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I asked "AI" to confirm metrics from other sources and it gave numerous formulas and dubious answers, none of which inspired confidence.

The goal is to set up a cable system to transport a load of no more than 200 lbs. from one floor to another.

Figure 1 Top View

Figure 2 Side View

The load will be attached to a 2" 'pulley'.

The 3/16" cable is represented by #1, Purple (90") and 2, Green (112"). Figure 1

Segment 2 is angled downward where it meets segment 1 which is horizontal/level.

Ideally I would install the cable in a single run that would loop through an eyelet (B") to create the 90 degree angle.

If doing it that way is not advisable, I would use two sets of thimbles, wire rope clips etc. to create the 90 degree leg. And "A" would then require a turnbuckle (connected to an eyebolt).

In all cases, "C" would consist of an eyebolt and a turnbuckle attached to a 5.7mm thick u shaped bracket made of Q235B carbon steel, mounted to a brick wall using (2) 3/8" x 3" Hex Nut Sleeve Anchor Expansion Bolts.

Because the angle of segment 2 (green) is somewhat steep, I will use an electric hoist to control its descent (or in a rare instance, pull the load up to a higher floor).

To measure the cable tension I can either use a Digital Force Gauge Push and Pull Meter (and somehow measure the tension after install) or a Digital Hanging Scale rigged to a wire rope clip to temporarily bypass the turnbuckle.

FYI the electric hoist is only available at point C and the wall along the #1 purple segment cannot be used for mounting.

My questions are:

1 Is it advisable to use a single eyebolt to create the 90 degree turn?

2 What are the minimum ratings for the mounting hardware and rigging?

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1 Answer 1

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As a rough estimate The tension in a cable suspended between two supports, T, is $$ T= w/2*l/2\delta $$

  • T =tension in cable
  • w = weight, 200 lbs
  • $\delta$ = deflection at the middle of the cable, say 2 inch
  • L = the distance between to supports, 112" $$200/2*112"/4"=2800lbs$$ $$2800*safety \ factor: \ 2=5600\ lbs $$

If there is a sudden start or stop in the loading or rolling of the cable, the dynamic safety factor could reach 5-6.

The wall, supports, pulleys, and anchors should be designed for such a load!

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  • $\begingroup$ 1st, thanks for answering--that's the formula I was looking for. With regard to pulleys, do you mean the pulleys that may be "in line" with the cable or are you saying that the trolley pulley which rides atop the cable and directly supports the 200 lb load also has to be at 5600 lbs? @kamran $\endgroup$ Apr 3 at 16:23
  • $\begingroup$ I have to add: all the concern that AI will render S.E. obsolete--the answers I have gotten from AI were at best, a good replacement for search engines. I can see AI being used here as a proctor to help find the appropriate forum and maintaining Q & A structures, but until it has the full spectrum of Theory of Mind, it can't replace human experts (IMHO). My thanks to S.E. and all its contributors. $\endgroup$ Apr 3 at 16:35
  • $\begingroup$ @Captain-Science, the pulleys that are online and don't act as terminal support just carry the vertical load. For instance, if you place a pulley between A and C, it will carry 200/2 = 100 lbs. and you will have a new L=l/2. As a rule of thumb, the more slack on the cable, the less tension at the support. In an extreme case where the cable hangs at 45 degrees between the supports, the tension will be w/2*sqrt2. Conversely, if the deflection is small, ultimately the tension will be very large, approaching infinity when the cable is a straight line. $\endgroup$
    – kamran
    Apr 3 at 18:17

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