# Tension on a rope from jumping tightwalker

1. Mar 3, 2009

### fanman791

A 65.0 kg tightrope walker stands at the center of a rope. The rope supports are 10 m apart and the rope sags 10.0^\circ at each end. The tightrope walker crouches down, then leaps straight up with an acceleration of 7.70 {\rm m}/{\rm s}^{2} to catch a passing trapeze. What is the tension in the rope as he jumps?

F=ma
Forces, basically.

So, based on the information, I think it's correct to assume because of the tightwalker jumping upwards then the Force from the tightwalker is slackened to 9.8(gravity) - 7.70(acceleration upwards).
From there: F=Fysin($$\theta$$)+Fysin($$\theta$$) - ma (from the tightwalker) = 0

I plug it in and get 180.71 and it's not the correct answer. I'm doing something wrong here.

2. Mar 3, 2009

### Imperitor

"because of the tightwalker jumping upwards then the Force from the tightwalker is slackened"

They want the tension as he jumps, before he leaves the rope.
Hint: The tightrope walker pushes down to jump up, putting more tension on the rope.

3. Mar 3, 2009

### BishopUser

Maybe I am wrong, but I think the 7.7 m/s^2 that is given is the absolute acceleration of the person. So, the acceleration and mass are both given, which makes it a simple calculation. Using those numbers do you get the correct answer?

4. Mar 3, 2009

### fanman791

ty Imperitor i got it

it doesn't slacken as he jumps - it adds more force to the equation
therefore, it = ma(jumper) + m(7.7)(jumper adding force)

and if i plug those in it yields the correct answer :D