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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.
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This experiment consists of three parts.
Principles of Physics students need do only the first part.
General College Physics students need do only the first and second parts.
University Physics students should do all three parts.
The three parts are:
Rod supported by doubled rubber band, pulled down by two rubber bands
Simulating Forces and Torques on a Bridge
Torques Produced by Forces Not at Right Angles to the Rod
For this experiment you will use four of your calibrated rubber bands, a printed copy of the 1-cm grid (grid, a .gif file, or grid_1cm, a PDF), the threaded rod, 4 push pins and eight paper clips.
Rod supported by doubled rubber band, pulled down by two rubber bands
Setup
The setup is illustrated in the figure below. The large square represents the one-foot square piece of plywood, the black line represents the threaded rod, and there are six crude-looking hooks representing the hooks you will make by unbending and re-bending paper clips. The red lines indicate rubber bands. The board is lying flat on a tabletop. (If you don't have the threaded rod, you can use the 15-cm ramp in its place. Or you can simply use a pencil, preferably a new one because a longer object will give you better results than a short one. If you don't have the plywood and push pins, you can use the cardboard and 'staples' made from paper clips, as suggested in the Forces experiment.)
The top rubber band is attached by one hook to the top of the plywood square and by another hook to the approximate center of the rod. We will consider the top of the square to represent the upward direction, so that the rod is considered to be suspended from the top rubber band and its hook.
Two rubber bands pull down on the rod, to which they are attached by paper clips. These two rubber bands should be parallel to the vertical lines on your grid. The lower hooks are fixed by two push pins, which are not shown, but which stretch the rubber bands to appropriate lengths, as specified later.
The rubber band supporting the rod from the top of the square should in fact consist of 2 rubber bands with each rubber band stretched between the hooks (each rubber band is touching the top hook, as well as the bottom hook; the rubber bands aren't 'chained' together).
The rubber bands will be referred to by the labels indicated in the figure below. Between the two hooks at the top the rubber band pair stretched between these notes will be referred to as A; the rubber band near the left end of the threaded rod will be referred to as B; and the rubber band to the right of the center of the rod as C.
In your setup rubber band B should be located as close as possible to the left-hand end of the threaded rod. Rubber band C should be located approximately halfway, perhaps a little more, from the supporting hook near the center to the right-hand end of the rod. That is, the distance from B to A should be about double the distance from A to C.
Rubber band C should be stretched to the length at which it supported 10 dominoes (in the calibration experiment), while rubber band B should be adjusted so that the rod remains horizontal, parallel to the horizontal grid lines.
(If there isn't room on the plywood to achieve this setup:
First be sure that the longer dimension of the plywood is directed 'up-and-down' as opposed to 'right-and-left'.
Be sure you have two rubber bands stretched between those top hooks.
If that doesn't help, re-bend the paper clips to shorten your 'hooks'.
If the system still doesn't fit, then you can reduce the length to that required to support a smaller number of dominoes (e.g., 8 dominoes and if that doesn't work, 6 dominoes).
Data and Analysis: Mark points, determine forces and positions
Mark points indicating the two ends of each rubber band. Mark for each rubber band the point where its force is applied to the rod; this will be where the hook crosses the rod. Your points will be much like the points on the figure below. The vertical lines indicate the vertical direction of the forces, and the horizontal line represents the rod.
Disassemble the system, sketch the lines indicating the directions of the forces and the rod (as shown in the above figure). Make the measurements necessary to determine the length of each rubber band, and also measure the position on the rod at which each force is applied.
You can measure the position at which each force is applied with respect to any point on the rod. For example, you might measure positions from the left end of your horizontal line. In the above figure, for example, the B force might be applied at 3 cm from the left end of the line, the A force at 14 cm from the left end of the line, and the C force at 19 cm from the left end.
indicate the following:
In the first line, give the positions of the three points where the vertical lines intersect the horizontal line, in order from left to right.
In the second line give the lengths of the rubber band systems B, A and C, in that order.
In the third line give the forces, in Newtons, exerted by the rubber band systems, in the same order as before.
In the fourth line specify which point was used as reference point in reporting the three positions given in the first line. That is, those three positions were measured relative to some fixed reference point; what was the reference point?
Starting in the fifth line, explain how the forces, in Newtons, were obtained from your calibration graphs.
Beginning in the sixth line, briefly explain what your results mean and how you obtained them.
----->>>>>>>> (note A doubled) intersections B A C, lengths B A C, forces B A C, reference point, how forces determined
******** ******** Your answer (start in the next line):
1.90 cm, 7.85 cm, 10.80 cm ( from left end of horizontal line )
9.15 cm, 8.75 cm, 12.75 cm
1.14 N, 1.14 N, 1.14 N
the left end of the horizontal line ( the horizontal line made by the threaded rod used as a reference point )
I obtained the force, 1.14 N, for each of the rubber bands by taking the corresponding force calculated from the rubber band calibration experiment exerted by rubber band C, which was rubber band two in the original experiment. I had to use a piece of cardboard in lieu of the plywood because the system's length exceeded that of the plywood. The force exerted was for six dominoes.
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Analyze results:
Vertical equilibrium: Determine whether the forces are in vertical equilibrium by adding the forces to obtain the net force, using + signs on upward forces and - signs on downward forces.
Give your result for the net force in the first line below.
In the second line, give your net force as a percent of the sum of the magnitudes of the forces of all three rubber band systems.
Beginning in the third line, briefly explain what your results mean and how you obtained them.
----->>>>>>>> Fnet, Fnet % of sum(F)
******** Your answer (start in the next line):
-1.14 N
33.3%
My net force is on the first line, -1.14 N. The result is negative because the two lower bands, whose magnitudes exceed that of the double upper bands, are pulling down on the threaded rod, making the overall direction is negative. The top, where the two rubber bands are hooked by the paperclips, represents the positive, upward direction. The net force, reported in the second line, has a magnitude that is 33.3% of the total sum of the forces, 3.42 Newtons.
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Rotational equilibrium: We will regard the position of the central supporting hook (the hook for system A) to be the fulcrum around which the rod tends to rotate. Determine the distance from this fulcrum to the point of application of the force from rubber band B. This distance is called the moment-arm of that force. Do the same for the rubber band at C.
report the moment-arm for the force exerted by the rubber band system B, then the moment-arm for the system C. Beginning in the second line, briefly explain what the numbers mean and how you obtained them.
----->>>>>>>>
******** Your answer (start in the next line):
5.95 cm, 2.95 cm
These are the moment-arms, the distance from the fulcrum to each of the respective exerted forces by rubber bands B to A. The first number is the distance horizontally from B to A. And the second reported number is the distance from A to C horizontally.
