I strongly suggest Steven H. Strogatz' book Nonlinear Dynamics and Chaos, which has several simple examples of all basic bifurcations in a few different fields. An interesting one in biology is that of fireflies synchronizing their light-flashing.
As a 1D physical example (for which the term saddle-node bifurcation seems a bit odd because saddles and nodes are really higher-dimensional fixed points but the mechanism is exactly the same in 1D), Strogatz turns to the problem of an overdamped pendulum driven by a constant torque $\Gamma$. (so a pendulum immersed in some viscous fluid like oil or honey and connected to some motor applying a constant torque to it) If $L$ is the length of the pendulum, $m$ its mass and $\theta$ the angle between the pendulum and the vertical direction, then Newton's law yields
$$mL^2\ddot{\theta} + b\dot{\theta} + mgL\sin{\theta} = \Gamma$$
where $b$ is a viscous damping factor. Now, in the overdamped limit of large $b$, the first term (the inertia term) can be neglected in comparison to the others and we get the equation
$$b\dot{\theta} + mgL\sin{\theta} = \Gamma.$$
We can simplify the analysis by transforming the problem into dimensionless variables. We can do this by dividing by a torque. A good choice here is to divide by $mgL$, yielding the following differential equation:
$$\frac{b}{mgL}\dot{\theta} = \frac{\Gamma}{mgL} - \sin{\theta}.$$
Subsequently substituting $\tau = \frac{mgL}{b}$ and $\gamma = \frac{\Gamma}{mgL}$ yields the dimensionless expression:
$$\theta' = \gamma - \sin{\theta}$$
where $\theta' = \frac{d\theta}{d\tau}$. Now it's easy to see that this system undergoes a saddle-node bifurcation as $\gamma$ varies.
- For $\gamma > 1$, $\theta'$ is never zero, meaning the pendulum keeps overturning continuously, with no fixed points around. Physically, since $\gamma$ is the ratio of the constant applied torque to the magnitude of the gravitational torque, this means gravity is never able to cancel out the applied torque completely. So we should expect no fixed points.
- For $\gamma = 1$, $\theta'$ is identically zero for $\theta = \pi/2$, meaning there is a fixed point for the pendulum hanging horizontally.
- For $\gamma < 1$, $\theta'$ has two zero's symmetrically located around $\theta = \pi/2$, meaning there are two fixed points now, one stable and one unstable. To find out which one is stable, you can consider the sign of $\theta'$ in either one, but on physical grounds it is already clear that the lower one (below $\pi/2$ so below the horizontal) is the stable one. Especially if we consider what happens if $\gamma$ decreases even further toward $0$.
- For $\gamma = 0$, $\theta'$ is just a sinusoidal function, so it has a zero at $\theta = 0$ and one at $\theta = \pi$ (inverted pendulum). Obviously, the inverted pendulum is unstable, so our conclusion for the stability of the fixed points was correct.
From the above analysis, it is clear that the saddle-node bifurcation occurs at $\gamma = 1$, where two fixed points are born (or, equivalently if we approach $\gamma \rightarrow 1^-$, where a stable and an unstable fixed point collide and mutually annihilate).