Why is it easier to push down a rope from the middle compared to the side? Intuitively, if I push down on a rope at the middle, I know that it will sink down farther than if I were to press it at the sides.
Below I drew a diagram of a person standing on a tightrope to give an example of this.

Why does this happen?
 A: This is the tremendous power of (pick of your favorite) trigonometry, free-body diagrams, force components, and leverage. There’s not a stretched rope or string or cable in this world that won’t yield substantially for a slight push near the middle, no matter how high the tension is cranked. In turn, given a sufficiently light, long, stiff, and strong rope, with no other equipment, I could easily move a freight train up an incline—and I'd certainly want to push near the rope's midpoint.
The reason is that flexible ropes can’t sustain a torque; they bend and fold easily. Consequently, the only load they can really bear is tensile. In addition, all applied forces and moments on a static object must sum to zero; otherwise, the object would start to accelerate.
So we have a stretched rope that we apply a lateral load to. The rope must deflect somewhat; there’s no other way for the reaction forces at the walls to provide a counteracting lateral force. What’s more, the small push is boosted by a factor related to $1/\sin\theta$ applied to the tension as a result of static equilibrium and trigonometry, providing enormous amplification of the force on the inward loads on the walls.
The amplification is largest if the displacement occurs in the middle, so this is where a lateral load is most effective.
On to the mathematics. Consider a rope of unit length attached to rigid supports, and assume that we push on it with force $F$ at position $l$ (where $l$ can range from 0 to 1) to obtain deflection $d$. The included angles on the left and right sides are $\theta_\mathrm{L}$ and $\theta_\mathrm{R}$, respectively, and the rope stretches a little on the two sides by $e_\mathrm{L}$ and $e_\mathrm{R}$.
The Pythagorean theorem tells us that
$$l^2+d^2=(l+e_\mathrm{L})^2\approx l^2+2le_\mathrm{L};$$
$$ (1-l)^2+d^2=(1-l+e_\mathrm{R})^2\approx (1-l)^2+2(1-l)e_\mathrm{R},$$
where the approximations are Taylor series expansions for small $e$.
Trigonometry tells us that $$\sin\theta_\mathrm{L}=\frac{d}{l}\qquad \sin\theta_\mathrm{R}=\frac{d}{1-l}.$$
Hooke's Law tells us that $$T_\mathrm{L}=\frac{Ee_\mathrm{L}}{l}\qquad T_\mathrm{R}=\frac{Ee_\mathrm{R}}{1-l},$$
where $T$ is the tension and $E$ is the rope stiffness (in terms of force per unit strain).
Equilibrium requires that the forces balance, so in the lateral direction, $$F=T_\mathrm{L}\sin\theta_\mathrm{L}+T_\mathrm{R}\sin\theta_\mathrm{R}.$$
Put this all together, and we obtain
$$d=\left[\frac{2F}{E}\left(\frac{l^3(1-l)^3}{l^3+(1-l)^3}\right)\right]^{1/3},$$ which is zero at $l=0$ and $l=1$ and maximized for $l=\frac{1}{2}$. Does this make sense?
A: The simple answer is torque. Exerting a downward force on the middle creates the maximum sum of torques on both ends, thus causing the rope to bend more.