#$&* moment arms for B, C
Make an accurate scale-model sketch of the forces acting on the rod, similar to the one below. Locate the points of application of your forces at the appropriate points on the rod. Use a scale of 4 cm to 1 Newton for your forces, and sketch the horizontal rod at its actual length.
Give in the first line the lengths in cm of the vectors representing the forces exerted by systems B, A and C, in that order, in comma-delimited format.
In the second line give the distances from the fulcrum to the points of application of the two 'downward' forces, giving the distance from the fulcrum to the point of application of force B then the distance from the fulcrum to the point of application of. force C in comma-delimited format, in the given order.
Beginning in the third line, briefly explain what the numbers mean and how you obtained them.
----->>>>>>>> (4 cm to 1 Newton scale) lengths of force vectors B, A, C, distances of B and C from fulcrum:
******** Your answer (start in the next line):
9.15 cm, 8.75 cm, 12.75 cm
5.95 cm, 2.95 cm
The numbers in the first line represent the actual lengths in centimeters of the respective vectors that are exerting force on the threaded rod. The first vector, vector B, is 9.15 cm, and has a projected force of 2.29 N, according to the 4 cm to 1 N scale. Vector C of 8.75 cm has a projected force of 3.19 N on this same scale. And vector A has a projected force of 2.19 N. The numbers in the second line show the actual length in centimeters from the fulcrum to the force of vector B and the distance from the fulcrum to vector C respectively.
You indicated previously that each system exerted a force of 1.14 N, which was not consistent with your reports of rubber band lengths. However I'm not sure about your 2.29 N, 3.19 N and 2.19 N forces reported here.
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The force from rubber band C will tend to rotate the rod in a clockwise direction. This force is therefore considered to produce a clockwise torque, or 'turning force', on the rubber band. A clockwise torque is considered to be negative; the clockwise direction is considered to be the negative direction and the counterclockwise direction to be positive.
When the force is exerted in a direction perpendicular to the rod, as is the case here, the torque is equal in magnitude to the product of the moment-arm and the force.
What is the torque of the force exerted by rubber band C about the point of suspension, i.e., about the point we have chosen for our fulcrum?
Find the torque produced by rubber band B about the point of suspension.
Report your torques , giving the torque produced by rubber band B then the torque produced by the rubber band C, in that order. Be sure to indicate whether each is positive (+) or negative (-). Beginning in the next line, briefly explain what your results mean and how you obtained them.
----->>>>>>>> torque C, torque B
******** Your answer (start in the next line):
-.0939 J ( N-m ), +.136 J ( N-m )
The first number is the torque of rubber band C, which was found by multiplying the force from the sketch, 2.2875 N ( 9.15 cm/ 4 cm/ 1 N ), by the centimeters divided by 100 to obtain the result in Newton-meters or Joules. The torque is negative in the first case, as the rod tends to be turned clockwise. The second number is the torque of rubber band B, which was found as the product of the force 3.1825 N and .0295 m. In this second case, the torque is positive, as the rod tends to be turned counter clockwise.
torques are not measured in Joules, though in the energy context a N * m is a Joule. Torque and work are two different things, having very different effects, though the units are the same.
I use m * N when referring to torque, to distinguish this unit from the unit of work.
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Ideally the sum of the torques should be zero. Due to experimental uncertainties and to errors in measurement it is unlikely that your result will actually give you zero net torque.
Express the calculated net torque--i.e, the sum of the torques you have found--as a percent of the sum of the magnitudes of these torques.
Give your calculated net torque in the first line below, your net torque as a percent of the sum of the magnitudes in the second line, and explain starting at the third line how you obtained this result. Beginning in the fourth line, briefly explain what your results mean and how you obtained them.
----->>>>>>>> tau_net, and as % of sum(tau)
******** Your answer (start in the next line):
.0421 J ( N-m)
18.3%
I obtained the percentage by taking the net torque, .0421 J ( N-m ), and dividing it by the total of the magnitudes, .2299 J (N-m ) of torque, and finally multiplying it by 100. I calculated the net torque by adding the negative torque from rubber band C, -.0939 J ( N-m ), and the positive torque of rubber band B, .136 J ( N-m ).
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Physics 121 students may stop here. Phy 121 students are not required to do the remaining two parts of this experiment, but may do so if they wish.
Simulating Forces and Torques on a Bridge
The figure below represents a bridge extended between supports at its ends, represented by the small triangles, and supporting two arbitrary weights at arbitrary positions (i.e., the weights could be anything, and they could be at any location).
The weights of the objects act downward, as indicated by the red vectors in the figure. The supports at the ends of the bridge hold the bridge up by exerting upward forces, represented by the upward blue vectors.
If the bridge is in equilibrium, then two conditions must hold:
1. The total of the two upward forces will have the same magnitude as the total of the two downward forces. This is the conditional of translational equilibrium. That is, the bridge has no acceleration in either the upward or the downward direction.
2. The bridge has no angular acceleration about any axis. Specifically it doesn't rotate about the left end, it doesn't rotate about the right end, and it doesn't rotate about either of the masses.
Setup
We simulate a bridge with the setup indicated below. As in Part I the system is set up with the plywood square, and with a 1-cm grid on top of the plywood.
The threaded rod will be supported (i.e., prevented from moving toward the bottom of the board) by two push pins, and two stretched rubber bands will apply forces analogous to the gravitational forces on two weights supported by the bridge.
Stretch one rubber band to the length at which it supported 8 dominoes in the calibration experiment, and call this rubber band B. Stretch the other to the length that supported 4 dominoes and call this rubber band C. Rubber band C should be twice as far from its end of the rod as rubber band B is from its end, approximately as shown below.
Use push pins (now shown) to fix the ends of the hooks and keep the rubber bands stretched.
Note that the length of the threaded rod might be greater than the width of the board, though this probably won't occur. If it does occur, it won't cause a serious problem--simply place the push pins as far as is easily feasible from the ends and allow a little overlap of the rod at both ends.
Be sure the rubber bands are both 'vertical'--running along the vertical lines of the grid. It should be clear that the push pins are each exerting a force toward the top of the board.
Place two more rubber bands, with the hooks at the positions of the push pins, as indicated below. Stretch these rubber bands out simultaneously until their combined forces and torques just barely begin to pull the rod away from the push pins supporting it. Fix push pins through the free-end hooks, so that the two new rubber bands support the rod just above the push pins supporting it, as close to the supporting pins as possible.
Remove the supporting pins. This should have no effect on the position of the rod, which should now be supported in its original position by the two new rubber bands.
