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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Assignment 16 Seed Question 1
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: To find the tension, which is .7 Newtons for every additional cm of length, we first find using the Pythagorean theorem, that the vector is equal to the square root of the length ( 5cm ) squared plus the height ( 8 cm ) squared, which is 9.43 cm. We can then use the fact that at 7.5 cm there is no tension, and with each increasing cm there is .7 Newtons of tension. We can extrapolate from 8.5 cm to 9.5 cm there is 1.35 Newtons of tension. (Because 9.43 is .07 cm short of 9.5 cm).
• What is the vector from the first point to the second?
answer/question/discussion: Since a vector is a measure like velocity, which is a rate, we can find the vector by comparing tension to displacement. Thus, we can say 1.35 Newtons / 9.43 cm = .143 N / cm.
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 magnitude of this vector?
answer/question/discussion: By the discussion above, the magnitude should be 9.43 cm using Pythagorean theorem.
• What vector do you get when you divide this vector by its magnitude?
answer/question/discussion: When you divide this vector by its magnitude, you get .143 N/cm / 9.43 cm = .015 N/cm/cm.
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: When you multiply this new vector by tension, you get .015 N/cm/cm * 1.35 N = .021.
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: We find cos (theta) = tan (8 / 5) = 58 degrees. The x and y coordinates of the new vector are vx = v cos (theta) = 9.43 cm * cos 58 degrees= 4.99 cm. The sin (58 degree) = 0.85, so .85 * 9.43 cm = 7.99 cm.
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This took me 30 minutes.
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This problem should have been given after the Introductory Problem Sets problems, which introduce vectors. You did well overall but a couple of the vector calculations weren't correct. See my notes and let me know if you have questions.