- #1

danshawvassar

- 1

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## Homework Statement

A 66.0 kg tightrope walker stands at the center of a rope. The rope supports are 10 m apart and the rope sags 9.00* at each end. The tightrope walker crouches down, then leaps straight up with an acceleration of 8.40 m/s

^{2}to catch a passing trapeze.

What is the tension in the rope as he jumps?

## Homework Equations

F=ma

## The Attempt at a Solution

I literally have no idea how to solve this problem whatsoever. I was thinking that the y-component of the tension (T*sin(9) N) minus the weight of the man (646.8 N) is equal to the man's mass times his acceleration.

So: Tsin9 - 646.8 = 66 * 8.4

Or: T=7678.6 N

I don't think this is right though.