A very Inquisitive question - Finding Force of Tension

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SUMMARY

The discussion focuses on calculating the forces of tension in a rope system involving a person jumping onto a tight rope. Participants clarify that the problem requires solving for two separate tensions: T1 and T2, corresponding to the segments of the rope measuring 5m and 12m, respectively. The key equations referenced include Newton's second law (F = ma) and the equilibrium of forces in both the y-direction (T1y + T2y = mg) and the x-direction. The confusion arises from the wording of the problem regarding the number of ropes and the distances 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 because 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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