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Question about the effect of acceleration on the force of a mass

  1. Oct 23, 2014 #1
    Ive got a query which im thinking should be simple but I cant for the life of me think how to solve it.

    I have a mass of 500 kg resting on a beam. It is subjected to vibration of 2m/s2 directly down, how would I calculate the extra force subjected on the beam due to this acceleration?

    Is it as simple as working out the force of the mass while stationary then working out the force applied due to acceleration (m*a) and then adding them together?
     
  2. jcsd
  3. Oct 23, 2014 #2

    mfb

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    A constant acceleration is not a vibration. For the momentary force: if all parts of the system are completely rigid and incompressible, then it is as simple as you described. Otherwise it gets more complicated.
     
  4. Oct 24, 2014 #3
    Cheers for the reply. Ive probably been a bit vague in the question so ill give a bit more info.

    Its a mass on a fabricated bracket attached a moving piece of machinery. Im treating it as a simple steel beam to avoid over complicating it. As the machine moves along the ground, imperfections in the surface cause the mass to bounce. 2m/s2 is the peak acceleration so I was just taking this as a worst case scenario. All im wanting to do is determine the maximum force this mass exerts on the beam so I can use it for FEA analysis on the beam and use it to design a new bracket as the current one is failing.
     
  5. Oct 24, 2014 #4

    mfb

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    If you measured the acceleration directly at the object, the corresponding force is just F=m*a.
     
  6. Oct 25, 2014 #5
    Thanks, I guess im just over thinking it :)
     
  7. Oct 25, 2014 #6

    A.T.

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    Is this acceleration relative to free fall or relative to rest?
    If the mass sits on the the beam, you should look for the maximal upwards acceleration.
     
  8. Oct 26, 2014 #7
    The accelerometer i had access to didnt give me both direction, only the maximum vibration along that axis. Im getting access to some better equipment next week which may give me better info. I was treating the peak as the maximum either way and its relative to rest
     
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