Basically...
No, there might be fundamental forces other than we know (however none have been detected so far at the LHC) but they would neither "explain" nor are required for explaining the wave-particle duality.
The behavior that came to be known as the wave-particle duality in the early days of Quantum Mechanics comes directly from the framework of Quantum Mechanics and doesn't require any further explanation, especially not at all in terms of any fundamental forces.
Details
In Quantum Mechanics, the state of a particle, in position-representation, is represented via a complex-valued function $\psi(x)$ which gives (is) the probability amplitude of finding the particle at position $x$.
Now, the momentum is represented via the operator $-i\hbar\frac{\partial}{\partial x}$. A state can have a definite momentum $p$ if and only if the state is an eigenstate of the operator $-i\hbar\frac{\partial}{\partial x}$ with the eigenvalue $p$, i.e., iff $-i\hbar\frac{\partial}{\partial x}\psi(x)=p\psi(x)$ where the real number $p$ would be the momentum of the state. Such eigenstates of the momentum operator with some momentum $p$ have the general form of $\psi_p(x)=e^{\frac{ipx}{\hbar}}=\cos\Big(\frac{px}{\hbar}\Big)+i\sin\Big(\frac{px}{\hbar}\Big)$. As you can see, this is a wave in $x$ with wavenumber $k=\frac{p}{\hbar}.$
Similarly, the position is represented via the operator $x$. A state can have a definite position $x_0$ if and only if it is an eigenstate of the position operator $x$, i.e., iff
$x\psi(x)=x_0\psi(x)$. Such eigenstates of the position operator with some position $x_0$ have the general form of $\psi_{x_0}=\delta(x-x_0)$. As you can see, this is a particle localized at $x_0$.
Conclusion
This is what the wave-particle duality intends to tell us. A quantum state can behave as a particle well-localized at $x_0$ when we measure its position but the same can behave as a wave with a wave-number $k=\frac{p}{\hbar}$ when it has a momentum $p$. We don't need any fundamental forces to explain this as elaborated.