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?

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2 Answers 2


Yes. If the string were instead a rigid iron rod suspended from one end, then even without another mass attached at the lower end you would need to apply a force to hold the rod at an angle to the vertical.

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.


Find the center of mass of the bob and string combination along the length of the string. Then simply assume that the total mass was concentrated as a bob on that part of the string.

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.

$$W=(m+m')g$$ $$F=Wtan\theta$$ $$T=\sqrt {F^2+W^2}$$


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