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Why are the force vectors equal?
It’s fairly obvious to me that the diagrams are drawn to make the force vectors appear equal:
I have a few questions:
I’m not sure if it’s relevant to my post or not, but I’m not looking for an answer.
What would happen if the force vectors on each side of the light were not equal?
A:
It is a property of Newton’s laws, when written in the form
$$\begin{array}{cccc}
F &=& m\ddot{x} \\
\Delta L &=& F \cdot \Delta s \\
E_t &=& \frac12 m\dot{v}^2 + U(x)\end{array}$$
One can always translate a system in such a way that one sees the end forces equal, without affecting the trajectory of the system. You can even see the result $F = F’$, while $F
e F’$.
A:
If you write Newton’s second law
$$m \dot{\mathbf{v}} = \mathbf{F}$$
you will see that the two force vectors turn out to be the same even for different mass $m$ or different particle velocity $\dot{\mathbf{v}}$ (as long as the masses or velocities are normalized).
So in reality, it is just a conveneince to simplify the drawing.
On the other hand, if you start to “multiply by mass” the equation $\Delta L = F \Delta s$ and then use $m \dot{\mathbf{v}} = F$, you will see that $m\dot{\mathbf{v}}$ will be equal to $\mathbf{F}$.
For the OP question, any matter system (your body, or any pendulum) will be in mechanical equilibrium, with $\mathbf{F}=0$ for all the atoms.
A:
If you plot both sides of the second