Mark the ends of each of the four rubber bands, and also the position of the rod. Your marks should be sufficient to later construct the following picture:
Now pull down to increase the length of the rubber band C to the length at which that rubber band supported the weight of 10 dominoes, and use a push pin to fix its position.
This will cause the lengths of the rubber bands A, B and D to also change. The rod will now lie in a different position than before, probably at some nonzero angle with horizontal.
Mark the position of the rod and the positions of the ends of the four rubber bands, in a manner similar to that used in the previous picture. Be sure to distinguish these marks from those made before.
Analyze your results
The figure below indicates the first set of markings for the ends of the rubber bands, indicated by dots, and the line along which the force of each rubber band acts. The position of the rod is indicated by the horizontal line. The force lines intersect the rod at points A, B, C and D, indicated by x's on the rod.
From your markings determine, for the first setup, the length of each rubber band and, using the appropriate calibration graphs or functions, find the force in Newtons exerted by each.
Sketch a diagram, to scale, depicting the force vectors acting on the rod. Use a scale of 1 N = 4 cm. Label each force with its magnitude in Newtons, as indicated in the figure. Also label for each force the distance along the rod to its point of application, as measured relative to the position of the leftmost force.
In the figure shown here the leftmost force would be the 2.4 N force; its distance from itself is 0 and isn't labeled. The 5 cm, 15 cm and 23 cm distances of the other forces from the leftmost force are labeled.
For the first setup (before pulling down to increase the force at C), give the forces, their distances from equilibrium and their torques, in comma-delimited format with one torque to a line. Give lines in the order A, B, C and D. Be sure your torques are positive if counterclockwise, negative if clockwise. Beginning in the following line, briefly explain what your results mean and how you obtained them.
----->>>>>>>> (ABCD left to right, position wrt A) four forces, four dist, four torques
******** Your answer (start in the next line):
3.025 N ( 12.10 cm/ 4 cm/ 1 N ), 12.10 cm, -.3660 J ( N-m )
3.3125 N ( 13.25 cm/ 4 cm/ 1 N ), 13.25 cm, +.4389 J ( N-m )
1.9375 N ( 7.75 cm/ 4 cm/ 1 N ), 7.75 cm, -.1502 J ( N-m )
3.025 N ( 12.10 cm/ 4 cm/ 1 N ), 12.10 cm, +.3660 J ( N-m )
The first number in each row is the force exerted by each rubber band, using the diagram; calculations are shown above. I multiplied each rubber band's distance from equilibrium, and divided it by the scale that 4 cm = 1 N of force. Finally, I multiplied the aforementioned calculated force by the length of each rubber band in meters to obtain the torque, accounting for positivity and negativity according to the way each turned the threaded rod ( clockwise versus counter clockwise ).
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In the figure shown above the sum of all the vertical forces is 2.4 N + 2.0 N - 3.2 N - 1.6 N = 4.4 N - 4.8 N = -.4 N. Is this an accurate depiction of the forces that actually acted on the rod? Why or why not?
In the first line give the sum of all the vertical forces in your diagram. This is the resultant of all your forces.
In the second line, describe your picture and its meaning, and how well you think the picture depicts the actual system..
----->>>>>>>> (from scaled picture) sum of vert forces, describe picture and meaning
******** Your answer (start in the next line):
.8 N
My picture displays the forces we exerted acting up the threaded rod; it is not a complete depiction of the scenario. It does not show the balanced forces of gravity supporting the cardboard and threaded rod on the table. In addition, friction is not accounted for. Like the depiction above, mine only shows the four forces of the rubber band, the two pulling the rod down below and the two pulling the rod above. The net force is .8 N, if upwards is the positive direction. This is illustrated as there is more total force acting from above than below.
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In the figure shown above the 1.6 N force produces a clockwise torque about the leftmost force (about position A), a torque of 1.6 N * 15 cm = 24 N cm. Being clockwise this torque is -24 N cm. The 2.0 N force at 23 cm produces a clockwise torque of 2.0 N * 23 cm = 26 N cm. Being counterclockwise this torque is +26 N cm.
In the first line below give the net torque produced by the forces as shown in this figure. Beginning in the second line describe your picture and discuss whether it could be an accurate depiction of torques actually acting on a stationary rod. Support your discussion with reasons.
----->>>>>>>> net torque from given picture, describe your picture
******** Your answer (start in the next line):
18 N-cm ( -24 N-cm + 26 N-cm + 16 N-cm ), .18 J ( N-m )
I calculated this net torque by calculating the sum of the all of the torques in the depiction above. Once again, my picture only shows a partial, yet sufficient diagram of the actual torque acting upon the threaded rod. What it does not show are the negligible forces acting upon the rod, as indicated in the last response. These include friction, gravitational pull, the support of the table, and the position of the system on the table. For our purposes, this figure represents what is happening in the system well. But, from a perfect, scientific point-of-view, my image is lacking in detail.
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Now calculate your result
What is the sum for your diagram of the torques about the point of action of the leftmost force (i.e., about position A)? This is your experimentally observed resultant torque about A. Give your result in the first line below.
For your diagram what is the magnitude of your resultant force and what is the sum of the magnitudes of all the forces acting on the rod? Give these results in the second line in comma-delimited format.
Give the magnitude of your resultant force as a percent of the sum of the magnitudes of all the forces. Give this result in the third line.
For your diagram what is the magnitude of your resultant torque and what is the sum of the magnitudes of all the torques acting on the rod? Give these two results, and the magnitude of your resultant torque as a percent of the sum of the magnitudes of all the torques, as three numbers in your comma-delimited fourth line.
Beginning in the fifth line, briefly explain what your results mean and how you obtained them.
----->>>>>>>> sum(tau) about A, Fnet and sum(F), Fnet % of sum(F), | tau_net |, sum | tau |, |tau_net| % of sum|tau|
******** Your answer (start in the next line):
26.57 N-cm or .2657 J ( N-m )
.8 N, 11.30 N,
7% ( est. )
.2887 J ( N-m ), 1.321 J ( N-m ), 21.85%
The number in the first line is the experimentally observed resultant force, which was calculated by summing the torques, obtained by multiplying the displacement from the leftmost force by the actual force exerted by each rubber band using the scale 4 cm to 1 N.
your first line reports torques, not forces; you've reported a resultant torque
The second number is the net force, the sum of all of the forces in the positive, upward direction and the negative, downward direction. It is followed by the sum of the absolute value of the magnitudes of each force, 11.30 N. The number in the third line is the absolute value of the percentage the net force is of the sum of the magnitude of all forces in the diagram. Finally, on the fourth line, is the torque, the net turning force, followed by the sum of the magnitude of the torque, and the percentage the net torque represents of the sum of the magnitudes.
