Where will friction act in rolling motion if the lines of external force pass through both COM and outside? It turns out that if the line of external force passing through center of mass, then friction acts backwards the force applied:

And if the line doesn't, then then friction acts in same direction:

My question: Suppose we had two external forces, one which had line passing through center and other not passing through center, what direction would friction act?
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 A: The idea is that the direction of the frictional force will be whatever it needs to be to enforce the rolling condition
$$a=R\alpha$$
where $a$ is the linear acceleration of the object, $R$ is the radius and $\alpha$ is the angular acceleration.
So let's set up a problem with two forces. $F_1$ is acting through the center as in your Fig 1, and $F_2$ is acting at the top of the object as in Fig 2. We will choose friction $f_s$ to be pointing as in Fig 1, which is fine because if it is really pointing the other way we will just find it to be negative.
So from the sum of the torques ($I$ is moment of inertia)
$$F_2 R + f_s R = I \alpha$$
and from the sum of forces ($M$ is total mass)
$$F_1+F_2-f_s = M a$$
so combining the two
$$F_1 + \left(1-\frac{MR^2}{I}\right)F_2 = \left(1+\frac{MR^2}{I}\right)f_s.$$
$I$ is some fraction times $MR^2$ (since not all the mass is at radius R), so we know
$$\left(1-\frac{MR^2}{I}\right) <0$$
This means that if there is no $F_1$ then $f_s$ must be negative, which is the premise of your question. If there is a $F_1$, then the result is if
$$F_1 + \left(1-\frac{MR^2}{I}\right)F_2 >0$$
then $f_s$ acts in the opposite direction as in Fig 1, and if it is less than zero then it acts in the same direction as in Fig 2
A: There is a middle ground location for force application called the center of percussion where the friction force will be zero. For a solid homogenous sphere it is located at a height $\frac{7}{10}D$, where $D$ is the diameter of the sphere.
