What is the tension in the stretched rope?

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The tension in a rope stretched horizontally between two supports, with a weight of 3160 N hung at its center, is calculated to be 7920 N. The rope sags by 35.0 cm when the weight is applied. To determine the tension, trigonometric principles are applied, resolving the downward force into components along the rope and perpendicular to it. The geometry of the situation involves a right triangle formed by the sagging rope, allowing for the calculation of the tension using the sine function.

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1. A rope of negligible mass is stretched horizontally between two supports that are 3.44 meters apart. When an object of weight 3160 N is hung at the center of the rope, the is observed to sag by 35.0 cm. What is the tension in the rope?



2. Does this have anything to do with torque??



3. ATTEMPTS:

I divided 3.44 by 2 (1.72) to find the horizontal length from one support to the object. From that I found that the length of the rope from the object to one of the supports is 1.72^2 + .35^2 = X^2 (it's a triangle), x=1.75

I can assume because the object is in the center of the rope that the tension on both sides of the rope is equal. Can I also assume the forces sum to 0 because its stagnant? BAH I'm so confused, I don't know what to do!
The correct answer is 7920. Help!
 
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You can solve this problem using trigonometry. You know the downward force. Draw a triangle resolving that force into a component along the string and one at right angles to the string. Then you can solve for the component of force along the string direction.

(BTW, future posts like these should go into Intro Physics...)
 
I'm sorry I still don't understand. I found the angle from the horizontal is 78 degrees. But when I do H*sin78=3160, I don't get the right answer.. I'm terribly confused.
 

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