Understanding Trig In Force Diagrams

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Discussion Overview

The discussion revolves around understanding the application of trigonometry in force diagrams, specifically in the context of calculating force components in a physics problem involving charges. Participants are reviewing their approaches to a homework problem and clarifying the reasoning behind the use of cosine in the force calculations.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant describes their approach to a problem involving forces acting on a charge and references a solution that includes a cosine term.
  • Another participant explains that the cosine term arises from the relationship between the resultant force and the individual forces acting on the charge, specifically in relation to the angle of the force diagram.
  • There is a discussion about the symmetry of the forces and how it justifies multiplying by 2 in the calculations.

Areas of Agreement / Disagreement

Participants generally agree on the reasoning behind the use of the cosine function and the symmetry in the problem, but the discussion does not resolve all aspects of the force diagram or the complete solution.

Contextual Notes

The discussion does not clarify all assumptions regarding the force diagram or the specific conditions under which the cosine term is applied, leaving some steps and definitions potentially unresolved.

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Hello,

I have the following problem part (b) which I already solved as you can see in the attached image. So I am not asking homework questions, I merely reviewing my homework for a better understanding for the test. I obtained the answer from a friend showing me his method. However, I am studying and a listed solution was the following:

F_y = 2*(k*(q^2/r^2))cos(30)


I've drawn out a force diagram but have no idea how, they have obtained the cosign. In the listed solution they state use oriented the y-axis such that it bisects charges q2 and q3.



2rxb9ef.png
 
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There are two forces acting on point 1. By symetry the result of these two forces will act along a line drawn through point 1 and a point mid way between 2 and 3. See diagram.

So work out the component of the two forces pointing in that direction.

Consider the triangle on the right.

Cos(30) = Fr/F

so

Fr = F Cos(30)

That's not the whole solution obviously, just where the cos(30) comes from. The angle doesn't come from the direction of the result per se, it comes from the direction of the result in relation to the forces. eg If you rotate the triangle/problem drawing 30 degrees so the resulting force is vertical (on the y-axis) the answer will still contain cos(30).
 

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Thank you very much! And because of symmetry is why it's multiplied by 2?
 
Yes in this case.
 

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