- #1
Bunny-chan
- 105
- 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 [itex]\vec{F_A}[/itex] of magnitude [itex]220~N[/itex], and Charles pulls with force [itex]\vec{F_C}[/itex] of magnitude [itex]170~N[/itex]. Note that the direction of [itex]\vec{F_C}[/itex] is not given. What is the magnitude of Betty's [itex]\vec{F_B}[/itex] force?
Homework Equations
[itex]F = ma[/itex]
The Attempt at a Solution
What I did was the following:
For Alex's angle, I did: [itex]137 - 90 = 47 \\ 180 - 47 = 133[/itex]
So Alex angle is [itex]\theta = 133^\circ[/itex] from the positive x-axis.
For x components, we have:[tex]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[/tex]And for y:[tex]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[/tex]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 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?