#$&*
Phy 121
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.
** Torques_labelMessages **
Submitted at 11:21 PM on April 30, 2013
** **
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.
******** ******** Your answer (start in the next line):
.39, 7.4, 11.3 all in cm
A is 6.7, B is 6.3, C is 8.7 cm
A is .21 N, B is .17 N, C is .31 N
The reference line or point is the line for the rod.. so the horizontal line is the rod and of course the vertical lines are the rubber band system A, B, or C....
The way I got the forces for each of the rubber bands was by getting the best accurate estimate by looking at the corresponding graph, which in this case was the 8 domino setup and saw that it was close to the forces that I have put above.. and what I was looking at was the the line on my graph to get the estimates..
In the first line it is the measurements on my reference line, which is where the rod was located, starting from left to right the points in which the vertical lines cross the horizontal line or the reference point..
The second line is the measurements of how long the rubber bands were stretched..
The third line is the forces that each are exerting according to my 8 domino setup calibrated graph...
#$&*
----->>>>>>>> (note A doubled) intersections B A C, lengths B A C, forces B A C, reference point, how forces determined
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.
******** Your answer (start in the next line):
The total net force of the system is -.27 N
The percentage for my net force as the sum of the magnitude of the forces of all 3 rubber band systems is 39.13%..
In the first line it shows the net force that is being done in the system... the A is a positive upward force at .21 N and B and C are negative downward forces at .17 + .31 = .48 N soo .21 - .48 is -.27 N net force and it is negative since there is more downward force than upward..
The second line is the percent and the way I got that was by .27 / .69 = .3913 * 100 = 39.13%... I divided .27, which is the net force I got by .69 which is the total of all three systems A, B, and C...
#$&*
----->>>>>>>> Fnet, Fnet % of sum(F)
Rotational equilibrium: We will regard the position of the central supporting hook (the hook for system A) to be the fulcrum around which the rod tends to rotate.
Determine the distance from this fulcrum to the point of application of the force from rubber band B. This distance is called the moment-arm of that force.
Do the same for the rubber band at C.
report the moment-arm for the force exerted by the rubber band system B, then the moment-arm for the system C. Beginning in the second line, briefly explain what the numbers mean and
how you obtained them.
******** Your answer (start in the next line):
The distance from the fulcrum to the point of application of the force from rubber band B is..
7.1 cm
The moment-arm for the force exerted by the rubber band system B is..
.04 N
The distance from the fulcrum to the point of application of the force from rubber band C is..
3.9 CM
The moment-arm for the force exerted by the rubber band system C is..
-.1 N
The way I got the distance from the fulcrum to the point of application from a particular force was by seeing the distance from point F (fulcrum) to point B...
Point F was at 7.4 cm on the reference line or the rod and point B was .3 cm.. so 7.4 cm - .3 cm is 7.1 cm...
The way I got the moment-arm force exerted by the rubber band system B was by..
taking the force that is being exerted by A and subtracting the rubber band force B which got me .04 N.. so .21 N - .17 N = .04 N is the moment-arm force exerted by the rubber band system...
#$&*
----->>>>>>>> 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.
******** Your answer (start in the next line):
The vector for system A
.84 cm
The vector for system B
.68 cm
The vector for system C
1.24 cm
The distance from the fulcrum to the point of application of the force from rubber band B is..
7.1 cm
The distance from the fulcrum to the point of application of the force from rubber band C is..
3.9 CM
For the first line, the numbers are the 4 cm to 1 newton scale.. so the for system A is .84 cm
For the second line, the numbers represent the distance from the fulcrum point to the point where the rubber band system is at point..
How did I get the 4 cm to 1 N vector scale for my systems?
For the first system A I got it by multiplying .21 N with 4 cm to get .84 cm.. because (4 cm / 1 N) * (.21 N) = .84 cm is the vector for system A or rubber band A
#$&*
----->>>>>>>> (4 cm to 1 Newton scale) lengths of force vectors B, A, C, distances of B and C from fulcrum
The force from rubber band C will tend to rotate the rod in a clockwise direction. This force is therefore considered to produce a clockwise torque, or 'turning force', on the rubber
band. A clockwise torque is considered to be negative; the clockwise direction is considered to be the negative direction and the counterclockwise direction to be positive.
When the force is exerted in a direction perpendicular to the rod, as is the case here, the torque is equal in magnitude to the product of the moment-arm and the force.
What is the torque of the force exerted by rubber band C about the point of suspension, i.e., about the point we have chosen for our fulcrum?
Find the torque produced by rubber band B about the point of suspension.
Report your torques , giving the torque produced by rubber band B then the torque produced by the rubber band C, in that order. Be sure to indicate whether each is positive (+) or
negative (-). Beginning in the next line, briefly explain what your results mean and how you obtained them.
******** Your answer (start in the next line):
The torque for C is..
-1.209
The torque for B is..
-1.258
The way I got the torque for C is..
The newton force that is being exerted by the rubber band C is -.31 (it is negative since it is going downwards or in the clockwise direction..
And the moment-arm distance is 7.1 cm..
So multiplying 3.9 cm * -.31 N gives me -1.209 cm * N
I did the same for Torque B but the force for B was .17 N and the moment-arm distance was 7.4 cm
#$&*
----->>>>>>>> torque C, torque B
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.
******** Your answer (start in the next line):
The sum of torque for the system that I have found is -2.467 cm * N
How would I get the percent of the sum of the magnitudes for these torques?
How did I get the net torque?
Added all the torques I have gotten for all systems which give sme -2.467 cm * N..
#$&*
*#&!*#&!
&#Your work on this lab exercise is good. Let me know if you have questions. &#