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course Phy 241
9/7 6I had some difficulty with the last few problems assigned. I understand the work but don't really know what I am doing. Probley doesn't make much sense saying I know how to do it but don't know what I'm doing. Anyway any advice on what I'm doing worng or an idea of a good place to start reviewing to get caught up would be greatly appericated. Sorry for any incovenice and as always thanks again.
1. On a coordinate plane sketch the points (5, 9) and (3, 12). You don't need to take a lot of time to meticulously mark off the scale or measure the points with a ruler, just make a reasonable estimate. You should be able to locate the origin and the two points in just a minute or so. Sketch a vector from the origin to the first point and call this vector B. Sketch another vector from the origin to the second point and call it A.
Sketch the projection of the A vector on the B vector, as we did in class. In case you need more detailed instructions:
* Sketch a dotted projection line from the tip of the A vector to the B vector. The projection line must make an angle of 90 degrees with the B vector.
* The 'projected point' is the point where the projection line meets the B vector.
* The projection of A on B is the vector from the origin to 'projected point' .
Estimate the coordinates of the tip of the projection vector and give them in the next line:
It appears to be approx. (6, 10.5)
??????I have not had any experience with a lot of this material, so please forgive me if my answers are ridiculous??????????????????????
The idea here is to gain experience. A lot of the vector operations should have been part of Mth 164, but they don't always get covered.
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What therefore is the length of the projection vector? Give the length and explain how you found it starting in the next line:
To find projection:
A*B = (3, 12) * (5, 9) = 15 + 108 = 123
B*B = (5, 9) * (5, 9) = 25 + 81 = 106
projB A = (123/106)*B = (123/106)*(5, 9) = approx. (5.8019, 10.4434)
To find length:
I don’t know formula and didn’t understand online instructions, so I used distance formula
From (3, 12) to (5.8019, 10.4434)
= `sqrt((3-5.8019)^2+(12-10.4434)^2) = `sqrt(10.2736) = approx. 3.2052
?????I found formula stating where C is magnatitude(or length) of projection vector, but wasn’t sure if this is what we needed to use?????????
You're getting there with the formulas, but if your estimate is (6, 10.5) then you would use those coordinates with the Pythagorean theorem to get the corresponding length. That's for the estimate.
We'll clarify the formulas tomorrow or Monday.
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2. Repeat the above exercise using the points (4, 2) and (8, 3). You might have to extend the line of the B vector so that the projection line will meet it. Give your results starting in the next line:
To find projection:
A*B = (8, 3) * (4, 2) = 32 + 6= 38
B*B = (4, 2) * (4, 2) = 16 + 4 = 20
projB A = (38/20)*B = (38/20)*(4, 2) = (7.6, 3.8)
To find length:
From (8, 3) to (7.6, 3.8)
= `sqrt((8-7.6)^2+(3-3.8)^2) = `sqrt(.8) = approx. .8944
????? Again I am not sure this is the correct method to use????????
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The idea here is to sketch and estimate. You're just about there with the formulas.
3. On a graph of velocity vs. clock time, with velocity in cm/s when clock time is in seconds, what does the point (3, 9) represent?
Velocity at the 3 sec mark being equal to 9 cm/sec.
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On this graph of velocity vs. clock time the points (3, 9) and (6, 13) represent the velocities of a moving object at the corresponding clock times. What does the 'rise' between these points represent? What does the 'run' represent? What does the slope of the straight line between (3, 9) and (6, 13) therefore represent?
‘rise’ represents `dv(change in velocity, or if for different fn then it would be difference in velocities)
‘run’ represents `dt(change in time)
slope of straight line represents the rate at which the velocity is changing from clock times 3 to 6, which happens to be 3 sec (I believe this is the same as saying the average acceleration)
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If you cut a triangle from this trapezoid, cutting in the horizontal direction starting at the midpoint of the line segment between the two given points, you can simply rotate this triangle 180 degrees and join it to the remaining part of the trapezoid and form a rectangle.
* What is the 'width' of this rectangle and what does it represent?
