Calculating the Tension in a Tightrope as a Trapeze is Caught

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In summary, the tightrope walker, weighing 68.0 kg, crouches down and then jumps with an acceleration of 8.10 m/s^2 to catch a passing trapeze. The rope supporting him sags 8.00 degrees at each end and is 10 m apart. To calculate the tension in the rope as he jumps, we use the equations y: 2Tcos(8)=F+Fg and x: Tsin(theta)-Tsin(theta)=0, where Fg is the force of gravity and F is the force necessary for the acceleration. By using the angle of sag and trigonometric functions, we can solve for T and find the tension in the rope.
  • #1
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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  • #2
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.
 
  • #3
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.
 
  • #4
is this right?

x: Tsin(theta)-Tsin(theta)

y: 2Tcos(8)=F+Fg
 
  • #5
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).
 
  • #6
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
 
  • #7
Who is not paying attention? "sags" means that it makes that angle with the horizontal.
 
  • #8
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! =]
 

1. How do you calculate the tension in a tightrope when a trapeze is caught?

The tension in a tightrope when a trapeze is caught can be calculated using the formula T = (m x v^2)/r, where T is the tension, m is the mass of the trapeze artist, v is the velocity of the trapeze artist, and r is the radius of the tightrope.

2. What factors affect the tension in a tightrope when a trapeze is caught?

The tension in a tightrope when a trapeze is caught is affected by the mass of the trapeze artist, the velocity of the trapeze artist, and the radius of the tightrope. Other factors that may also have an impact include the elasticity of the tightrope, wind conditions, and the strength of the trapeze artist's grip.

3. Can the tension in a tightrope ever be greater than the weight of the trapeze artist?

Yes, the tension in a tightrope can be greater than the weight of the trapeze artist. This can happen if the trapeze artist is moving at a high velocity or if the tightrope has a small radius. In these cases, the centrifugal force acting on the trapeze artist can increase the tension in the tightrope beyond their weight.

4. How does the tension in a tightrope change as the trapeze artist swings?

The tension in a tightrope will vary as the trapeze artist swings due to changes in velocity and radius. As the trapeze artist swings higher, the tension will increase due to the increase in velocity and centrifugal force. As they swing lower, the tension will decrease as the velocity and centrifugal force decrease.

5. What safety precautions should be taken when calculating the tension in a tightrope for a trapeze act?

When calculating the tension in a tightrope for a trapeze act, it is important to ensure that the tightrope is properly secured and able to support the weight and movements of the trapeze artist. Safety nets and harnesses should also be used as a precaution in case of any accidents. It is also important to regularly inspect and maintain the tightrope and all equipment to ensure they are in good condition.

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