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: **
could you check the boxes marked with a &&&&....I m not sure if I am doing this right.
** 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: **
&&&&&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 sum of the vertical forces on the rod, and your discussion of the extent to which your picture fails to accurately describe the forces: **
&&&&&.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: **
&&&&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.
** 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: **
&&&&&&&&&&&&&LINE 1 :
To calculate the sum of torques about point of action of the leftmost force I added up the torques 0.0165 + 0.0157 + 0.0173 + 0.0025 = 0.05199. to get the torques I (0.0088m * 1.88N)+ (0.0087m * 1.8 N)+(0.0091m * 1.8 N)+(0.001m * 2.5 N)
I converted the cm to M and then multipled by the Netwons that corresponded w/the length.
LINE 2 :
magnitude of force and what is sum of magnitude of all forces on rod?
1.88N +1.8N + 1.9N + 2.5 N = 8.08 N
sum of magnitude of all forces = 1.88 N + - 1.8 N + - 1.9 N + 2.5 N = .68 N.
LINE 3 :
magnitude of resultant force/sum of magnitudes
.68 N /8.08 N = 8.4%
LINE 4 :
magntidue of torque and what is sum of magnitude of all torques
To find the magnitude of torque and sum of magnitude of all torques acting on rod?
TO get the first number I added 0.016544 N + -0.01566 + -0.01729 + 0.0025. To get the second number I added all the torques together.
-0.013906/ 0.051994 = -.267
LINE 5 :
Explained
** In the second setup, were the forces all parallel to one another? **
&&&&&&&&no, they were not, because in the second setup we had to add additional force (a few cm) to C. This caused the rod to be slanted up in the left direction.
** Estimated angles of the four forces; short discussion of accuracy of estimates. **
My estimates...to make this estimate I I drew a straight horizontal line like the one my rod made in the initial setup. I then drew a line where the rod was now. and marked where each rubberband was, A, B, C, and D, before and after the movement.
I found the points where the initial mark would be if the rod was horizontal then marked where it was when they were slanted.
&&&&&&&&&&&&I then took those points and used the tan-1 of y/x to find the angle.
tan ^ -1 1cm / .5 cm = 63 degrees.
tan ^ -1 .4cm / .1 cm = 76
tan ^ -1 .4cm / .1 cm = 76
tan ^ -1 1cm / .5 cm = 63.
** 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: **
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: **
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
** Torque produced by B, torque produced by C: **
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
** Net torque, net torque as percent of the sum of the magnitudes of the torques: **
.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.
** Forces, distances from equilibrium and torques exerted by A, B, C, D: **
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
** 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: **
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.
** Net torque for given picture; your discussion of whether this figure could be accurate for a stationary rod: **
Net torque = Postion A + (-) Position B + (-) Position C + Position D
Net torque = (0.0088m * 1.88 N) + (-0.0087m *1.8 N) +(-0.0091m*1.9 N) +(0.01m*3.0) = 0.0795 N * m
For my initial setup the rod was parrallel to the lines on the grid of the board. And the rod was straight. This could be an accurate depiction of the torques acting on the rod because of the way the rod was positioned, it was not at an angle. Thus the torques canceled each other out.
** 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: **
LINE 1
sum of torques.
(1.9 cm * .76) + (-1.65 cm *1.8) + (-1.9 cm *.76) + (1.65 cm *1.8) = 0
LINE 2
magnitude of resultant force, sum of all forces
.76 N + - 1.8 N + - .76 N + 1.8 = 0 N
.76 N + 1.8 N + .76 N + 1.8 N = 5.12 N
LINE 3
magnitude of force as percent of sum
0/5.12 N = 0 %
LINE 4
Magnitude of torque, sum of magnitudes
(.76 N *8.3 cm ) + (- 8.7 cm * 1.8 N ) + ( -8.3 cm * .76 N ) + ( 8.7 cm * 1.8 N ) = 0 N * m
(.76 N *8.3 cm ) + (8.7 cm 1.8 N ) + ( 8.3 cm .76 N ) + ( 8.7 cm * 1.8 N ) = 0.043936 N * m
0 N * m / .043936 N * m = 0 %
** 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: **
LINE 1
sum of torques = (0.0088m * 1.88 N) + (-0.0087m *1.8 N) +(-0.0091m*1.9 N) +(0.01m*3.0) = 0.0795 N * m
LINE 2
magnitude of force, sum of magnitudes of all forces
1.88 + (-1.8) + (-1.9) + 3 = 1.18 N
1.88 + 1.8 + 1.9 + 3 = 8.58 N
LINE 3
1.18 N * / 8.58 N * = 13.8%
LINE 4
1.18 N * .001 m, 8.58 N * .0366 m,
.00118 N * m, .314 N * m, 3.76%
magnitude of resultant torque, sum of magnitude of all torques on rod.
