An archer's bow is drawn at its midpoint until the tension in the string is 0.842 times the force exerted by the archer. What is the angle between the two halves of the string

Consult the attached free body diagram.

If we take the direction of F to be the positive horizontal axis, and upward to be the positive vertical axis, then using Newton's second law we have net forces

• ∑ F [horizontal] = F [archer] + T cos(180° - θ) + T cos(180° + θ) = 0

• ∑ F [vertical] = T sin(180° - θ) + T sin(180° + θ) = 0

since the bow is held in place while it's drawn. T is the magnitude of the tension in the string, and it can be shown to be equal in both strings since they both make the same angle with the negative horizontal axis (the dashed line).

We only really need the first equation. Simplifying it, we get

F [archer] - T cos(θ) - T cos(θ) = 0

F [archer] - 2T cos(θ) = 0

F [archer] = 2T cos(θ)

cos(θ) = F [archer] / (2T)

We're given that the tension T in the string is 0.842 times the force exerted by the archer, which is to say

T = 0.842 F [archer]

and from this we have

cos(θ) = F [archer] / (2 • 0.842 F [archer])

cos(θ) = 1/1.684

cos(θ) ≈ 0.593

Solving for θ gives an angle of θ ≈ arccos(0.593) ≈ 53.6°. Then the angle between the two tension forces is twice this, or about 2θ ≈ 107°.

c. magnitude and direction

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form