Does a force have less translational effect if it applies torque? I was taking an edx course, and came upon the following question:
A cylinder of mass    and radius    is rotated in a V-groove with constant angular velocity  $\omega_0$ . The coefficient of friction between the cylinder and the surface is   . What external torque must be applied to the cylinder to keep it rolling at a constant angular speed?

I wasn't able to solve the problem, and when I saw the solution, I got kinda confused. As always, it starts by labeling all forces and applying Newton's second law along both axis and the torque law in the following way:
$\tau _{ext}=f_1\times R_{f1}+f_2\times R_{f2}$
$\frac1{\sqrt2}(N_1-f_1-N_2-f_2)=0$
$\frac1{\sqrt2}(N_1+f_1+N_2-f_2)=mg$
Where $f_1$ and $f_2$ are the frictional forces, while $N_1$ and $N_2$ are the normal forces. And now they just solve the system of equations.
What confuses me is the use of $f_1$ and $f_2$. I probably have some huge misconception, but how  come the friction forces are capable of both applying torque to the cylinder and counteract the other translational forces on the system? I'd think that the rotational effect they have on the cylinder would decrease the translational effect, and vice versa. Instead, friction is treated as if it was both uniquely applying torque and uniquely applying force. What am I missing?
 A: 
Can a force both affect translational and rotational motion?

Yes, it can. If the force does not point along a line that goes through the center of mass of the object then it can effect both its translational and rotational motion. It is not one or the other in general.
A: In classical mechanics, the angular quantities can be entirely derived from the linear quantities
$\vec{\tau} = \vec{r} \times \vec{F} = \vec{r} \times m\vec{a} = mr^2\vec{\alpha} = I\vec{\alpha}$
$\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m\vec{v} = mr^2\vec{\omega} = I \vec{\omega}$
The insight being, the angular quantities are another way of viewing the same set of linear quantities. We simply calculate them in a roundabout fashion.
Which is to say, force and torque doesn't exchange with each other or affect each other. They are each other.
This begs the question, how do they provide two independent sets of equations in your example?
Well, they don't. You are in fact using an implicit assumption of rigid bodies, which requires that every particle in that cylinder be fixed relatively to each other. That is the implicit additional equation used to solve the problem.
