#$&*
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.
** **
** **
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):
2.13, 9.6, 14.0
7.9, 9.7, 8.4
1.42, 3.9, 1.8
I used the left-most end of the line I sketched
to represent the rod.
I used the slope-intercept of the graphs for
each of the rubber bands to obtain the force in
Newtons.
The first line represents the hook points on
the rod in centimeters from an arbitrary
reference point. The second line is the length
of the rubber bands in centimeters. The third
line is the force in Newtons exerted by the
rubber bands.
#$&*
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):
+.68N
9.5%
The first line is the Fnet of the rubber band
system. I added all of the forces together
assigning negative values to the bottom rubber
bands and a positive value to the top rubber
band. The third line is the net force
percentage value of the sum of the magnitudes
of the forces of all three rubber bands.
#$&*
Rotational equilibrium: We will regard the
position of the central supporting hook (the
hook for system A) to be the fulcrum around
which the rod tends to rotate. Determine the
distance from this fulcrum to the point of
application of the force from rubber band B.
This distance is called the moment-arm of that
force. Do the same for the rubber band at C.
report the moment-arm for the force exerted by
the rubber band system B, then the moment-arm
for the system C. Beginning in the second
line, briefly explain what the numbers mean and
how you obtained them.
----->>>>>>>>
******** Your answer (start in the next line):
7.47, 4.4
Above are the moment arms for each of the lower
rubber bands in regards to the attachment point
of the middle upper rubber band. They were
obtained by measuring the distance from the
central fulcrum to the point of application of
force for rubber bands B and C.
#$&* 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):
2.84, 7.8, 3.6
7.47, 4.4
The first line is the length of the vectors I
constructed. They are in a 2:1 scale (2 cm for
each 1 Newton) rather than a 4:1 scale because
of the limitations of my paper size. The
second line lists the distances from the
fulcrum for rubber bands B and C and they are
in 1:1 scale. All measurements are in
centimeters.
#$&*
The force from rubber band C will tend to
rotate the rod in a clockwise direction. This
force is therefore considered to produce a
clockwise torque, or 'turning force', on the
rubber band. A clockwise torque is considered
to be negative; the clockwise direction is
considered to be the negative direction and the
counterclockwise direction to be positive.
When the force is exerted in a direction
perpendicular to the rod, as is the case here,
the torque is equal in magnitude to the product
of the moment-arm and the force.
What is the torque of the force exerted
by rubber band C about the point of suspension,
i.e., about the point we have chosen for our
fulcrum?
Find the torque produced by rubber band
B about the point of suspension.
Report your torques , giving the torque
produced by rubber band B then the torque
produced by the rubber band C, in that order.
Be sure to indicate whether each is positive
(+) or negative (-). Beginning in the next
line, briefly explain what your results mean
and how you obtained them.
----->>>>>>>> torque C, torque B
******** Your answer (start in the next line):
+10.6, -7.92
Listed above are the torques of rubber band B
and C respectively. They were obtained by
multiplying the moment arm of each rubber band
by the force that rubber band exerted on the
system. The units are in N*cm.
#$&*
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):
+2.68
14.5
I divided 2.68 by the sum of the two torques
then multiplied quotient by 100. (2.68/
(10.6+7.92))*100
The first line is the Fnet of the torques. The
second line is the Fnet percentage of the sum
of the torques.
#$&*
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):
1.56, 0, 0
1.43, 1.0, -1.43
.57, 12.68, -7.23
.58, 16.64, 9.65
The lines above are the force in Newtons, the
displacement from equilibrium in centimeters,
and the torque in N*cm for each of the rubber
bands A, B, C, D. Rubber band A was designated
the point of equilibrium. The other rubber
bands torque was determined by multiplying the
moment arm by the force they were exerting on
the system.
#$&*
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):
+ .14
My free body diagram depicts a rod of 19 cm
being supported by 4 rubber bands. The two
left most rubber bands are longer than the two
right most rubber bands. The forces were drawn
to a scale of 4 cm to 1 Newton. The sum of all
forces in y was +.14 N. This result is close
to the actual system which was in balance with
0 acceleration.
#$&*
In the figure shown above the 1.6 N force
produces a clockwise torque about the leftmost
force (about position A), a torque of 1.6 N *
15 cm = 24 N cm. Being clockwise this torque is
-24 N cm. The 2.0 N force at 23 cm produces a
clockwise torque of 2.0 N * 23 cm = 26 N cm.
Being counterclockwise this torque is +26 N cm.
In the first line below give the net torque
produced by the forces as shown in this figure.
