#$&*
Phy 201
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).
torque_experiment_setup.gif (2724 bytes)
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).
torque_experiment_labeling_of_rubber_bands.gif (2366 bytes)
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.
torque_experiment_lines_sketched.gif (1966 bytes)
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):
1.9cm, 8.5cm, 13.25cm
7.79cm, 8.51cm, 8.67cm
.52N, 3.3N, 1.90N
The reference point was the far left edge of the paper were I had my line drawn through.
The forces were obtained by looking at my graph.
The 1st line contains the distance the points B,A,C were from the reference point. The second line is the stretch measured on the bands.A
was doubled because 2 bands were used. The third line is the force corresponding to B A C.
#$&*
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):
0.88,
15%
I got fNet by adding the forces -.52N+ 3.3N+ -1.90N= 0.88N
(.88)/ (.52+3.3+1.90) * 100 = 15%
#$&*
Rotational equilibrium: We will regard the position of the central supporting hook (the hook for system A) to be the fulcrum around which
the rod tends to rotate. Determine the distance from this fulcrum to the point of application of the force from rubber band B. This
distance is called the moment-arm of that force. Do the same for the rubber band at C.
report the moment-arm for the force exerted by the rubber band system B, then the moment-arm for the system C. Beginning in the second
line, briefly explain what the numbers mean and how you obtained them.
----->>>>>>>>
******** Your answer (start in the next line):
6.6cm, 4.7cm
The moment arm for B is 6.6cm becuase that is how far the force is applied from the fulcrum. The moment are for C was 4.7cm
#$&* 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.
torque_experiment_force_vectors_first_setup.gif (1720 bytes)
* 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):
2.08cm, 13.2cm, 7.6cm
6.6cm, 4.7cm
If 4cm = 1N .52N * 4cm = 2.08cm 3.3N * 4cm= 13.2cm 1.9N * 4cm = 7.6cm
The length of B from the fulcrum was 6.6cm , while C was 4.7cm
#$&*
The force from rubber band C will tend to rotate the rod in a clockwise direction. This force is therefore considered to produce a
clockwise torque, or 'turning force', on the rubber band. A clockwise torque is considered to be negative; the clockwise direction is
considered to be the negative direction and the counterclockwise direction to be positive.
When the force is exerted in a direction perpendicular to the rod, as is the case here, the torque is equal in magnitude to the product of
the moment-arm and the force.
* What is the torque of the force exerted by rubber band C about the point of suspension, i.e., about the point we have chosen for our
fulcrum?
* Find the torque produced by rubber band B about the point of suspension.
Report your torques , giving the torque produced by rubber band B then the torque produced by the rubber band C, in that order. Be sure to
indicate whether each is positive (+) or negative (-). Beginning in the next line, briefly explain what your results mean and how you
obtained them.
----->>>>>>>> torque C, torque B
******** Your answer (start in the next line):
-2.44, +12.54
The moment arm for C=4.7cm B=6.6cm Force on C=.52N B= 1.90N Torque of C= 4.7cm*.52N = -2.44 B= 6.6cm *1.90N= 12.54
#$&*
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):
10.1, 67%
12.54-2.44 = 10.1 I feel that my experiment was set up correctly. The sum of the torques didn't equal out to 0 because the data I read from
my rubber-band callibration graph skewed it. 10.1/ (12.54+2.44) *100 = 67%
#$&*
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.
wpe1A.jpg (3628 bytes)
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.
wpe8.jpg (3903 bytes)
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.
wpe9.jpg (4203 bytes)
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:
wpe1.jpg (4717 bytes)
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.
wpe1.jpg (4717 bytes)
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.
wpe3.jpg (5662 bytes)
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):
2.20N, 2.00N, 0.80N, 0.63N,
0cm, .9cm, 10.1cm, 13.7cm
-15.07, 11.9, -2.6, 4.32
The first line contains the Force exerted on each band, this was found by using my rubber-band callibration graph. The second row contains
the distance of the points from A. A is at 0cm then B,C,D.
