Magnitude of the tension on the ends of the clothesline?

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Homework Help Overview

The problem involves a horizontal clothesline tied between two poles, with a mass causing it to sag. The subject area includes concepts from mechanics, specifically tension and forces in equilibrium.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the tension in the clothesline, questioning whether it can be simplified to just the weight of the mass. They explore the need to account for both horizontal and vertical components of the forces involved.

Discussion Status

There is an ongoing exploration of the relationships between the components of tension and the weight of the mass. Some participants have offered insights into the vector nature of tension, while others are clarifying assumptions about the setup and calculations.

Contextual Notes

Participants are navigating the complexities of the problem, including the geometry of the situation and the need for accurate component analysis. There is a mention of validation issues with an external website, indicating potential constraints on the problem-solving process.

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A horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 2 kilograms is tied to the middle of the clothesline, it sags a distance of 3 meters.

What is the magnitude of the tension on the ends of the clothesline?

Isnt the tension just 2g Newtons? or for just one end of the clothesline g Newtons? The website where i enter these problems is not validating these answers. I don't know why. I also did the x componets, the y componets, and the negative of all of that, the answers are never valid. I have no clue why.
 
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You must also account for the horizontal componet of force. Draw some triangles.
 
the horizontal componet and the vertical componet adds up to the weight, which is 2(9.81)N. isn't that the tension?
 
No, the vertical component of each of the two ends of the clothesline is equal to .5*1*9.8 N. Consider that the direction of the tension is not directly up, but up and sideways because it must be in the same direction as the line. The net tension at one end will therefore be the vector sum of the horizontal and vertical tensions.
 
shouldnt the vertical componet of each pole be: 1*9.8*sin[tan^(-1)(3/7)]?

how did you get .5*1*9.8?
 
The vertical components of each of the two must add up to the weight of the hanging mass, which is downward.
 
ok, i got it, thanks
 

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