The halves of the string make 35 degrees with the vertical. What is the tension in the string? What is the tension in the string?

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I think the top answer is wrong, the second way gets the correct answer but goes about it in a strange way (the arrow is being pulled back horizontally..?)

Here is my contribution:

The angles below are not correct but hey, a diagram always helps

Tension

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\_______ 90N

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/

Tension

Each part of the string has a Tension contributing to keeping the forces in equilibrium. Vertically, T cos35 = T cos 35 , which doesn't help.

Solving horizontally: T cos 55 + T cos 55 = 90

2 T cos 55 = 90

T cos 55 = 90 /2 = 45

T = 45 / cos 55

T = 78N

The tension in the string is the tension at every point in the string, so it is not necessary to assume that the arrow is notched at the center. However, we must assume it is notched at some particular point to solve the problem. Without the notch point being specified, there is not enough information to solve the problem. However, since the problem refers to "halves of the string", it is reasonable to assume that the arrow is notched in the middle of the bowstring.

Pulling back on the arrow and bowstring with a force of 90 Newtons and holding the arrow in that position, there is an equal and opposite vertical component of the tension in the bowstring. Each half of the bowstring contributes 45 N to the vertical component of the tension.

45/T = cosine (35 degrees) where T is the tension in the bowstring.

T = 45/cosine (35 degrees) = 45/.8194 = 54.91 N

Assume the arrow is notched at the center of the bowstring. Thatis where the forces are in equilibrium and all forces add up to zero.

The tensions in each half of the bow string each have a horizontal and a vertical component.

Since the arrow is in the center and both haves are mirror images of each other, the tensions in each half are equal.

Since the point of equilibrium is in the center it is also true that both horizontal components are equal, and since there are no other horizontal forces acting on the arrow and it is not accelerating they must add to zero.

∑ F hor = 0

All the forces in the vertical at that point must also add to zero since there is no acceleration in that direction either. I just assumed the arrow was still drawn and not loosed.

There are 3 forces there. The two equal vertical components of the 2 tensions (1 for each half of the bowstring) and the downward 90 N pull the shooter makes.

The 2 sides of the bowstring make a right triangle with the arrow and the hypotenuse is the tension in the string and is T sin(35). The vertical forces add to zero so you have:

∑ F vert = 2•T•sin(35) - 90 N = 0

solving for T you get:

2•sin(35) = 90 N /T ==>> T = 90 N / (2 sin(35) = 45 N / sin(35)

Just don't forget; what goes up straight, comes back down straight ... on your head;)

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