How Is Tension Calculated in an Acrobat's Wire?

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SUMMARY

The tension in the wire when an acrobat hangs from it at an angle of 9.15 degrees, with a mass of 88.4 kg, is calculated to be approximately 5518.09 N. The calculations involve resolving the forces in the vertical (Fy) and horizontal (Fx) directions using the equations of equilibrium: Sum of Fy = 0 and Sum of Fx = 0. The final tension is derived using the Pythagorean theorem, combining the vertical and horizontal force components.

PREREQUISITES
  • Understanding of basic physics concepts such as force and tension.
  • Knowledge of trigonometric functions, specifically sine and cosine.
  • Familiarity with Newton's laws of motion.
  • Ability to perform vector addition and resolve forces.
NEXT STEPS
  • Study the principles of static equilibrium in physics.
  • Learn about vector resolution and its applications in force analysis.
  • Explore free body diagrams and their importance in solving physics problems.
  • Investigate the effects of different angles on tension in similar scenarios.
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of tension and force analysis in real-world applications.

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Homework Statement


An acrobat hangs by his hands from the middle of a tightly strestched horizontal wire so that the angle between the wire and the horizontal is 9.15 degrees. If the acrobat's mass is 88.4kg, what is the tension in the wire? Answer in units of N.


Homework Equations


Sum of Fy =0
Sum of Fx =0
F= sqrt(Fx^2 + Fy^2)


The Attempt at a Solution



Fysin(9.15)-(88.4kg)(9.8m/s2)=0
Fy=5447.879N

Fxcos(9.15)-(88.4kg)(9.8m/s2)=0
Fx=877.486N

F= Sqrt(5447.879^2+ 877.486^2)
F=5518.0945N
 
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draw a free body diagram and add the forces as vectors.
 

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