TENSION IN THE BOOM LINE

we’ve got a 10# load hanging from the boom of the NS4 crane

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)

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