Question

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?

Answer 1

Answer:

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.