Beginning in the second line describe your
picture and discuss whether it could be an
accurate depiction of torques actually acting
on a stationary rod. Support your discussion
with reasons.
----->>>>>>>> net torque from given picture,
describe your picture
******** Your answer (start in the next line):
+6N*cm
The net torque of my diagram is +.99N*cm. This
is a close representation of the actual forces
acting on the rod. The actual forces in my
system were balanced with zero acceleration.
#$&*
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):
+.99N*cm
+ .14N, 4.14N
3.4%
+.99N*cm, 18.31N*cm, 5.4%
The first line is the resultant torque of my
diagram. The second line is the resultant
force in y of my diagram followed by a sum of
all forces in y disregarding positive and
negative. The third line is the percentage the
resultant vertical force was of the total
vertical forces. This was found by dividing
the resultant by the total vertical force and
multiplying the quotient by 100. The fourth
line is the resultant torque, the total torque
disregarding rotation, and the percentage the
resultant torque was of the total torque. This
percentage was found using the same method
described earlier.
#$&*
Perform a similar analysis for the second setup
(in which you increased the pull at C) and give
your results below:
For your diagram, what is the sum of
the torques about the point of action of the
leftmost force (i.e., about position A)? This
is your experimentally observed resultant
torque about A. Give your result in the first
line below.
For your diagram what is the magnitude
of your resultant force and what is the sum of
the magnitudes of all the forces acting on the
rod? Give these results in the second line in
comma-delimited format.
Give the magnitude of your resultant
force as a percent of the sum of the magnitudes
of all the forces. Give this result in the
third line.
For your diagram what is the magnitude
of your resultant torque and what is the sum of
the magnitudes of all the torques acting on the
rod? Give these two results, and the magnitude
of your resultant torque as a percent of the
sum of the magnitudes of all the torques, as
three numbers in your comma-delimited fourth
line.
Beginning in the fifth line, briefly
explain what your results mean and how you
obtained them.
----->>>>>>>> (pull at C incr) sum(tau) about
A, Fnet and sum(F), Fnet % of sum(F), | tau_net
|, sum | tau |, |tau_net| % of sum|tau|
******** Your answer (start in the next line):
-.07N*cm
+.66N, 6.36N
10.4%
-.07N*cm, 47.67N*cm, .15%
The first line is the resultant torque of my
diagram. The second line is the resultant
force in y of my diagram followed by a sum of
all forces in y disregarding positive and
negative. The third line is the percentage the
resultant vertical force was of the total
vertical forces. This was found by dividing
the resultant by the total vertical force and
multiplying the quotient by 100. The fourth
line is the resultant torque, the total torque
disregarding rotation, and the percentage the
resultant torque was of the total torque. This
percentage was found using the same method
described earlier.
#$&*
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 were very close to parallel with
one another. I used a protractor to estimate
the angles and there might have been 1 degree
difference between the forces.
#$&*
Estimate the angles made by the lines of force
with the rod in the second setup, and give your
angles in comma-delimited format in the first
line below. Your angles will all likely be
close to 90 degrees, but they probably won't
all be 90 degrees. The easiest way to estimate
is to estimate the deviation from 90 degrees;
e.g., if you estimate a deviation of 5 degrees
then you would report an angle of 85 degrees.
Recall that you estimated angles in the
rotation of a strap experiment.
Starting in the second line give a short
statement indicating how you made your
estimates and how accurate you think your
estimates were.
----->>>>>>>> angles of lines of force with rod
******** Your answer (start in the next line):
92, 88.5, 88.5, 92
I used a protractor to make my estimates. I
believe they are within 1 degree of accuracy.
#$&*
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):
1.18, 2.25, 1.2, 9.35
8.75, 12.95, 6.5, 20.17
19.36, 1.42, 15.65, 9.2
Listed above are the endpoint coordinates for
the rubber bands used in this setup. The first
is the left most rubber band B, The second is
the two rubber band setup A, and the third is
the right most rubber band C. The measurements
are in centimeters and were made with a plastic
ruler.
#$&*
Analysis
Using your coordinates and the Pythagorean
Theorem, find the length of rubber band system
B.
Do this by first finding the difference
in the x coordinates of the ends of this band,
then the difference in the y coordinates of the
ends.
This gives you the lengths of the legs
of a right triangle whose hypotenuse is equal
to the length of the band.
Then using your calibration information
find the force in Newtons exerted this system.
Do the same for systems A and C.