`dt( change in t)
* What is the 'height' of this rectangle and what does it represent?
vAve(average velocity between t’s)
* What is the area of this rectangle and what does it therefore represent?
total displacement from 3 sec mark to 6 sec mark. Which is 11(height or vAve) * 3(width or `dt) = 33 cm/s^2
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4. On a graph of velocity vs. clock time the points (t_0, v_0) and (t_f, v_f) represent the velocity and position of an object at the beginning and the end of some interval. The vertical lines from these points to the horizontal axis form what we will call the 'graph altitudes' of a trapezoid. The line segment from the first point to the second forms another side of the trapezoid, and the segment from t_0 to t_f on the horizontal axis forms the fourth side, which we call the 'graph width/ of the trapezoid. (in correct geometrical language this In terms of the symbols t_0, v_0, t_f and v_f, what are the following:
t_0 = initial time( or time at the start of the interval)
t_f = finial time( or time at the end of interval)
v_0 = initial velocity ( or velocity at start of interval)
v_f = finial velocity( or velocity at the end of interval)
5. For your data in the rubber band experiment, give the endpoints of each rubber band in the first trial, where the rubber bands were just barely beginning to exert a tension force. Report one rubber band in each of the next three lines, giving in each line the x and y coordinates of the first point, followed by the x and y coordinates of the second point, in that order. Separate each number from the next by a comma. You will have four numbers in each of three lines, separated by commas. In the fourth line give the units of your measurements. Start in the next line:
First Group (barley any force):
1st rubber band: (18 (9/10), 18 (3/10)) to (14 (3/5), 13 (7/10))
2nd rubber band: (12 (1/2), 11 (2/5)) to (9 (3/5), 6(1/2))
3rd rubber band: (12 (1/2), 11 (1/2)) to (7 (1/2),8 (1/5))
Second Group (medium force):
1st rubber band: (19 (1/10), 18 (3/5)) to (13 (1/10), 12)
2nd rubber band: (11, 9 (9/10)) to (5 (9/10), 5(7/10))
3rd rubber band: (11 (1/10), 9 (4/5)) to (7 (2/5),3 (9/10))
Third Group (Most force):
1st rubber band: (18 (3/10), 19 (1/10)) to (9, 10 (4/5))
2nd rubber band: (7 (1/5), 8 (3/10)) to (3 (4/5), (1/10))
3rd rubber band: (6 (4/5), 8 (4/5)) to (- (1/5),8 (9/10))
Both x and y axis are measured in centimeters, with (0, 0) being where measurements begin.
!!!!!!Graph is a little unusual, but was not aware of what we were going to do while taking measurements. Should work fine for calculating distances between points, which I believe is the idea!!!!!
Good so far.
The fractions are fine but decimals are quicker to type. I see that you started using them in the next set of calculations, which was pretty much inevitable.
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Find the lengths of the three rubber bands, based on the coordinates of the points you reported. Give the lengths in the first line below, separated by commas. Starting in the next line explain how you calculated the lengths:
First Group (barley any force):
1st rubber band: (18 (9/10), 18 (3/10)) to (14 (3/5), 13 (7/10))
Distance = `sqrt((18.9 – 14.6)^2 + (18.6 – 13.7)^2) = approx. 6.5192
2nd rubber band: (12 (1/2), 11 (2/5)) to (9 (3/5), 6(1/2))
Distance = `sqrt((12.5 – 9.6)^2 + (11.4 – 6.5)^2) = approx. 5.6939
3rd rubber band: (12 (1/2), 11 (1/2)) to (7 (1/2),8 (1/5))
Distance = `sqrt((12.5 – 7.5)^2 + (11.5 – 8.2)^2) = approx. 5.9908
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Now report the coordinates of the points you observed in the second trial, using the same format as before. Report the three different rubber bands in the same order as before:
Second Group (medium force):
1st rubber band: (19 (1/10), 18 (3/5)) to (13 (1/10), 12)
Distance = `sqrt((19.1 – 13.1)^2 + (18.6 – 12)^2) = approx. 8.9196
2nd rubber band: (11, 9 (9/10)) to (5 (9/10), 5(7/10))
Distance = `sqrt((11 – 5.9)^2 + (9.9 – 5.7)^2) = approx. 6.6068
3rd rubber band: (11 (1/10), 9 (4/5)) to (7 (2/5),3 (9/10))
Distance = `sqrt((11.1 – 7.4)^2 + (9.8 – 3.9)^2) = approx. 6.9642
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Report the lengths of the three rubber bands, using the same order as before:
Third Group (Most force):
1st rubber band: (18 (3/10), 19 (1/10)) to (9, 10 (4/5))
Distance = `sqrt((18.6 – 9)^2 + (19.1 – 10.8)^2) = approx. 12.6905
2nd rubber band: (7 (1/5), 8 (3/10)) to (3 (4/5), (1/10))
Distance = `sqrt((7.2 – 3.8)^2 + (8.6 – .1)^2) = approx. 9.1548
3rd rubber band: (6 (4/5), 8 (4/5)) to (- (1/5),8 (9/10))
Distance = `sqrt((6.8 – (-.2))^2 + (8.8 – 8.9)^2) = approx. 7.0007
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Assuming that each rubber band exerts a force of .5 N for every centimeter of length in excess of its 'barely-exerting-a-force' length, what are the three forces? Report the three forces, separated by commas, in the first line below. In the second line below, explain how you determined your forces.
