Newton's Third Law Problem - Tightrope

[danshawvassar]

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/s² 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.

 

[tiny-tim]

welcome to pf!

hi danshawvassar! welcome to pf!

(award yourself a degree: ° )

when he straightens his legs, his centre of mass accelerates upward, so there must be a downward force by his feet on the rope, and an equal upward force by the rope on his feet

that is the force you need to find, and then find the (extra) tension that will produce that upward force