&
got stuck at
**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: **
I have the rest of the lab completed just need help w/calculations.
** 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: **
LINE 1 1.25, 7.75, 11 LINE 2 Length of rubber bands B 8.4 cm A 7.9 cm C 9.1 cm LINE 3 Forces B 1.0 N A .19 N C 2.0 N LINE 4 What I did was I drew the rod exactly according to scale, which was 15.5 cm. I placed the B rubberband system at 1.25 cm from the edge on the left. The A system was located in the middle which was at 7.75 cm. System C was located 3.25 cm beyond the A system, @ 10.55 cm. LINE 5 to obtain the Forces in Newtons I used the information gathered in my cm v. Force calibration curve that I constructed on July 12 for the rubberband calibration experiment. LINE 6 To obtain my results I measured the distances and then converted them to Newtons for my experiment. I then checked my work using the formula force * distance = force * distance. For this formula to work there has to be a mid point, kind of like a seesaw and then weights distrubuted on each side of the rod. So to check my work I used the equation f1*d1=f2*d2 1 N * 6.5 cm = 2 N * 3.25 cm 6.5 N * cm = 6.5 N * cm
** Net force and net force as a percent of the sum of the magnitudes of all forces: **
Fnet = Force in A + B + C
.19 N + 1 N + 2 N = 3.19 N
** Moment arms for rubber band systems B and C **
6.5 cm, 3.25 cm
The moment arm is the displacement from axis to the point where the line of the force crosses the axis. Since the forces are perpendicular to the rod, the moment arms are equal to the displacement from axis to the point of application.
** Lengths in cm of force vectors in 4 cm to 1 N scale drawing, distances from the fulcrum to points B and C. **
Distance from A to B on rod = 6.50 cm
Distance from A to C on rod = 3.25 cm
To determine the distance from the fulcrum to the points of application I had to use Pyth Theorey. I'm not sure if this is the right formula for this problem but I will elaborate on my thinking as much as possible so you can see where I am coming from.
The fulcrum is the support or point of support on which a lever turns in raising or moving something. So I took the top of where rubberband A stretches and measured to the rod. For this measurement I had 4.8 cm for the paperclip and the rubberband has a measurement of 7.9 cm (4.8 cm + 7.9 cm = ) 12.7 cm horizontal. I took this number and used the pyth theory to solve.
For A to B
12.7 cm ^ 2 + 6.50 cm ^ 2 = 14.27 cm from fulcrum to B
12.7 cm ^ 2 + 3.25 cm ^ 2 = 13.11 cm from fulcrum to C
N from fulcrum to B = 14.27 cm / 4 N / cm = 3.57 N
N from fulcrum to C = 13.11 cm / 4 N / cm = 3.28 N
** Torque produced by B, torque produced by C: **
Distance from A to B on rod =6.5 cm
6.5 cm * 4 N /cm = 26 N
distance from A to C on rod = 3.25 cm
3.25 cm * 4 N / cm = 13 N
** Net torque, net torque as percent of the sum of the magnitudes of the torques: **
26 N + 13 N = 39 N
sum of magnitudes = 6.5 cm + 3.25 cm = 9.75 cm
0.0975 m /39 N = 0.0025 N * m
** Forces, distances from equilibrium and torques exerted by A, B, C, D: **
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This was my initial set, where the rod was completely parallel and level.
To determine the Torque I multipled the force * how far they were from the end of the rod.
.76 N , 1 cm, .76 N * cm
1.52 N (force for 8 dominos), 2.75 cm, 4.18 N * cm
0.76 N (force for 4 dominos), 12.75 cm, 9.69 N * cm
1.52 N, 14.5 cm 22.04 N * cm
** 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: **
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.76 N + 1.52 N - .76 N - 1.52 N = 0 N
Once I had my rod situated on my wall (used alot of duck tape to get wooden board to stay up there), I began inserting the rubberbands and paperclips in the respected positions. After using pins to mark the length of the rubberbands on the bottom's location (8 domino length and 4 domino length) I measured the distance of each rubberband.
I can only assume my picture accurately represents the forces acting on the rod, because the Net force should be 0.
** Net torque for given picture; your discussion of whether this figure could be accurate for a stationary rod: **
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Took force and multiplied it by the distance from one side of the rod to the other based on its location.
.76 N * 1 cm = .76 N cm clockwise = + .76 N cm
1.52 N * 2.75 cm = 4.18 N cm clockwise = - 4.18 N cm
.76 N * 12.75 cm = 9.69 N cm clockwise = - 9.69
1.52 N * 14.5 cm = 22.04 N cm clockwise = + 22.04 N cm
** 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: **
8.93 N cm
The magnitude of each was .76cm + 1.52cm + .76cm + 1.52cm = 4.6 cm = 1.15 N (4.6 cm / 4 N / cm )
sum of magnitudes = 33.97 cm (added up distance each rubberband was stretched) 33.97 cm / 4 N / cm = 8.5 N
.76 N / 8.5 N = 8.9% of total.
Confused here...could you give me some pointers here...I already have the second setup for this lab ready and calculated so I am just waiting so some feedback to show me how to do this problem.
I'm not sure specifically what you are asking about. Assuming it's related to the second setup:
Find the angle of each force with the horizontal rod, using the arctan(y / x) idea.
Find the component of each force perpendicular to the rod. Perpendicular to the rod is perpendicular to the x axis, which is in the direction of the y axis. The y component of a force is F sin(theta).
** 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: **
I'm not sure what your question was, but I've made my best guess and hope my response is helpful.
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