Help with a tug of war problem

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SUMMARY

The discussion revolves around solving a two-dimensional tug-of-war problem involving three participants: Alex, Betty, and Charles. Alex exerts a force of 211 N, and Charles exerts 181 N, while the direction of Charles's force is crucial for determining Betty's force. For part (a), the calculated magnitude of Betty's force is 321.60 N when Charles pulls at a specific angle. The challenge lies in part (b), where the direction of Charles's pull is reversed, necessitating a reevaluation of the equilibrium conditions to find Betty's force in this alternate scenario.

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  • Understanding of vector forces and equilibrium in physics
  • Familiarity with trigonometric functions and their application in force resolution
  • Knowledge of Newton's laws of motion
  • Ability to interpret and analyze force diagrams
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  • Review vector addition and equilibrium conditions in physics
  • Learn how to resolve forces into components using trigonometric identities
  • Study the implications of changing force directions on equilibrium
  • Practice similar tug-of-war problems with varying angles and forces
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This discussion is beneficial for physics students, educators, and anyone interested in understanding force equilibrium in two-dimensional systems, particularly in applied mechanics scenarios.

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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 of magnitude 211 N, and Charles pulls with force of magnitude 181 N. Note that the direction of is not given. What is the magnitude of Betty's force if Charles pulls in (a) the direction drawn in the picture or (b) the other possible direction for equilibrium?

** I got part a correct but i just do not understand what part b is looking for diagram link attached if doesn't work email.


http://i1148.photobucket.com/albums/o578/bettingw15/ScreenShot2012-06-04at95053AM.pn


Homework Equations


for part a i used Fb= -fa-Fc
found angle of fa = 180-42= 138
fby=-fay-fcy
fbsin(-90)=-fasin138-fcsintheta
then did same thing with x components to get angle theta= 29.97 then plugged that angle into formula -fb=-(211N0(sin138)-181N(sin 29.97)
= 321.60 N

I understand that but how do i do the other possible direction? The question does not even make sense to me


The Attempt at a Solution

 
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