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Two Supports and Three Forces

  1. Feb 6, 2007 #1
    I'm most likely complicating this, but since I can find no examples of a related problem, I'm forced to bring it here.

    What I have is a 70 lb beam supported evenly by two supports that are 10 feet apart. Looking at the beam side on, support 1 is on the left, and support 2 is on the right.

    A downward force of 50 lbs is acting 3 feet to the left of support 1 and a downward force of 120 lbs is acting 3 feet to the left of support 2.

    What I need to find is the force exerted by each support on the beam.

    F1---------------------------------F2
    |-----------------------------------|
    v(50 lbs)---------------------------v(120 lbs)
    ============================================= (70 lb beam)
    --------------^---------------------------^
    --------------^-(S1)----------------------^-(S2)

    |-----3ft------|------------10ft------------|-----3ft-----|


    I understand moment and balance, but the two supports are screwing me up.

    I thought it made sense to first divide the 70 pounds of the beam evenly onto the two supports, so each needed to react 35 lbs against the beam, plus whatever forces F1 and F2 contributed to the individual supports. Is this the correct thinking?

    I know that I can't just add up all the downward forces and distribute them evenly to both supports, but I'm totally confused how to account for F1 and F2 on the separate supports. Every problem I've seen like this so far involves only a single support, and there's no explanation how two supports would change the system.

    This is probably a really simple thing, but I'd appreciate any feedback - especially just concerning the whole "two support" system.
     
    Last edited by a moderator: Feb 6, 2007
  2. jcsd
  3. Feb 6, 2007 #2

    radou

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    Why do you think you should divide anything? Simply calculate the resultant force of the weight of the beam. Where does that force act?
     
  4. Feb 6, 2007 #3

    AlephZero

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    If you take moments about the position of S1, the force S1 has zero moment about that point, so you will get an equation for S2.

    As radou said, the weight of the beam acts at its centre of mass. That has nothing to do with where the supports are.
     
  5. Feb 6, 2007 #4

    PhanthomJay

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    before you get too far, your problem statement of the location of the 120 lb. force is not consistent with your diagram. Does it act to the left or right of support 2 ?
     
  6. Feb 6, 2007 #5
    How can weight act as location?

    If I understand you correctly, though, you are just saying that the beam's force of weight acts equidistant from both supports in this case...

    Is that correct?
     
  7. Feb 6, 2007 #6
    It seems to look consistent to me. But for clarification, it is 3 feet to the left of support two. The three feet marked on the far right and bottom of the diagram is just showing that the beam is placed symmetrically on the supports. Sorry if that was confusing.
     
  8. Feb 6, 2007 #7

    PhanthomJay

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    Well that makes a difference, in which case due to symmetry you are correct that the beams weight is equall;y distributed between the 2 supports. But in the more general case, you must apply the beams weight thro the cenetr of mass as noted. Now sum moments etc to get the reaction forces from the other applied forces.
     
  9. Feb 6, 2007 #8
    I think I'm a little confused again. Maybe because my question was misunderstood the first time.

    So, here's just a general question...

    The 50 lb force to the left of S1 - does that affect the reaction force of S2 in any way?

    My estimate is yes, since, given enough force to the left of S1, the beam would become unbalanced.
     
  10. Feb 6, 2007 #9

    PhanthomJay

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    Yes, it does. As noted bove, sum momnts about S1 --- watch plus an minus signs----and solve for S2, noting that the vector sum of the momnts must equal__________?
     
  11. Feb 6, 2007 #10

    PhanthomJay

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    When you summed moments about S2, you forgot to add the moment from the beam weight, You've got to be consistent, since when you summed abou S1, you included the beam weight also. The reaction forces must equal the downward forces of 240N. Alternatively, you may have confused yourself in the beginning by finding the 35 pound reactions from the beam weight only. This is perfectly valid by the superposition principle, but then when you sum moments, you do not consider the beam weight , but then you must add the reactions from each case together to get the same result. It's probably best to do it all at once rather than in several steps.
     
  12. Feb 6, 2007 #11
    Moments about S1:

    (-3ft)(50lbs) + (7ft)(120lbs) + (5ft)(70lbs) + (-10)(S2) = 0

    S2 = 104 lbs

    Moments about S2:

    (-13ft)(50lbs) + (10ft)(S1) + (-5ft)(70lbs) + (-3ft)(120lbs) = 0

    S1 = 136 lbs


    S1 + S2 = 240 lbs = Total downward force on system.

    This seems to work. If I did something out of place, let me know, but thanks for all your help! I'm sorry I made something so trivial such a chore. Anyway, it all makes much more sense now.
     
  13. Feb 6, 2007 #12
    Yeah, whoops. I caught that mistake too, but I guess I didn't delete my post fast enough.

    Thanks for your help again!
     
  14. Feb 6, 2007 #13

    PhanthomJay

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    Yep, nice work.
     
  15. Feb 7, 2007 #14

    AlephZero

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    Weight is a force. The weight of each particle of the beam is a force acting at the position of that particle. When you add up all those small forces, you get a force equal to the total weight, that acts at the centre of mass.

    No, the centre of mass is half way along the length of the beam, for a uniform beam. It is a property of the beam. It doesn't depend where the supports are.
     
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