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Perform a similar analysis for the second setup (in which you increased the pull at C) and give your results below:
For your diagram, what is the sum of the torques about the point of action of the leftmost force (i.e., about position A)? This is your experimentally observed resultant torque about A. Give your result in the first line below.
For your diagram what is the magnitude of your resultant force and what is the sum of the magnitudes of all the forces acting on the rod? Give these results in the second line in comma-delimited format.
Give the magnitude of your resultant force as a percent of the sum of the magnitudes of all the forces. Give this result in the third line.
For your diagram what is the magnitude of your resultant torque and what is the sum of the magnitudes of all the torques acting on the rod? Give these two results, and the magnitude of your resultant torque as a percent of the sum of the magnitudes of all the torques, as three numbers in your comma-delimited fourth line.
Beginning in the fifth line, briefly explain what your results mean and how you obtained them.
----->>>>>>>> (pull at C incr) sum(tau) about A, Fnet and sum(F), Fnet % of sum(F), | tau_net |, sum | tau |, |tau_net| % of sum|tau|
******** Your answer (start in the next line):
.1125 J ( N-m )
-.325 N, 13.55 N
2.40%
.019 J ( N-m ), 1.84 J ( N-m ), 1.03%
The number in the first line is the experimentally observed resultant force, which was calculated by summing the torques, obtained by multiplying the displacement from the leftmost force by the actual force exerted by each rubber band using the scale 4 cm to 1 N. The second number is the net force, the sum of all of the forces in the positive, upward direction and the negative, downward direction. It is followed by the sum of the absolute value of the magnitudes of each force, 13.55 N. The number in the third line is the absolute value of the percentage the net force is of the sum of the magnitude of all forces in the diagram. Finally, on the fourth line, is the torque, the net turning force, followed by the sum of the magnitude of the torque, and the percentage the net torque represents of the sum of the magnitudes.
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For the second setup, the forces were clearly different, and the rod was not completely horizontal. The angles of the forces were therefore not all 90 degrees, though it is likely that they were all reasonably close to 90 degrees.
Look at your diagram for the second setup. You might want to quickly trace the lines of force and the line representing the rod onto a second sheet of paper so you can see clearly the directions of the forces relative to the rod.
In the first setup, the forces all acted in the vertical direction, while this may not be the case in this setup.
In the second setup, were the forces all parallel to one another? If not, by about how many degrees would you estimate they vary? Include a brief explanation of what your response means and how you made your estimates.
----->>>>>>>> (incr pull at C) variation of forces from parallel
******** Your answer (start in the next line):
The forces in the second set-up, although close, were not all quite parallel to each other, but they were very close, give or take 5 degrees. In the second set-up, the increased tension in rubber band C lead to a slight change in the direction of the threaded rod because the angle between it and the x-axis slightly changed from the net force of the rubber bands. The lengths of the other rubber bands increased, causing the orientation of the threaded rod to change even more. I say by about five degrees because the change, although very little, was noticeable, especially in the rubber bands over the threaded rod. The rubber bands did not quite fully abandon their parallelism to the grid lines, but they shifted slightly their orientation along them.
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Estimate the angles made by the lines of force with the rod in the second setup, and give your angles in comma-delimited format in the first line below. Your angles will all likely be close to 90 degrees, but they probably won't all be 90 degrees. The easiest way to estimate is to estimate the deviation from 90 degrees; e.g., if you estimate a deviation of 5 degrees then you would report an angle of 85 degrees. Recall that you estimated angles in the rotation of a strap experiment.
Starting in the second line give a short statement indicating how you made your estimates and how accurate you think your estimates were.
----->>>>>>>> angles of lines of force with rod
******** Your answer (start in the next line):
85 degrees, 87.5 degrees, 91 degrees, 95 degrees
In the order of rubber bands, I estimated the degrees to which the force reacted when I stretched rubber band C to 14 centimeters. I made these estimates in relation to how the rubber bands adjusted with the added tension of rubber band C. The top two rubber bands, A and D, seemed to have the biggest adjustment, at no more than 5 degrees, while the other two were changed, yet even to a slighter degree.
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Torques Produced by Forces Not at Right Angles to the Rod
Setup and Measurement
Set up a system as illustrated below.
As in our very first setup, the 'top' rubber band will in fact consist of two rubber bands in parallel.
The leftmost rubber band will remain vertical, while the rightmost rubber band will be oriented at a significant angle with vertical (at least 30 degrees).
The rightmost rubber band will be stretched to a length at which it supports the weight of 10 dominoes, and its point of attachment will be at least a few centimeters closer to that of the center rubber band than will the leftmost rubber band.
The leftmost rubber band will be stretched to the length at which it supports 8 dominoes.
Mark the ends of the rubber bands, the points at which the forces are exerted on the central axis of the rod, and the position of the central axis of the rod.
Measure the positions of the ends of the rubber bands:
Disassemble the system and draw an x and a y axis, with the origin somewhere below and to the left all of your marks.
Measure the positions of the ends of the rubber bands. Measure both the x and y coordinate of each of these positions, and measure each coordinate in centimeters.
Give in the first line below the x and y coordinates of the ends of the leftmost rubber band, which we will call rubber band system B. Give four numbers in comma-delimited format, the first being the x and y coordinates of the lower end, the second being the x and y coordinates of the upper end. All measurements should be in cm.
In the second line give the same information for the two-rubber-band system above the rod, which we will call system A.
In the third line give the same information for the rightmost rubber band which we will call system C.
Beginning in the fourth line, briefly explain what your results mean and how you obtained them.
----->>>>>>>> (BAC) endpts of B, endpts of A, endpts of C
******** Your answer (start in the next line):
( -6.60 cm , -11.95 cm ), ( -6.60 cm, 0 cm )
( 0 cm, 8.15 cm ), ( 0 cm, 19.80 cm )
( 12.85 cm, -10.40 cm ), ( 3.45 cm, 0 cm )
I obtained this data by using my singly-reduced ruler to measure the endpoints of where the rubber band was stretched.
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Analysis
Using your coordinates and the Pythagorean Theorem, find the length of rubber band system B.
Do this by first finding the difference in the x coordinates of the ends of this band, then the difference in the y coordinates of the ends.
This gives you the lengths of the legs of a right triangle whose hypotenuse is equal to the length of the band.
Then using your calibration information find the force in Newtons exerted this system.
Do the same for systems A and C.
