Hakim, It's a bit more complicated than that... in fact, it's a LOT more complicated. In the first place, although we often call it the nuclear force, in quantum mechanics it's really the potential V(r) that's important. Force as such is not used. Secondly, we don't typically measure things in Newtons! (Or weigh particles in kilograms.) Nuclear physics has its own set of units, much more convenient on a subatomic scale.
The attraction between a proton and neutron can be described roughly by a square well potential, however the size and depth depend on the orientation of the particle spins. If the spins are parallel ("triplet state") the well depth is roughly 36 MeV with a radius of 2 fm (fermis = femtometers). This is deep enough to have a bound state, the deuteron. On the other hand if the spins are antiparallel ("singlet state") the well depth is less, more like 18 MeV and radius 2.5 fm. In this case there is no bound state.
But that's only the beginning. There's a spin-orbit force present, which varies depending on the relative motion of the proton and neutron. Plus a tensor force, which is noncentral. And everything I've said relates only to isolated particles. Protons and neutrons within a nucleus are affected by forces considerably different.
