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Torque of a Tree Limb or Pole

Date: 05/28/2008 at 18:29:08
From: Sayer
Subject: Trying to figure out leverage weight

My stepdad wants to know the formula for leverage weight when cutting 
dangerous limbs off trees.  He explains it to me like this: "If I
remove 1/3 of the growth from the end of a 20' limb and I piled what I
pruned on a scale and it weighed 300 lbs., the weight of that growth
on the end of a 20' limb would be much more than 300 lbs. because of
leverage."  I guess the weight is the 'pull weight' of the limb with
the leaves on the end.  When he cuts that off he is taking off a lot
of weight that might rip the limb from the tree.

The weight of a tree limb is guess work so it's all approximate 
weights to begin with.  What is confusing is that if a limb weighed 
500 lbs. and was 20' long and you took off 30% of the limb because 
of the leverage factor you cannot do simple math. 

Date: 05/29/2008 at 21:18:48
From: Doctor Rick
Subject: Re: Trying to figure out leverage weight

Hi, Sayer.

I think that what your stepdad calls "leverage weight" is what 
physicists and engineers call "torque".  If you have a force of 100 
lb acting at a distance 10 feet from the fulcrum or pivot point, the 
torque is 100 lb times 10 feet = 1000 foot-pounds.  If you double 
either the force (weight) or the distance ("lever arm"), you double 
the torque.

A tree limb gets thinner as you go out, but also has branches that 
may well increase the proportion of the weight that is farther out. 
Let's simplify things by replacing the limb with a straight pole of 
constant diameter.  The weight acts at the center of gravity of the 
pole, which is halfway out, under my assumption of constant 
diameter.  Thus, if the pole is 20 feet long, its weight acts at a 
point 10 feet out.  If the pole weighs 500 pounds, then the torque is 
10 feet * 500 lb = 5,000 ft-lbs.

Now, what if you remove the outermost 30% of the pole?  The length of 
the remaining pole is 70% of 20 feet = 14 feet, so the new center of 
gravity is 14/2 = 7 feet out.  The remaining weight is 70% of 500 lb 
= 350 lb.  Thus the torque is 7 feet * 350 lb = 2450 ft-lb.  Removing 
30% of the weight removed about 50% of the torque. (The torque is
reduced by 5,000 ft-lb - 2450 ft-lb = 2550 ft-lb; expressed as a
percentage of the original torque, this is 2550/5,000*100% = 51%.)

If the weight is distributed farther out on the limb, the center of 
mass will be farther out, so the torque will be greater to begin 
with.  But removing the outermost 30% of the weight will require 
taking off less than 30% of the length, so I don't think that the 
torque will be reduced quite as much.  Still, the principle holds 
true, even if the numbers aren't quite right.

- Doctor Rick, The Math Forum

Date: 05/30/2008 at 17:11:04
From: Sayer
Subject: Thank you (Trying to figure out leverage weight)

Wow, this is cool, THANK YOU!  I can't wait to give this to my stepdad
and tell him that he's talking about "torque"!

Associated Topics:
High School Physics/Chemistry

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