Lumberjack physics

The pdf solution is posted at this link along with the figure

https://imgur.com/BzIZFph

We calculate the angle of displacement for the trunk. Using a .5-12 legged right triangle we solve for this angle. Using this angle and looking at the trunk, we know the tree will tip iff the center of mass lies past the furthest point of support so we can calculate L from the triangle by projecting a vertical line up from this point of support. L = 10/tan(2.4) ~ 240 so the answer is 2.4 meters.

Once the tree is tilting just B4 being toppled, the C.M. must be exactly above the hinge-point at the right of the drawing. At this position the tree is listing at an angle which has a tangent of 0.5/12(Pic.). The axe of the tree(on which the C.M is) makes too the same angle with a vertical line. In that situation it is not hard to show, that condition of the C.M. above the hinge point will require Lxtg(A)=0.1m Where A is the listing angle with tg(A)=0.5/x12

So from here L=0.1/tg(A)=0.1/(0.5/12)=2.4m

The perfectly symmetrical cylindrical tree has a uniformly distributed density in which the center of mass is in the middle of the tree. The chainsaw is forming an angle with the horizon in which this angle can be represented by $tan(\theta)=\frac{opp}{adj}$ . The $opp$ is the height of chainsaw, $0.5cm$ and the $adj$ is the width of the tree minus the length of the chainsaw hence is $12cm$ . Therefore $tan(\theta)=\frac{0.5}{12}$ , the length $L$ is the distance from the centre of the tree trunk to the center of mass, as the tree tips over the angle formed from the vertical will be equivalent to the angle that is lifting it. Hence the angle $\theta$ can be shown in terms of the length $L$ and the distance from the center to the edge seeming as once this center of mass passes the point in which it is balancing, it will fall. Therefore $tan(\theta)=\frac{10}{L}$ , merging the two equation found and cancelling $tan(\theta)$ we get $\frac{0.5}{12}=\frac{10}{L}$ . Solving for $L$ gives us the answer of $240cm$ , hence our answer in meters is $L=2.4m$ .

Before the tree topples, the net torque on it is 0. Let $\theta$ be the acute angle between the slanted axis and the horizontal.

On the right side, the torque is mgL $\cos\theta$

while on the left side, the torque is mg(0.1) $\sin\theta$

cancel and balance the equation, we will get L=(0.1) $\tan\theta$

From the given diagram, we get $\tan\theta$ = $\frac{20-8}{0.5}$ =24

Substitute it into the equation and we get L=0.1(24)=2.4 metres