# A very Inquisitive question - Finding Force of Tension

• Weebay
In summary, the conversation is about a problem involving finding the two forces of tension in a string, with the expectation that the y-components will add up to equal the force of gravity. The person attempting to solve the problem used simple trigonometric ratios and Newton's second law, and provided a link to their attempt. However, there is some confusion about the wording of the question and the person asking for help is trying to clarify the details. They mention that the question involves a person jumping onto a tight rope 5m away from its start, and that there are two separate tensions to solve for. The conversation ends with the person expressing frustration with the problem.

## 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 [Broken]

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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.

:grumpy: this is really getting to me

Ok, according to your diagram, you have the data for the y-direction: $$T_{1y}+T_{2y}=mg$$