Equilibrium: An archer's bow is drawn at its midpoint until

[soccerball3211]

An archer's bow is drawn at its midpoint until the tension in the string is 0.963 times the force exerted by the archer. What is the angle between the two halves of the string?

I started by analyzing the forces present.

I have ma = F - 2Tcos(theta)

Since there is no acceleration ma goes to zero

Therefore we have F = 2Tcos(theta)

F = 2(0.963F)cos(theta)

F's will cancel and we have 1/(2*0.963)=cos(theta)

Inverse cosine leaves us with theta = cos^(-1)(1/(2*0.963))

Can someone help me find my mistake?

 

[skeeter]

let a = angle between the two halves of the string

F = 2Tcos(a/2)
T = 0.963F

F = 2(0.963F)cos(a/2)

1 = 2(0.963)cos(a/2)

a = 2arccos[1/(2(0.963))] = approx 117 degrees

 

[soccerball3211]

Thanks skeeter I believe I found half the angle between the two pieces of string