Finding an angle with Newton's Second Law

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Homework Help Overview

The discussion revolves around finding an angle θ in a problem involving a free-body diagram, where a body is in equilibrium. The context is rooted in Newton's Second Law and the analysis of forces acting on the body.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law in the vertical direction and express uncertainty about the correctness of their calculations. There is a focus on the relationship between the forces and the angle θ, with some questioning the vector nature of the forces involved.

Discussion Status

Some participants have offered insights regarding the relationships between the forces and the angle, suggesting that the forces may have equal magnitudes but differ as vectors. There is ongoing exploration of the implications of these relationships on the value of θ.

Contextual Notes

Participants note that the angle θ is expected to be acute, which raises questions about the validity of their current calculations and assumptions regarding the forces involved.

Mr Davis 97
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Homework Statement


Given the free-body diagram below, and that the body is in equilibrium, find θ.
ee3VjBf.jpg

Homework Equations


F = ma

The Attempt at a Solution


[/B]
Basically, I used Newton's second law in the y direction.

Here is my work:
o4KjcyS.jpg

This does not seem like the correct answer, because the angle is acute. Thus, what am I doing wrong?
 
Last edited:
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Mr Davis 97 said:

Homework Statement


Given the free-body diagram below, and that the body is in equilibrium, find θ.
[ IMG]http://i.imgur.com/ee3VjBf.jpg[/PLAIN]

Homework Equations


F = ma

The Attempt at a Solution


[/B]
Basically, I used Newton's second law in the y direction.

Here is my work?
[ IMG]http://i.imgur.com/o4KjcyS.jpg[/PLAIN]
This does not seem like the correct answer, because the angle is acute. Thus, what am I doing wrong?
First of all, the three forces may have equal magnitude, but they are not equal as vectors.

That should be cos(θ/2) or equivalently sin(90° - θ/2)
 
SammyS said:
First of all, the three forces may have equal magnitude, but they are not equal as vectors.

That should be cos(θ/2) or equivalently sin(90° - θ/2)
So θ = 120 degrees?
 
Last edited:
Mr Davis 97 said:
So θ = 120 degrees?
Yes.
 

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