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Two ropes hold a 200N weight. Find the tension in each rope.

  1. Jan 31, 2015 #1
    1. The problem statement, all variables and given/known data
    The question is;
    Two ropes hold a 200N weight as shown here; http://i.imgur.com/BgeoIVq.png

    Find the tension in each rope using a vector diagram.
    The forces are in equilibrium in the diagram btw.

    2. Relevant equations

    Well they gave us 200N so no need for F = MA
    Rx = R * cos(theta)
    Ry = R * sin(theta)
    The above are resolution to component equations.

    Other equations the teacher gave us were:
    tan(theta) = Ry/Rx
    R = SQRT[(Rx)^2 + (Ry)^2]
    The above are component to resolution equations.

    3. The attempt at a solution

    So I fished around on this site and there were some solutions to this question, but I never quite got it, partially because (theta) for R2 is 90.

    R1y = R1 * sin(53)
    R2y = R2 * sin(90)
    Downwards force = 200N

    Therefore; because of equilibrium;
    R1sin(53) + R2sin(90) = 200N [First Equation ( Y Axis )]

    ---
    R1x = R1 * cos(53)
    R2x = R2 * cos(90)

    Therefore; because of equilibrium;
    R1cos(53) = R2(cos90) [Second Equation (X Axis)]

    ---
    Okay, the problem here is cos(90) = 0,
    So does that mean that R1 = 0? because 0* R2 = 0.....

    So the answer I got is
    R1 = 0
    R2 = 200

    Im pretty sure that im wrong, but then again i only just started physics in year 11 (australia)
     
  2. jcsd
  3. Jan 31, 2015 #2

    Simon Bridge

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    You have applied the equations first, and then checked using phsics, and observed that these do no match. It follows there is something wrong with the equations or how you have applied them.

    ... so you have the wrong angles.

    You are better just to use the physics first, with your understanding of forces and vectors, to construct the correct equations first go.
    Less memorizing to do too. So throw away the equations you were given:

    So look at the picture - can you see that the horizontal components of ##\vec R_1## is just going to be ##R_2##?
    What is the role of rope 2 anyway? If you cut it, what would happen?

    For R1, draw the triangle that shows the horizontal and vertical components adding up to make the vector ##\vec R_2##.
    Draw in the 53deg angle you are given.
    Now use trigonometry to work out the relationships.

    (It may help you to draw in the x and y axes too.)

    The other way is to draw the vectors for R1+R2 ... you know what they have to add up to, the weight 200N straight up. This gives you a triangle. You know how to deal with triangles.
     
    Last edited: Jan 31, 2015
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