Physics Asked by Bunji on January 21, 2021
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
$$
F_{mb} = mg sintheta = mgfrac{Delta x}{L},
$$
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?
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 : $$FLcostheta=mg(frac12Lsintheta)+Mg(Lsintheta)$$ 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.
Correct answer by sammy gerbil on January 21, 2021
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.
$$W=(m+m')g$$ $$F=Wtantheta$$ $$T=sqrt {F^2+W^2}$$
Answered by Sam on January 21, 2021
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