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Answer :
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.
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Final answer:
The tension in the string of the bow when the archer pulls it back with a force of 29 lb and an angle of 128 degrees is approximately 125.4 N.
Explanation:
The archer pulling back the string of a bow with a force of 29.0 lb represents a common problem in the field of Physics dealing with forces and tension. The tension in the string is equally distributed on both sides of the string from the pull point, given symmetry in the problem setup. The effective force on each half of the string is thus half the force applied by the archer, or 14.5 lb. However, since this is a 128 degree angle (not a straight line or 180 degrees) we have to adjust for this. The force acting on the string is not perfectly aligned with the string’s tension direction – it is off by a few degrees. The tension in the string is then 14.5 lb / cos(64 degrees). Converting lb to N (since 1 lb = 4.448 N), we get the tension as approximately 125.4 N.
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