Give the length and force exerted by rubber band system B in the first line below, in comma-delimited format, then in the second and third lines give the same information for systems A and C. Starting in the fourth line give a brief description of what your results mean and how you obtained them.
----->>>>>>>> length and force of B, of A, of C
******** Your answer (start in the next line):
11.95 cm, .4255 N
11.65 cm, .1826 N
14.02 cm, 1.964 N
I found the length of rubber bands A and B by simply finding the difference in the y-component of the two endpoints. With the length of rubber band C, I had to use Pythagorean theorem by finding the differences in both the x and y components, which were the legs of the right triangle. Then, by adding the sum of the squared legs, I calculated the square root of the result to obtain rubber band C's length of approximately 14.2 cm. I found the corresponding forces by putting the lengths into the equations of the respective rubber bands from the calibration experiment.
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Find the sine and the cosine of each angle with horizontal:
You earlier found the lengths of the x and y legs of the triangle whose hypotenuse was the length of rubber band system A.
The magnitude of the sine of the angle for the system the y component divided by the hypotenuse, i.e., the ratio of the y component to the hypotenuse. The sine is negative if the y component downward, positive if the y component is upward.
The magnitude of the cosine of the angle for the system the x component divided by the hypotenuse, i.e., the ratio of the x component to the hypotenuse. The cosine is negative if the x component is to the left, positive if the x component is to the right.
Find the sine and cosine for this system.
Using the same method find the sine and the cosine for system B and system C. Ideally system B will be acting vertically, so the cosine will be 0 and the sine will be 1; your measurements might or might not indicate a slight divergence from this ideal.
Report your results , giving in each line the sine and the cosine of the angle between the line of action of the force and the horizontal. Report lines in the order B, then A, then C. Beginning in the fourth line, briefly explain what your results mean and how you obtained them.
----->>>>>>>> sin and cos of angle w horiz of B, A, C
******** Your answer (start in the next line):
sine = -1, cosine = 0
sine = 1, cosine = 0
sine = -.7418, cosine = .6706
I found the cosine and sine of each angle of the first two systems, A and B, by simply taking the sine and cosine of their respective vertical forces, 90 degrees and 270 degrees. For the last one, system C, I had to use the inverse sine function, using the legs, to find the angle of the vector. Then, by finding the angle, I input it into the sine and cosine function for their respective values.
#$&*
Find the angles of the force vectors with the horizontal, and the angles of the force vectors in the plane:
The angle of the force vector with horizontal is arcTan(y / x): the arctangent of the magnitude of the quantity you get with you divide the y component of the triangle used in the preceding, by the x component.
The arctangent is easily calculated using the 2d fn or inverse key on your calculator, along with the tan function.
The angle of the force vector in the plane is measured from the positive x axis, in the counterclockwise direction.
Give for each system the magnitude (i.e., the force in Newtons as you calculated it earlier), the angle with the x axis and the angle in the plane for each of the force vectors, reporting three comma-delimited lines in the order B, A and C. Starting in the fourth line briefly explain how you determined these values and how you obtained them:
----->>>>>>>> magnitude and angle of B, of A, of C
******** Your answer (start in the next line):
270 degrees
90 degrees
312.115 degrees
I found the corresponding angle of systems B and A as 90 easily because they are simply vertical forces. For system C, I used my knowledge of the angle from the inverse sine function explained in the previous problem.
#$&*
Sketch a force diagram showing the forces acting on the rubber bands, using a scale of 1 N = 4 cm. Label the positions at which the forces act on the rod, the magnitude in Newtons of each force and the angle of each force as measured counterclockwise from the positive x axis (assume that the x axis is directed toward the right).
Find the components of each force:
Sketch the x and y components of each force vector, measure them and using the scale of your graph convert them back to forces. Then using the magnitude of the force and sine and cosine as found earlier, calculate each x and y component.
In the second line below you will report the x and y components of your sketch of vector A, the x and y components of the force of this system as calculated from the x and y components on your sketch, and the x and y components as calculated from the magnitude, sine and cosine. Report six numbers in this line, in comma-delimited format.
In the first line report the same information for vector B, and in the third line the same information for vector C.
Beginning in the fourth line, briefly explain what your results mean and how you obtained them.
----->>>>>>>> comp of sketch, implied comp of force, comp calculated from mag and angle B, A, C
******** Your answer (start in the next line):
( 0 , 2.9125 ), ( 0 cm, 19.80 cm ), ( 1 ( sine ), 0 ( cosine ) )
( -1.65, 3.3125 ), ( -6.60 cm , -11.95 cm ), ( -1 ( sine ), 0 ( cosine ) )
( 3.21, -2.6000 ), ( 12.85 cm, -10.40 cm ), ( -.7418 ( sine ), .6706 ( cosine ) )
As before, I calculated the x and y components of each on my new sketch by reducing each magnitude by 4. All other coordinates were from before.
#$&*
Calculate the sum of the x components and of the y components, as determined by the magnitude, sine and cosine.
What is the sum of all your x components? What should be the sum of all the x components? How close is your sum to the ideal? Report as three numbers in comma-delimited format in line 1.
What is the sum of all your y components? What should be the sum of all the y components? How close is your sum to the ideal? Report as three numbers in comma-delimited format in line 2.
Beginning in the third line, briefly explain what your results mean and how you obtained them.
----->>>>>>>> sum of your x comp, actual sum, how close to ideal x, then y
******** Your answer (start in the next line):
7.0682, 0, 7.0682 more
1.7456, 0, 1.7456 more
These sums should equate closer to zero. Thus, this tells me that my calculations have error in them, when it comes to calculating force and torque. Some of this was induced in rounding and estimation. But, I think even more of this was induced when my upper-most line was straight, instead of slanted. I did not see that this line had to be at an angle, until the very end. It did not seem to note that in the beginning of the section three instructions.
#$&*
The torque produced by a force acting on the rod is produced by only the component perpendicular to the rod. The component parallel to the rod has no rotational effect.
give in comma-delimited format a line for each force, indicating the distance of its point of action from that of the leftmost force, its component perpendicular to the rod, and its torque. The order of the lines should be B, A then C. Remember that torques should be reported as positive or negative.
Beginning in the fourth line, briefly explain what your results mean and how you obtained them.
----->>>>>>>> (about B) dist from ref, perpendicular comp, torque for B, for A, for C
******** Your answer (start in the next line):
0 J ( N-m )- Leftmost
.1922 J ( N-m )
-.3020 J ( N-m )
I calculated each by multiplying the force from the sketch in Newtons by the distance between the point of action ( the leftmost force ), divided by 100.
