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How to get tis relation any hints:

  1. Oct 30, 2009 #1
    [​IMG]

    plz help me :) to get tis relation.
     
  2. jcsd
  3. Oct 30, 2009 #2

    Doc Al

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    Assuming those vectors add to zero, just add the horizontal components.
     
  4. Oct 30, 2009 #3
    :confused: still not clear :confused:
     
  5. Oct 30, 2009 #4

    Doc Al

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    Do you have a specific question? Where did this relationship come from? I assume it's part of a larger problem?

    Do you know how to find components of a vector? Read this: Resolving a Vector Into Components
     
  6. Oct 31, 2009 #5

    HallsofIvy

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    The statement [itex]T_1= T_2+ T3cos(\theta)[/itex] is only true if, "T1", "T2", and "T3" are the lengths of the vectors shown and the horizontal components sum to 0. (If both T1 and T2 are horizontal and T3 isn't, as shown, the vectors them selves cannot sum to 0.)

    Since T1 and T2 are both horizontal, you need to look at the horizontal component of T3. Drop a perpendicular from the tip of T3 to the horizontal. You get a right triangle with angle [itex]\theta[/itex] and hypotenuse of length T3. The horizontal component of T3, Tx then satisifies [itex]cos(\theta)= T_x/T_3[/itex] so [itex]T_x= T_3 cos(\theta)[/itex].
     
  7. Oct 31, 2009 #6
    [​IMG]

    yes u r right, i was solving a problem on surface tension. For the equilibrium of the drop, the horizontal components must balance each other.
     
  8. Oct 31, 2009 #7
    yes, its a part of a larger problem; problem on surface tension.

    :approve: yes i know to find components, BTW i appreciate ur help u always replies to my posts; thanks for ur valuable time :smile:
     
  9. Oct 31, 2009 #8

    HallsofIvy

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    Doc Al's point was "Apply that method (of finding components) and you will answer your own question!"
     
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