Here's a simulated crane boom supported by a green cable

that goes over a sheave
(pulley) mounted in the tower.

The end of the cable is attached to a load cell

that is hooked to a turn-buckle
which is anchored to the base.

GIVEN; the boom is 60" long, weighs 6 lbs. ,

and makes a 25 deg angle
with the horizontal.

FIND: the tension in the boom line.

In
the crane industry they use terms like

Overturning moment (OTM) and Righting moment.(RM)

Engineers use the formula for the sum of the moments.

We assume that since the boom is not moving,
it is in

equilibrium.

That's the basis for our formulas.

With the boom stationary, the moment pulling
the boom

down, and the moment keeping it up at a 25 deg angle must

be equal- or the sum of the moments must equal zero.

The Moment pulling the
boom down acting in a Clock-wise

direction (CW) added to the opposite moment

acting Counter Clock-wise (CCW) must equal zero.

We can simplify this by saying:

the CW Moment = CCW Moment.

Remember Moment = Force x Distance

The CW moment
has 2 parts:

1-the moment that is causing the boom to fall-i.e. the lead cylinder

hanging from the tip of the boom. The cylinder weighs 6lbs. 13oz. or 109 oz.

The
FORCE is the force of gravity acting on that 109 oz. load

The DISTANCE is NOT
60". With moment we're interested in

the distance from the line of force
to the pivot point.

The force of gravity pulls straight down on the boom. The direction

of that force ( the line of force ) is straight down.

We
need to find the shortest distance from somewhere along

that line to the pivot point.

You can see from the picture below that the perpendicular

distance from the line
to the pivot is the shortest.

That will be D(load).

In this rt. triangle, we have an hypotenuse of 60",

and a base angle
of 25 deg. To find D(load) we'll

use the cosine function.

Cos 25 = D (load) / 60 cos 25 = .9063

solving we get

D(load) = .9063 x 60 or

D(load) = 54.378 inches

now back to the basic equation M = F x D

M = 109 oz. x 54.378"

M = 5927.202 inch ounces

2- the next part of the moment is
the boom itself.

Don't forget it weighs almost as much as the load (6lbs.)

and it is also pulling down on the cable. So

M(boom) = F x D(boom)

The boom weighs 6lbs. so that's the F, but what's the D?

Remember we're looking for the line of force which goes through

the center of gravity (CG) of the object. For the load (cylinder) the

CG went through
the measured center of the cylinder and it's the

same for the boom. The CG of the boom is 30"
from the pivot point.

M = F x D

the boom weighs 6lbs. or 96 oz. so that's the F.

To
find the D we use cosine again.

cos 25 = D(boom) /30"
cos 25 = .9063 so

D(boom) = .9063 x 30 = 27.189 inches

M = F x D so

M = 96oz. x 27.189" = 2610.144 inch ounces

finally

M(load) = 5927.202

M (boom) = 2610.144

TOTAL MOMENT = 8537.346 inch ounces

That's the total moment trying to pull the boom down .

Now we need an equal moment to KEEP the boom

stationary at 25 deg.

In other words the CW Moment 8537.346
must be equal to

the CCW moment.

Now we work backwards.

8537.346 (CW M) = F x
D (CCW M)

We want to find F (the tension in the boom line) but first

we need to compute
D, and for that we have to add

more info to our diagram.

If the tower is at a rt. angle to the horizontal, then the

complementary
angle to 25 is 65 deg. Also the cable is

55" long and the distance from the pivot point
to where

the cable attaches to the tower is 40".

The LINE OF FORCE in many of these examples is the

line of force of gravity pulling down on the load, the boom
or ??

In this example, the line of force is along
the cable. Imagine

you're pulling on the end of the cable to keep the boom from

dropping. Your pulling force is exactly in line with the cable,

so the line
of force IS THE CABLE, and the Distance is measured

at a rt. angle from the line
of force to the pivot point.

First we have to find the dimensions of triangle ABC

and for that we'll
need the Law of Sines.

a/sina = b/sinb
= c/sinc

We already have side a =40"
side c =55" and

angle c =65 deg. so,

40"/sin a = 55"/ sin 65 and sin
65 = .9063 so

40/sin a = 55/ .9063

sin a = 40 x .9063/55 sin a = .6591 which gives us

angle a = 41.2 deg.

and since there are 180 deg in every triangle, then

angle
b = 73.8 deg

Now all that's left is to find d or side CE of triangle ACE.

sin
41.2 = d/60" and sin 41.2 = .6591 so

.6591 = d/60 ..................d = .6591
x 60 = 39.546"

now back to M = F x D

and we
already know the value of M 8537.89 inch ounces so

8537.89 = F x 39.546 that gives us

F = 215.897 ounces ...convert back to lbs.

F = 13.493 lbs. or just under 13lbs.8 oz.

We can
also use triangle BEC to check our answer.

sin 73.8 = d/40"
you compute the answer.

I got 13.89lbs. or 13lbs. 14 oz.

The load
cell reading was .....................13lbs. 12oz.

So we were only off by a few ounces,
and considering that

the measurements of the sides of the triangles were not

very exact, and the load cell is also not perfectly

calibrated, a few ounces
discrepancy is pretty good.