Tension of a rope between two trees

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SUMMARY

The tension in a rope of mass m hanging between two trees, with ends at the same height and forming an angle θ, is calculated as T = (mg)/(2cos(θ)). To determine the tension at the midpoint, a free body diagram should be utilized, summing forces in both x and y directions. The discussion emphasizes that while integrals and complex calculus may provide deeper insights, they are not required for solving this specific problem.

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Elphaba123
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1. Suppose a rope of mass m hangs between two trees. The ends of the rope are at the same height and they make an angle ! with the trees.



I believe the tension at either end of the rope is T=(mg)/(2cos(theta)) but I don't know how to solve the for the middle. The only way I can think of doing it is by using integrals but my teacher told me we won't be using integrals till later in the year.
 
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How did you get the answer at the ends? The same working may help for the middle.

If you haven't already, try drawing a force diagram for just, say, the left half of the rope.
 
To find the rope at the end, I drew a free body diagram. From that I found the sum of the forces to be Tcos(theta)-(1/2)mg=0 because there is no movement. From this I found the tension. The problem is if I use this way I end up with the force in the middle being zero which I don't think is true.
 
Elphaba123 said:
To find the rope at the end, I drew a free body diagram. From that I found the sum of the forces to be Tcos(theta)-(1/2)mg=0 because there is no movement. From this I found the tension. The problem is if I use this way I end up with the force in the middle being zero which I don't think is true.
As Modulated suggests, draw a free body diagram of the left half of the rope, and sum forces in the x and y directions to solve for the unknown tension at the middle of the rope.
 

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