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Requirements of Equilibrium

  1. Dec 13, 2007 #1
    In the figure below, a rigid beam is attached to two posts that are fastened to the floor. A small but heavy safe is placed at the six positions indicated, in turn. Assume that the mass of the beam is negligible compared to that of the safe.
    (a) Rank the positions according to the force on post A due to the safe, greatest compression first, greatest tension last (use only the symbols > or =, for example, 1>2>3=4>5=6).
    (b) Rank the positions according to the force on post B due to the safe, greatest compression first, greatest tension last (use only the symbols > or =, for example, 1>2>3=4>5=6).


    I was easily able to get the answer to a which is 1>2>3>4>5>6, but using the same method i was not able to come up with a correct answer to b. I know that 3 is the greatest compression and 6 is the greatest tension. I was pretty sure that 2=4 and 1=5 but that does not seem to be the case. can you please help me get the answer. Thank you.
     

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  3. Dec 13, 2007 #2

    Doc Al

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    Staff: Mentor

    What method are you using? Show what you did.
     
  4. Dec 13, 2007 #3
    for part a i understood that the closer the block was to A the greater compression I tried to do that for part b but it did not work. Since 3 was on B i assumed that the was the greatest and 6 is the furthest away so that would be the least compression. 2 and 4 where equal distances away from B and so were 1 and 5. Is there a real method i should be using? If so please help me to figure it out.
     
  5. Dec 13, 2007 #4

    Doc Al

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    Solve it systematically, using the conditions for equilbrium: The net force must be zero & the net torque must be zero.

    Just call the support forces F_A & F_B. Call the weight of the safe W. Call the distance from A to B L. Now set up torque equations (using A & B as the pivot points) and figure out what F_A must be for each safe position. Give it a try.
     
  6. Dec 13, 2007 #5
    I dont understand becuase wouldn't the mass and w alawys be the same and how would i do the cross product for torque. When i do the cross product i get for 1-3 Lmg is that correct.
     
  7. Dec 13, 2007 #6

    Doc Al

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    I'll analyze position #1:

    Forces: F_A + F_B = W

    Torques about B: LW = (F_A)L

    That second equation tells us that F_A = W, so F_B = 0.

    We could also have used torques about A: (F_B)L = 0, so F_B = 0.
    (cause there's no torque from W about point A)

    OK, that's position #1. Now you do the rest in the same manner.
     
  8. Dec 14, 2007 #7
    thank you Doc Al for your help i was able to get the answer.
     
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