Portlandjib.com
Law of Sines and Sum of Moments
 

Enter subhead content here

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

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

DSC00357.JPG

The end of the cable is attached to a load cell

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

DSC00359.JPG

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

and makes a 25 deg angle with the horizontal.

boom25.jpg

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).

 

 

 

forceline.jpg

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.  

cgboom.jpg

                         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.

40tower.jpg

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.  

cablelineofforce.jpg

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 

deg783.jpg

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.

DSC00393.JPG

I made some improvements to the basic crane model.

 

 

 

. 

goodboom.JPG

I added a windlass with an automatic brake,and the load cell is connected

to the line going to the windlass.

As the boom is raised the load cell will show the Tension in the line. 

  

modelwindlass.JPG

The load can be placed anywhere along the length of the boom.

inclinebrickJPG.JPG