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Solving for unknown tensions

  1. Apr 11, 2008 #1
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    The weight of the engine is known and so are angles B and C.
    I need to determine the tension in the chain AB and the compressive force acting in the spreader bar BC.

    What I want to know is:

    1. Is it possible to make a force triangle of the system to determine AB? If so how is this done?

    2. Is it possible to set up an equation for the horizontal and vertical forces and have BA's x and y components as the variables?

    3. Is the compressive force in BC determined in the same way that the tension force in AB is determined?
     
  2. jcsd
  3. Apr 11, 2008 #2

    Hootenanny

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    Yes to all of the above. However, (1) requires you to do (2) anyway.
     
  4. Apr 11, 2008 #3
    Can someone please confirm whether the resultant divides the triangle into 2.
    i have 2 reasons for this.
    1. the force in the chain connected to the roof would have the same force as the resultant of the triangle and would be in the same direction.
    2. the angles B and C are equal therefore they would be an equal distance from the resultant.
    so if angle b and c are 40 degrees do 180-40x2 to determine angle A then divide angle A by 2 to get the angle of the resultant from B and C.
     
    Last edited: Apr 11, 2008
  5. Apr 11, 2008 #4
    when resolving the x and y components of the acting forces do I need to take BD into consideration?
     
  6. Apr 11, 2008 #5
    I have three variables and cant solve. what am i doing wrong?
    0=ABsin40+ACsin40-4kN
    0=ABcos37-ACcos37+BC
    engine = 4kN
     
  7. Apr 12, 2008 #6

    Hootenanny

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    You're doing nothing wrong, however, It may be useful to note that,

    [tex]AB\sin40 = AC\sin40[/tex]

    Hence,

    [tex]2\cdot AB\sin40 = 2\cdot AC\sin40 = 4 \;\left[kN\right][/tex]
     
  8. Apr 12, 2008 #7
    What direction is the BC x component moving?
     
  9. Apr 12, 2008 #8

    Hootenanny

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    BC exerts a force away from the centre of BC.
     
  10. Apr 12, 2008 #9
    how do i determine if it is negative or positive?
     
  11. Apr 12, 2008 #10

    Hootenanny

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    Since BC acts in both directions (left and right) it doesn't make sense to define a direction, a magnitude is all that is needed.
     
  12. Apr 12, 2008 #11
    From my equation
    0=ABcos37-ACcos37+BC
    Bcos37 and ACcos37 would obviously cancel each other out. BC therefore makes this equation incorrect and BC can therefore not be found using this method can it?

    Because the triangle is symmetrical cant i just divide it into two right triangles and solve for BC/2 then multiply by 2?
     
  13. Apr 12, 2008 #12

    Hootenanny

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    Consider the forces acting at B. You know that since the chain BD hangs vertically the sum of the horizontal components at B must be zero. That is,

    [tex]|CB| = |AB\cos37|[/tex]

    Similarly for point C.
     
  14. Apr 12, 2008 #13
    could you explain it a bit further?
    also can you confirm which of these is correct
    1. 0=ABcos37-ACcos37+BC
    2. 0=ABcos37-ACcos37
     
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