How to get tis relation any hints:

  • Thread starter sphyics
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  • #1
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t1.jpg


plz help me :) to get tis relation.
 

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  • #2
Doc Al
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Assuming those vectors add to zero, just add the horizontal components.
 
  • #3
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Assuming those vectors add to zero, just add the horizontal components.

:confused: still not clear :confused:
 
  • #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].
 
  • #6
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t2.jpg


horizontal components sum to 0.

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.
 
  • #7
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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?

yes, its a part of a larger problem; problem on surface tension.

Do you know how to find components of a vector? Read this: http://hyperphysics.phy-astr.gsu.edu/HBASE/vect.html#vec5"

:approve: yes i know to find components, BTW i appreciate ur help u always replies to my posts; thanks for ur valuable time :smile:
 
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  • #8
HallsofIvy
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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:
Doc Al's point was "Apply that method (of finding components) and you will answer your own question!"
 

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