What Is the Tension in a Clothesline with a 46 N Load at Its Center?

AI Thread Summary
The discussion centers on calculating the tension in a clothesline with a 46 N load at its center, where each half of the line forms a 25° angle with the horizontal. The equation used is ΣY = FT - FG, leading to the conclusion that the tension in each half of the line is approximately 54 N. However, clarification is sought regarding whether this value represents the tension in each half or the entire line. It is explained that the tension on both sides is the same, but the overall line tension is an internal force and should not be summed. Understanding the distinction between the tension in each segment and the total line tension is crucial for accurate calculations.
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Homework Statement


A load on which the force of gravity is 46 N hands from the centre of a frictionless clothesline, pulling its centre down so that each half of the line makes an angle of 25° with the horizontal. What is the force of tension in the line?

Homework Equations


\SigmaY = FT - FG
= 2FTsin25° - FG

The Attempt at a Solution


I found that FT is equal to 46 N / 2sin25° \approx 54 N. However, I don't understand if this is the force of tension in each half of the line, or through the whole line. Can someone help me understand this?

Thanks.

Michael
 
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I think you've calculated the tension in one of the lines.
You can calculate both of them by removing the 2 from the denominator.

I approached it like this: If the weight is 46N, then each side of the line is contributing 23N to supporting it. That means that the vertical component of the tension in one side of the line is 23N, so 23/sin(25) = F_T in one side. This is clearly the same as your expression.
 
The tension on both sides of the hanging weight are the same. But you can't add them to get the line tension. The line tension is an internal force. At any point on the clothesline, the line tension is___________??
 
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