Converting Effective Mechanical Load to Newtons (Capybara)

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SUMMARY

The effective mechanical advantage (EMA) of a capybara is determined to be 0.71, with a mass of 55 kg and an average speed of approximately 3.0 km/h. To calculate the force output in Newtons, the formula EMA = Fload/Feffort is utilized, leading to a force output of around 700 N per capybara. To overcome a frictional force of approximately 125,000 N, a specific number of capybaras must be calculated based on their individual force output. This discussion highlights the need for clarity in converting EMA to force output and emphasizes the importance of maintaining coherent threads for effective communication.

PREREQUISITES
  • Understanding of effective mechanical advantage (EMA)
  • Basic knowledge of force calculations in Newtons
  • Familiarity with the physics of friction and normal force
  • Ability to rearrange and manipulate algebraic equations
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  • Research the relationship between EMA and force output in mechanical systems
  • Learn about calculating friction and normal force in physics
  • Explore the concept of mechanical advantage in animal locomotion
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Students of physics, animal behavior researchers, and anyone interested in mechanical advantage calculations in biological systems.

enigmaticbacon
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TL;DR
I have the EMA of the average capybara, and I'm trying to figure out the maximum force one can exert given that EMA.
Hello again!
I've found the capybara's EMA to be 0.71. Their mass to be 55kg. And their average speed to be ~3.0km/h.

I want to figure out how many capybaras it would take to overcome Friction * Normal force of ~125,000N. How would I go about doing that?

https://www.quora.com/How-many-capy...wheeled-chariot-across-the-Bolivian-altiplano

The answer of a Quora question states:
EMA of a capybara is around 0.7. That would give us a force output of around 700N.

But I have no idea how they converted from EMA to a force output in Newtons. Can anyone help me?
 
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I know that EMA is = Fload/Feffort

I thought I could rearrange the variables to solve for the Force of Effort. But that obviously didn't work. A capybara cannot pull 88750N.
 
Don't you already have a thread going on this set of questions?
 
I do, but I figured out some stuff since and it felt like a different question? I assumed for a different question, I'd make a different post :) Sorry, new to this forum.
 
No worries. Please keep this discussion in your original thread. Otherwise it gets too confusing and fragmented for others to keep up. Thanks. :smile:
 
Oh, and please define the acronym EMA unless it's obvious in your other thread (I don't remember), and if EMA has units, please include those. Thanks. :smile:
 
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