Calculating Force Needed to Raise Steel Tower

[old fart]

How did you find PF?: looking for simple formula to calculate force needed to raise steel tower. I am not on social media and have no idea how to use a forum.

at almost 80, I find having to hand crank a ham radio tower up and down to be challenging. I must lay it down to work on the antennas. (It is a self supporting, 4 section steel tower that hand cranks up and down.) To lay it down, one must un-crank until it is horizontal. It has 1/4" steel cable and if I "pluck" the cable, it has a tone somewhere above middle C. Very tight. I would like to convert to a winch but have no idea as to the strength needed to raise and lower the tower.

It is 21' long (tall), weighs between 1040 and 1200 lbs. One of your commentators said that if you pick up one end, you are lifting 1/2 the weight. The tower is hinged at the bottom and has a pulley at 7' above the ground. The cable comes from the hand crank over a pulley. to the tower pulley and back to the top of the pulley post and hooks with a hook. (if I am correct, one doubling strength)

Common sense is telling me that if lifting from the top is 1/2 the weight, the closer one gets to the bottom of the tower, the more force is needed to move the tower. (I am also guessing it is a log progression.)

Is there a formula or can someone help me out. Thanks.

 

[berkeman]

Welcome to PF (fellow HAM here). Your description is pretty good, but can you Attach a PDF or JPEG diagram of the setup? Thanks.

 

[old fart]

Tower height is 21' and point where lifting cable attaches to tower is 7' or 1/3 the way up the tower.

 

[jrmichler]

looking for simple formula to calculate force needed to raise steel tower.

The calculation procedure is relatively simple if you are reasonably proficient at algebra and trigonometry. It starts with a free body diagram (FBD). A partial FBD for your antenna is shown below.

A FBD is a schematic that shows all forces on an object. Start with the weight of the antenna. The entire weight of the antenna is shown as being at one point - the center of gravity (CG). The CG is the balance point if you laid it horizontal and disconnected it from the base.

The force from the cable is shown as two forces, one for each length of cable from the pulley. The forces start at the center of the pulley, and extend parallel to the two lengths of cable.

There is also a force on the pivot, but we can ignore that for now.

At this point, we know the direction of each force, and where it pulls on the antenna. What we need to find next is the perpendicular distance from each force to the center of the pivot. You can calculate these distances using trigonometry, or you can measure from the antenna.

Now for the algebra. The two forces due to the cable are equal, so we can call both of them $F$. The force due to gravity is 1200 lbs. In this case, the gravity force is trying to rotate the antenna in one direction, and the forces from the cable are trying to rotate the antenna in the opposite direction. The antenna is not moving, so the net effect of all those forces is zero.

The rotational force, or moment due to a force, is the magnitude of the force times the perpendicular distance from the force to the pivot. That distance is shown as dashed lines in the FBD. The moment due to the gravity force is roughly 10 feet times 1200 lbs, $10 ft * 1200 lbs = 12000 ft-lbs$.

For lack of real numbers, I will assume that one cable force has a perpendicular distance of 2 feet, and the other cable a perpendicular distance of 3 feet. Then the moment from the cable is $F lbs * 2 ft + F lbs * 3 ft = F * (2 + 3) = F * 5 ft-lbs$.

Since the total moment in one direction is equal to the total moment in the other direction, we take the above moments and make them equal: $12000 ft-lbs = 5 * F ft-lbs$. Now we use algebra to solve for $F$: $F = 12000 / 5 = 2400 lbs$. That is the force that the winch sees. The post holding the winch sees two forces, each 2400 lbs, for a total of 4800 lbs.

The force will be highest when the antenna is in the lowest position because the CG is farthest from the pivot and the cables are closest to the pivot. You can verify this by doing the calculations with the antenna near vertical, at 45 degrees, and horizontal.

MTA to add: And welcome to PF.

 

[berkeman]

Tower height is 21' and point where lifting cable attaches to tower is 7' or 1/3 the way up the tower.

You can reduce the peak tension required in the cable if you add a rest to your setup so the tower does not go all the way to horizontal on the ground. If you can have the tower stop at around 20 degrees or so, you still should be able to reach the top part with a short ladder for any servicing or antenna additions. $21ft~sin(20^o) = 7ft$

 

[DaveC426913]

I have a similar mast-raising system on my sailboat - though it uses ropes rather than cable. There are a few ways that you might improve your setup - if you are of-a-mind.

1. If your winch doesn't already have one, a brake on the cable (cam cleat or clam cleat) will allow you to stop at any point in the raising - whether to catch your breath or to unfoul something.
2. Depending on how flexible your cable is to go around pulleys, a 4:1 or even 6:1 block system will make lifting the tower single-handed far easier.

I can show you examples if you are interested. But the responsibility remains with you to ensure your solution is safe and sufficient.