#$&*
course Phy 241
September 7 around 2:45pm
The first two problems deal with projections of one vector onto the line of another vector. The third deals with trapezoids on the v vs. t graph. The fourth deals with the rubber band lab exercise. If you don't understand the projections and/or the trapezoids, you can start with #4 and whatever you do understand about the first three, then wait for class notes before completing those questions.
1. On a coordinate plane sketch the points (5, 9) and (3, 12). You don't need to take a lot of time to meticulously mark off the scale or measure the points with a ruler, just make a reasonable estimate. You should be able to locate the origin and the two points in just a minute or so. Sketch a vector from the origin to the first point and call this vector B. Sketch another vector from the origin to the second point and call it A.
Sketch the projection of the A vector on the B vector, as we did in class. In case you need more detailed instructions:
* Sketch a dotted projection line from the tip of the A vector to the B vector. The projection line must make an angle of 90 degrees with the B vector.
* The 'projected point' is the point where the projection line meets the B vector.
* The projection of A on B is the vector from the origin to 'projected point' .
Estimate the coordinates of the tip of the projection vector and give them in the next line:
Approx. (6,3)
from the slope you can tell that this point isn't on or particularly close to the line of either of the given vectors.
#$&* (the 'next line' of the instruction is the blank line above this line)
What therefore is the length of the projection vector? Give the length and explain how you found it starting in the next line:
Length of the projection vector: sq. rt. of 45
I found it by taking the magnitude of projection point (6,3)=
Sq. rt of (6^2+3^2)=sq rt of 45
#$&* (the 'next line' of the instruction is the blank line above this line; usually when you see #$&* you should be entering answers in the lines above the prompt)
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:
Approx. (7,4)
Length of the projection vector: sq. rt of 65
I found it by taking the magnitude of projection point (7,4)=
Sq. rt of (7^2+4^2)=sq rt of 65
#$&* (the 'next line' of the instruction is the blank line above this line; usually when you see #$&* you should be entering answers in the lines above the prompt)
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?
The point (3,9) represents that at 3 seconds, the velocity will be at 9 cm/s.
#$&* (enter your answers in the line above this prompt)
On this graph of velocity vs. clock time the points (3, 9) and (6, 13) represent the velocities of a moving object at the corresponding clock times. What does the 'rise' between these points represent? What does the 'run' represent? What does the slope of the straight line between (3, 9) and (6, 13) therefore represent?
The “rise” between the points represents the change in velocity (going from 9cm/s to 13cm/s=4cm/s). The “run” represents the change in clock time in seconds (3 seconds to 6 seconds=3 seconds). The slope of the straight line between the two points is (4cm/s / 3 seconds= 4/3 cm) the distance between the two points or the length. It also tells us how quickly the velocity is changing and the sign of the change, whether it’s increasing or decreasing.
#$&* (enter your answers in the line above this prompt)
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? Tf-t0=delta t, represents the final incline minus the initial incline, which will give you the width of the rectangle.
* What is the 'height' of this rectangle and what does it represent? Average of initial velocity and the final velocity ((vf+v0)/2), which will give you the height or the altitude of the rectangle.
* What is the area of this rectangle and what does it therefore represent? Average altitude times the width “(vf+v0 / 2) * delta t” This will show how far and therefore will give you “delta s” which is the change in position. Therefore, represents the product of the average velocity and the change in clock time during the interval.
#$&* (enter your answers in the line above this prompt)
4. On a graph of velocity vs. clock time the points (t_0, v_0) and (t_f, v_f) represent the velocity and position of an object at the beginning and the end of some interval. The vertical lines from these points to the horizontal axis form what we will call the 'graph altitudes' of a trapezoid. The line segment from the first point to the second forms another side of the trapezoid, and the segment from t_0 to t_f on the horizontal axis forms the fourth side, which we call the 'graph width/ of the trapezoid. (in correct geometrical language this In terms of the symbols t_0, v_0, t_f and v_f, what are the following:
The average velocity v_Ave.
(vf+v0)/2
#$&*
The change in velocity `dv.
Vf-v0=’dv
#$&*
The time interval `dt.
