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Statics: Internal forces at a point along a beam

  1. Jul 15, 2013 #1
    1. The problem statement, all variables and given/known data
    z0VQCtH.png


    2. Relevant equations
    ƩFx=0
    ƩFy=0
    ƩM=0

    3. The attempt at a solution

    FBD of the entire thing:
    JveTSR9.png

    ƩMB = 780(0.3) + (5/13)T*(0.72) - (12/13)T*(0.6) = 0
    T = 845N

    ƩFx = Bx + (5/13)*845 = 0
    Bx = 325N ←
    ƩFy = By + (12/13)*845 = 0
    By = 780N ↓

    Now here is where I tried 2 different methods for the forces at J:


    Method 1:

    FBD of CDJ:
    Jrk8kfg.png

    ƩFx = (5/13)*845 - (12/13)V + (5/13)F = 0
    ƩFy = (12/13)*845 - 780 + (5/13)V + (12/13)F = 0
    θ = arctan(5/13) = 22.6°
    F = 125N 67.4° down and to the left
    V = 300N 22.6° up and to the left

    horizontal distance from D to J = 0.24/tan(67.4°) = 0.1m
    ƩMJ = (5/13)*845*(0.24) - (12/13)*845*(0.4) + 780*(0.1) + M = 0
    M = 156 N*m counterclockwise

    This method gives me the correct magnitudes, but the directions are supposed to be opposite of what I got, which I don't understand.

    Method 2:

    FBD of BJ:
    5IOevDl.png

    ƩFx = (12/13)V - (5/13)F - 325 = 0
    ƩFy = -(5/13)V - (12/13)F - 780 = 0

    F = 845N 67.4° up and to the right
    V = 0N

    horizontal distance from J to B = 0.48/tan(67.4°) = 0.2m
    ƩMJ = 325*(0.48) -780*(0.2) + M = 0
    M = 144 N*m counterclockwise

    I am not sure why this method gives me incorrect answers.
     
  2. jcsd
  3. Jul 15, 2013 #2

    TSny

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    Homework Helper
    Gold Member

    Did you leave out one of the vertical forces here?

    I agree with your answers for Method 1 including your directions.
     
    Last edited: Jul 15, 2013
  4. Jul 15, 2013 #3
    There is no By force. The horizontal force in the AC rope is 780. It looks like your forgot to add that in on your Sum of y-direction forces.
     
  5. Jul 15, 2013 #4
    Oh yeah, thanks for the correction. I have the right numbers and directions now, using the second method.

    Doing method 2 I get the "correct" directions according to the book, which are opposite those in method 1.
     
  6. Jul 15, 2013 #5
    Your answers for both methods should have opposite directions. If you were to put those two sections back together the forces at J would cancel out and the whole system would be in static equilibrium. I guess your book just used method two but either way would be correct.
     
  7. Jul 16, 2013 #6
    Ah, ok, this makes sense. Thanks!
     
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