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For example, for the following scenario:

A $m \;\text{kg}$ person is walking across a level bridge and stops three-fourths of the way from one end. The bridge is uniform and weighs $M \;\text{kg}$. What are the values of the vertical forces exerted on each end of the bridge by its supports?

We can solve it by considering the net torque on the bridge about the end closest to the person is $$\frac{mL}{4} + \frac{ML}{2}−FL= 0$$ which has a solution for the supporting force on the far end of F.

What's the intuition for why the exact length of the bridge is unnecessary when solving questions such as the above one?

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Imagine stretching or contracting the bridge. This changes the total length $L$ by some factor $L \to \lambda L$. Intuitively the bridge will still be in equilibrium even after this rescaling.

This implies that the equilibrium condition cannot depend on the total length of the bridge, but only on the relative position of where the various forces are applied.

EDIT (to answer the comment):

My reasoning works because the system is symmetric under scaling along the length of the bridge. This is the statement that you can only measure things that do not depend on that $\lambda$ parameter I said at the beginning. In your case this scaling also causes the position of where the force is applied $x$ to scale in the same way, thus everything only depends on the $\lambda$ independent ratio $x/L$.

This only works however given the assumptions that the bridge is rigid (can't be deformed) and so $L$ and $x$ are the only length scale of the problem. Introducing the deformations in the mix will introduce other parameters such as the stress-tensor of the bridge and the deformations that will transform with some powers of $\lambda$ and make the whole thing more complicated.

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  • $\begingroup$ Thank you! Just a quick follow-up question: as mentioned in Vincent Fraticelli's answer to this question, "it feels like a short bridge is stronger than a long bridge". I am wondering how would your method of visualization take this into account? $\endgroup$
    – Alexia.
    Jul 5 at 17:44
  • $\begingroup$ @Cheryl added a comment $\endgroup$
    – FrodCube
    Jul 5 at 18:58
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You could use a dimensional argument: the only lengths available in the problem are $x$ (position on the bridge) and $L$. The Pi theorem (Vaschy Buckingham) requires that only the ratio $x/L$ appears in the final result: $T=mg\varphi(x/L)$

This is not intuitive because it feels like a short bridge is stronger than a long bridge. But bond strength cannot be measured simply with the total force alone. If you had calculated the bending moments within the bridge, the length of the bridge would be involved. For example, imagine a rigid diving board attached to only one end: the global force at the attachment point is the opposite of the weight of the diving board $T=mg$ But the moment of actions at the point of fixation is $-Mgl/2$, the greater the longer the diving board. And intuitively, this moment plays a critical role in the risk of rupture.

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