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Plant Operations/Maintenance/Questions per Rigging and Hoisting via Millwright Study Guide


Hello Matt,
On the Rigging and Hoisting section of the Millwright Study Guide, I am having trouble finding the proper information to plug in for Questions # 2., 19.,35., and 38.respectively.
If you can help with these questions it would be so awesome as I am kind of stumped for now! I also noticed that your study guide is geared for Canada is there a whole lot of difference between a North American Millwright test and Canadian Information???
Any help will be most appreciated.Thank You very much,

C.J. Redding

P.S. Is there a chart that will tell you what size rope will support how much weight safely or is it just a formula?? Thanks again!! Happy Thanksgiving!!

Good morning.

The answer to question number two can be found using the formula (DxD)x8 giving an answer in tons.  SO, in this question, we can take the diameter of 1/2" and plug it into the formula.  (.5x.5)x8.  ultiplying .5x.5 gives you  (.25)x8 =2 tons.

This formula is a very good rule of thumb to use to calculate a wire ropes lifting strength for vertical lifts.

Question number 19 requires a little bit of geometry (or trigonometry) knowledge.  All triangles have a total of 180 degrees when you add up the three angles.  If we have a given angle, and two given lengths of sides, we can determine what type of triangle this will be (Isosceles, scalene or equilateral).  In this case, we know that if the given angle is 60 degrees, and those two leg lengths are the same, it has to be an equilateral triangle.  In an equilateral triangle, all three angles are always the same, and all three sides are the same length too.

For question 35, we need to first determine the SWL of the wire ropes if used vertically (the same formula as in question number 2).  So, we have SWL of the wire rope in vertical position = (DxD)x8.  Plug in the rope diameter.  (.375 x .375) x 8 = 1.125 tons (or 2250 lbs).  Now we need to use the SWL formula.  For this, we again need to know some basic trigonometry as well as the formula for two part lifting bridal.  The formula for this is SWL of the two part bridal = SWL (of rope in vertical lift) x H/L x 2.  This means the safe working load of the rope in vertical position, multiplied by the (height of lift/sling length) times two ropes. The tricky part here is that we do not have a lift height or a sling length.  However, because we know that it is a 60 degree angle we also can determine that regardless of wire rope sling length, the ratio of the height and length will remain constant.  Now we can simply pick any length at all, and use some more geometry (Pythagorean theory)to determine a distance between lifting points.

I know this is confusing, but I will break it down a bit more.  Lets assume that we have a sling length of 6'.  Using Pythagorean theory, we know that  to solve a right angle triangles side length, we need to use the formula A squared x B squared = C squared, and the length of C will = the square root of your answer.  We also need to remember that on a right angle triangle, "a" and "b" are always the sides that join at the right angle.

So, we can lay it out with a sling length of 6'.  We can also then determine that the distance from the sling to the center point is going to be 3'.  Now we can determine the height. (a squared plus b squared = c squared.)  So we already have A = 3'.  We do not have B (the height), but we have C (Sling length).  3 squared plus "B" squared = 6 squared.  9 + B squared = 36.  We can now determine B.  36 - 9 = B squared.  B squared is 27.  The square root of 27 is the height (5.19 feet).

Now that we have determined our sling SWL, our height and our sling length, we can determine the SWL of the 3/8 rope at 60 degrees in a two part bridal.  SWL of two part bridal = 2250 x (5.19/6) x 2 = 3900 lbs or closest answer = a.

For 38, there are a few ways to do this.  The easiest is to determine the weight of a cubic inch of steel.  Since a cubic foot is 12" x 12" x 12" = 1728 cubic inches in a cubic foot.  Now take 500 lbns, and divide it by 1728 to determine how many lbs a cubic inch weighs.  This equals .289 lbs per cubic inch.  Now we need to figure out how many cubic inches are in the plate.  72" long x 68" wide x 1.25" thick = 6120.  Multiply this by .289 lbs per cubic inch = 1768 lbs.  Or the approximate answer as asked in the question would be "D".

As for the difference between Canada and the US, there really is not much.  All of the forumals and principals are the same.

There are tons of charts available on the internet too.  Just google wire rope SWL.

I hope this helps, and please let me know if you have any more questions!  Contact me through my email address below.



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Matt Bolger


I can answer questions you may have regarding Millwright trade exams such as power transmission, safety, fluid power, some welding questions, bearings, and rigging and hoisting, as well as a variety of questions about equipment maintenance. Troubleshooting questions are ok, but without physically looking at and testing the equipment (as there are infinite numbers of industrial equipment applications), these questions may be difficult to accurately answer.


I am a licensed industrial mechanic/millwright (Canada - C of Q red seal). I have 13 years experience as a Millwrigt, maintenance supervisor, maintenance manager and operations and engineering manager.

Millwright license (Red Seal). I have also written a millwright study guide and shop reference manual that was designed to assist mechanics in passing the millwright trade exam.

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