Give the length and force exerted by rubber
band system B in the first line below, in
comma-delimited format, then in the second and
third lines give the same information for
systems A and C. Starting in the fourth line
give a brief description of what your results
mean and how you obtained them.
----->>>>>>>> length and force of B, of A, of C
******** Your answer (start in the next line):
7.1, 1.3
7.56, 3.37
8.62, 2.14
Listed above are the lengths and forces for the
rubber bands in the system B, system A, and C.
The lengths were found by using the Pythagorean
Theorem and are measured in centimeters. The
forces were found by using the slope intercept
formula equation for each rubber band. For the
two rubber band setup A I found the force
exerted on the system by each rubber band then
added these forces together for the Fnet.
#$&*
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):
-1, .002
.95, -.29
-.9, .43
Listed above are the sine and cosine for each
of the rubber bands in order B, A, C. For the
sine I divided the y component of the length of
the rubber band by the length and for the
cosine I divided the x component of the length
by the length.
#$&*
Find the angles of the force vectors with the
horizontal, and the angles of the force vectors
in the plane:
The angle of the force vector with
horizontal is arcTan(y / x): the arctangent of
the magnitude of the quantity you get with you
divide the y component of the triangle used in
the preceding, by the x component.
The arctangent is easily calculated
using the 2d fn or inverse key on your
calculator, along with the tan function.
The angle of the force vector in the
plane is measured from the positive x axis, in
the counterclockwise direction.
Give for each system the magnitude (i.e., the
force in Newtons as you calculated it earlier),
the angle with the x axis and the angle in the
plane for each of the force vectors, reporting
three comma-delimited lines in the order B, A
and C. Starting in the fourth line briefly
explain how you determined these values and how
you obtained them:
----->>>>>>>> magnitude and angle of B, of A,
of C
******** Your answer (start in the next line):
1.3, -90, 270
3.37, 72.7, 107.3
2.14, -64.5, 295.5
Listed above are the forces in Newtons, the
angle from the x axis in degrees, and the angle
in the plane for each of the three rubber
bands. The force was calculated using the
slope intercept formula, the angle was
calculated using arctangent and the x and y
components of the vectors, and the angle in the
plane was calculated using the angle from the x
axis and the orientation of the vector.
#$&*
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, -4.6, 0, -1.15, 0, -1.15
-4.1, 12.73, -1.03, 3.18, -.96, 3.17
3.7, -7.72, .92, -1.93, .92, -1.93
The first two numbers in each line listed
above are the scaled x, y components for each
rubber band. The scale is 4 cm = 1 N. The
following two numbers are the x and y forces as
determined by the scale. The last two numbers
are the x and y forces found by multiplying the
resultant by the sine and cosine found earlier.
#$&*
Calculate the sum of the x components and of
the y components, as determined by the
magnitude, sine and cosine.
What is the sum of all your x
components? What should be the sum of all the x
components? How close is your sum to the ideal?
Report as three numbers in comma-delimited
format in line 1.
What is the sum of all your y
components? What should be the sum of all the y
components? How close is your sum to the ideal?
Report as three numbers in comma-delimited
format in line 2.
Beginning in the third line, briefly
explain what your results mean and how you
obtained them.
----->>>>>>>> sum of your x comp, actual sum,
how close to ideal x, then y
******** Your answer (start in the next line):
-.04, 0, -.04
.09, 0, .09
Listed in the first line is the sum of all x
components followed by the ideal sum and the
distance from the ideal sum. The second line
is the same for the y components. They were
found by adding all x and y components as
determined by multiplying the vectors by their
respective sine and cosine values. The ideal
values should be 0 because the system is not
accelerating.
#$&*
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, 0
8.21, 26.1
13.72, -26.4
Listed above are the moment arms and torques
for each of the rubber bands in the system.
The moment arm is determined by measuring the
distance from the arbitrary left most point.
The torque is determined by multiplying this
distance by the y component of the force
vector. The units are centimeters and N*cm
respectively.
#$&*
Finally report the sum of your torques:
What is the sum of the torques about the point
of action of the leftmost force? What should
this sum be? How close is your sum to the
ideal? Report as three numbers in comma-
delimited format in line 1. Beginning in the
second line, briefly explain what your results
mean and how you obtained them.
----->>>>>>>> sum of torques, ideal sum, how
close to ideal:
******** Your answer (start in the next line):
-.3, 0, -.3
This is the sum of the torques in N*cm,
followed by the ideal and the distance from the
ideal.
#$&*
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):
8 hours
#$&*
*#&!
&#Very good data and responses. Let me know if you have questions. &#