Phy 231
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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: ->->->->->->->->->->->-> :
By finding the difference of the magnitude of vectors (5,9) and (10,17), this gives me the distance between the points. The difference between the magnitudes is 9.43. Subtracting 7.5 cm from 9.43 you get 1.933 cm. Since the tension increases by 0.7 N for every cm, we multiply 0.7 by 1.933 and we get the tension to be 1.35 N.
• What is the vector from the first point to the second?
answer/question/discussion: ->->->->->->->->->->->-> :
The vector from the first point to the second is (10-5, 17-9) which is (5,8).
• What is the magnitude of this vector?
answer/question/discussion: ->->->->->->->->->->->-> :
The magnitude of the vector is sqr rt(5^2 + 8^2) which is 9.43.
• What vector do you get when you divide this vector by its magnitude? (Specify the x and y components of the resulting vector).
answer/question/discussion: ->->->->->->->->->->->-> :
You get (0.53, 0.848). I believe this is called the vector projection.
This is the unit vector. Vector projection involves multiplying the dot product by a unit vector and dividing by the length of one of the vectors. (proj of u on v is u dot v * unit vector in direction of v / magnitude of v, or less intuitively but more compactly u dot v * v / | v |^2, or u dot v / v dot v * v)
• What vector do you get when you multiply this new vector by the tension?
answer/question/discussion: ->->->->->->->->->->->-> :
You would get (.7155, 1.15). I'm not sure what this represents.
• What are the x and y coordinates of the new vector?
answer/question/discussion: ->->->->->->->->->->->-> :
(.72, 1.15)
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20 min
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Please compare your solutions with the expanded discussion at the link
Solution
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