We need the perpendicular distance (90 degree).
So we draw a right triangle- with a 30 degree base angle and a hypotenuse of 4ft.
To find DISTANCE/load we'll use the cosine function.
Cos 30 = Dload / Hyp.
30 = .866
.866 = Dload / 4ft
Next we plug in the numbers
= F x D
M = 10# x 3.464 ft
M = 34.64ft pounds or 34.64 ft#
Now for the boom. It
weighs 11# so we have to find M/boom
M = F x D
We know the force is 11#, and the distance
is measured at right angles from the line of force to the pivot. So where's the line of force?
For the bucket, the line of force went through the center of mass, or the center of gravity (CG)
of the bucket- that makes sense.
Is the CG of the boom in the middle (halfway
up the length)? It would be if the weight of the boom were evenly distributed along
the length of the boom- but the tip of the boom is heavier ( it has all those pulleys and steel bolts, etc.
plus the jib.)
We found the CG of the boom to be 32" ( 2.6ft) from
Now we have a similar triangle with a 2.6ft. hyp.
To find DIST/boom
Cos 30 = Dboom / 2.6ft
Cos 30 = .866
.866 = Dboom /
D/boom = 2.25ft
M = F x D
M = 11# x 2.25ft
TOTAL M 24.75ft# + 34.64ft# = 59.39ft#
So the turning force(moment) that's trying to tip the crane over -the OTM- is almost 60ft#.
What's preventing this? - the righting moment- RM.