# How to derive the wave equation with external forces?

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?

• 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. – sammy gerbil Jul 13 '17 at 19:58

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.
• 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? – Guacho Perez Jul 13 '17 at 23:03
• That would have the wrong unit compared to the rest of the equation: look at the left-hand side, its unit is $[\text{mass}][\text{length}][\text{time}]^{-2}\equiv [\text{force}]$. – user154997 Jul 13 '17 at 23:28