- #1

aliciaw0

- 17

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What is the tension in the rope as he jumps?

i guess i just don't understand the question in general.

y: 2Tcos(8)-Fg=68.0kg*8.10m/s/s

x: 2Tsin(8)=ma=0

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- Thread starter aliciaw0
- Start date

- #1

aliciaw0

- 17

- 0

What is the tension in the rope as he jumps?

i guess i just don't understand the question in general.

y: 2Tcos(8)-Fg=68.0kg*8.10m/s/s

x: 2Tsin(8)=ma=0

- #2

EnumaElish

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The acrobat is weighing down due to gravity which applies a downward force on the rope and creates tension. Then as he jumps, he applies additional force to accelerate himself, which creates additional tension. You are asked to calculate total tension at the moment of his jump.aliciaw0 said:i guess i just don't understand the question in general.

- #3

HallsofIvy

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- #4

aliciaw0

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is this right?

x: Tsin(theta)-Tsin(theta)

y: 2Tcos(8)=F+Fg

x: Tsin(theta)-Tsin(theta)

y: 2Tcos(8)=F+Fg

- #5

HallsofIvy

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aliciaw0 said:is this right?

x: Tsin(theta)-Tsin(theta)

y: 2Tcos(8)=F+Fg

x: T sin(θ)- T sin(θ)= 0 of course- since the two sides of the rope make the same angle the horizontal forces due to tension must be the same but oppositely directed.

y: 2Tcos(8)= F+ Fg is the equation you want to solve for T.

OF course, Fg= 68(9.81), the weight of the person. The F is the force necessary to give an acceleration of 8.10 m/s

- #6

mukundpa

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- #7

HallsofIvy

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Who is not paying attention? "sags" **means** that it makes that angle with the horizontal.

- #8

aliciaw0

- 17

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so my final equation did turn out to be

2Tsin(8)=F+Fg and that worked. Thanks! =]

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