Tf-f0=’dt
#$&*
The displacement `ds.
((vf+v0)/2)*(tf-t0)
The area beneath the slope.
#$&*
The average rate of change of velocity with respect to clock time.
(vf-v0)/(tf-t0)
#$&*
The average acceleration.
(vf-v0)/’dt
(Slope of the trapezoid) represents avg. acceleration on corresponding interval.
#$&*
5. For your data in the rubber band experiment, give the endpoints of each rubber band in the first trial, where the rubber bands were just barely beginning to exert a tension force. Report one rubber band in each of the next three lines, giving in each line the x and y coordinates of the first point, followed by the x and y coordinates of the second point, in that order. Separate each number from the next by a comma. You will have four numbers in each of three lines, separated by commas. In the fourth line give the units of your measurements. Start in the next line:
The ENDPOINTS of the r.b in the 1st trial (barely stretched out): (1.5, 13.5), (7, 8), (12, 18)
(6.5, 13.5, 1.5, 13.5)
(7, 13.5, 7, 8)
(9, 15, 12, 18)
Used graph paper “units”; close to cm.
#$&*
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:
1st rubber band: 5 units
2nd rubber band: 5.5 units
3rd rubber band: sq. rt of 18 units
I found the MAGNITUDE of each of the two points (sq. rt. of (x2-x1)^2+(y2-y1)^2). This gave me the length from the start of rubber band to the end. I also used the Pythagorean Theorem to find the length and it gave me the same answer.
#$&*
Now report the coordinates of the points you observed in the second trial, using the same format as before. Report the three different rubber bands in the same order as before:
The ENDPOINTS of the r.b in the 2nd trial (stretched out): (0.5, 13) (5.5, 6.5) (13.5, 19)
(7, 13, 0.5, 13)
(7, 12.5, 5.5, 6.5)
(9, 14, 13.5, 19)
Measured in graph paper “units”
#$&*
Report the lengths of the three rubber bands, using the same order as before:
1st rubber band: 6.5 units
2nd rubber band: (1.5 units^2+6 units^2=c^2)….sq. rt. of 38.25 units approx. 6.18 units
3rd rubber band: (4.5 units^2+5 units^2=c^2)….sq. rt. of 45.25 units approx. 6.73 units
I could also have found the MAGNITUDE of the two points in each, which would give me the length. However, the first rubber band had the same “y” coordinate so I just measured the units between the two points for the length. The second and third rubber band, I used the Pythagorean theorem, making a triangle, finding the units of the two sides, and then finding the hypotenuse, which would be the length between the two points.
#$&*
Assuming that each rubber band exerts a force of .5 N for every centimeter of length in excess of its 'barely-exerting-a-force' length, what are the three forces? Report the three forces, separated by commas, in the first line below. In the second line below, explain how you determined your forces.
(0.75N*cm, 0.3425N*cm, 1.25N*cm)
I found the forces by taking the difference in the 2nd trial lengths by the 1st trial lengths to get the excess:
6.5-5=1.5cm EXCESS
Sq. rt of 38.25-5.5=approx. 0.685cm EXCESS
Sq. rt of 45.25-sq. rt of 18=approx. 2.5cm EXCESS
-------------
Then multiplied the EXCESS by the force of 0.5N:
1.5cm*0.5N=0.75N*cm
0.685cm*0.5N=0.3425N*cm
2.5cm*0.5N=1.25N*cm
#$&*
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:
<6.5cm, 0cm> or 6.5i
<1.5cm, 6cm> or 1.5i+6j
<-4.5cm, -5cm> or -4.5i-5j
Using the “stretched out” rubber band data, I got these displacement vectors by using the point from where the rubber band was attached to the paper clip in the middle (x1, y1) and the point where it was stretched out on the paper (x2, y2) and did to get the vectors.
#$&*
The vectors you reported above are displacement vectors. It should be clear that the magnitude of each is the length of the corresponding rubber band. If you divide both components of a displacement vector by its length, you will get a unit vector which represents the direction of the displacement vector. In the first three lines below, give the x and y components of the three resulting unit vectors. Starting in the fourth line give a brief description of how you obtained your results and what they mean.
