# Tension prob

A 68.0 Kg tightrope walker stands at the center of a rope. The rope supports are 10 m apart and the rope sags 8.00 degrees at each end. The tightrope walker crouches down, then leaps straight up with an acceleration of 8.10 m/s^2 to catch a passing trapeze.

What is the tension in the rope as he jumps?

i guess i just dont understand the question in general.

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

x: 2Tsin(8)=ma=0

EnumaElish
Homework Helper
aliciaw0 said:
i guess i just dont understand the question in general.
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.

HallsofIvy
Homework Helper
Draw a picture, look at the triangles formed by the two sides of the rope and the horizontal. Since the person is not moving right or left the horizontal forces due to the two sides of the rope must balance and the vertical forces must be equal to the force of gravity plus the force necessary for him to accelerate upward.

is this right?

x: Tsin(theta)-Tsin(theta)

y: 2Tcos(8)=F+Fg

HallsofIvy
Homework Helper
aliciaw0 said:
is this right?

x: Tsin(theta)-Tsin(theta)

y: 2Tcos(8)=F+Fg

x: T sin(&theta;)- T sin(&theta;)= 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/s2,which is, of course, (68)(8.10).

mukundpa
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Pay attention...... the rope sags 8 degree at each end. Whether this angle will be with horizontal or with vertical, according to that the components will be sin theeta of cos theeta

HallsofIvy