[John]

The sum of the moments around the pivot must be zero. The forward friction on the wheel and the sideways friction on the wheel are the only two forces with a component perpendicular to the rod. So

0 = forward friction * length * cos(theta) - sideways friction * length * sin(theta)

the forward friction would be F= tau/r, so solving for the sideways friction gets:

0 = tau/r *length * cos(theta) - sideways friction * length * sin(theta)

sideways friction = tau/r *cos(theta)/sin(theta)
sideways friction = tau/r * cotan(theta)

The total magnitude of the total friction force is just
F = sqrt ((tau/r)^2 + (tau/r * cotan(theta))^2)
F = tau/r * sqrt(1+cotan(theta)^2) = tau/r * (1/sin(theta))^2 = tau/r * cosec(theta)^2

Edit: should be F= tau/r * cosec (theta) (forgot to take square root)

The angle from the horizontal is
arctan((tau/r)/(tau/r * cotan(theta))) = arctan (theta) =theta

This makes sense because the friction force has to act exactly opposite to the tension force, or the wheel won't be in static equilibrium.