How is torque transmitted between inclined surfaces? In the picture below, in a), a body K1 is pivotably attached to a bearing. My question is about the torque that results from a force exerted onto a surface of the body K1.
A first force F1 applied orthogonally onto the surface should result in a torque M1 in clockwise direction.
Is it correct that a second force, F2, applied almost parallel to the surface will result in a torque M2 in counterclockwise direction?
My thoughts are, F2 is split into F2t and F2o (transversal and orthogonal components) by the surface of the body K1. To get a torque, F2o is multiplied by the lever b and F2t is multiplied by the lever a (M2 = F2t * a - F2o * b > 0). As a>b and F2t>F2o, the torque from the force F2 results in counterclockwise direction.
Applying these thought to the two bodies K1, K2 in b), a torque of M3 applied to the body K2 will result in a torque M4 in the body K1. (The bodies won't move because they are in each others movement path)
Is this correct or am I forgetting something? What is the job of friction in this case? From looking at b), K2 should push K1 away by applying a clockwise torque, but that is wrong then, right?
Suppose there is enough friction so that no slippage occurs.

 A: I think friction is required for any torque to be applied CCW, which be definition works against torque being applied CW. So with no friction it would be net torque CW, but with "infinite" friction (i.e. no slipping) it would be net torque CCW (and also locked up and not spinning). I don't think this question can be solved without some assumption or knowledge about the friction involved.
A: Don't wrap yourself around in pretzels. Even for planar cases, assume they are defined in 3D (with z-axis out of plane) and use the cross product to define torque
$$ \vec{\tau} = \vec{r} \times \vec{F} $$
which expands to
$$ \pmatrix{ 0 \\ 0 \\ \tau_z} = \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix} \pmatrix{F_x \\ F_y \\ 0} $$
and projects into 2D as
$$ \tau_z = -y F_x + x F_y $$
The nature of force does not matter. Use the combined normal and frictional forces to find the net torque, or just an individual component to gauge the effect of on the body.
The same goes for moment of impulse in case you have contacts.
$$ \vec{\gamma} = \vec{r} \times \vec{J} $$
