Pulling the string on a bow back with a force of 29.0 lb, an archer prepares to shoot an arrow. if the archer pulls in the center of the string, and the angle between the two halves is 128 degrees , what is the tension in the string?

The tension in the string 33.08 lb.

Explanation:

Step 1

Consider the forces acting where the arrow is held. The string has a tension T on both sides of the arrow inclined 64 degrees to the horizontal.

The sum of all the forces in the x-directions is:

∑F_x = T cos(Ф₁) + T cos(Ф₂)

Since Ф₁ = Ф₂ = 64°, the sum of all the forces in the x-direction will be

∑F_x = 2T cos(64°)

The sum of all the forces in the y-directions is:

∑F_y = T sin(Ф₁) + T sin(Ф₂)  

Since Ф₁ = Ф₂ = 64°, the sum of all the forces in the y-direction will be

∑F_y = T sin(64°) - T sin(64°)

         = 0

Step 2:

The tension in the string:

The net force acting at the midpoint is 29.0 lb

Thus,

F_net = ∑F_x

F_net = 2T cos(64°)

29.0 = 2T cos(64°)

⇒ T = 29.0 / (2 cos(64°))

    T = 33.08 lb

Therefore, the tension in the string 33.08 lb.

50 inches should be the resultant vector

6/12=1/2

step-by-step explanation: