Portlandjib.com
CENTER OF GRAVITY
 


This wood beam is 22" long and weighs 4 pounds  (4#), and
the steel block (3" wide) weighs 7.375#
We attach slings with the hook centered over the middle of
the beam and lift----oops!!!

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We have to find the combined Center of Gravity--- CG.
First the wood beam-If we assume the density of the beam
is uniform throughout its length, then the CG will be in
the geometric center

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You knew that already, just like the CG of a 12" ruler is
centered at 6" 

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Same with the steel block. It's 3" wide so the CG is 1 1/2" in

from the edge.

 But when we put the 2 together, everything changes and we'd better

know where the combined CG is or our load will tip and could injure someone.

Here's how- let's assume the beam and the block are balanced on a pivot point

at 1 end and would therefore try to rotate- like a see-saw--  only with just 1 side.

Remember forces that cause rotation can be computed using the basic formula for Moment-

M = Force x Distance

For the pivot point we'll use one end of the wood beam-the end opposite the steel block.

To compute the Moment of the beam    M = F x D

The force is the weight (4#) and the distance is measured from the line of force to the pivot

(at a right angle). The line of force (gravity remember) goes through the geometric

center of the beam- so the Dist is 11"

 M = F x D

M = 4# x 11"

M = 44inch pounds-44"#

 Now the steel block- wt = 7.375#.and the distance from the pivot is 20 1/2" 

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M = F x D

M = 7.375# x 20.5"=151.18"#

Here's the formula to find the combined CG

MOMENTbeam  + MOMENT block divided by combined weight of

beam and block.

44"# + 151.18"# divided by 7.375# + 4# =

195.18"# /11.3745# =

17.1"

So now we place the hook directly above the CG (17.1" from pivot) and lift.

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