How to get tis relation any hints:

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Discussion Overview

The discussion revolves around understanding a specific relationship involving vectors, particularly in the context of a problem related to surface tension. Participants explore how to derive this relationship and clarify the conditions under which it holds, focusing on vector components and equilibrium.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related

Main Points Raised

  • Some participants suggest that if the vectors add to zero, one can add the horizontal components to derive the relationship.
  • One participant clarifies that the statement T_1 = T_2 + T_3 cos(θ) is valid under specific conditions regarding the vectors' orientations and their horizontal components.
  • A participant emphasizes the need to consider the horizontal component of T3 by constructing a right triangle based on its angle θ.
  • Another participant mentions that the problem is related to surface tension and that the horizontal components must balance for equilibrium.
  • There is a suggestion to apply the method of finding vector components to resolve the question at hand.

Areas of Agreement / Disagreement

Participants generally agree on the importance of vector components and their roles in the problem, but the exact relationship and conditions remain somewhat unclear, indicating that multiple views and interpretations exist.

Contextual Notes

Some assumptions about the vectors' orientations and the specific context of the surface tension problem are not fully detailed, which may affect the clarity of the discussion.

sphyics
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t1.jpg


please help me :) to get tis relation.
 
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Assuming those vectors add to zero, just add the horizontal components.
 
Doc Al said:
Assuming those vectors add to zero, just add the horizontal components.

:confused: still not clear :confused:
 
Last edited by a moderator:
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].
 
t2.jpg


HallsofIvy said:
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.
 
Doc Al said:
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.

Doc Al said:
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:
 
Last edited by a moderator:
sphyics said:
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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