For Second Group (medium force):
1st rubber band,
= (8.9196 - 6.5192)(difference between little force distance and medium force distance, in cm) = 2.4004(diff. in distance)*.5(force in Newton’s needed per cm) = 1.2002
2nd rubber band,(Same setup as prior problem in regards to diff. in length and force needed)
= (6.6068 - 5.6939)*.5 = approx. .4565
3rd rubber band,(Same setup as prior problem in regards to diff. in length and force needed)
= (6.9642 - 5.9908)*.5 = approx. .4867
For Third Group (Most force):
1st rubber band,
= (12.6905- 6.5192)(difference between little force distance and medium force distance, in cm) = 6.1713(diff. in distance)*.5(force in Newton’s needed per cm) = approx. 3.0857
2nd rubber band,(Same setup as prior problem in regards to diff. in length and force needed)
= (9.1548- 5.6939)*.5 = approx. 1.7305
3rd rubber band,(Same setup as prior problem in regards to diff. in length and force needed)
= (7.0007- 5.9908)*.5 = approx. .503
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Everyone should be able to do the above. At some point below I would expect most College Physics students to get a little lost, so if you are a College Physics student and do get lost, it's OK. We'll go through some further explanation based on the above and you'll be able to answer the questions after the next class.
For your second trial, each rubber band can be represented by a displacement vector, a vector from one of its endpoints to the other. More specifically we will represent each rubber band by the displacement vector from the point closest to the middle to the point furthest from the middle (the middle is the center of that paperclip on which all three rubber bands are pulling).
In the first line below, report the x and y components of the vector representing the first rubber band, using the notation . For example a vector with x and y components 3 cm and -7 cm would be reported as <3 cm, -7 cm>. In the second and third lines, report the components of the vectors representing the second and third rubber bands. Report the rubber bands in the same order as before. Starting in the fourth line give a brief explanation of how you got your results:
First Group (bareley any force):
<4.3, 4.6>, <-2.9, -4.9>, <-5, -3.3>
Second Group (medium force):
<6, 6.6>, <-5.3, -4.2>, <-3.7, -5.9>
Third Group (Most force):
<9.3, 8.3>, <-3.4, -8.2>, <-7, -.1>
!!!!I calculated these vectors under the assumption that the end of the rubber band attached to the center paper clip was to be treated as the origin. From the instructions that is what I think you are asking us to do. However this is not exactly correct I believe due to the fact that my paper clips did not all meet in the center of the paper clip. So instead of my origin being a single point it is the area of a paper clip hope this does not throw off my data!!!!!!!!
It would, mostly by distorting the length of the paper clip.
You should have marked both ends of each rubber band. Actually it appears you did, so I think you did account for everything.
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The vectors you reported above are displacement vectors. It should be clear that the magnitude of each is the length of the corresponding rubber band. If you divide both components of a displacement vector by its length, you will get a unit vector which represents the direction of the displacement vector. In the first three lines below, give the x and y components of the three resulting unit vectors. Starting in the fourth line give a brief description of how you obtained your results and what they mean.
First Group (barley any force):
<4.3, 4.6>/6.5192 = approx. <.6596, .7056>
<-2.9, -4.9>/5.6939 = approx. <-.5093, -.8606>
<-5, -3.3>/5.9908 = approx. <-.8346, -.5508>
Second Group (medium force):
<6, 6.6>/8.9196= approx. <.6727, .7399>
<-5.3, -4.2>/ 6.6068= approx. <-.8022, -.6357>
<-3.7, -5.9>/6.9642= approx. <-.5313,- .8472>
Third Group (Most force):
<9.3, 8.3>/12.6905= approx. <.7328, .654>
<-3.4, -8.2>/9.1548= approx. <-.3714, -.8957>
<-7, -.1>/7.0007= approx. <-.9999, -.0143>
!!!Divided displacement vector by magnitude of vector, which we solved for earlier!!!!!
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If you now multiply the unit vector for each rubber band by the force it exerts, you get the force vector for the force exerted on the 'middle' paperclip by that rubber band. Do this and report your force vectors below, one to each line, reporting the forces of the three rubber bands in the same order as before. Starting in the fourth line give a brief description of how you obtained your results and what they mean.