#$&*
Finally report the sum of your torques:
What is the sum of the torques about the point of action of the leftmost force? What should this sum be? How close is your sum to the ideal? Report as three numbers in comma-delimited format in line 1. Beginning in the second line, briefly explain what your results mean and how you obtained them.
----->>>>>>>> sum of torques, ideal sum, how close to ideal:
******** Your answer (start in the next line):
-.1098 J ( N-m ), 0 J ( N-m ), -.1098 J ( N-m ) less
I calculated the sum of the torques by adding the torques about the point of action together. The total should have been 0 J, but I was very close to the theoretical prediction, only off by -.1098 J.
#$&*
Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
Approximately how long did it take you to complete this experiment?
******** Your answer (start in the next line):
3 Hours, 45 Minutes
#$&*
Please copy your document into the box and submit:
Whatever you use (plywood, cardboard, etc.), you'll want to support it above your tabletop, in order to avoid damaging the tabletop. A cardboard box works very well for this purpose; so would a couple of books to support the ends of the board.
Set up the experiment:
1. Print out the grid, which is a .gif file and should give you a printout with 1-cm squares. (NOTE: If the 'grid' link doesn't work, or if you prefer to use a PDF, download and/or print grid_1cm). Place this grid on the plywood square. (Place an accurate ruler on the grid and verify that the squares indeed line up with the 1-cm marks. If they don't email the instructor and include the number of millimeters, to the most accurate fraction of a mm you can estimate, by which the grid is off on a 10-cm measurement in the horizontal direction, and the number of mm of error on a 10-cm measurement in the vertical direction.)
You may also print out a page of rulers. Be sure to right-click on the page with the rulers, then click on Print. Otherwise you might print out the top frame of the page, which is not what you want here. These rulers should be accurate to within a fraction of a millimeter over their 22-cm length. Measure them and be sure they are indeed accurate. Note that the markings on these rulers might not be perfect, due to limitations of pixel placement, and that this may introduce an additional small element of uncertainty in the results of your measurements.
2. Using a paper clip, three rubber bands and three push pins (or paperclip 'staples') set up the system as shown in the figure below. In the figure the dark thick lines represent the rubber bands, all three of which are attached directly to the paper clip. The small circles at the other end of each rubber band represents a push pin (or a 'staple'). The setup should have the following characteristics:
One rubber band (the one lowest in the figure below) should run along one of the vertical lines of the grid. In the figure the lowest rubber band isn't quite parallel to the vertical line. Make sure yours is parallel and that it's on the line.
Don't let any rubber band exceed its original length by more than 30% (e.g., a rubber band with unstretched length 8 cm should not exceed 10.4 cm in length). The rubber band most likely to exceed this length will be the one parallel to the vertical grid lines. Keep an eye on that one as you position the other two. It would be a good idea to check your calibration graphs to see which rubber band is capable of exerting the most force at the 'plus-30% length', and use that for the 'vertical' rubber band.
The second rubber band will be directed toward the upper left and will be angled at least 45 degrees to the left of vertical. This rubber band should stretched to a length that will cause it to exert a force of about 1 Newton.
The third rubber band should exert enough force to cause the lower rubber band to stretch to at least 20% beyond its maximum unstreched length, but not more than 30% (standard #32 or #33 rubber bands with an initial length of about 8 cm stretch to about 10 cm when supporting about 2 Newtons of weight; if your rubber bands are shorter or stiffer, the lengths may be shorter). The third rubber band will probably end up closer to vertical than the second rubber band. Remember that the lower rubber band has to be parallel to the vertical lines.
It's not shown that way in the picture but you should use paper clips bent into hooks to attach the rubber bands to the push pins. That will prevent the push pins from 'crushing' the rubber bands and preventing the ends from stretching, and it will also make it much easier to accurately mark the positions of the ends of the rubber bands. However the 'hooks' can't be too long, or the system won't fit on the board. (If you use paperclip 'staples', as described near the beginning, there is no need to use these 'hooks').
Be very sure that the 'downward' rubber band is stretched to between 10.0 and 10.5 cm, and that it is parallel to the vertical gridlines.
Mark the positions of the ends of the rubber bands, measure their lengths and determine the force exerted by each:
1. Using a pencil with a good point or a reasonably fine-tipped pen make an x under the end of each rubber band, with the x crossing directly under the very end of the rubber band. Locate this mark as accurately as possible. The first figure below indicates the positions of the x's (note that the x's under the circles representing push pins will in fact be under hooks at the ends of the rubber bands).
2. Once you have marked both ends of each rubber band all your information will be on the grid, and you may remove the pins, the rubber bands and the paper clip. The second figure above shows the grid as it would be with just the x's marked.
3. Using a ruler measure the length, in cm, of each rubber band, based on the distance between your marked points.
4. Using the appropriate calibration graph or calibration function determine the force exerted by each rubber band.
In the space below give the length of each rubber band and the corresponding force in Newtons, in this order and separated by a comma:
In the first line give the information for the 'vertical' rubber band.
In the second line give the information for the rubber band which slopes up and to the left..
In the third line give the information for the rubber band which slopes up and to the right.
Starting the fourth line give a brief explanation, in your own words, of what the numbers you have reported mean and how you determined your forces.
Your answer (start in the next line):
10.1, 3.6 N
7.5, 1 N
9.3, 2.8 N
The force is determined using Hookes Law stating that the amount of elongation is proportional to the force exerted on the object.
This means the most stress is exerted on the vertical rubber band, the second most on the right rubber band, and the least on the left rubber band.
#$&* _ lengths and forces in Newtons _
We will now proceed to calculate the components of the net force using three different methods. The methods are:
Graphical determination of force components using sketches of the vectors and projection lines.
Graphical determination of the resultant vector using the polygon method of vector addition.
Calculation of components using basic trigonometry (methods of Introductory Problem Set 5).
Graphically determine the components of your forces, using sketches and projection lines:
Set up coordinate axes on the grid, select a point on each line and draw projection lines to determine coordinates; use coordinates to determine angles:
1. Find the point at which the lines of force meet. For each rubber band use a straightedge to draw a line along the center of the rubber band's position. Extend all three lines to their point of intersection. The first figure below illustrates the lines corresponding to the above figures.
2. Construct a y vs. x coordinate system whose origin is at the point where the three lines of force intersect, as shown in the second figure below.
3. Locate on each line of force the point which is 10 cm from the intersection (i.e., the origin of the coordinate system), as indicated by the heavy red x's in the first figure below. For each rubber band, just measure 10 cm from the origin, along the line of force.
