Calculating the Tension in a Tightrope as a Trapeze is Caught

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The problem involves calculating the tension in a tightrope as a tightrope walker jumps to catch a trapeze. The scenario includes a tightrope walker with a mass of 68.0 kg, a rope that is 10 m long with a sag of 8 degrees at each end, and an upward acceleration of 8.10 m/s² during the jump.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the forces acting on the tightrope walker, including gravitational force and the additional force during the jump. There are attempts to set up equations based on vertical and horizontal force balances, with some participants questioning the correct interpretation of the angles involved.

Discussion Status

Several participants are actively engaging with the problem, sharing their interpretations and equations. There is a mix of agreement and questioning regarding the setup of the equations. Some have provided guidance on visualizing the problem through diagrams and considering the balance of forces.

Contextual Notes

Participants note the importance of understanding the angle of sag in relation to the horizontal, which affects the tension calculations. There is also mention of the need to account for both gravitational force and the force required for upward acceleration in the tension equations.

aliciaw0
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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 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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aliciaw0 said:
i guess i just don't 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.
 
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
 
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/s2,which is, of course, (68)(8.10).
 
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
 
Who is not paying attention? "sags" means that it makes that angle with the horizontal.
 
yepp yepp i had to use sin to get it right

so my final equation did turn out to be

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

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