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HELP Static Equilibrium

  1. Oct 30, 2005 #1
    HELP!!! Static Equilibrium

    |_______/
    |........./|
    |....../ 0 400N
    |.../
    |/_37?
    |
    Uniform pole of 6.0 m w/ weight 800N is attached to a vertical wall. A load of 400N hangs from other end of pole. The pole it held by a wire with an angle of 37?.
    WHAT is the TENSION of the wire and the HORIZONTAL and VERTICAL components of the reaction forces at the wall support!!
    HELP PLEASE! NEED EXPLANATION!:surprised
     
  2. jcsd
  3. Oct 30, 2005 #2

    cepheid

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    :rolleyes: What do you mean, "HELP, PLEASE, NEED EXPLANATION?" Do you not understand what the question is asking for? If not, that's fine, we can explain it to you. But if you do understand what is being asked, then I don't quite get what the melodrama is about. Bear in mind, it's static equilibrium. So, did you start by drawing a free body diagram?
     
  4. Oct 30, 2005 #3
    Alright I don't know how to solve the problem. All I know is that sum of forces in xa nd y are 0 and sum of torque is 0. I don't know how to draw the FBD.
     
  5. Oct 30, 2005 #4

    cepheid

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    Alright, I'll get you started. Keep in mind that, as the name implies, a FREE body diagram considers only one element at a time, ignoring all the others. In this case we want to draw a free body diagram for the pole. Why? Because all of the forces we are asked to solve for act on the pole. What are they? Make a list of them:

    Forces that act on the pole:

    - its own weight
    - the reaction force(s) of the vertical wall
    - the load
    - the force on the pole due to the tension of the wire

    Next, ask yourself, where does each of these forces act, and in what direction? If the force is not nicely horizontal or vertical, resolve it into horizontal and vertical components. The problem gives you enough information to do so.

    Now, draw the picture of the pole (just the pole), with all the force vectors in the correct places and orientations. Now, keep in mind, everything is in equilibrium. There can't be a net force acting on the pole, otherwise it would be accelerating. Thus, all the horizontal components of all the forces must balance each other out:

    [tex] \sum F_x = 0 [/tex]

    and all the vertical components of all the forces must cancel each other out:

    [tex] \sum F_y = 0 [/tex]

    Substitute all the relevant forces into these two equations, and ask yourself whether you have enough information to solve for the unknowns (the forces the problem asks you to calculate). If not, you may have to consider torques as well. Again, because this is an equilibrium problem, the net torque on the pole must be zero. Otherwise, the pole would be rotating.

    So you see what equilibrium problems are all about? Just be methodical, go through things step by step, in this way.
     
  6. Oct 30, 2005 #5
    Alright, could someone give me an answer to see if I applied the method correctly?
     
  7. Oct 30, 2005 #6

    cepheid

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    No. We don't do your homework for you here. You don't learn anything that way, it's a pointless, futile exercise. We have no way of knowing if you actually did the problem or not. POST your work here, and I'd be more than happy to correct it. :smile:
     
  8. Oct 30, 2005 #7
    Alright

    Sum of forces in y = 0 = Ay - 800N - 400N
    Sum of forces in x = 0 = Ax - Tr
    Sum Torque = 0 = Tr(6.0m) - 800N(6.0m) - 400N(cos(38))(3.0m)

    Tr = 960N

    Ax = 960N
    Ay = 1200N

    I think that is right?
     
  9. Oct 30, 2005 #8

    cepheid

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    Well, I'm assuming Ay and Ax are the vertical and horizontal reaction forces at the wall, respectively? Ok. What is Tr?

    At first glance, it doesn't look correct. Remember, we listed 4 forces acting on the pole? Yet, your sums of forces in each direction have only 3 forces each. What happened to to the vertical and horizontal components of the tension force in the wire?
     
  10. Oct 30, 2005 #9
    Here is my FBD
    ........T---./
    .........../.|
    AY...../...M
    |.../ W
    |/____AX

    Tr represents tension in the rope
     
  11. Oct 30, 2005 #10

    cepheid

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    *sigh*. Please READ my second post if you haven't already, and it seems like you haven't. Like I said, we're drawing a free body diagram *for the pole*. It should consist of only the pole and the forces acting on it.

    Note, the tension force in the rope has both horizontal and vertical components. It is not horizontal only, as your equations suggest.
     
  12. Oct 30, 2005 #11
    I don't understand how the rope has a vertical component
     
  13. Oct 30, 2005 #12

    cepheid

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    Maybe it's my bad. Maybe I misunderstood your original diagram? I thought the pole was the horizontal line at the top, and the rope was the diagonal line, at an angle of 37 degrees from the horizontal. Did I switch the two around? Is it the rope that's horizontal, and the pole is the diagonal line?
     
  14. Oct 30, 2005 #13
    yes, There is a vertical wall. A diagnol beam is attached at one end to the bottom of the vertical wall and attached at the other end with a ropw that is connected to the top of the vertical wall. It is a confusing diagram becuase of the periods.
     
  15. Oct 31, 2005 #14

    cepheid

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    Lol, that does seem to be a more sensible arrangement too, from the point of view of securing the pole to the wall. The sum of forces equations look right. The answer for Ay is pretty obvious and correct.

    The lever arms don't look right. The weight can be considered to act at the centre of mass. The load is at the end of the rod. And these distances should all be multiplied by the appropriate trig ratios, since the lever arms are perpendicular (horizontal or vertical) distances.
     
  16. Oct 31, 2005 #15
    OK.

    Sum Torque = 0 = Tr(6.0m) - 800(6.0m)(cos(37)) - 400(3.0m)(cos(37))
     
  17. Oct 31, 2005 #16

    cepheid

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    It looks like you're summing the torques about the point where the pole joins the wall. Ok.

    The cos37's are correct, because those two forces (the weight and the load) are vertical forces, so their lever arms will be horizontal (which you get by multiplying by the cos of the angle).

    but I question the individual values. Like I said, the weight can be considered to act at the centre of mass, which is in the middle of the rod. Isn't that 3m along the rod, not 6? The load acts at the end of the rod. Isn't that 6m along the rod, not 3?

    what happened to the lever arm for the tension? It's a horizontal force, so it has a vertical lever arm (multiply by sine).

    I don't have any more time to spend on this, so here are the values I get, I think they are right, hopefully you will agree.

    Tr(6sin37) - 800(3cos37) - 400(6cos37) = 0
     
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