Horizontal Force on a trapeze pivot

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SUMMARY

This discussion focuses on calculating the horizontal forces acting on a trapeze bar and determining the necessary tension in guy wires to counteract these forces. The key formula for tension in a pendulum swing is established as Tension = m*g*cosθ + m*v^2/L, where v is derived from kinetic energy as v = √[2*g*L*(cosθ)]. The analysis reveals that the maximum horizontal force occurs at a 45-degree angle, and importantly, the tension in the rope remains constant regardless of rope length, depending solely on the mass of the bob and the release angle.

PREREQUISITES
  • Understanding of pendulum physics and dynamics
  • Familiarity with tension and centripetal force concepts
  • Basic knowledge of trigonometry and angles
  • Experience with Excel for data analysis and calculations
NEXT STEPS
  • Research the effects of varying angles on pendulum motion using "Pendulum Motion Simulation" tools
  • Explore "Dynamics of Rigid Bodies" to understand forces in more complex systems
  • Study "Tension in Cables and Ropes" for applications in rigging and structural engineering
  • Examine "Kinetic Energy and Work" principles in more depth to enhance understanding of energy transformations
USEFUL FOR

This discussion is beneficial for riggers, trapeze artists, physics students, and anyone involved in the design and safety of aerial performances or similar applications requiring an understanding of forces and tensions in dynamic systems.

Nthaoe
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-sorry if I got the prefix wrong.

I'm working on calculating the forces on a trapeze bar, specifically, what tension I would need on guy wires to resists the horizontal forces on a swinging trapeze. If I could get the horizontal force generated by the the trapeze/pendulum I can do the rest.
Looking on various websites, and trying to remember my physics, I got the following formulas:

Tension on a pendulum swing = m*g*cosθ + m*v^2/L

Looking for a formula for velocity at any angle from the vertical up to 90°, I find, through kinetic energy, (and on this board),

v = √[2*g*L*(cosθ)]

When I set this up and started playing with values, I got numbers that were close to experience for speed & our experience of heaviness at the lowest point. And by setting this up on excel, and calculating the tension every 5 degrees, it seems as if the maximum horizontal force is at 45 degrees, no big surprise.

What did surprise me was when I changed the rope length, and the tension didn't change. It makes sense mathematically, if I put the velocity expression into the Tension equation, the lengths cancel out.

I've never seen this expressed in that way, that the tension on the string is not dependent on the length of the string, just on the mass of the bob, and the angle you release it at. (assuming no change in gravity). Am I getting that right?

Thanks for any help, riggers & trapeze artists will be grateful.
 
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Hello Nthaoe, :welcome:

This is a difficult situation for PF: we can't assume any responsibilities for this kind of work.

But your physics/mathematics is just fine: ##v^2## is proportional to ##L## and the required centripetal force is proportional to ##v^2/L##.
 
Thanks, people in the circus world have been guying trapeze bars for a century, this wouldn't be used to create a minimal tie down. I was just surprised when I first looked into it that I couldn't find anyone who'd done the math.
-just out of curiosity, it's fairly common to be taught that the period of a pendulum is not dependent on the weight of the bob, just on the length of the swing, but I've never seen it expressed that the tension on the string doesn't increase when you increase the length of the string, creating a higher maximum velocity for a given angle. For some reason I find that even less intuitive.

Thanks for the confirmation,
 

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