I figured the equilibrium point would be the halfway point of the rail. This point came to be 6.85cm. To find the torques I found the
distance from this point A B C D were then multiplied by the Newtons. A= 2.2N * 6.85cm= -15.07N B= 2.00N * (6.85cm-.9cm) = 11.9
C= .80N * (10.1cm-6.85cm) = -2.6 D= .63N * (13.7cm-6.85cm)= 4.32
The clockwise torques are negative, counter-clockwise +
#$&*
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):
0.03N
2.2N+ .63N-2.0N-.8N= 0.03N My picture looks almost exactly like the one shown. All of the torques seem to balance out the system so that
the center line is level.
#$&*
In the figure shown above the 1.6 N force produces a clockwise torque about the leftmost force (about position A), a torque of 1.6 N * 15
cm = 24 N cm. Being clockwise this torque is -24 N cm. The 2.0 N force at 23 cm produces a clockwise torque of 2.0 N * 23 cm = 26 N cm.
Being counterclockwise this torque is +26 N cm.
In the first line below give the net torque produced by the forces as shown in this figure. Beginning in the second line describe your
picture and discuss whether it could be an accurate depiction of torques actually acting on a stationary rod. Support your discussion with
reasons.
----->>>>>>>> net torque from given picture, describe your picture
******** Your answer (start in the next line):
0Ncm -16Ncm -24Ncm + 26Ncm = -14Ncm
I believe my picture shows a pretty accurate depiction of the torques acting on the rod. I say this because the torques balance the rod out
and if my calculations were totally right I would assume the Net force would be equal to O.
#$&*
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):
-1.97Ncm
0.03N, 5.63N
5.31%
-1.97Ncm, 18.511
10.6%
2.2N+.63N-2.00N-0.80N= 0.03N 2.2+ 2.00 + .63 + .80 = 5.63N 0.3/5.53 * 100= 5.31%
(0cm*2.2N) + (.9cm*-2.00N) + (10.1cm * -.80N) + ( 13.7cm * 0.63N) = 0Ncm -1.8cNcm -8.80Ncm + 8.631Ncm = -1.97Ncm
#$&*
Perform a similar analysis for the second setup (in which you increased the pull at C) and give your results below:
* For your diagram, what is the sum of the torques about the point of action of the leftmost force (i.e., about position A)? This is
your experimentally observed resultant torque about A. Give your result in the first line below.
* For your diagram what is the magnitude of your resultant force and what is the sum of the magnitudes of all the forces acting on the
rod? Give these results in the second line in comma-delimited format.
* Give the magnitude of your resultant force as a percent of the sum of the magnitudes of all the forces. Give this result in the third
line.
* For your diagram what is the magnitude of your resultant torque and what is the sum of the magnitudes of all the torques acting on
the rod? Give these two results, and the magnitude of your resultant torque as a percent of the sum of the magnitudes of all the torques,
as three numbers in your comma-delimited fourth line.
* Beginning in the fifth line, briefly explain what your results mean and how you obtained them.
----->>>>>>>> (pull at C incr) sum(tau) about A, Fnet and sum(F), Fnet % of sum(F), | tau_net |, sum | tau |, |tau_net| % of sum|tau|
******** Your answer (start in the next line):
#$&*
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):
#$&*
Estimate the angles made by the lines of force with the rod in the second setup, and give your angles in comma-delimited format in the
first line below. Your angles will all likely be close to 90 degrees, but they probably won't all be 90 degrees. The easiest way to
estimate is to estimate the deviation from 90 degrees; e.g., if you estimate a deviation of 5 degrees then you would report an angle of 85
degrees. Recall that you estimated angles in the rotation of a strap experiment.
Starting in the second line give a short statement indicating how you made your estimates and how accurate you think your estimates were.
----->>>>>>>> angles of lines of force with rod
******** Your answer (start in the next line):
#$&*
*#&!
&#Very good data and responses. Let me know if you have questions. &#