Dynamics problem about two-dimensional tug-of-war

AI Thread Summary
In a two-dimensional tug-of-war problem, Alex pulls with a force of 220 N and Charles with 170 N, while the tire remains stationary. The calculations for Betty's force, F_B, were performed using vector components, resulting in F_B being approximately 240.7 N. There was confusion regarding the vector sum of the forces, leading to questions about the validity of the calculations. It was clarified that the forces must be treated as vectors to ensure their sum equals zero. The discussion also suggested using Lami's theorem for a potentially simpler solution.
Bunny-chan
Messages
105
Reaction score
4

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?
 
Physics news on Phys.org
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
 
Last edited by a moderator:
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top