In this part of the world, we have oak trees. Technically they are called live oaks – but I don’t get it. Of course they are alive. I was at a soccer game and this is the tree I always look at.

Look how far those limbs extend horizontally. That branch is about 12 meters long. Why is this amazing? Have you ever tried to hold an 8 foot 2 x 4 board horizontally by holding one end? Pretty tough. How about I calculate the forces needed to hold that branch in place? I will do a simple model and then maybe later I can make it more complicated. Suppose I replace that limb with one straight uniform limb that looks like this:

In this replacement limb, I am going to say it is a cylinder that is 9 meters long and 30 cm in diameter. Let me assume that this limb is connected at two points to the tree (the white dots). So, for this limb to stay there, the following must be true:

The first two equations say that the total force must be zero. The last one says that the torque about any point must be zero (since it is in rotational equilibrium about any point). First for the forces. There is the gravitational force. This pulls on all parts of the limb, but I can represent this as one force pulling on the limb at the center of mass (long ago, I said I would explicitly show this – but I haven’t yet). Then there are two other forces. Let me pretend like there are two pins that hold the limb to the tree. Each of these pins can exert a force in the vertical and horizontal direction I will call these F_{1-y} F_{2-y} etc…where the top pin will be 1. That is 5 forces.

For these forces and the first two equations, I get:

So, already I have some constraints. The horizontal components of the forces from the two pins must be equal and opposite. The vertical components of the forces from the pins have to add up to the weight of the log. Now for the torque, I am going to add up the torques about the lower pin. Let me draw a distorted view of the log so that the important distances can be seen.

What is torque? Torque is like a rotational force. Here is an example, what if you try to open a door by pushing near the hinges? It is much harder than pushing near the handle, right? When rotating about some axis, the torque is:

Here F is the applied force, r is the distance from the point where the force is applied to the axis. Theta is the angle between F and r. I will call torques that would make a rotation counterclockwise positive (really, torque is a vector). So, what is the some of torques about axis O (that passes through point O)? First, there are some forces that have zero torque. Both of the vertical pin forces have either theta = 0 or r = 0 so that the torque is zero. The same is true for the horizontal force on the bottom pin. This just leave two forces that have non-zero torques:

Now that I have the horizontal force on the top pin, the bottom pin has the same value (but in the opposite direction). I don’t have an expression for the two vertical pin forces. Let me just say that each has a force equal to have the weight of the limb.

How about some values? First, I need the mass of the limb. If this is a cylinder of wood (with a density rho) then the mass is:

So the magnitude of the two horizontal forces on the pins would be:

If I use my values from above and an estimation of the density of wood, I get:

Wow. Oh, I know I made some estimations but even 50,000 Newtons would be huge. Impressive, most impressive. I salute you mighty tree.