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.
