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This question is relatively easy and I know you use trig mainly for the triangle and solve for the tension. I worked it out and got 223 N for the tension. Can anyone confirm this please?

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- Thread starter Windwaker2004
- Start date

- #1

- 34

- 0

This question is relatively easy and I know you use trig mainly for the triangle and solve for the tension. I worked it out and got 223 N for the tension. Can anyone confirm this please?

- #2

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- #4

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wait 1 sec, I'm trying to figure it out...

- #5

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Anyways, I got 392N for the tension force. I drew a vector diagram (Fg pulling down, one side of the string pulling down and to the right, and the other side of the strind pulling down and to the left.

Then I found the angle, which is equal to about 1 degree, and made a triangle with a hypotenuse of x (the tension force), an angle of 1 degree and an adjacent side of half the Fg (7.84N /2). I got 392N for x.

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Ok i'll try to redo it and see what i get

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- #11

Doc Al

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- #12

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- #13

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196 is the tension in each side of the cord.

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yea that 88.85 degrees would be to the vertical so it still works.

- #16

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Alright thanks again!

- #17

Doc Al

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The tension is uniform throughout the rope. So there is no meaning to "total" tension. You are confusing the tension (a property of the rope) with the force that the rope exerts on the shirt: which gets a contribution from both sides of the rope that help support the shirt.Windwaker2004 said:

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We never assumed the length of the cord changed.

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True but if u calculate the hypotenuse using the sides 4 m and 0.08 m there is a little difference.

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I guess it is negligible because it is such a small difference.

- #22

Doc Al

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I make the reasonable assumption that the rope stretches before the poles move closer. But something's got to give.Windwaker2004 said:

For small sag it won't make much difference.

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- #24

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Yea I agree. It makes sense now lol. OK thanks to the both of you!

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no problem

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