What is the magnitude of force with which Betty pulls?

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SUMMARY

In the tug-of-war problem involving Alex, Betty, and Charles, the forces are balanced, resulting in a stationary tire. Alex exerts a force of 220 N, while Charles pulls with a force of 170 N at an unknown angle. To determine the magnitude of the force with which Betty pulls, denoted as B, one must set up a coordinate system where Alex's force vector is <220, 0> and Betty's force vector is . The total force vector must equal <0, 0>, leading to two equations that can be solved for B and the angle θ of Charles's pull.

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Homework Statement



In a two-dimensional tug-of war, Alex, Betty, and Charles pull horizontally on an automobile
tire. The angle between Alex and Betty is 137degrees. The tire remains
stationary in spite of the three pulls. Alex pulls with a force of magnitude 220 N, and Charles
pulls with a force of magnitude 170 N. Note that the direction of the force applied by Charles
is not given. What is the magnitude of force with which Betty pulls?

Homework Equations



F1 + F2 + F3 = 0 ?

The Attempt at a Solution



I do not know how to even approach this problem. Charles could be pulling with Betty, or with Alex, or in some other random direction. What formula could I use that would compensate for Charles?
 
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Hint: What must be the magnitude of the sum of Alex's force + Betty's force?
 
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Set up a coordinate system. let's set it up so that Alex is pulling along the positive x-axis: his force vector is <220, 0>. Then Betty is pulling at an angle of 137 degrees has force vector <Bcos(137), Bsin(137)> where "B" is the magnitude of the force (which is what you are asked). Finally, we know that Charles is pulling with magnitude 170 but don't know the angle: let's call the angle \theta degrees. His force vector is &lt;170 cos(\theta),170 sin(\theta)&gt;. The total force vector is &lt;220+ Bcos(137)+ 170cos(\theta), Bsin(137)+ 170sin(\theta&gt;= &lt;0, 0&gt;. That gives you two equations to solve for B and \theta.
 

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