On the left-hand side of the image I drew below, a pendulum bob hangs from a pendulum string of length $L$. A magnetic force of magnitude $F_{mb}$ pulls the bob to the left such that the bob equilibrates at an angle $\theta$; the bob is a horizontal distance $\Delta x$ from its equilibrium point. The magnitude of the force of the string on the bob is $F_{sb}$, and the magnitude of the weight force due to the Earth on the bob is $W_{eb}$.
Assuming I know the mass of the bob $m$ and the length of the pendulum, I can use this device to find $F_{mb}$. Using a simple 2D force-balancing approach, I get
$$ F_{mb} = W_{eb} \tan \theta = mg \tan \theta. $$ Assuming the angle $\theta$ is small, we can approximate $\sin \theta \approx \tan \theta$, and thus $$ F_{mb} = mg \sin\theta = mg\frac{\Delta x}{L}, $$ which is exactly what I need.
Now, what if the mass of the string is a significant fraction of the mass of the pendulum bob? Does this affect my expression for $F_{mb}$?
As an attempt to answer my problem, I drew the new extended free body diagram on the right-hand side of the figure below. The forces acting on the string are $F_{ps}$ (the force of the pivot on the string), $W_{es}$ (the weight force of the earth on the string, which acts at the COM), and $F_{bs}$ (the force of the pendulum bob on the string, which should be equal in magnitude to $F_{sb}$. It seems that what I want to do is to get an expression for $F_{bs}$, and then use that in the original free body diagram to solve for $F_{mb}$. The problem is that when I try to do this, everything is in terms of the unknown pivot force $F_{ps}$. How do I overcome this challenge to get a more accurate expression for the magnetic force?
