Phy 121
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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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?
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When I read this question, I personally think of two different ways it could be solved and I’m not sure if either way is correct. The first way that I thought about this problem was that if the rubberband has no tension until it reaches 7.5 cm that for 5cm there would be 0 tension and for 9 cm there would be a tension of 1.05 Newtons. Continuing to solve in this manor, I had a number of 1.75 Newtons for 10 cm and 6.65 Newtons for 17 cm.
If we look at this as coordinates and the cm that we count are the 9 cm and the 17 cm then our respective Newtons would be 1.75 Newtons and 6.65 Newtons.
Not bad thinking, but there is no rubber band with length 5 cm and no rubber band with length 9 cm. There is a single rubber band whose length is the hypotenuse of a triangle with legs 5 cm and 9 cm.
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• What is the vector from the first point to the second?
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Reading this question makes me think that my second assumption of the tension for the coordinates was 1.75 Newtons and 6.65 Newtons. I am going to assume that the vector for this would be the difference between the tension at each y coordinate value. In that case, I would say that the vector would be 4.9 Newtons.
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• What is the magnitude of this vector?
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To find the magnitude of the vector I am going to use the Pythagorean Theorem. So the sq root of 1.75N^2 +6.65N^2 = magnitude of vector.
= sq root 3.0625 N + 44.2225
=sq root 47.285N
= 6.87N
=Approximately 7 N
Very good, but you should have used the Pythagorean Theorem earlier, to find the length of the rubber band, then used this length to find the magnitude of the force.
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• What vector do you get when you divide this vector by its magnitude? (Specify the x and y components of the resulting vector).
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When we divide this vector by its magnitude, we get 4.9 N/7 N. This equals .7 N.
This is equal to my tension force from the rubberband. Is this supposed to be this way???
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• The new vector should have magnitude 1. When you divide a vector by its magnitude the result is a vector with magnitude 1. We call a vector of magnitude 1 a unit vector. What vector do you get when you multiply this new vector (i.e., the unit vector) by the tension?
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• What are the x and y components of the new vector?
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I am not sure if this is correct but I think the x coordinate is 0 and the y coordinate is my .7.
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This new vector is called the tension vector. It is a force vector which represents the tension. A force vector can be specified by its components, or equivalently by its magnitude and direction.
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35 - 40 minutes
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You've got most of the right ideas and procedures. Check the link, and my notes, to see how to put them together.
&#Please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified. &#