Your 'cq_1_16.1' 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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Seed question 16
A rubber band has no tension until it reaches a length of 7.5 cm. Beyond that length its tension increases by .7 Newtons for every additional centimeter of length.
What will be its tension if its endpoints are at the points (5 cm, 9 cm) and (10 cm, 17 cm) as measured on an x-y coordinate system?
answer/question/discussion:
Magnitude of vector A (5cm,9cm)
5^2 + 9^2 = 106
Sqrt 106 = 10.3
Magnitude of vector B (10cm,17cm)
10^2 + 17^2 = 389
Sqrt 389 = 19.7
19.7-10.3 = 9.4cm
9.4cm * .7 = 1.33N
Very good work, except that you didn't find the length of the rubber band correctly. Coincidentally (as pointed out below) the length you found turns out to be about the same as the correct length, but your method will not work in general.
I wrote this before I read the rest of your solution. You did a great job working this out. See my note here in any case, since it addresses why the approach you used on this part doesn't quite work.
The rubber band runs from (5cm, 9cm) to (10 cm, 17 cm). Its length is equal to that of a line segment from the first point to the second.
This line segment is the hypotenuse of a right triangle with a 'run' of 10 cm - 5 cm = 5 cm, and a 'rise' of 17 cm - 9 cm = 8 cm. (The 5 cm leg of this triangle is parallel to the x axis and the 8 cm leg is parallel to the y axis; the right angle is formed at the point (10 cm, 9 cm); if this isn't completely clear you should sketch this triangle on a graph).
The length of the segment, by the Pythagorean Theorem, is therefore sqrt( (5 cm)^2 + (8 cm)^2 ) = sqrt (25 cm^2 + 64 cm^2) = sqrt( 89 cm^2) = 9.4 cm, approx..
It is just coincidence that this turns out to be equal, to the nearest .1, to the difference between the distances of the two points from the origin. The coincidence occurs because the origin and the two points all very nearly lie on a straight line.
Had the points been (15 cm, 9 cm) and (20 cm, 17 cm) the distance between them would still be sqrt( 89 cm^2), about 9.4 cm. However the distance of the first point from the origin is about 17.6 cm, and the distance of the second is about 26 cm; the difference between their distances from the origin is less than 9 cm.
The length of the segment from (5 cm, 9 cm) to (10 cm, 17 cm) can be understood in terms of vectors as the length of the vector from the first point to the second.
A vector has magnitude and direction. The vector from (5 cm, 9 cm) to (10 cm, 17 cm) is <5 cm, 8 cm>, indicating a vector with components 5 cm in the x direction and 8 cm in the y direction.
The magnitude of this vector is sqrt( (5 cm)^2 + (8 cm)^2 ) = sqrt( 89 cm^2) = 9.4 cm, approx..
Its angle is arctan(8 cm / (5 cm) ) = arctan(1.6) = 58 degrees, approx..
What is the vector from the first point to the second?
answer/question/discussion:
I’m not sure how to answer this. The vector is in quadrant 1, it’s origin is point (5cm,9cm) and it’s endpoint is point (10cm, 17cm). It is moving in a NE direction or in an increasing or positive direction. Or is the vector the tension? Which would be 1.33N
New thought…I I treat each point as an endpoint to two different vectors the slopes are different: 9/5 = 1.8 and 17/10 = 1.7 so, maybe I am supposed to treat the two endpoints as the endpoints of two separate vectors. If I do that then I don’t know what good that is going to do me. I can add them together and get a new vector, but it wont tell me from endpoint A to B, it will tell me from endpoint (0,0) to a new endpoint that we might call C….Never mind
New thought…How about the angle? I draw my points and connect them, that’s the hypotenuse. I draw a vertical line from point (10,17) down to my x axis and I draw a horizontal line from point (5,9) over to my y axis and over to the vertical line. This forms a right triangle with the third point at (10,9). I want to know the angle at point (5,9). Inverse tangent (8/5) = 58 degrees. So the vector is 9.4cm at 58degrees counterclockwise from positive x axis?
Very, very good thinking. Your final thought nails it down perfectly.
What is the magnitude of this vector?
answer/question/discussion:
I can figure it out the way I did above, in order to figure out the tension, or I can subtract the points first and then figure out the hypotenuse.
X = 10cm-5cm = 5cm and Y= 17cm-9cm= 8cm
5cm^2 + 8cm^2 = 89
Hypotenuse = sqrt 89 = 9.4
The magnitude of the vector is 9.4cm
What vector do you get when you divide this vector by its magnitude?
answer/question/discussion:
Once again, I’m not sure. The vector is a magnitude with direction, so the direction divided by the magnitude? The magnitude divided by the magnitude? 9.4cm/9.4cm = 1 Or if the vector is labeled as the tension then 1.33N/9.4cm = 0.14 (Do I have to change the cm to m in order to divide it?) Or do I divide the degrees 58 by the magnitude 9.4? 58/9.4 = 6.17
You're right about the magnitude, which is 1. You also have to specify the direction, or equivalently find its x and y components.
If we divide the vector <5 cm, 8 cm> by its magnitude we get <5 cm, 8 cm> / sqrt(89 cm^2) = < 5 cm / sqrt(89 cm^2), 8 cm / sqrt(89 cm^2) > = <.53, .83>, approximately. That is, we get a vector with x component .53 and y component .83. Note that both these components are unitless, since dividing cm by sqrt(cm^2) yields cm/cm so that the units 'cancel out'.
This vector is 1 unit in length (therefore called a 'unit vector'), and directed at the same angle as the original vector.
What vector do you get when you multiply this new vector by the tension?
answer/question/discussion:
I don’t know which of my above guesses to use. I will guess 6.17 * 1.33 = 8.2
The tension is about 1.35 N. Multiplying the vector <.53, .83> by 1.35 N we obtain the new vector
<.53, .83> * 1.35 N = <.71 N, 1.13 N>, approximately.
This represents a force vector with x and y components .71 N and 1.1 N, respectively. The magnitude of this vector is the original 1.35 N, and it is directed at angle arcTan (1.13 N / (.71 N) ) = 58 degrees, approximately.
What are the x and y coordinates of the new vector?
answer/question/discussion:
X = 8.2 cos (58) = 4.3
Y = 8.2 sin (58) =6.9
(4.3, 6.9)
I’m curious, if I go back to treating the two endpoints as endpoints of two separate vectors and I add them together will I get the same new endpoint?
This is a great question. The answer is that unless everything lies along the same straight line through the origin, you won't. However the closer to this straight line, the more nearly this will be so. The reasons are similar to those outlined in my first note.
X component of new vector is 5 + 10 = 15
Y component of new vector is 9 + 17 = 26
No, not even close.
If I find the angle of this vector it is 60 degrees…
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3 hours
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Well, I don’t know if I came anywhere close on any of this, but I feel like I understand some of it, which is amazing since 5 hours ago ( yes, between the problem set and reteaching myself trig which took 2+ hours and this seed question which took almost 3 hours, I have been working for 5 hours this evening!) I didn’t remember ANY trig.
You didn't get it all, but you got most of it, and I think you'll understand the rest from my notes. You did some outstanding reasoning here.
&#Let me know if you have questions. &#