Portlandjib.com Tension in the boom line
 Home about Roosevelt HS STEM PROGRAM BRIDGES CRANES BRIDGE BEAMS ENGINEERING middle school projects crane projects Engineer's kit Portland Transformer Junior builder's kit montessori bridge beams Montessori Children's House scissors lift new crane new bridges contact FAQ

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

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) Enter content here

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

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

.819 = Dload /4ft

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.