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  1. #1
    Join Date
    Aug 2005
    Posts
    80

    Equilibrium calculation

    Hi,

    I'm doing a calculation for the normal force needed to keep a device clamped to a table with only frictional forces.

    And i'm getting high numbers at the end of my calculations I think i might have a problem in my equilibrium calculations and would appreciate it if someone here could take the time to check them for me

    oh, I know that the last picture is drawn wrong, but i only need the force not the direction for subsequent calculations.
    Attached Thumbnails Attached Thumbnails Calc1.jpg   Calc2.jpg   Calc3.jpg   Calc4.jpg  


  2. #2
    Join Date
    Apr 2007
    Posts
    777
    You have labeled the reaction force at the pivot as zero but the correct equation is Fr+Fx_2=0 => Fr=-Fx_2. If there were no reaction force, the part in the drawing would be translating to the left. . .

    Looking back to the first drawing however, I think that what you need to do is resolve the force into normal and tangential directions around the center of the hole and then use the tangential force to compute the torque around the hole. This is dependent on the distance between the hole and the edge. Once you have the torque around the hole, you know that the distance between the hole center and the clamp times the force is equal to the previously computed torque.

    Since this is a statics analysis, you want to sum the forces and torques around the pivot point rather than the center of mass to make the problem easier!

    Regards,

    Cameron

  3. #3
    Join Date
    May 2007
    Posts
    8
    The big red flag is in your third sheet where you calculate Fx_2 as non-zero, and then as zero.

  4. #4
    Join Date
    Apr 2007
    Posts
    1955
    Hi, my suggestion would be to first start by putting everything in a single common unit system, such as SI. For instance, the Newton is 1 Kg x meter / sec2.

    http://en.wikipedia.org/wiki/Newton_(unit)

    It has been a long time since I did mechanics / statics, so don't take this for being correct, but I suspect that the 14mm offset is just extra information to confuse you. In other words, for a static case, it does not matter and can be ignore. Someone please double check my logic on this.

    500 N is a considerable force. In the real world, this is very close to the force from a college age woman's weight hanging onto that rope.

  5. #5
    Join Date
    Aug 2005
    Posts
    80
    Thanks for all the help, i'll get back to this problem when i have time over.

    Now that i know the problem of the problem it might go easier the next time around

    Thx again

  6. #6
    Join Date
    Mar 2004
    Posts
    1661
    Quote Originally Posted by harryn View Post
    ...

    500 N is a considerable force. In the real world, this is very close to the force from a college age woman's weight hanging onto that rope.

    And to me it all looks like an exercise made in college (or, "tekniskt gymnasium" it might be), and the OP is trying to get the answer from us...


    Am I right, am I right?

  7. #7
    Join Date
    Aug 2005
    Posts
    80
    Quote Originally Posted by svenakela View Post
    And to me it all looks like an exercise made in college (or, "tekniskt gymnasium" it might be), and the OP is trying to get the answer from us...


    Am I right, am I right?
    And to me it looks like the mod is calling me a cheater and is trolling a thread on the forum that he is moderating.

    And no, it's for a fixture at work. The 500N force comes from using a high safety factor.
    I eventually used pitbull clamps instead though.

  8. #8
    Join Date
    Mar 2004
    Posts
    1661
    Quote Originally Posted by MIKEL12 View Post
    And to me it looks like the mod is calling me a cheater and is trolling a thread on the forum that he is moderating.

    And no, it's for a fixture at work. The 500N force comes from using a high safety factor.
    I eventually used pitbull clamps instead though.

    Well, good then.

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