Question

Could you consider work against friction as being torque with the radius of the earth as being the lever arm?

Answer 1

http://www.physics.wisc.edu/undergrads/courses/spring10/207/groupproblems/HW8-S10.pdf
This could probably help you; unless you already checked it out.

Answer 2

You could look at forces from friction as torques on the earth, sure (Work doesn't quite equate to torque, but you can relate them through the angle that a torque is applied through).

If we look at a box sliding on the earth's surface with friction, the frictional force is directly opposite to the direction of motion of the box. According to Newton's third law, this means that the box is pushing the earth by the same amount as the friction, but in the opposite direction as the frictional force (that is, in the direction of travel for the box).

If we approximate the earth to be a sphere, then finding the torque on the earth caused by the box stopping is relatively easy. T = F x r. Multiply this by the angle the to get the work done.

In the small angle regime (which, on Earth, can be decent distances), we can see that they are the same. Let's have a box of some mass sliding to a stop over 500m from a 50N frictional force.

Work from friction, the old fashioned way:\[50N \cdot 500m = 25000Nm\]
Let's try it with using torque on the earth:
\[\tau = r \times F = 6378100m \times 50N\]\[W = \tau\theta \]
Since 500m is way, way less than 6378100m, I'll use a few small angle tricks:
\[Tan(\theta)=\frac{Sin(\theta)}{Cos(\Theta)} \approx Sin(\theta) \approx \theta\]
and use the approximation that the earth is flat over that 500m, which allows me to define Tan(Theta) =500/6378100

Then, I just plug it all in:
\[W = 50N \times 6378100m \times \frac{500m}{6378100m} = 50N \times 500m = 25000Nm\]The same as the "old-fashioned" way.

It'll work in general, as far as I know. This was just an easy example :)

Answer 3

Woah awesome, I really didn't think it would work (or torque? =P ), just sort of one of those ideas that popped into my head. I guess now I'm a little bit confused. It seems like in a way we could consider all work to be torque done on objects from really distant points, like stars.

Answer 4

You can just use s = r theta to get rid of the "flat" approximation. I didn't think about that while writing it :)

Answer 5

Actually I wonder if it is at all related to Mach's Conjecture which is supposed to be one of the ideas behind relativity

http://en.wikipedia.org/wiki/Mach's_principle

Answer 6

You could, but you wouldn't really gain much by doing it.

It's similar to talking about linear motion in angular terms. It's true that a particle moving in a straight line has some angular momentum with respect to whatever point you wish to use, but it's not always convenient to think of it that way.

Answer 7

No, I don't think so. The earth is not that rigid of a body. If you kick sand, then the only thing really moving is he sand.