** In the second setup, were the forces all parallel to one another? **
No the forces were not all parallel to one another.
I'd say the angles varied by about 10 degrees or less.
The reason I say this is because after marking the initial position when the rod was straight and then comparing them with where they are now it was slightly off vertically and horizontally.
** Estimated angles of the four forces; short discussion of accuracy of estimates. **
No they were not, the rod was slanted up to the left.
What I did to determine the angle was use the tan - 1 of y/x components.
To determine the y/x components I placed a mark on my paper where the points were initially and then a mark on my paper where the points ended up. For A and D on my rod they both differed from there initial point by 1 cm vertically and .5 cm horizontally. For B and C on my rod, they both differed from there initial point by .4 cm vertically and .1 cm horizontally.
I used negative values for when the magnitude shifted down vertically and to the left horizontally.
A= (tan-1)1/.5 =63 degrees
B= (tan-1).4/.1 =76 degrees
C= (tan-1)-.4/-.1 =75 degrees
D= (tan-1)-1/-.5 = 63 degrees
** x and y coordinates of both ends of each rubber band, in cm **
&&&&&
System B
x and y coordinates of the ends of the leftmost rubberband (bottom)
-6.5 cm, -14 cm
x and y coordinates of the ends of the leftmost rubberband (top)-5.5 cm, -5 cm
System A
x and y coordinates of the ends of the vertical rubberband (bottom)
1 cm, 5
x and y coordinates of the ends of the vertical rubberband (top)
0 cm, 16
System C
x and y coordinates of the ends of the rightmost rubberband (bottom)
9.5 cm, -13.5 cm
x and y coordinates of the ends of the rightmost rubberband (top)
5.5 cm, -5 cm
Here is what I did to make a coordinate system. I took the rod and found the center of it and drew a line perpendicular to it dividing the page into 1/2s. this was my y axis. I then used the rod straight rod the x axis. I then marked off the line in intervals of 1 cm each so I could find the x and y intervals for each respected point.
** Lengths and forces exerted systems B, A and C:. **
&&&&&
Difference in x coordinates of the ends of band B = -6.5 - - 5.5 = -1
Difference in y coordinates of the ends of band B = -14 - - 5 = -7
length of band = 9cm = 1.8 N
Difference in x coordinates of the ends of band A = 1 - 0= 1
Difference in y coordinates of the ends of band A = 16 - 5 = 11
length of band = 10 cm = 3.3 N
Difference in x coordinates of the ends of band C = 9.5 - 5.5 = 4
Difference in y coordinates of the ends of band C = -13.5 - - 5 = - 8.5
length of band = 8 cm = .4 N
** Sines and cosines of systems B, A and C: **
&&&&&&&&&&
system B
hypotenuse = 9 cm
cosine = .1
sin = 1
system A
hypotenuse = 10 cm
cosine .1
sin 0
system C
hypotenuse = 8 cm
cosine .44
sin .86
** Magnitude, angle with horizontal and angle in the plane for each force: **
&&&&&
B, 1.8 N,
A, 3.3 N,
C, 0.4 N,
** 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): **
&&&&
B =9 cm = 2.25 N
A = 10 cm = 2.5 N
C = 8 cm = 2 N
used the formula 4 cm = 1 N
** 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? **
8 hours
** Optional additional comments and/or questions: **