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<JPS>
posted
I am very bad at math, but know that to find the force of a dropped object you multiply the mass times the aceleleration. Now it is finding the "a" part of the equation where I fall short.

If a 10lbs chainsaw is dropped on a five foot lanyard, how much force is place on it when it stops?
 
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<Chris Rose Licensed Arborist>
posted
Reply to post by jps, on November 22, 2000 at 20:52:35:

To figure the equation you need to know a physics "constant", which I can't recall from the top of my head. It has to do with a falling object reaching constant velocity (or reaching constant velocity). All falling objects ultimately reach a similar speed no matter what their weight, given enough time. What you are looking for is the "shock load" force being delivered to the climber, usually expressed in kilo-newtons (kN). Rock climbers/ice climbers, tree climbers, etc. should be familiar with these quantities. My guess would be that the object in question could produce a shock-load force of 100 or more pounds on the climber in question. I am at my brothers house right now, but when I am home I can figure out the quantity exactly. E-mail me @ islandgarden@acadia.net, I'd be happy to "discuss" details!
 
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<joe>
posted
Reply to post by Chris Rose Licensed Arborist, on November 22, 2000 at 20:52:35:

To figure the force of a falling object which starts from rest, multiply the force times the distance fallen.

F=ma=10lbs.
distance=5ft.

10 lbs. times 5 ft.= 50 lb.ft. of force

The constant of acceleration, 32.2 ft./sec^2
(read thirty-two feet divided by seconds squared or to the power of 2) is already figured into the force measurement the pound.

Joe
 
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<joe>
posted
Reply to post by Joe, on November 24, 2000 at 13:49:37:


I want to make a correction to my reply here someone pointed out to me at the I.S.A. discussion board. The 50 lb.ft of force I calculated should be 50 lb.ft of "energy", not force. Sorry for the confusion.

Joe
 
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<kc>
posted
Reply to post by Joe, on November 24, 2000 at 13:49:37:

I believe for short distances that don't achieve full acceleration we can use this formula............but I thought it would have 10# of force sitting still; and pick up one equivalent measure of force per unit of drop.

Whereby:

If we are speaking of foot pounds (we could do psi then we'd measure the drop in inches add 1 and multipy!):

WEIGHT X (DROP + 1) = FORCE
10# x (still)(0 + 1) = 10#
10# x (4' + 1) = 50#
10# x (5' + 1) = 60#

This has to be true because 10# without being dropped still has enough force if slid over to trip a 10# trigger etc.

Now if the lanyard performs any de-acceleration function this can be lessened.

Of course if you and the saw fall and you are suddenly stopped that force can increase just as if some other force reacts on the saw adding force ie. severe kickback; snatched from you etc.
 
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<wes>
posted
Reply to post by Joe, on November 25, 2000 at 20:33:48:

It goes like this:

Potentential energy = Kinectic energy
PE = KE

PE = mass x accelleration x height = mgh
KE = the integration of mv wrt velocity:
KE = integral of mv/dv = 1/2 mv^2
therefore
mgh = 1/2 mv^2

if you want the velocity at the point of impact, use this equation.
v = sqrt(2gh)
Bottom line, the kinecitc energy (not the force) measured in foot-pound, joules, calories, BTUs or whatever is mgh
 
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<joe>
posted
Reply to post by wes, on December 02, 2000 at 19:59:29:


...v=sqrt(2gh)

v=sqrt(2 à 32.2 ft/s² à 4.5 ft)

v=sqrt(289.8 ft²/s²)

v=17 ft/s

Using v(f)²=v(i)²+2a(x-x(0)) to find the rate of deceleration where the distance of deceleration is assumed to be 1/2 foot. This gives a freefall of 4.5 feet and .5 foot to decelerate. This equals a total of 5 feet of fall. The final velocity (v(f)) is 0 since the saw is no longer falling.

0= 289.8 ft²/s² Ã2a(.5 ft)

(289.9 ft²/s²)/ (2Ã.5 ft)= 289.8 ft/s² = a, the rate the saw decelerates.

Using f=ma

f=(10 (slugsÃft/s²)/32.2 (ft/s²))Ã(289.8 (ft/s²))

or f=(.311)(289.8)=90

90 lbs of force or tension in the lanyard.

Joe
 
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