For Second Group (medium force):
1st rubber band,
1.2002(force needed calculated earlier)* <.6727, .7399>(unit vector just calculated)
= <.8074, .888>
2nd rubber band,
.4565(force needed calculated earlier)* <-.8022, -.6357> (unit vector just calculated)
= <-.3662, -.2902>
3rd rubber band,
.4867(force needed calculated earlier)* <-.5313,- .8472> (unit vector just calculated)
= <-.2586, -.4123>
For Third Group (Most force):
1st rubber band,
3.0857(force needed calculated earlier)* <.7328, .654> (unit vector just calculated)
= <2.2612, 2.018>
2nd rubber band,
1.7305(force needed calculated earlier)* <-.3714, -.8957> (unit vector just calculated)
= <-.6427, -1.55>
3rd rubber band,
.503(force needed calculated earlier)* <-.9999, -.0143> (unit vector just calculated)
= <-.5029, -.0072>
!!!!Resulting vectors are force vectors, which I calculated by mult. force (which we found by mult. .5 for every cm. over least amount of tension) by unit vectors which we just solved for!!!!!!!!!
????We were not to calculate for 1st group correct???
The 1st group was your basis for calculating forces. Presumably all forces were close to 0 for that group.
?????Are these force vectors supposed to be the absolute value or does the neg sign show which way force is exerted???????
The ratio of the components tells you the direction of a vector. The components are the main thing.
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Sketch your force vectors, using an appropriate scale. You should select a scale such that your sketch takes up at least half of a sheet of paper.
* Pick one of the three vectors and sketch the dotted line which extends this vector (this will be the dotted line you will need in order to sketch the projections on this line of the remaining vectors).
* Sketch the projections of the other two vectors onto your selected vector.
* Measure the lengths of your projections.
Give the lengths of your projections in the first line, separated by a comma. In the second line give the length of the vector you picked.
To find projection:
A*B = <.8074, .888> * <2.2612, 2.018> = 1.8257 + 1.792 = 3.6177
B*B = <2.2612, 2.018> * <2.2612, 2.018> = 5.113 + 4.0723 = 9.1853
projB A = <3.6177/9.1853>*B = <3.6177/9.1853>* <2.2612, 2.018> = <.8906, .7948>
To find projection:
A*B = <-.3662, -.2902> * <-.6427, -1.55> = .2354 + .4498 = .6852
B*B = <-.6427, -1.55>* <-.6427, -1.55> = .4131 + 2.4025= 2.8156
projB A = <.6852/2.8156>*B = <.6852/2.8156>* <-.6427, -1.55> = <-.1564, -.3772>
To find projection:
A*B = <-.2586, -.4123>* <-.5029, -.0072> = .13 + .003 = .133
B*B = <-.5029, -.0072> * <-.5029, -.0072> = .2529 + .0001 = .253
projB A = (.133/.253)*B = (.133/.253)* <-.5029, -.0072> = (-.2644, -.0038)
?????I am not sure I am doing this correctly; I only have the 2 separate vectors. I projected 2nd group onto 3rd group?????????
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In the first line below give the forces corresponding to your projections, based on the scale of your sketch. In the second line give the force of the vector you picked previously. In the third line explain how you got the forces from the measured lengths:
?????I don’t believe I am doing this correctly, On my graph my vectors extend into the 1st and 3rd quadrants. I know I should know this but I have never worked with vectors before. We are just now starting to cover them in my linear algebra class. Everything I have done has been through me researching this subject on-line and in my different text books. Please let me know how I have set this up incorrectly and I will make the necessary corrections and get it back to you a.s.a.p. I apologize for not being better prepared!???????
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University Physics Students only: Find the magnitude of the projection of each of the force vectors on each of the others, based on the coordinates of the force vectors. (You will use the dot product, as explained in Chapter 1 of your text. If you're in multivariable calculus you'll be familiar with this. If you're not in that course, you might have some questions about this, but take my word for it, it's easy with just a little practice. So if you don't understand it from the text, ask. The magnitude of the projection of vector A on vector B is just the || A || cos(theta), where theta is the angle between the two vectors. Using the dot product it's easy to see that this is just the magnitude of A dot B / || B || ). Using the order in which you have been reporting the three rubber bands, give the projection of the first on the second, the first on the third, the second on the first, the second on the third, the third on the first and the third on the second. In the fourth line give the details of one of your calculations, selecting one that represents the general process:
?????Any advice on this problem as well would be greatly appreciated. I know I can do this work I am just having trouble seeing what it is exactly I am doing??????????
you're doing fine. You'll understand fully when we do this in class.
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&#Your work looks good. See my notes. Let me know if you have any questions. &#
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