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A weight on a beam held up by two columns - force problem

  1. Jul 25, 2011 #1
    1. The problem statement, all variables and given/known data
    A weight (mass given) is sitting on a beam that is supported by two columns. Given the location of the weight on the beam, calculate the force on each column.

    2. Relevant equations
    not sure

    3. The attempt at a solution
    not sure
  2. jcsd
  3. Jul 25, 2011 #2


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    You haven't given nearly enough detail !
  4. Jul 26, 2011 #3


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    Start by drawing a diagram and labelling what you know. There is a beam with length L, let's say, and there is a weight W sitting somewhere on the beam. The beam is held up by two supports, one at each end. Assume for the time being that the reaction forces are RL and RR (or R1 and R2). If the weight W is located a distance a from the left support, use the equations of statics to determine what RL and RR are in terms of W, L, and a.

    After all, this is how beam tables are developed.
  5. Jul 26, 2011 #4
    Thank you Steam King. This is the type of information I'm looking for. One hint I was given is to use a coordinate system that makes the problem easier to attack. I'm not sure what to do with that. Also, can you point me in the right direction with the basic statics equations?
  6. Jul 26, 2011 #5
    You could also attack this from a static equilibrium point of view, realizing that both net force and net torque must be zero for the thing to stay still.

    Your first step is to find the center of mass of the weight and horizontal beam as a distance from a column.

    Once you've done that, set the net torque and net force equal to zero and solve.

    Also, choosing the left pillar as the origin would be easiest in my opinion.
  7. Jul 26, 2011 #6
    oh, this is it Firestorm! can you help me out with some equations? I just need to be pointed in the right directions with some equations. any help you can provide would be great.
  8. Jul 26, 2011 #7
    Sure; first, realize that FL + FR = m * g
    and that torque is [itex]\tau[/itex]= r x f

    That should get you going.
  9. Jul 27, 2011 #8
    I tried this problem again, and am still stuck. Please see the attachment for my work thus far.
    I know that I need to be able to pick an origin that will make the problem easier and cancel some stuff out, but I don't know what this is.
    any other help/ideas/hints?
    very much appreciated.

    Attached Files:

  10. Jul 27, 2011 #9


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    I don't think the centre of mass is necessary to calculate.

    For an overlaying coordinate system, have the origin at the top of one of the columns - eg the left one.
    The mas is then at (4,0) and the top of the other column is at (5,0)

    Does the beam have a mass??

  11. Jul 27, 2011 #10
    the beam does not have a mass...
  12. Jul 27, 2011 #11


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    Well that makes calculating the torques more simple.

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