A little integral calculus helps you out here. You can add up small contributions to the total required torque from small portions of the wheel's width depending on the frictional force and the radius, take the limit, and compute the integral.
Spoiler for I guess I need a title:
We'll split the wheel in half to make it easier, starting from distance r=0 to r=W/2 from the wheel pivot point, with W the wheel width (assuming wheel pivot is in the middle, it's easy to compensate otherwise).
At each dr chunk, the normal force there is the total normal force (equal to the load, unless you're moving in and out of the ground...) on the wheel scaled to the size of dr, or L*dr/W, where L is the load on the wheel, dr is the width of the small, and W is the total wheel width.
The frictional force then is the normal force times u, the coefficient of friction, or uLdr/W. The torque is the force crossed with the radius, so we have the torque for each chunk as uL/W * rdr, which we integrate from 0 to W/2 and multiply by 2 to get our final expression, T = uLW/2.
This assumes the contact area between the wheel and the floor can be described as a line. You could integrate over a rectangular contact if you knew roughly how big that was and would find a slightly larger answer, due to the larger moment. This also assumes uniform distribution of the wheel load over the contact area.
Any calculation you might do on this has a good chance of being off by such a large margin that it is not practical to actually use it in sizing your system. You would probably get a far more usable result by measuring a physical test if you can arrange it. Build a pivot, load the wheel, torque it with a lever, measure the force, calc the torque.
Just put it in the dryer for a few extra cycles. That should shrink it down nice and small. - notmattlythgoe [more] 
13-07-2012, 12:24
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