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Tension/Compression in Joints

  1. Jun 5, 2012 #1
    Statics - Forces in a Truss

    1. The problem statement, all variables and given/known data
    http://i.imgur.com/NbZlc.png


    2. Relevant equations
    Equations of equilibrium (sum of the forces in all directions = 0, sum of the moments about an point = 0)



    3. The attempt at a solution
    My attempt: http://i.imgur.com/qlLCY.jpg
    Hopefully you can read it. What I did is that I did the FBD of the entire system neglecting the internal forces, and I found the reaction forces. Then I did a FBD for joint A alone and solved for the two forces. Me and my friends are getting different answers so I just want to make sure. What I'm unsure about is my signs and the whole tension vs. compression thing. I don't understand it at all. Let me know where I went wrong (if I went wrong). Thanks!

    EDIT: I'm not really sure why this was moved to the engineering section, this problem is from my Statics course which is an introductory physics course (right?).
     
    Last edited: Jun 5, 2012
  2. jcsd
  3. Jun 5, 2012 #2

    SammyS

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    Re: Statics - Forces in a Truss

    Your solution looks fine to me. I didn't check the numbers, but they appear to be in the right ballpark.

    From my experience, a Statics course is generally Engineering, even if it's taught by the physics department. (No, I wasn't the one who moved it.)

    attachment.php?attachmentid=48077&stc=1&d=1338955228.jpg
     

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  4. Jun 5, 2012 #3
    Thank you :)
    A lot of people in my class seem to be getting 23.7N for the axial force in AE, and they found that the joint is under tension. I don't see anything wrong with my solution so I'm keeping it.

    I'm confused about one thing: how exactly do you know if a member is under compression or tension? For example, for my free-body diagram at joint A, both axial forces point towards the joint, so I call that compression. Is that how it works? Also, when you get a negative answer for a force or reaction force, do you carry that negative sign in the rest of your equations? I know that you have to change the direction on the FBD after.
     
  5. Jun 6, 2012 #4
    In calculations compression units are negative and tension are positive

    if you have 2 forces going away from the beam you would draw arrows inside the beams going in [ -> <- ] (opposite to reaction) - that beam would be in tension

    this is how i was taught to do it, hope it helps.
     
  6. Jun 6, 2012 #5

    SammyS

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    They would get 23.7N by adding 17 + 17cos(60°) . But those forces are in opposite direction, so you're solution is correct.
     
  7. Jun 6, 2012 #6

    PhanthomJay

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    Always, when a force of a member on a joint points toward the joint, it is in compression; if the force in a member on a joint points away from the joint, it is in tension. As Sammy S has noted, your answer is correct. And yes, if you get a negative answer, you assumed the wrong direction, and must reverse the direction on your FBD.
     
  8. Jun 6, 2012 #7
    Thanks so much guys, I appreciate it =)
     
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