Here again we have to solve another question submitted to us. This is regarding how to solve for the size of manila rope. Here it is...
What size of manila line is required to hold a weight of 932 lbs., if you use a safety factor of six? Your Answer :: 3.0? Correct Answer :: 2.5? what is the solution of this
What size of manila line is required to hold a weight of 932 lbs., if you use a safety factor of six? Your Answer :: 3.0? Correct Answer :: 2.5? what is the solution of this
Load = 932 pounds
Safety Factor = 6
Find:
Size of the Manila rope
Solution:
Since the unit of weight is pounds, the formula for Manila rope's breaking strength would be C² × 900, where C is the circumference. So, BS = C² × 900.
We can solve for the Manila rope's size by using this formula:
SWL = BS
SF
So by transposition we can have BS = SWL x SF
C² × 900 = SWL x SF
C² × 900 =932 x 6
C = √932 x 6
900
C = √5592
900
C = √6.21333
C = 2.49265 or simply put 2.5" is the final answer.
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Close link chain of not less the 3/4" (or the wire rope equivalent) is required for lashing deck cargoes of timber. What size flexible wire rope would provide the strength equivalent to 3/4" chain, using a safety factor of 5?
ReplyDeletehow about in metric ton unit ?
ReplyDeleteBS= C2 x 900 not C2 / 900, thank you
ReplyDelete