Oak Trees are awesome

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.

i-2fed948d5ad88836d3eba1ef912fb8c2-2010-03-30_i_photo.jpg

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:

i-d0142e349421546acaa5967d79764df2-2010-03-30_untitled.jpg

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:

i-419073cb81199f696d7043b6a43e764c-2010-03-30_la_te_xi_t_1_12.jpg

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 F1-y F2-y etc...where the top pin will be 1. That is 5 forces.

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

i-012da5fbf1ab33373bcf5508754093b9-2010-03-31_la_te_xi_t_1.jpg

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.

i-b594a2212739ded07e13824a140429ec-2010-03-31_untitled.jpg

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:

i-b7f138f0c9d89ac15838a97b46040a05-2010-04-01_la_te_xi_t_1_1.jpg

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:

i-9309d5ed088ac03170bcc6c01e707432-2010-04-01_la_te_xi_t_1_2.jpg

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:

i-35527118f76b93738777d53a00c751d9-2010-04-01_la_te_xi_t_1_3.jpg

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

i-0c19a3934323b0cbcbda003370fae2b8-2010-04-01_la_te_xi_t_1_4.jpg

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

i-50ae30186211d595ea8b926574f4a448-2010-04-01_la_te_xi_t_1_5.jpg

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

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They are called "live" oaks because they keep their leaves
during the winter, thus still looking alive when their
northern cousins look dead, dead, dead.

Your density is way too low for green live oak. Even seasoned live oak has a density of nearly 90 g/cm3 - it's one of the densest woods around. I'd expect its green density to be nearly double that. This gives the green density as 90 pounds per cubic foot, whatever the hell that is in real units.

Oh, and it's called "live" oak because it's evergreen.

@Ahcuah, @Dunc,

You might be right about the density - I estimated the density from a stick I found near the tree. Who knows how old it was. If the density was higher, then the forces would be higher.

Thanks for the info about "live" - that makes sense.

I estimated the density from a stick I found near the tree

Yeah, I looked - a stick that size is practically all sapwood, but the heartwood is much more dense.

This non-physics major thanks you for this post. I am really going to be looking at trees through new eyes, ones more appreciate of the way in which they are an impressive example of natural engineering. (On a literary note, the photograph of the live oak makes me think of the Party Tree in Tolkien's The Lord of the Ring.)

@Dunc,

thank you tree expert. I am glad that I underestimated the density rather than overestimate.

I'm nowhere near an expert, but thanks anyway. ;)

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

Is there a typing error in the line for the net y-force?

I know it's off-topic, but how about a post on kitesurfing and the forces on a kitesurfer. What can a kitesurfer do to travel faster? What role does his mass play?

Trees are amazing. We just had some 65mph wind gusts last eevening and the tops of the trees in my backyard swayed as much as 20 or so feet at the top. Over the last year we had winds in the neighborhood of 106 mph and I didn't lose one tree. My backyard gate didn't fare so well.

By Eric Juve (not verified) on 02 Apr 2010 #permalink

I love trees. I ued to climb them and sit in their branches and listen to them in the breezes. I'd hug them. I don't have as much of my innocent connection to them but in being a gardener I care for the trees in my yard ad have planted/nurtured many a smaller shrub, brush trees for visiting songbirds. But I do love to visit and walk through parks with old growth stands of oaks and walnuts, maples etc.

I don't think the density could be higher than 64 lbs/ft^3, because I'm pretty sure oak floats and water is 64 lbs/ft^3.

Hmmm. From my experience, live oak wood will float, but barely. If you really want to see a tree that looks like it is physically impossible to stay upright, check out the mesquite. They tend to grow sideways farther than they grow up, and sometimes all limbs on one side.

By CherryBomb (not verified) on 03 Apr 2010 #permalink

Unit conversions and decimal points ...
90g/cm^3 is 5 times denser than anything on earth, I guess 0.9 g/cm^3 is more realistic? That gives 56.25 lb/cft.

Probably find that the tree isn't so much "strong" as cleverly dare I say it? "designed"
I.e the load is spread much farther than at those 2 points, possibly opposite branches are counterbalanced and are effectively self anchored around the trunk? Load could be spread
in the trunk, suspended above and butressed below the branch line with the tree fibers acting as cables and supports but running deep into the vertical trunk, Sorry no equations, just a day dream thought experiment? :)