Chopping Down A Tree

The image below shows the limiting case

mm

Torque about the vertex which contains $\theta$ is zero.

So, $mg cos(\theta) \times 10 = mg sin(\theta) \times L \implies tan (\theta) =\frac{10}{L} =\frac{0.5}{12} \implies L=240cm= \boxed{2.4m}$

The tree begins to topple when the center of mass is not aligned vertically on the base of the tree. The critical angle $\theta_{crit}$ is the angle in which toppling hapens. To find the critical angle, extend an imaginary line vertically upwards from the outermost part of the base, until it intersects the axis of the tree trunk. The intersection point forms a right triangle with the base's right side and the midpoint of the top cut of the base. Let the angle formed by the cut be $\theta$ The angle of inclination of the tree trunk is also equivalent to $\theta$ . Applying geometry concepts, we find that $\theta_{crit} = \theta$ , hence, the right triangle formed between the cut and the edge of the chainsaw blade is similar to the right triangle formed earlier, with $\theta_{crit}$ as one of its angles. We treat $L_{min}$ as the length of the latter triangle's legs. Solving for the ratio $0.5:12 = 10:L_{min}$ , from the similar triangles proportion, we arrive with $L_{min} = 240 cm = \boxed{2.4 m}$ .