# Physics of a TreadWheel [Abstract Concept?]

1. Oct 5, 2013

### 012anonymousx

1. The problem statement, all variables and given/known data
[EDIT] Cleaned up the problem.

A tread wheel crane is an ancient device used to lift heavy objects. It consists of a large circle (like a hamster wheel) for a person to walk inside. The large circle connects to the middle where there is a small circle that is attached a rope that goes over a pulley and hooks onto some mass.
http://www.lowtechmagazine.com/2010/03/history-of-human-powered-cranes.html

Givens:
My weight: 500N
My speed: unknown.
I stand 1.5m away from the center of both circles

What is the maximum weight I can lift?

2. Relevant equations
F = (m)(v^2)/r
v(tang) = (r)(w)

3. The attempt at a solution
I draw the two circles and placd a dot 1.2m from the center on the edge of the outer circle. The distance from the edge to the center is the radius of the big circle.

You can work out that the tangential force is mgsin(theta) given the geometry. Similarily, you can get a force directed outward mgcos(theta)

I'm thinking that F(Radial) is equal to F(outward). But I also thought perhaps you can solve v(tangential) given the tangent force...

500cos(theta) = m(v^2)/r

That is where I was able to get to, because I do not know what mass is in this context. Nothing is really spinning around. Maybe my whole approach is wrong. I would greatly appreciate insight and help!

Last edited: Oct 5, 2013
2. Oct 5, 2013

### rcgldr

v^2/r is normally associated with centripetal acceleration. What is wanted in this case is the tangental force at the radius of the small circle.

3. Oct 5, 2013

### 012anonymousx

Yeah, I figured that after I thought about it a lot.

So F = ma, and we have the tangental force at the radius of the bigger circle.

But f = ma, and once again, we don't have m and we shouldn't even have a. The circle should not be accelerating. The angular velocity anyway.

4. Oct 5, 2013

### rcgldr

You have the weight that is supposed to be lifted by a rope attached at the radius of the inner wheel.