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Math Help - Tensions & Vectors

  1. #1
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    Tensions & Vectors

    To carry a 100 lb cylindrical weight, two people lift on the ends of short ropes that are tied to an eyelet on the top center of the cylinder. Each rope makes an angle of theta degrees with the vertical

    1. Find the tension in the ropes if theta = 30 degrees
    2. Write the tension T of each rope as a function of theta
    3. Determine the domain of the function
    4. Explain why the tension increases as theta increases

    Thank you thank you thank you!!

    Jennifer
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  2. #2
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    Quote Originally Posted by jmoran View Post
    To carry a 100 lb cylindrical weight, two people lift on the ends of short ropes that are tied to an eyelet on the top center of the cylinder. Each rope makes an angle of theta degrees with the vertical

    1. Find the tension in the ropes if theta = 30 degrees
    2. Write the tension T of each rope as a function of theta
    3. Determine the domain of the function
    4. Explain why the tension increases as theta increases

    Thank you thank you thank you!!

    Jennifer
    1. Draw a sketch.

    2. You are dealing with a right triangle: Each rope has to carry 50 lbs measured vertically. This force is drwn in blue.

    50\  lbs = t \cdot \cos(30^\circ)~\implies~\boxed{t = \dfrac{50\ lbs}{\cos(30^\circ)}} \approx 43.3\ lbs

    ...

    to #4:

    If \theta approaches 90 the \cos(\theta) approaches zero and thus the quotient approaches infinity.
    Attached Thumbnails Attached Thumbnails Tensions & Vectors-zweiseile.png  
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  3. #3
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    So is force the same thing as tension?

    Also when I calculated (via my graphing calculator) the equation I came up with 57.7 lbs not 43.3 lbs, why is that?

    One more, if I have to draw a graph of the function I am trying to figure out how to do that on the graphing calculator and not having any luck, any advice would be appreciated...I am trying to graph the function which I am assuming is y1=50/cos(theta)

    Jennifer
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  4. #4
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    Quote Originally Posted by jmoran View Post
    So is force the same thing as tension?

    Also when I calculated (via my graphing calculator) the equation I came up with 57.7 lbs not 43.3 lbs, why is that?
    That's the correct value. I hit the \times button by accident. Sorry for the confusion.

    One more, if I have to draw a graph of the function I am trying to figure out how to do that on the graphing calculator and not having any luck, any advice would be appreciated...I am trying to graph the function which I am assuming is y1=50/cos(theta)

    Jennifer
    That depends which calculator you use. I assume you have a TI83/84. If so:
    1. Go to MODE and switch to degree (3rd row, 2nd entry)
    2. Go to Y= editor and type: 50/cos(x)
    3. go to WINDOW and use the following settings:
    Xmin = -1
    Xmax = 90
    Xscl = 10
    Ymin = -1
    Ymax = 180
    Yscl = 30
    Xres = 1
    Xmin = -1 and Ymin = -1 are necessary if you want the axes to be drawn.
    4. Go to GRAPH

    I've attached screenshots of the production of the graph.
    Attached Thumbnails Attached Thumbnails Tensions & Vectors-jmoran.png  
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  5. #5
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    Thank you!! I never know what to put the window settings at.

    Jennifer
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