<1, 0>
<0.25, 1>
<-0.67, -0.74>
I got these results by taking the displacement vectors in the answer above and dividing those by the length of the “stretched out rubber bands.” These unit vectors represent the DIRECTION OF the displacement vectors above. A unit vector is a dimensionless vector with a magnitude of 1.
<6.5, 0>/6.5
<1.5, 6>/sq. rt of 38.25
<-4.5, -5>/sq. rt of 45.25
#$&*
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.
<0.75, 0>
<0.085, 0.3425>
<-0.8375, -0.925>
I obtained these results by taking the unit vectors in the answer above, and multiplied them by the forces I got in 3 questions before this (0.75, 0.3425, 1.25) in order to get these force vectors.
<1, 0>*0.75N*cm
<0.25, 1>*0.3425N*cm
<-0.67, -0.74>*1.25N*cm
???Was I supposed to multiply the unit vectors by the forces of the “stretched out rubber band” (taking 0.5N*the lengths of 2nd trial r.b) instead of what I got in the earlier question? Because I just used the forces I got in the earlier question to get these force vectors. ?????
Looks like you did the right thing. You would want to actually do the multiplications to get the components of the force vectors.
#$&*
Sketch your force vectors, using an appropriate scale. You should select a scale such that your sketch takes up at least half of a sheet of paper.
* Pick one of the three vectors and sketch the dotted line which extends this vector (this will be the dotted line you will need in order to sketch the projections on this line of the remaining vectors).
* Sketch the projections of the other two vectors onto your selected vector.
* Measure the lengths of your projections.
Give the lengths of your projections in the first line, separated by a comma. In the second line give the length of the vector you picked.
Lengths of projections: 0.085, 0.8375
Length of vector: 0.75
I found the lengths by finding the magnitude of the projection points.
(0.085, 0)=sq rt. of (0.085^2)=0.085
(0.8375, 0)=sq rt. of (0.8375^2)=0.8375
#$&*
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 understand this question. I sketched out my graph and projections of the force vectors and used <0.75, 0> as my selected vector. My graph had one vector that was in the 3rd quadrant and I didn’t know how to project that onto the x-axis (0.75, 0) point, so that it would make 90 degrees. So, I extended that vector into the 1st quadrant, with the same points, just different signs and then drew my projection down onto the x-axis. Is this right? So, is this question asking for the “projection points?” How do I go about solving this question?
We can take a look at this in class. If necessary, remind me, but you won't be the only one with questions at this point.
#$&*
University Physics Students only: Find the magnitude of the projection of each of the force vectors on each of the others, based on the coordinates of the force vectors. (You will use the dot product, as explained in Chapter 1 of your text. If you're in multivariable calculus you'll be familiar with this. If you're not in that course, you might have some questions about this, but take my word for it, it's easy with just a little practice. So if you don't understand it from the text, ask. The magnitude of the projection of vector A on vector B is just the || A || cos(theta), where theta is the angle between the two vectors. Using the dot product it's easy to see that this is just the magnitude of A dot B / || B || ). Using the order in which you have been reporting the three rubber bands, give the projection of the first on the second, the first on the third, the second on the first, the second on the third, the third on the first and the third on the second. In the fourth line give the details of one of your calculations, selecting one that represents the general process:
DOT PRODUCTS divided by MAGNITUDES:
1st onto 2nd: 0.06375/0.353=0.18……1st onto 3rd: -0.628/1.25=-0.5024
2nd onto 1st: 0.06375/0.75=0.085……2nd onto 3rd: -0.388/1.25=-0.3104
3rd onto 1st: -0.628/0.75=-0.837……3rd onto 2nd: -0.388/0.353=-1.1
MAGNITUDE OF FORCE VECTORS:
<0.75, 0>=0.75
<0.085, 0.3425>=0.353
<-0.8375, -0.925>=1.25
If I understand this question correctly, I found the dot products of all 3 force vectors’ coordinates, then took that and divided it by the magnitude of the one you’re projecting it onto. right
(A dot B)/ ||B||
#$&*
"
Very good work. See my notes.
#$&*