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A common statement for Newton's 3rd law reads as "for every action there is always an equal and opposite reaction". The important part which is generally missed out is that it concerns the forces exerted by two bodies on each other. This demo shows that forces exerted by two ring magnets are equal and opposite.

You need a weighing machine, two ring magnets and a PVC stand.

- Take a PVC stand and two ring magnets. Weigh the two ring magnets, say \(\mathrm{M_1}\) and \(\mathrm{M_2}\), and the stand separately. Let the weights be \(W_1\), \(W_2\) and \(W_3\), respectively.
- Put the ring magnet \(\mathrm{M_2}\) in the stand and place this stand on weighing machine pan, display of the weighing machine will show \(W_3+W_2\).
- After this put the other magnet \(\mathrm{M_1}\) in the same stand in such a way that it will be in the repulsive mode with magnet \(\mathrm{M_2}\). The magnet \(\mathrm{M_1}\) will be floating in the air having no vertical contact force with anything. Still the dial reading will be \(W_1 + W_2 + W_3\).

Although the magnet \(\mathrm{M_1}\) is floating in air i.e., it is not on weighing pan and is stationary in air (i.e., net vertical force acting on \(\mathrm{M_1}\) is zero), but the scale reading has increased by \(W_1\). That means \(\mathrm{M_1}\) is pushing \(\mathrm{M_2}\) downwards by the force \(W_1\). Now, \(\mathrm{M_1}\) is not falling so some force acts on it upwards to hold it there. This force is from magnet \(\mathrm{M}_2\) only. So, magnet \(\mathrm{M}_2\) is pushing \(\mathrm{M}_1\) by a force \(W_1\) upwards. Thus the two forces exerted by the two magnets on each other are equal and opposite.

Force diagram makes concept more clear. Since \(\mathrm{M}_2\) and stand are together, we consider them as a single body of mass \(W_2+W_3\). The free body diagram of \(\mathrm{M}_1\) and $\mathrm{M}_2$ + stand is shown in the figure. Here \(F_{12}\) is the magnetic repulsion force acting on \(\mathrm{M_1}\) due to \(\mathrm{M_2}\) and \(F_{21}\) is the magnetic repulsion force acting on \(\mathrm{M_2}\) due to \(\mathrm{M_1}\). The reaction force on $\mathrm{M_2}$+stand is \(R\). This reaction is the force measured by the weighing balance. Since \(\mathrm{M_1}\) and $\mathrm{M_2}$+stand are at rest (non-accelerating), net force on each should be zero i.e.,

\begin{align} &F_{12}=W_1, \nonumber \\ &R=W_2+W_3+F_{21} \nonumber\\ &R=W_{1}+W_{2}+W_3 \quad (\text{measurement}).\nonumber \end{align}Solve these equations to get \(F_{12}=F_{21}\). Thus force acting on \(\mathrm{M}_1\) due to \(\mathrm{M}_2\) is equal and opposite to the force acting on \(\mathrm{M}_2\) due to \(\mathrm{M}_1\).

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