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Force Calculation to balance arm

  1. Dec 5, 2011 #1
    Hello,

    I am after a little help on an applicaition, I need to calculate the force require to hold a pivoted arm in a position using a pneumatic ram. Basically the arms C of G is 740mm up the arm, and the arm weights 100kg. Roughly 1000N working at an offset of 147mm. The arm is 16 degrees from vertical, with the ram at 62 degrees to the arm, so it is pulling to hold the arm in position.

    I attach a diagram to give all the information

    I had a friend take a look and he calculated 676N, I am not sure if this is correct. As I could not follow is method or workings. I attach them as well.

    Thanks

    Mr Howell
     

    Attached Files:

  2. jcsd
  3. Dec 5, 2011 #2

    gneill

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    Staff: Mentor

    The arm is rotating around the pivot at its lower end, right? The pivot at the end of ram's shaft is there just to allow the ram / arm angle to change freely.

    Moments should be taken about the pivot at the base of the arm. I see a larger ram force being required than what you (and your friend) calculated.
     
  4. Dec 6, 2011 #3
    yes the arm is free to rotate around the pivot at the base of it. The ram will slightly change the angle of the arm when it moves in and out.

    My calculation were, to first calculate the torque at the pivot:

    100kg x 9.81 x 0.147m = 203.07 Nm

    so 203.07 / .06m = 3384.5N to support at the pivot point of 215mm up the arm. If the ram was at 90degrees to the arm.

    So need to calculate the angle from 90 degrees to the arm to the real position of the ram.

    90 - 62 = 28 degrees.

    3384.5 / (cos 28) = 3833.18N

    But I am not happy that my method is correct as it is vastly different end figure then my friends.
     
  5. Dec 6, 2011 #4

    gneill

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    Staff: Mentor

    Calculating torques requires that you use the components of the forces that are perpendicular to the moment arm. So, for example, the torque about the pivot point due to gravity would use the component [itex] mg \; sin(16°)[/itex].
     
  6. Dec 6, 2011 #5
    Ok I have had another look at the problem and tried is a different way, please could some let me know if this is correct:

    981N*0.74m*Sin16 = ? * 0.215m*Sin62

    (981N*0.74m*Sin16) / (0.215m*Sin62)

    = 1054N
     
  7. Dec 6, 2011 #6

    gneill

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    Staff: Mentor

    Now THAT looks good!
     
  8. Dec 6, 2011 #7
    thank you
     
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