Converting Effective Mechanical Load to Newtons (Capybara)

[enigmaticbacon]

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?

 

[enigmaticbacon]

I know that EMA is = F_load/F_effort

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.

 

[berkeman]

Don't you already have a thread going on this set of questions?

 

[enigmaticbacon]

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.

 

[berkeman]

No worries. Please keep this discussion in your original thread. Otherwise it gets too confusing and fragmented for others to keep up. Thanks.

 

[berkeman]

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.