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Static Equillibrium. [T=0, [F=0? Problem finding force using component

  1. Mar 3, 2013 #1
    Taking a physics 12 class in night school and he only gave us an example of a Beam and Wire question where the Wire was holding a Beam against a wall. The wire was at the end of the beam and holding a sign.

    Using the methods taught, I have been able to understand the Torque = 0, but Forces get me stuck

    This is a nifty site and my next class (and unit test) is on Tuesday. Teacher hardly checks emails!

    1. The problem statement, all variables and given/known data
    Find the cord tension and force from hinge on this beam.

    Hinge is at the wall. Beam is 4.50m long. Object at end is 20 kg heavy. The beam is 80kg. Also, the wire this time is 1.0m from the end (**star** is the wire) with an angle of 25 degrees.

    H.______*__


    2. Relevant equations

    Sum of T = 0
    Sum of F = 0

    3. The attempt at a solution

    First using the sum of Torques:
    Tcw = Tccw


    Clockwise applications:
    Mb*G (Beam)
    Mo*G (Object)

    Counter-cw:
    Ty

    Using all these applications and their distances relative to the pivot (Pivot @ H)
    Mb*G*d + Mo*G*d = Ty*d
    (80)(9.8)(2.25m) + (20)(9.8)(4.5m) = Tsin25 * (3.5m)
    1764 + 882 = (3.5) * Tsin25

    Divide by 3.5

    756 = Tsin25

    Divide by sin25

    T = 1788.8N --> 1790N (The answer is rounded in my book)


    Now for forces.

    Fx = 0
    FHx - Tx = 0
    FHx = Tx
    FHx = Tcos25
    FHx = 1790cos25
    FHx = 1622.3N

    But the answer is 1600N. Using pythagorean theorem, the x is larger than the resultant?
     
  2. jcsd
  3. Mar 4, 2013 #2

    CWatters

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    I can't see an error in your working.

    The vertical component at the hinge isn't zero so it's not just 1622 rounded down to 1600N. I make the resultant about 1800N so could be a typo ?
     
  4. Mar 4, 2013 #3
    I understand that there is a Y component but I stopped after I saw FHx = 1622N

    From my understanding... The answer should be 1637N (I had a friend who is into math, calculus and taking engineering in university see if I made an error after posting this?)
    He got the same 1637N as me

    [Fy=0
    FN+Ty(up)+F1g (down)+F2g(down)=0
    FN + 1790sin25 + (-784) + (-196) = 0

    Flipping over some question to the other side

    Fn = 784 + 196 - 756
    Fn = 224

    R= Sq. Root {1622^2 + 224^2)
    R= 1637N

    Tan-1(224/1622)
    Angle = 7.86 degrees (the angle is 7.87 in answer?)

    So I am assuming that they just rounded down to 1600N?
     
  5. Mar 4, 2013 #4

    CWatters

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    Yes looks like misscalculated. I agree with your figure of 1637N.
     
  6. Mar 4, 2013 #5
    I emailed my teacher it is correct, rounded to 2 sig figs. Thanks for helping out! I'm on to beams at angles now so if I need any more help I shall ask on here :)

    Just wondering, this is a physics forum with some advanced applications all over mathematics. Would physics 12 fit in this introductory physics all the way through?
     
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