Dynamics problem about two-dimensional tug-of-war

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Homework Help Overview

The problem involves a two-dimensional tug-of-war scenario where three individuals are pulling on an automobile tire at various angles. The tire remains stationary despite the forces applied by Alex, Betty, and Charles. The magnitudes of Alex's and Charles's forces are given, but the direction of Charles's force is unspecified, leading to a need to determine the magnitude of Betty's force.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculations related to the angles and components of the forces involved. There are attempts to verify the correctness of the original poster's calculations and reasoning regarding the forces' equilibrium. Questions arise about the vector nature of the forces and whether the sum of the forces should equal zero.

Discussion Status

Some participants affirm the original poster's calculations, while others raise questions about the implications of the results, particularly regarding the equilibrium condition of the forces. There is an acknowledgment of the need to consider the vector addition of the forces to ensure they balance correctly.

Contextual Notes

Participants note the lack of information regarding the direction of Charles's force, which is critical for resolving the problem accurately. The original poster expresses uncertainty due to conflicting information found online and the absence of textbook answers.

Bunny-chan
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Homework Statement


In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on an automobile tire at the angles shown in the picture. The tire remains stationary in spite of the three pulls. Alex pulls with force \vec{F_A} of magnitude 220~N, and Charles pulls with force \vec{F_C} of magnitude 170~N. Note that the direction of \vec{F_C} is not given. What is the magnitude of Betty's \vec{F_B} force?

Homework Equations


F = ma

The Attempt at a Solution


334da32a-a449-48c7-a35b-49dd73829d4e.gif


What I did was the following:

For Alex's angle, I did: 137 - 90 = 47 \\ 180 - 47 = 133
So Alex angle is \theta = 133^\circ from the positive x-axis.

For x components, we have:F_{Bx} + F_{Ax} + F_{Cx} = 0 \\ \Rightarrow F_B\cos -90 + 220\cos 133 + 170\cos \theta = 0 \\ \Rightarrow 0 - 150 + 170\cos \theta = 0 \Rightarrow -150 = -170\cos \theta \Rightarrow \cos \theta = \frac{150}{170} = 0.88235 \\ \Rightarrow \cos^{-1} 0.88235 = 28.0^\circAnd for y:F_{By} + F_{Ay} + F_{Cy} = 0 \\ \Rightarrow F_B\sin -90 + 220\sin 133 + 170 \sin 28 = 0 \Rightarrow -F_B + 160.9 + 79.8 = 0 \Rightarrow -F_B = -240.7 \\ \Rightarrow F_B = 240.7~NAnd this is my result. I'm posting this here because I feel like I did everything right, but my textbook doesn't have answers, and when I searched through the web, I came across some different answers and ways of solving it, and now I don't know if I'm missing something. So, is there any mistake on my answer?
 
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Your work and results look fine.
 
gneill said:
Your work and results look fine.
Really? OK. Thank you!
 
gneill said:
Your work and results look fine.
One question though... Shouldn't F_B + F_A + F_C be equal to 0? If F_B = 240.7~N, that couldn't happen.
 
Bunny-chan said:
One question though... Shouldn't F_B + F_A + F_C be equal to 0? If F_B = 240.7~N, that couldn't happen.
They are vectors so you need to add them as vectors. If you do so you should find that the sum has a magnitude of zero.
 
Bunny-chan said:

Homework Statement


In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on an automobile tire at the angles shown in the picture. The tire remains stationary in spite of the three pulls. Alex pulls with force \vec{F_A} of magnitude 220~N, and Charles pulls with force \vec{F_C} of magnitude 170~N. Note that the direction of \vec{F_C} is not given. What is the magnitude of Betty's \vec{F_B} force?

Homework Equations


F = ma

The Attempt at a Solution


View attachment 199302

What I did was the following:

For Alex's angle, I did: 137 - 90 = 47 \\ 180 - 47 = 133
So Alex angle is \theta = 133^\circ from the positive x-axis.

For x components, we have:F_{Bx} + F_{Ax} + F_{Cx} = 0 \\ \Rightarrow F_B\cos -90 + 220\cos 133 + 170\cos \theta = 0 \\ \Rightarrow 0 - 150 + 170\cos \theta = 0 \Rightarrow -150 = -170\cos \theta \Rightarrow \cos \theta = \frac{150}{170} = 0.88235 \\ \Rightarrow \cos^{-1} 0.88235 = 28.0^\circAnd for y:F_{By} + F_{Ay} + F_{Cy} = 0 \\ \Rightarrow F_B\sin -90 + 220\sin 133 + 170 \sin 28 = 0 \Rightarrow -F_B + 160.9 + 79.8 = 0 \Rightarrow -F_B = -240.7 \\ \Rightarrow F_B = 240.7~NAnd this is my result. I'm posting this here because I feel like I did everything right, but my textbook doesn't have answers, and when I searched through the web, I came across some different answers and ways of solving it, and now I don't know if I'm missing something. So, is there any mistake on my answer?
I am not solving but you may use lami's theorem to get easy and faster results.

https://en.wikipedia.org/wiki/Lami's_theorem
 
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