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Equilibrium of coplaner force systems-help!

  1. Oct 14, 2005 #1
    prob 3-19.JPG
    A 100-kg mass is suspended by a boom hinged to the wall at one end and tied to the wall at the other end by a cable.Determine the tension in cable AB and the axial force in boom AC.
    Could you please give me a detailed explanation of how to do this and the answer. Thanks
     
  2. jcsd
  3. Oct 14, 2005 #2

    Doc Al

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    Staff: Mentor

    To get help, show us what you've done so far. (Hint: What are the rotational and translational conditions for equilibrium?)
     
  4. Oct 14, 2005 #3
    equilibrium of coplaner forces systems

    My professor gave us this for homework and has not really gone over it yet.
    The answer is in the book but I really need to know how to work the problem.


    I think you may multiply 100 kN by 9.8m/s, but besides that i have no clue what to do. The answer in the back of the book is Tension AB=1430 N
    and F AC=2200 N. I worked on this problem for a while but that is all i can figure out to do.

    Thanks for your help.
     
  5. Oct 14, 2005 #4

    hotvette

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    Homework Helper

    Are you familiar with the methodology of solving statics problems?

    - draw free body diagram of each component and label the forces
    - sum of forces in each coordinate direction must equal zero
    - sum of moments about any point must equal zero

    Does any of this sound familiar?
     
  6. Oct 15, 2005 #5
    Equilibrium of coplaner forces

    yes, I already know that but I don't know how to calculate the forces in each direction. Is there any way you can help me?
     
  7. Oct 15, 2005 #6

    hotvette

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    Ok, here's the methodology. A little hard to tell from the picture, but I assume B and C are pinned joints (i.e. can't support a moment) and A is a frictionless pulley (same thing, can't support a moment)

    1. Draw each member (AB, BC, AC) separately without touching each other. In other words, explode the picture apart keeping the orientation of each member intact

    2. At the ends of each member, draw forces in x and y directions and label them F1, F2, F3, etc. It doesn't matter which direction (+ or -) you place each force, as the correct direction will come out of the equations. Hint: at point A, you already know the magnitude and direction of the vertical force

    3. In each coordinate direction (x,y), create an algebraic equation summing the forces. The sum must equal zero. You'll end up with 2 equations in the form of F1 - F3 + F6 + .... = 0

    4. For moments, pick a point (A, B, or C) and write the moment equations about that point, making sure to properly keep track of the direction of the moment. You'll need the geometry to calculate the moment arms. I presume you know at least one of the distances AB, BC, CD. It is pure trig to find the moment arms. The moment equation also must equal zero. It will be in the form of F1*d1 - F2*d2 + .... = 0, where F's are the forces you labeled in step 2 and d's are the moment arms you calculate from the geometry.

    5. In general, you'll end up with n equations in n unknowns, the unknowns being the F's. You may need to write moment equations at additional points if you don't have enough equations.

    6. Solve the equations for the F's by combining equations to eliminate some of the unknowns and by substitution (system of linear equations).

    This is tedious work, but straightforward. It is critical that you keep consistent with +/- directions for the F's.
     
    Last edited: Oct 15, 2005
  8. Oct 15, 2005 #7
    equilibrium of coplaner forces

    I still cannot get the answer in the back of the book. It is not even close. I am really lost. Please help me
     
  9. Oct 15, 2005 #8

    Pyrrhus

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    You know how i would solve this problem, i would use the trigonometry :smile:
    Make a triangle, and the sides length will be the force magnitude, Use sine Law.
     
  10. Oct 15, 2005 #9
    equilibrium of coplaner forces

    I already did it that way using an example from the book and it does not come out right. I need this one done step by step, because it does not come out right. thanks
     
  11. Oct 15, 2005 #10

    Pyrrhus

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    Great!, show me you attempt. It'll be better if you could draw it somehow.
     
  12. Oct 16, 2005 #11
    can anyone please help me with this?
     
  13. Oct 19, 2005 #12
    Some hints

    Hi! In solving this problem, I believe you need to make an important assumption: that the boom is massless....

    Also, you will need to come to the realization that there are actually 2 pieces of cable present here, so the tension in the cable AB may not be equal to the tension in the vertical cable (which is connected to the 100 kg mass).

    Now the tension in the vertical cable can be found very easily. Since the 100 kg mass is in equilibrium, how is the tension in the vertical cable related to the weight of the mass?

    Knowing that, you can begin...

    1) Generally, for an equilibrium question, we can choose to take moments about a point or resolve forces in 2 perpendicular directions. We normally start by taking moments, as this enables us to eliminate some of the forces. For example, if we take moments about a point, and a force passes exactly through that point, the moment exerted by the force about that point is zero. So now, let us consider the forces acting on the boom. About which point should you take moments so that you can determine the tension in cable AB? Remember, the only force whose magnitude you know is the tension in the vertical cable...

    2) Once you have determined the tensional forces in both cables, you can proceed to find the axial force in boom AC. Resolve forces vertically and horizontally, then calculate the horizontal and vertical components of the axial force. Finally, use Pythagoras' Theorem to calculate the actual value of this force... This step may involve numbers with many decimal places, but do not give up! Equilibrium questions normally involve non-exact numbers due to the involvement of trigonometry.

    Hope my explanation makes things clearer for you!
     
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