A very Inquisitive question - Finding Force of Tension

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The discussion revolves around calculating the forces of tension in a rope system, specifically addressing a scenario where a person jumps onto a tight rope. Participants express confusion about the problem's wording, particularly regarding whether there is one rope or two and the correct interpretation of distances. It is clarified that the problem involves solving for two separate tensions in the rope segments of 5m and 12m, as the person is not positioned in the middle. The conversation emphasizes the need to consider both the y-components and x-components of the forces to apply Newton's second law effectively. Overall, the complexity of the problem lies in its non-central configuration, which requires careful analysis of the forces involved.
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Homework Statement



Problem in Picture, I'm looking for the two forces of tension in the string, I expect the y-components to add to equal the force of gravity

Homework Equations


I used simple trig ratios to get the distances and angles.
Mainly using Newtons second law F = ma

The Attempt at a Solution


here is a link at my attempt, I'm simply stumped beyond this point.

http://ioj.com/v/ytyqh

website is an image uploader
 
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help is greatly appreciated !

I'm helping out others in the meantime here
 
Sorry, I am confused. The question first says "the rope", and then it says "two ropes". Are there just one or two? And it also says "5 m from on the rope from one side of the building". Shouldn't it be the horizontal distance of the person from the building and not the length of the rope from the building to the point where the person stands?
 
benhou said:
Sorry, I am confused. The question first says "the rope", and then it says "two ropes". Are there just one or two? And it also says "5 m from on the rope from one side of the building". Shouldn't it be the horizontal distance of the person from the building and not the length of the rope from the building to the point where the person stands?

Sorry for the wording, it must have caused the confusion. I don't have the question on hand so I am paraphrasing it as best as i Can. It says that a person (unrealistically) jumps onto a tight rope 5m away from its start ( total distance of rope is 17m, hence the 5m and 12 m sides ).

So yes i believe you should be solving for two separate tensions ( of one rope ) , the tension in the 5m length and the tension on the 12 m length
This question is a little tricky only becuase it is not in the middle.

this is really getting to me
 
Ok, according to your diagram, you have the data for the y-direction: T_{1y}+T_{2y}=mg

How about the x direction?
 
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