4. Sketch the projection line from each of these points to the x and y axes, as shown in the second figure below. Since one point is already on the y axis it isn't necessary to sketch its projection lines.
5. Determine the x and y coordinate of the point selected on each line of force. For example in the picture the x and y coordinates of the line in the first quadrant are about 2.3 cm and 8.5 cm, so the point is (2.3, 8.5). Note that the point (2.3, 8.5) isn't really 10 cm from the origin; when you located your points you hopefully did much better.
Report your results in the space below, x and y in centimeters in comma-delimited format, one rubber band to a line. Report with the rubber bands in the same order you reported previously.
Starting in the fourth line, give a brief explanation of how you constructed this graph and how you obtained your results.
Your answer (start in the next line):
0, 10
-7.5, 7.5
5, 8.5
The graph was constructed following the directions above, and looks similar to the example provided in this lab. Measurements were taken with a ruler, in cm, and verified using the 1 cm squares on the paper.
#$&* _ coordinates of points 10 cm from origin along three lines of force _
6. Find the angle of each line of force by calculating the inverse tangent of the y coordinate divided by the x coordinate (this means inverse tangent of (y coord / x coord), making sure your calculator is in degree mode. For example, for the point (2.3, 8.5) we would get an angle equal to arctan(8.5 / 2.3) = 74.9 degrees, approximately.
On a calculator you can use the inverse tangent function (usually INV and TAN keys, or 2d function and TAN). Be sure you are in degree mode if you want the angle in degrees, or in radian mode if you want the angle in radians.
If you use Excel you can use the ATAN function (for this angle you would type =ATAN(8.5 / 2.8) into a cell). Excel gives angles in radians, which you can convert to degrees if you multiply by 180 / `pi (you could just type in = ATAN(8.5 / 2.8) * 180 / PI(), which will give you the angle in degrees. Note that pi() is Excel's way of denoting `pi).
7. Find the angle made by each force vector with the positive x axis, as measured in the counterclockwise direction from the positive x axis. Note that the angle you found in the preceding instruction is the angle of the line of force with the x axis.
The actual force vector might point in either direction along this line.
For example, a line at angle 60 degrees might indicate a force vector into the first quadrant, at 60 degrees as measured counterclockwise from the positive x axis; or the force vector might also point into the third quadrant, making its angle 240 degrees.
Similarly a line at angle -30 degrees could indicate a vector into the fourth quadrant, at -30 degrees or equivalently at 330 degrees; or it could indicate a vector into the second quadrant, at angle 150 degrees.
The rule for the angle of the line with the x axis is arctan(y component / x component). The angle of a vector (which as indicated above can point in either of two directions along the line) is arctan ( y component / x component), plus 180 deg if the x component is negative.
Report your angles in the space below, one to a line, reporting the rubber bands in the same order as before. This will require the first three lines.
Starting in the fourth line explain briefly, in your own words, how you obtained your angles.
Your answer (start in the next line):
0
135
59.5
These angles were obtained by taking the arctan (Y/X) and then compensating to make them from the positive x axis. The 0 degrees doesnt make any sense, it should be 270 from the positive x axis.
#$&* _ angles of three forces
8. Verify that all your points were really 10 cm from the origin by using the Pythagorean Theorem; i.e., by calculating `sqrt( x^2 + y^2 ), where x and y are the coordinates of the points. For the point (2.3, 8.5), for example, you would get `sqrt( 2.3^2 + 8.5^2) = 9.2, approx.. The result should be 10. Obviously whoever located the point on the above graph didn't do a very good job. Hopefully your results will be better.
Give the results of the Pythagorean theorem in the space below, reporting in each line, in comma-delimited format, the result you obtained and its deviation from the ideal 10 cm. Report three lines, one for each rubber band, reporting in the same order as previously.
Starting in the fourth line explain what the numbers you have reported mean and how they were obtained.
Your answer (start in the next line):
9.9 cm
10.6 cm
10 cm
These numbers were obtained through the Pythagorean theorem, but clearly show thet my estimate of mm might be off a little. Im close, but only 1 vector is truly 10 cm.
#$&* _ pythagorean theorem applied to reported coordinates of three points
Draw to scale the force vectors corresponding to the rubber band forces, determine their components and find their sum. Note carefully that you are drawing force vectors here, NOT the lengths of the rubber bands. You are now done with the lengths of the rubber bands. Those lengths were only used to find the forces and they are no longer important.
1. Using a scale of 4 cm per Newton sketch the vector representing the force exerted by each rubber band (the force, not the length of the rubber band). Use a different color or style for the vector corresponding to each rubber band; you might use pens of different colors, or you might use pencil for one, pen for another, and another pen with a different color; or you could use a line which is thicker or thinner than the other lines..
2. Each vector should have its initial point at the origin, should lie along the line of force and be directed in the direction of the force exerted on the paper clip by the rubber band. The first figure below illustrates three vectors representing forces of roughly 1.5 Newtons, .6 Newtons and 1.8 Newtons.
3. Now draw the projection lines, sketch the component vectors, and validate the components as calculated using the sines and cosines of the angles: From the tip of each vector sketch the projection lines back to the x axis and to the y axis. Be sure the projection lines run parallel to the grid lines and use a straightedge to locate the projection lines as accurately as possible.
4. Using the same colors and/or line styles you used to sketch the vectors, sketch the component of each of the force vectors along the x and y axes, as shown in the second figure below.
5. Measure the length of each of these components, and using the 4 cm to 1 Newton scale determine as accurately as possible how many Newtons are represented by each component. In the figure below, for example, it looks like the red vector in the first quadrant has components measuring 1.4 cm in the x direction and about 5.8 cm in the y direction. These forces would correspond to .35 Newtons and 1.45 Newtons.
Enter your measurements in the space below. In each line enter 4 numbers, the first and second being the x and y components of your sketches, in cm, and the second and third being the corresponding x and y forces. Enter one line for each rubber band, using the same order of rubber bands as before. Note each quantity you report will be positive or negative, depending on whether that quantity is in the direction of the corresponding axis or opposite the direction of that axis (i.e., right and left are + and -, respectively; up and down are also + and - respectively).
Starting in the fifth line explain briefly what you numbers mean and how you obtained them.
Your answer (start in the next line):
0 cm, -10 cm, 0, 3.6 This would be -3.6 N
-4.9 cm, 4.4 cm, .7, .7 the x component would be -.7 N, not .7 N
5.9 cm, 10 cm, 1.42, 2.4
The x and y components were found by measuring using the grid, while the forces were found with the x and y component equations of N * Cos (angle), and N*Sin (angle).
