There are a whole pile of issues in your question. There is a fundamental difference between binding energy and forces, and usually, thinking about the forces between nucleons is not particularly helpful. The concept of binding energy is much more powerful.
In a collection of nucleons, the force experienced by one nucleon must be the summation of the forces acting on it due to all the other nucleons. This will be a combination of the residual strong nuclear force, which to first order is the same for protons and neutrons, and the coulomb repulsion felt only between protons. These two types of force have radically different separation dependencies. Whilst the proton-proton repulsion is an inverse square law and comparatively weak, the strong-force is much stronger at small separations, is attractive and decays exponentially with a scale length of 1.4 fermis, but becomes extremely repulsive at separations less than about 0.7 fermis.
What this means is that all nuclei tend to be about as small as they can be, with the nucleons separated by about 1 fermi. They are like little hard, close-packed spheres. Each of these spheres will be bound to its nearest neighbours by the attractive strong nuclear force, but the protons will also feel a repulsive force due to all the protons in the nucleus. For a given set of neutrons and protons, the separation between the nucleons could be adjusted to achieve a force balance. If you tried to squish the nucleus, the strong nuclear force would become repulsive and it bounces back. If you try to expanded the nucleus the coulomb repulsion would get weaker bringing the nucleons back towards an equilibrium. What you imagine is that the nucleons sit in a potential well. The equilibrium point between the forces is at the bottom of the well where the gradient in potential (i.e. the net force) is zero. The depth of the well corresponds to the binding energy and is the sum of the binding energy terms due to the strong nuclear force, the coulomb repulsion (and also other items like whether the nucleons are paired in energy states or whether there is symmetry between the numbers of protons and neutrons).
These considerations basically apply to all nuclei. There is usually some sort of equilibrium to be found between the forces, but this does not equate to stability and this where I think your book is leading you astray. Instability arises because there is a feasible way of rearranging the nucleons (i.e. changing a proton to a neutron or vice-versa) or ejecting a nucleon (or alpha particle) in order to reach a configuration that has an even deeper potential well. A general principle in physics is that systems will seek to minimise their energy density if there is a means of doing so. Similarly, if there are nucleons or particles in the vicinity of the nucleus, it may be possible to capture or interact with these in order to produce a new nucleus with an even deeper potential well.
So radioactive decay is really not best thought of in terms of a competition between forces. For instance, a nucleus will beta decay (a process governed by the weak force, a much shorter range force which is not a factor in the forces between the nucleons), because by changing a neutron to a proton the resulting nucleus has a larger binding energy (deeper potential well).