# Tug of war in two demensions.

1. Feb 9, 2008

### AnkhUNC

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 an angle of 141d (Alex and Betty's angle). The tire remains stationary in spite of the three pulls. Alex pulls with force Fa of magnitude 208 N, and Charles pulls with force Fc of magnitude 178 N. Note that the direction of Fc is not given. What is the minimum magnitude of Betty's force Fb ?

Sorry don't have a picture :(

2. Relevant equations

3. The attempt at a solution

OK so for this I get
x = -Fa cos(39)[141-180]+0+Fc cos (Theta)
y = Fa sin(39)-Fb+Fc sin (theta)

Solving for Theta I get 24.75240251.
So when I try to plug this into the equation for y to find Fb I find the magnitude to be 56.37042988 (There is no magnitude in the x direction). But this is incorrect. Any idea where I went wrong?

2. Feb 9, 2008

### Shooting Star

Apply the cosine formula directly:

Fb^2 = Fa^2 + Fc^2 + 2 FaFcCos B. Now find the min value of Fb.

3. Feb 9, 2008

### AnkhUNC

So I get Fb^2 = 142192.8858 -> sqrt = 377.0847196 is that correct?

4. Feb 9, 2008

### Shooting Star

I am terribly sorry, but apply the formula for the resultant of the force opp the angle which is given. The angle which is given is between Fa and Fb, i.e., angle C=141 deg . (Draw a diagram of three concurrent forces, to be on the safe side.)

Fc^2 = Fb^2 + Fa^2 + 2FbFaCos C.

Now you have a quadratic expression in Fb, and can minimize it. All the other terms are known.

(How did you get the value you have given? Cos B was unknown.)

5. Feb 9, 2008

### AnkhUNC

Can't I just use the fact that acceleration is 0 to make this an easier problem? Was the way I was trying to solve completely incorrect?

6. Feb 10, 2008

### Shooting Star

The method I've given should be the easiest.

Write properly in which direction you are choosing the x-axis etc and the components of the forces along the axes. Ultimately any method should lead you to the same eqn.