Tension in the boom line

We've got a 10# load hanging from the boom of the NS4 crane


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The boom is at a 35 degree angle and the boom line makes a  20 degree angle with the horizontal


First we have to find the OTM-  see "Don't tip over the crane"

   M =F x D

 We know the force is  10#, and D is the shortest distance from the pivot point to the 

line of force (the force of gravity acting straight down)  


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Here's our triangle- now we need to find Dload

Cos 35 = Dload /hyp.         Cos 35 = .819     Hyp. = 4ft

.819 = Dload /4ft     

Dload = 3.27ft

M =F x D

M = 10# x 3.27ft

M = 32.7ft#

Don't forget the boom, It weighs 11# and acts just like any other load that's pulling down on the boom.

The hypotenuse of our next triangle is 2.6ft- the distance from the pivot to the CG of the boom.


Cos 35 = Dboom /hyp.       cos 35 = .819       Hyp. = 2.6ft

.819 = Dboom /2.6ft           Dboom = 2.13ft

M = F x D     M = 11# x 2.13ft             M = 23.43ft#

TOTAL MOMENT  23.43ft# + 32.7ft# =  56.13ft#

For this exercise we're not interested in how much RM is needed but rather how much

Tension is in the boom line- the black wire. The boom line is a thin steel wire (3/32") covered with a

black plastic sheath. We used wire here instead of string as we did everywhere else, because there's

a lot more tension (stretch) in the boom line than in all the other lines.

How much tension?


We've got a total of 56.13ft# of moment (OTM) that's trying to tip the crane over Counter clockwise CCW.

Since the crane is still upright, there must be at least 56.13ft# of moment (clock wise-CW) preventing this.

Let's call the CCW moment positive and the CW moment negative.

Engineers would explain this equilibrium by saying the sum of the moments = 0

(+56.13ft#)  + (-56.13ft#) = 0

That 56.13ft# is the moment exerted by the boom line.Remember that's moment NOT FORCE.

To find the force we have to go back to our formula  M = F x D

The boom is @ 35 degrees and since there are 180 degrees in a straight line , and 180 in every triangle then;

- the base angle of the new triangle must be 145 deg.

-which makes the last angle 15 deg.


M = F x D

Where's the line of force?  This time it's NOT GRAVITY-but the force in the boom line. Imagine you're

holding the end of the boom line- you have to pull with a certain force to counter the OTM of 56.13ft#.

So the line of force IS THE BOOM LINE.

Remember D is the shortest distance (perpendicular distance) from the line of force to the pivot


Now we have another triangle with a hyp. of 4ft and an acute angle of 15 deg.

This time we need the sine function. Sine = opposite/ hypotenuse

 sin 15 = D/4ft                           sin 15 = .259

.259 = D/4                             D = 1.03ft

Now back to the basic formula

M = F x D    M = 56.13ft#

 56.13ft# = F x 1.03ft

F = 54.4 pounds of force

That's why we didn't use string for the boom line- it could break with 54 pounds of  force (tension) in it.