# Homework Help: Dynamics problem about two-dimensional tug-of-war

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1. May 6, 2017

### Bunny-chan

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
$F = ma$

3. The attempt at a solution

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:$$F_{Bx} + F_{Ax} + F_{Cx} = 0 \\ \Rightarrow F_B\cos -90 + 220\cos 133 + 170\cos \theta = 0 \\ \Rightarrow 0 - 150 + 170\cos \theta = 0 \Rightarrow -150 = -170\cos \theta \Rightarrow \cos \theta = \frac{150}{170} = 0.88235 \\ \Rightarrow \cos^{-1} 0.88235 = 28.0^\circ$$And for y:$$F_{By} + F_{Ay} + F_{Cy} = 0 \\ \Rightarrow F_B\sin -90 + 220\sin 133 + 170 \sin 28 = 0 \Rightarrow -F_B + 160.9 + 79.8 = 0 \Rightarrow -F_B = -240.7 \\ \Rightarrow F_B = 240.7~N$$And 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 accross 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?

2. May 6, 2017

### Staff: Mentor

Your work and results look fine.

3. May 6, 2017

### Bunny-chan

Really? OK. Thank you!

4. May 6, 2017

### Bunny-chan

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.

5. May 6, 2017

### Staff: Mentor

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.