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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?
I am not sure how to find tension when there are two dimensions of endpoints,
answer/question/discussion:
• What is the vector from the first point to the second?
30.0162 ( I added x and y coordinates and took the square root of their sums squared)
, 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..
answer/question/discussion:
• What is the magnitude of this vector?
30.0162
answer/question/discussion:
• What vector do you get when you divide this vector by its magnitude?
I do not know what is the difference between the rubber band’s vector and its magnitude.
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.
answer/question/discussion:
• What vector do you get when you multiply this new vector by the tension?
Again I do not know what is the difference between the rubber band’s vector and its magnitude.
answer/question/discussion:
• What are the x and y coordinates of the new vector?
I cannot solve this because I do not know what is the difference between the rubber band’s vector and its magnitude.
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.
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10 minutes
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I do not understand how to solve tension. It has been with several assignment seed questions and I am having great difficulty. Also this one was confusing to me.
You should be sure you understand this solution, then review the part of the introductory chapter that applies to vectors.
Let me know if you have questions.