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Without external forces the derivation is quite straightforward, but when external forces are involved, it becomes harder. I've been stuck on this problem for a while now so I turned to this derivation for help. Unfortunately, the only parts I do not understand are probably the most important.
I think the gist of those notes is to break up the string into many tiny "rigid" intervals and then use Newton's Law to derive a relation between the mass, acceleration and sum of all forces on that rigid piece of wire. However, when they say "we shall denote by $F(x,t)\Delta x$ the sum of the external forces on the string" I do not get why we would multiply a force by a length.
The formula given in the linked notes is, $$\overbrace{\rho(x)}^{\text{density}}\underbrace{\sqrt{\Delta x^2+\Delta u ^2}}_{\text{length}}u_{tt}=\overbrace{T(x+\Delta x,t)\sin \theta (x+\Delta x,t)-T(x,t)\sin \theta (x,t)}^{\text{sum of tension forces on ends of string}}+\underbrace{F(x,t)\Delta x}_{\text{external forces}}$$ Where $u(x,t)$ is the distance at time $t$ from the equilibrium point to the point at position $x$ on the string and $\theta(x,t)$ is the angle between the horizontal and the tangent line at position $x$ at time $t$. What I just do not understand is why the external forces get represented as $F(x,t)\Delta x$ and not simply as $F(x,t)$. Is it just to avoid problems when dividing by $\Delta x$ and letting $\Delta x \to 0$? Why is this justified, or what is the reasoning behind it?

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  • $\begingroup$ That is the notation which the author has chosen to use. $F(x,t)$ is not a force but a force per unit length. This is a question about why a particular author uses a particular notation, it is not a question about physics. $\endgroup$ – sammy gerbil Jul 13 '17 at 19:58
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The left-hand side is infinitesimal, right? So is the difference between tension forces on the right-hand side. So if the author had just written $F(x,t)$, then that would have been an infinitesimal quantity too, i.e. $dF(x,t)$ in fact. But necessarily $dF(x,t) \propto \Delta x$, and the author has chosen to denote by $F(x,t)$ that proportionality factor instead.

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  • $\begingroup$ Thank you, I did not even know this was a notational issue, but would it be problematic to interpret $F(x,t)\Delta x$ as force per unit length? $\endgroup$ – Guacho Perez Jul 13 '17 at 23:03
  • $\begingroup$ That would have the wrong unit compared to the rest of the equation: look at the left-hand side, its unit is $[\text{mass}][\text{length}][\text{time}]^{-2}\equiv [\text{force}]$. $\endgroup$ – user154997 Jul 13 '17 at 23:28

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