Say we consider a simple experiment of balancing a wooden rod on two fingers. The finger on the left, (1), will remain stationary, whereas the finger on the right, (2), will be moved toward the left. For the sake of discussion, let’s assume that the coefficient of static friction between each finger and the rod is the same, μ_S.

As finger (2) is moved to the left, the rod will need to slip with respect to one of the two fingers. At which finger with the rod slip?

We know that slipping will occur when a friction force reaches its maximum value, the value of the μ_S times the normal force. So, the answer is: the finger with the smallest normal force on the rod is where slipping will occur. The analysis below shows that the finger furthest away from the center of mass, G, will slip on the rod. The finger at which slipping occurs has a friction force of μ_KN where μ_K  is the coefficient of kinetic friction. Note that typically  μ_S > μ_K.

Once the distance between each finger and G approaches the same value, then what happens? Here it gets a little more complicated. The friction force at each finger must be the same. The normal force at the sticking finger will continue to decrease until the maximum friction force is reached, causing it to now slip, and therefore will stick at the other finger.

Animation
Consider the animation from a simulation of this problem shown below. Here the moving finger on the right starts out closer to the center of mass. As this finger moves to the left, the distance between the left finger and the center of mass decreases, eventually to the point where the normal force at the left finger has increased enough for sticking. At this point the finger on the right slips.

This cycle of sticking and slipping at the two fingers continues until the two fingers come together with the center of mass half way between the two fingers. This will ALWAYS occur, regardless of the initial positioning of the two fingers! Try it on your own.