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Math Help - Express the height of the crate over the ground in terms of distance to the person

  1. #1
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    Express the height of the crate over the ground in terms of distance to the person

    A 20m ropes passes over a pulley 4m above the ground and a crate on the ground is attached at one end. The other end of the rope is held at a level of 1m above the ground and is drawn away from the pulley. Express the height of the crate over the ground in terms of the distance the person is from directly below the crate. Sketch the graph of distance and height.

    I am having a really tough time with this. I have assumed that I need to use the tangent of the angle created to solve this but I am really not finding any solutions that work.
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  2. #2
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    Re: Express the height of the crate over the ground in terms of distance to the perso

    What "angle" are you talking about? There is nothing said about the person pulling on the rope standing a fixed distance from the pully so unless there is a picture or something else you haven't told us, it seems to me we have to assume the person pulling on the rope is directly below the pulley.The rope has length 20 m. That is equal to the distance from the pulley to the crate (4 meters minus the height of the crate) plus the length of the rope from the pulley to the person pulling on the rope, 4- 1= 3 meters.
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    Re: Express the height of the crate over the ground in terms of distance to the perso

    It states that the rope is drawn away from the pulley, I assume in the x direction.
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    Re: Express the height of the crate over the ground in terms of distance to the perso

    Express the height of the crate over the ground in terms of distance to the person-pulley.png

    From the picture:

    x + h = 4
    x + z = 20
    Therefore, z + 4-h = 20

    z^2 = 3^2 + d^2
    z = \sqrt{3^2 + d^2}

    Substituting in z you get: \sqrt{3^2 + d^2} -h = 16

     h = \sqrt{3^2 + d^2} - 16
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