In fact, in a an environment dominated by friction, the answer might not be obvious. To find the answer analytically, recall that a rigid body is particular collection of particles. Let us decompose the force acting on particle $i$ into $\vec F_i^{ext}$ which is due to particles external to the system and $\sum_j\vec F_{ij}$, where $\vec F_{ij}$ is the force on $i$ due to $j$. Hence
$$m_i\ddot{\vec r}_i=\vec F_i=\vec F_i^{ext}+\sum_j{\vec F_{ij}}.$$
Sum over all particles,
$$\sum_im_i\ddot{\vec r}_i=\sum_i\vec F_i^{ext}+\sum_{ij}{\vec F_{ij}},$$
use the fact that internal forces cancel up (third Newton's law), $\sum_j\vec F_{ij}=0$, and use the definition of the centre of mass vector, $\vec R_{cm}$,
$$\sum_im_i\ddot{\vec r}_i=\frac{d^2}{dt^2}\sum_im_i\vec r_i=M\ddot{\vec R}_{cm},$$
where $M$ is the total mass of the system. Therefore
$$M\ddot{\vec R}_{cm}=\vec F^{ext},$$
where $\vec F^{ext}$ is the total external force acting on the system. As you can see, the centre of mass has acceleration in the same direction as the (total) external force. This video shows clearly how force and centre of mass translation are in the same direction even though the force was not applied into the centre of mass.
