Is there an easier way to solve for theta in this equilibrium problem?

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Homework Help Overview

The problem involves finding the angle theta in an equilibrium scenario, where forces are balanced. The original poster attempts to derive an equation involving theta using trigonometric identities and relationships related to a stretched length.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the manipulation of trigonometric identities, with the original poster expressing difficulty in simplifying the equation. Some suggest using a substitution involving the tangent half-angle identity, while others question the validity of this approach.

Discussion Status

The discussion is ongoing, with participants exploring different methods to express sin(theta) and cos(theta) in terms of tan(theta/2). There is no clear consensus on the best approach yet, but several lines of reasoning are being examined.

Contextual Notes

Participants are navigating through complex trigonometric relationships and identities, indicating potential confusion or lack of clarity in the transformations being discussed.

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Homework Statement


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The Attempt at a Solution


2Fsin(theta)=45 N

Where F=ks=(75N/m)(s-1.2m).

Looking at the triangle, we can state that s, the stretched length is equal to 1.2m/cos(theta). Therefore,

2(75N/m)[(1.2m/cos(theta))-1.2m]sin(theta)=45 N
sin(theta)[(1/cos(theta))-1]=0.25

From here, the math gets tricky and complicated for me. Is there any easier way to solve theta?:S
 
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hi quantum_enhan! :smile:

(have a theta: θ :wink:)
quantum_enhan said:
sin(theta)[(1/cos(theta))-1]=0.25

From here, the math gets tricky and complicated for me. Is there any easier way to solve theta?:S

You have sinθ(secθ - 1) = 0.25.

Try substituting t = tan(θ/2). :wink:
 
I don't see how you can get tan(θ/2) from sinθ(secθ - 1). Don't you need (1-cosθ)/sinθ. I can't seem to get it to equal the half angle identity..
 
You can get sinθ from tan(θ/2), and you can get cosθ from tan(θ/2) …

so just build it up! :smile:
 

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