#$&* _measured lengths of components of vector _ 4 cm / Newton scale
In the space below, report in the first line the total of the x components of all your forces, and in the second line the total of the y components of all your forces. In the third line explain how you obtained these totals and show the numbers you added to obtain each.
Your answer (start in the next line):
2.12
6.7
X = 0+.7+1.42, Y = 3.6+.7+2.4
see my note above on the signs of these quantities
Im getting more skeptical as this goes on that Im doing it correctly. This makes sense in the homework and the Open QAs, but I think Im missing something here.
#$&* _ total of x and of y components
Graphically determine the vector sum by constructing a polygon
Add the three vectors by the polygon method:
1. Again sketch your three force vectors, this time arranged to form a polygon. Let the terminal point of each vector be the initial point of the next. That is, the vectors should be connected head to tail, as illustrated in the second figure below.
In this figure he polygon runs along the 'red' vector from (0, 0) to (1.4, 6), then from this point along the 'green' vector to about (-.3, 8), then from this point along the 'purple' vector to about (-.3, 1).
At a scale of 4 cm per Newton, which can also be expressed as 1/4 Newton per cm or .25 Newton per centimeter, the resultant vector has x component -.3 cm * .25 N / cm = -.075 N and y component 1 cm * .25 N / cm = .25 N.
2. Now sketch a vector from the initial point of the first vector to the terminal point of the last vector. This vector is the short blue vector in the second and third figures below. The 'blue' resultant vector originates at (0, 0) and terminates at the third point of the polygon, which we have estimated in the figure below to be at (-.3, 1).
The fourth figure below indicates the components of this resultant vector, about -.3 cm in the x direction and about 1 cm in the y direction..
What are the coordinates of the resultant vector in your sketch?
Give the x and y coordinates of the points on your actual graph, in cm, then the x and y coordinates of the corresponding force components, in Newtons, in the first line of the space below, in comma-delimited format.
Starting in the second line give a brief description of your polygon, including the initial and terminal points of each of the vectors in the polygon, and explain how you calculated the force components of your resultant vector using the scale of the graph.
Your answer (start in the next line):
(.1,0), 2.8, (5.9,10), 1, (1, 14.8), 10, and (1,0)
Im definitely sure at this point that Im doing something incorrectly here. The vectors were drawn like the ones above, and connected head to tail to form a polygon. The force vector has an angle of 95 degrees, and components a y component of 1 and an x component of 0.1.
It could be that you got some of your signs wrong.
#$&* _ describe polygon
3. Are the components of your resultant vector reasonably consistent with the results you got in the preceding activity when you added the components of the vectors? Answer in the space below.
Your answer (start in the next line):
Not at all. I definitely made a mistake somewhere along these lines and have carried that mistake along the lab.
#$&* _ consistency between polygon and adding components
4. What is the magnitude of the force vector indicated by your sketch? Use the pythagorean theorem to calculate the magnitude of the resultant force vector, as indicated by the components you gave in the preceding space . Give your magnitude in the first line, the magnitude calculated by the Pythagorean Theorem in the second, and explain starting in the third line how you calculated it:
Your answer (start in the next line):
The force vector is about 1 cm in length, and has coordinates of .1,0 and 0,1. Using the Pythagorean Theorem, the force vector was found to have a magnitude of 1.
#$&* _magnitude of force vector indicated by measurement _ then by pythagorean theorem on components
Use sines and cosines to find the components of the resultant vector and compare with the results of graphical methods:
Add the three vectors by calculating components:
1. For each rubber band use the magnitude and the angle of the force it exerts to determine the x and a y component of that force.
Use the methods of Problem Set 5, using magnitude with the sine and cosine of each angle; i.e., the x component of a force is F cos(theta) and the y component is F sin(theta), where F is the magnitude and theta the angle of the force; you listed these magnitudes and angles early in your report.
List in the space below, using comma-delimited format, the magnitude, angle, x component and y component of each vector. List one vector per line, in the order of rubber bands used throughout this experiment. Starting in the third line explain in detail how you calculated the components of the second of these vectors. Also include a statement comparing the components you calculated with the components you obtained from the scale drawing of the vectors (the one where you used a scale of 4 cm to 1 Newton).
2. What is the sum of all the x components you calculated in the preceding step? What is the sum of all your y components?
You previously obtained x and y components graphically, first using a sketch and projection lines, then using the polygon method of addition. Compare your results with the results you obtained before.
In the first line below give, in comma-delimited format, the sum of your x components, as found here, then the sum as found using the first sketch and projection lines, and finally the x component you obtained by the polygon method. Give in the second line the same information for the sum of the y components. Beginning in the third line explain what you have done and comment on the relative accuracy of the various methods:
Your answer (start in the next line):
At this point Im definitely lost, but seem to have gotten the best results, or the only ones that made sense, from the Pythagorean theorem method.
#$&* _ sum of x comp using sin, cos _ sum using first sketch _ x comp of polygon _ then same for y _discuss relative accuracy
Compare results with ideal result:
Compare what you got with the ideal result:
1. What is the actual sum of the x components of the forces that actually occurred, in the real world, when you perform this experiment? What is the actual sum of all the y components?
2. What is the resultant vector corresponding to these components? What is the magnitude of this resultant?
Report the answers to these two questions in the space below, and explain how you know that your answers must be so:
Your answer (start in the next line):
The sum of the x components should be the x component of the force vector, while the sum of the y components should be the y component of the force vector
The magnitude of the resultant is the average force, or the force and direction put on the paper clip by thye three rubber bands in the three different directions.
#$&* what is actual sum of x comp, y comp _ what is ideal resultant vector
3. For each of the three methods of calculating the resultant of the three forces, your error is the magnitude of the difference between the expected result and the actual result, your actual result being the resultant you obtained. Express the magnitude of your error as a percent of the largest of the three force vectors. Give in the first line the magnitude of your error, and the percent error, for the first graphical method. In the second line give the same information for the polygon method of addition. In the third line give the information for your calculations based on magnitudes and angles of the three vectors.
Your answer (start in the next line):
The Pythagorean theorem made sense to me, the graphing went well, and somewhere along the way I got lost and confused. I understand this was a way to achieve hopefully the same results using three different methods, all of which make sense to me on paper, but I got really weird results when I sketched and tried to make sense of it that way.
#$&* _ magnitude of percent error each way as percent of largest force vector _
Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
Approximately how long did it take you to complete this experiment?
Your answer (start in the next line):
A little over 3 hours
#$&*
See my notes and see if you can patch this up. Your data aren't bad. However you did get some signs wrong when first calculating the resultant. Up to that point you were doing fine.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#