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Equilibrium 3

  1. Mar 16, 2009 #1
    1. The problem statement, all variables and given/known data

    4990612.gif
    A 4.2 m long uniform post is suspended by a cable having a tension of 1 700 N.

    2. Relevant equations
    What is the mass of this post?


    3. The attempt at a solution

    I figure there are 2 forces acting in a clockwise direction (the mass of the log on each side of the rope. And that each of those masses should be multiplied by sin60 not sure about the tension though. I figure that the tension of the rope is acting in a counter clockwise direction, and that all torques are = 0.)

    the answer is 530kg.
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Mar 16, 2009 #2

    LowlyPion

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    Is that your answer? Or the book's?

    For instance I appreciate the hints, but I think I know how to work it.

    What are you stuck on is what I'm asking?
     
  4. Mar 16, 2009 #3
    No I dont know how to solve it. The booklet I have has the answer of 530 kg but I dont understand how to get to that solution.

    I gave the info on how I think I should do it but im stuck/confused.
     
  5. Mar 16, 2009 #4

    LowlyPion

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    OK. If the sum of the torques is 0, what are the torques about the pivot?
     
  6. Mar 16, 2009 #5
    thats whats confusing me, What about the tension of the rope? Is does that have torque?

    I think its like
    [tex]\tau[/tex]1+[tex]\tau[/tex]2-Tr=0
    m(9.8)sin60(2.1)+m(9.8)sin60(.5)-(1700)(4.2)
    17.8m+4.24m=7140
    22.04m=7140

    22.04m=7140
    [tex]\overline{22.04=22.04}[/tex]

    m=323.96kg...thats a stab at it.
     
  7. Mar 16, 2009 #6

    LowlyPion

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    What about the tension in the rope? Isn't that a force? Isn't it acting over a moment arm about the pivot? And how far away from the pivot is it acting ? So what is the torque from the tension in the rope?
     
  8. Mar 16, 2009 #7
    thats why I did the force of tension 1700xdistance from pivot (4.2)

    But I assume that you need to do cosine60 or something?
     
  9. Mar 16, 2009 #8

    LowlyPion

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    The log is 4.2m. How far from the pivot is the Tension acting?
     
  10. Mar 16, 2009 #9
    darn I read it wrong so 3.2m
     
  11. Mar 16, 2009 #10
    that gives me an even lower number though (246kg)
     
  12. Mar 16, 2009 #11

    LowlyPion

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    Now what other Torque is that acting against?
     
  13. Mar 16, 2009 #12
    wouldnt that make the weight even less?

    I cant think of any others other than maybe the ground or the angle of the log?
     
  14. Mar 16, 2009 #13

    LowlyPion

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    Where is the center of mass acting?
     
  15. Mar 16, 2009 #14
    thats one of the things thats confusing me is it 2.1 metres?
    or something like 2.1(sin60)
     
  16. Mar 16, 2009 #15

    LowlyPion

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    It is at 2.1m along the log. But what distance is the projection of m*g acting from the pivot?

    Draw a line straight down. That distance from the pivot along the ground is the projection for a ⊥ moment the weight is acting through.
     
  17. Mar 16, 2009 #16
    See how much weight would provide torque in this case?
    Its not all of it.
     
  18. Mar 16, 2009 #17
    isnt it just 2 torques the log clockwise and the rope counterclockwise

    mgsinθr=1700r
    m(9.8)sin60(2.1)=1700(3.2)
    17.8m=5440
    m=305kg?
     
  19. Mar 16, 2009 #18
    Very close. Resolve the vectors again and see which component counters torque due to tension in the thread.
     
  20. Mar 16, 2009 #19
    so its cosine. right? thanks!
     
  21. Mar 16, 2009 #20
    If it wasn't the sine function, it had to be the cosine function obviously. But please understand why its the cosine function. And try more problems related to resolution of vectors to get that idea clear.
     
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