Why we rip the way we do

We first simplify by assuming the natural length of the springs is 1 m and that we stretch the springs slowly towards the breaking point. In this case, one can treat the ends of the springs as moving at a constant velocity and hence with zero acceleration. For the case of the horizontal forces, the force balance on the right node is $F_{Net}=0=F_1-k\Delta_m-k\Delta_m/2$ where $\Delta_m$ is the amount the middle spring has stretched (NOT the length of the spring). Hence $F=3k \Delta_m/2$ .

Similarly, one can do a force balance for the vertical force case. However, one must consider both x and y directions and do a little geometry first. Center a coordinate system on the middle spring (so the origin is in the center of the middle spring) and denote the horizontal position of the right end of the middle spring by $1/2-x$ . The y displacement will be denoted by y. Since the middle spring will snap if it only stretches by 0.01 m, the displacements are small relative to the length of the springs. The angles the springs make with the horizontal are also small and we can use the small angle approximation ( $\cos~\theta \approx 1, \sin~\theta=\theta$ ). The total stretch of the middle spring (remember that it's stretching on both 'sides') is $\Delta_m=2y^2-2x$ and the total stretch of the right spring is $x+y^2/2$ .

The horizontal force balance and the small angle approximation allows us to relate x and y, yielding $2x=y^2$ . The vertical force balance allows us to find N as $N=3k \Delta_m^{3/2}$ . $F/N$ is therefore $F/N=(2 \Delta_m^{1/2})^{-1}$ . At snapping $\Delta_m=0.01$ , which yields $F/N=5$ .