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Mass hanging by two ropes, find rope tensions.

  1. Sep 22, 2010 #1
    1. The problem statement, all variables and given/known data
    There is a steel beam of 1400kg hanging from the ceiling by two ropes. The first rope (R1) is 20 degrees to the left of the steel attach point, the second rope is 30 degrees to the right of the attach point.

    Here is the diagram: http://img202.imageshack.us/img202/7654/06p29.jpg [Broken]


    2. Relevant equations

    The relevant equations would be newton's second law, F = ma

    Fx = F cos(theta)
    Fy = F sin(theta)

    This are all the relevant equations as far as I know.

    3. The attempt at a solution

    First, I found W (weight, the force of gravity on the object) by simply using newton's second law: F = ma where m = 1400kg and a = -9.8; with that I get W = -13720N

    As the steel beam is not moving side to side, R1x = R2x Fnetx = 0

    Because the tension is holding the steel beam in the air, the x-components of R1 and R2 must be equal to W. So: W = R1 cos 20 + R2 cos 30

    Is this correct so far?

    This is the part I get stuck. I know neither R1 or R2 and am not sure where to go.


    Any help would be appreciated so much.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 22, 2010 #2

    PhanthomJay

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    yes, but you seem to be considering the vertical axis as the x axis, which is OK, but uncoventional
    Put R1x and R2x into their vector component form , just like you did in the vertical direction, and solve the 2 equations with 2 unknowns.
     
    Last edited by a moderator: May 4, 2017
  4. Sep 22, 2010 #3
    Oh, I meant to be doing y.. Our teacher (she's confusing) said that because the angle is not along the x axis and is on the other side of the vector that the y component would be cos instead of sin..

    Sorry, I'm not trying to do it in an unconventional way, I want to do it the simplest and most straightforward way possible.
     
  5. Sep 22, 2010 #4
    What I mean is, sin(30) == opposite/hypotenuse; so in the diagram (using a triangle) this would be Fy = F sin 30 == (F hyp)*(parallel to the x-axis opp) /hyp

    Does this make sense? Like am I understanding why the component equation is what it is?

    We use sin because is opp/hyp and when we multiply that by the vector magnitude(which is the hypotenuse) it results in opp; which is the component we were looking for in the first place?
     
  6. Sep 22, 2010 #5

    PhanthomJay

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    Your teacher is correct..the y (vertical) component of Rope 2 is R2(cos 30) or R2(sin 60)...same result. Proceed as you were doing, just keep your components straight....and axes....
     
  7. Sep 22, 2010 #6
    Thanks, I got it figured out now. I just had to get each component in the form Fx/y = F cos/sin(theta) and then set each equaling both W and zero and plug one into the other.

    I'm not sure why I was so confused in the first place, hindsight--blah.


    Thanks again.
     
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