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Weight supported by two supports

  1. Jul 31, 2013 #1
    Hi all,

    I am working on a problem that I can't even seem to figure out where to start. I have a horizontal beam of known weight (1100lbs, or 4893N). It is 1714.5mm long, and is supported by two vertical supports, which are at known location along the beam (Support 1 @ 200mm; support 2 @ 1150mm). I am trying to figure out how much mass (in Newtons) is carried by support 2, and then I want to add additional 5G of force to the mass supported by support 2 straight down in vertical direction.

    I know the starting point would be:
    sum of moments = sum of forces = 0
    [itex]\Sigma[/itex]M = [itex]\Sigma[/itex]F = 0

    I know I do something with the reaction forces at the support, but not sure.

    Any help to get started would be appreciated.

    Thanks!
     
  2. jcsd
  3. Jul 31, 2013 #2

    nvn

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    abe_cooldude: Yes, start by performing a summation of moment about support 1, then solve for the reaction force at support 2.

    Is the additional 5 g acceleration applied only to the mass supported by support 2? Or is an additional 5 g also applied to the mass supported by support 1? If the latter, then just multiply the beam weight by 6, then solve for the reaction force at support 2 again, as you did before.
     
  4. Aug 3, 2013 #3

    E_Q

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    First, draw a diagram! Like you say, for the system to remain in equilibrium the forces (so the system doesn't accelerate) and the moments (so the system doesn't rotate) need to sum to 0.

    Like nvn said I would start by summing moments about support 1 (remembering to consider the reactive force at support 2 as an unknown). If the horizontal beam is uniform it's entire weight will act at the centre of gravity, which will be in the middle. A moment is the magnitude of force * perpendicular distance from line of action of force.

    Obviously for forces Reaction 1 + Reaction 2 = Weight of beam

    Not entirely sure about the second part of your question, but hopefully you can work it out now :) Hope this helps.

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