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Simple Statics Problem

  1. Oct 23, 2011 #1
    1. The problem statement, all variables and given/known data
    Determine the tension in cables BC and BD, and reactions at the ball-and-
    socket joint at A for the mast shown in the figure.

    Answers are given by the teacher for checking purposes: TC=707 N; TD=1500 N, Ax=0 N,
    AY=0 N, AZ=1500 N


    2. Relevant equations
    ƩMx=0
    ƩMy=0
    ƩMz=0
    ƩFx=0
    ƩFy=0
    ƩFz=0


    3. The attempt at a solution
    Taking clockwise to be the positive moment direction about the x-axis. In other words, if you are standing at the positive side of the x-axis and look down towards the negative side, the moment will be spinning clockwise.
    ƩMx=0= -1kN(6m) + TBDcos(tan-1(3/6))cos(45)[kN]*(6m)
    ....this attempt at a solution is way wrong.

    The way I thought about it was, take the cos(tan-1(3/6)) to get your TBD in the yz-plane. Next, I took the cos(45) to get it going strictly in the y direction, and then multiplied by the perpendicular distance from the x-axis, 6 m. The problem I'm having is knowing what to do with the given force of 1 kN. There is no specified direction or any information stating where this force is going to. I know how to use the moment and force equations if I can get these things broken up into components correctly.
     

    Attached Files:

  2. jcsd
  3. Oct 23, 2011 #2

    gneill

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    From your rough figure it looks as though force F is anti-parallel to the y-axis (that is, it is parallel to the xy plane and directed towards the -y direction).

    If you project all the forces onto the xy-plane then you'll have a pretty straightforward exercise in statics.
     
  4. Oct 23, 2011 #3

    rude man

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    Right now, without knowing the direction of F I'm dubious as to whether the problem can be solved.

    If we assumed point D is along the y axis, just for the sake of argument, then the force summations at point B would be

    Fx + Tc/√2 = 0
    Fy + Td/√2 = 0

    (Summing moments about A gives no new information.)

    But √(Fx^2 + Fy^2) = C = 1kN

    So we have

    Fx + Tc/√2 = 0
    √(C^2 - Fx^2) + Td = 0

    I can pick any Fx < C, that determines Tc, the Td is also determined. So Fx and Fy are not uniquely defined. If I reorient F in the (presumably) x-y plane, Tc and Td will be correspondingly redistributed to satisfy ƩF = 0.

    I guess the question is, since point D is offset from the y axis into the -x direction, does this impose another condition on F such that its direction is then defined? Stay tuned, gotta think about it some more.
     
  5. Oct 23, 2011 #4

    rude man

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    Looking at it more, I don't think the fact that D is offset into the -x direction changes the aforesaid.

    Now we write
    ƩFx = 0 → Fx + Tc/√2 + Tdi = 0
    ƩFy = 0 → Fy + Tdj = 0
    Fx^2 + Fy^2 = 1 kN

    where

    i,j are unit x and y vectors,
    Tc = Tc*ûc
    and Td = Td*ûd
    Vectors in bold. ûc and ûd are unit vectors pointing from the flagpole top to C and D respectively. Tc and Td are positive scalars.

    ûc and ûd are known (check your analyt geometry or calculus book).

    Again, I don't see that Ʃmoment about point A will give more info. The moments are just the x and y forces about the y and x axes respectively, times L, the length of the pole.

    This means again that, since there are 4 unknowns and only 3 equations, we can choose Fx arbitrarily < 1 kN.

    Maybe someone else can look at this and refute my reasoning. Woukldn't be the first time ...
     
  6. Oct 23, 2011 #5
    Rudeman, you had the right idea. The 1kN force is in fact in the y direction. I used the j unit vector for the -1kN force. I used the moment about A, and then I used the i, j, and k components of that to find T(bd) and T(bc). Then I summed up the force components to find the A forces. Thank you all for the helpful repiles.
     
  7. Oct 23, 2011 #6

    PhanthomJay

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    There are just 3 external unknowns at A, C, and D...all members whether tension wires or compression columns are 2-force members, so the problem is statically determinate using just 3 of the 6 equations (the other 3 are superfluous, as you noted).
     
  8. Oct 24, 2011 #7

    rude man

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    There were 4 unknowns before it was determined that F was in the -y direction: Fx, Fy, Tc and Td. Defining F = -1kn*j resolved the indeterminacy.
     
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