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Homework Help: Billboard Problem

  1. Jan 4, 2010 #1
    1. The problem statement, all variables and given/known data

    The supporting structure of a billboard is attached to the ground by a pin at B, and its rear leg rests on the ground at A. Friction may be neglected. Point G is the center of gravity of the billboard and structure, which together weigh 2,800 lb. To prevent tipping over in high winds, a 2,370-lb weight is placed on the structure near A, as shown. (a) Compute the magnitudes of the reactions at A and B if the wind load on the billboard is q = 120 lb/ft. (b) Find the smallest wind load q that would cause the structure to tip over.

    physprob.jpg

    2. Relevant equations

    M = Fd
    I don't know the others..

    3. The attempt at a solution

    interp.jpg

    I totaled Moment of Weight and CG and minus it to Moment of wind to get Rxn on Point A... and Add it to get Rxn on Point B.

    Am I doing it right? I already answered it using this method and got approx. 15 kNm on point A and 41 kNm on point B, and having 260 lb/ft for the billboard to tip over....
     
  2. jcsd
  3. Jan 4, 2010 #2

    PhanthomJay

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    Don't mix up your SI and USA units! The problem is given in units of pounds and feet, so stick with those units.

    I'm not sure why you flattened out the vertical support and billboard and made it horizontal. Keep it as shown when summing moments. And the moment arm of the wind force moment is 13 feet, not 8 feet (arm = distance from cg of wind load to pivot point). Please show your calculations. The problem is in 2 parts: first get reactions (in pounds) at A and B, then, for part b, get min q for tipover, which occurs when A = ???
     
  4. Jan 4, 2010 #3
    sorry but it's 11:30 pm here already.. gotta sleep, have class tomorrow.

    Thanks for replying though, and thanks for correcting that 13 ft thingy.

    Ill try to post my calculations tomorrow.
     
  5. Jan 4, 2010 #4
    Forces in -> direction
    lbs from wind = 120 x 10 @ 13ft (need to take centre of the force which, asuming it is constant over the area will be 5ft from the top.
    Forces in <-
    lbs from sign
    2800lbs @ 1.5ft from pivot
    w @ 7ft from pivot
    so moments ->
    1200 x 13 = 15,600mlbs
    so to stop it tiping
    7w +(1.5x2800)=15600
    7w+4200=15600
    7w=11400
    w=1628.57lbs

    since it is now 5 hours later and too late i though i would at least show an answer. trying to warm my brain up.
     
  6. Jan 4, 2010 #5

    PhanthomJay

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    You should please refrain from giving solutions to problems; but in any case, the problem, in part b, is not asking for what 'w' should be to prevent tipover; it is asking what minimum wind load 'q' will tip the structure over with the given 'w' of 2370 lbs.
     
  7. Jan 5, 2010 #6
    sorry for the late reply, been busy...

    ok here's what I did.

    For Reaction on point A:

    I replaced the 2 forces (W and CG) with a single force positioned at point A using this equation..

    W(7) + CG(1.5) = F(8.5) F = 2,440 something.

    Then I reduce the Moment of that F by the Moment of the wind load (pivot is point B)..

    F(8.5) - WindLoad(13) = Moment at A.

    Rxn on A = Moment at A / 8.5 = 610 lb

    is this correct?
     
  8. Jan 5, 2010 #7

    PhanthomJay

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    Yes, although that's a bit of an unorthodox way of doing it. Why not sum moments of all forces about B directly, to solve for the unknown force reaction at A. Then what's the reaction at B? then move on to part (b).
     
  9. Jan 5, 2010 #8
    The problem is... I don't know how should I get Rxn on point B...

    but anyways, I tried it using this calculations..

    First I set point A is the pivot. Then I totaled the 3 moments (Weight, CG, and wind).

    W(1.5) + CG(7) + Wind(21.5) = Moment A

    Then I divided it by 8.5 (the distance from A to B) to get Rxn on B.

    Moment A / 8.5 = 5759 lb = Rxn on Point B

    is this correct?

    --------------------------

    Now for smallest wind to tipover the billboard... here's what I did.... :

    I set point B as pivot, then Totaled the 2 moments (Weight and CG).

    W(7) + CG(1.5) = Moment WCG

    Then I set this Moment WCG as Moment Wind, because the wind needs just about this force to tip over the billboard.

    Moment WCG = Moment Wind

    Force = Moment Wind / 13
    Force = 1600lb

    or 160lb/ft
     
    Last edited: Jan 5, 2010
  10. Jan 5, 2010 #9

    PhanthomJay

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    No. You can always check sum of forces in y direction = 0 to solve for the reaction at B. In solving moments about A, you again made the mistake of making the vertical member a horizontal one, which gives you the wrong moment arm for the wind force. The arm is not 21.5 feet, it's______feet. By making the vertical member horizontal, that messes up the sum of forces in the y direction also. Don't change the problem.
     
  11. Jan 5, 2010 #10
    hmm.. ok lemme try again :D

    For Reaction on Point B:

    I'll set A as pivot, then total the Moment of W and CG.

    W(1.5) + CG(7) = Moment WCG

    then I'll divide it by 8.5 to get Reaction on point B? then B = 2724 lb

    so I will ignore wind load completely on calculating Rxn on B?

    or I will use that Wind load to produce a Resultant force?
     
  12. Jan 5, 2010 #11

    PhanthomJay

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    Firstly, you got part (b) correct. But part (a) is wrong. You can't ignore the moment from the wind force! Moment is force times perpendicular distance from line of action of the force to the pivot. For the moment from the wind force about A, the perpendicular distance from the line of action) of that force to A is ????????????.. Now check your work by summing forces in the y direction = 0.
     
  13. Jan 5, 2010 #12
    ah.. ok I get it! lemme try again

    For Reaction on B:

    Set A as pivot, then total 3 moments (W CG and Wind)

    W(1.5) + CG(7) + Wind(13) = Moment WCG

    then to find RXN on B

    Force B = Moment WCG/8.5

    = 4559 lb

    am I right?
     
  14. Jan 5, 2010 #13

    PhanthomJay

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    Yes! Which is a good thing, since it's near 11 pm here and i've got to get some sleep. :wink: Now just check your work by noting that the sum of forces in y direction equals 0,,that is, 4559 + 610 equals 2370 + 2800 , yes, good night or good morning, wherever you are.
     
  15. Jan 5, 2010 #14
    hehe thanks Jay! :D good night
     
  16. Jan 6, 2010 #15

    PhanthomJay

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    Morning. Let's not forget that there is a horizontal force acting also, the wind force of 1200 lbs acting to the right, in part A. So there must also be, in addition to the vertical reaction force at B, a horizonal reaction force at the pin at B, of 1200 pounds , acting to the left., this in order to fulfill the requirement of sum of forces in x direction = 0. Sorry about that, FairyChatMan.:redface:
     
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