# Does the mass of a pendulum string affect the amount of force required to hold it at a specific location?

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?

The horizontal force $$F$$ that you need to apply to hold the pendulum in static equilibrium can be found from balancing moments around the suspension point : $$FL\cos\theta=mg(\frac12L\sin\theta)+Mg(L\sin\theta)$$ where $$m, M$$ are the masses of rod and bob.
A flexible string which has non-negligible weight will hang in a curve called a catenary, not in a straight line. The same method can be used to find $$F$$. However the position of the centre of mass of the string is not so easy to calculate as for a straight rod.
Let the original bob have mass $$m$$, string have mass $$m'$$, tension be $$T$$ and magnetic force be $$F$$. The constant in this question is the angle made by the string.
$$W=(m+m')g$$ $$F=Wtan\theta$$ $$T=\sqrt {F^2+W^2}$$