Problem: Standard Size Cylinder 3
Find Out The Solution
Calculate the angle that the cylinder is mounted to the horizontal frame: Use the tangent of ø = 5.1962 / 3 = 1.732
Now do Tan -¹ to find degrees = 60°
Step 2: The 400 lbs. has a 2:1 mechanical disadvantage, so the true load on the end of the arm is 800 lbs.
Step 3: Load x load distance = effort x effort distance (800 x 8 x 3 = 2133.33 lbs. = vertical load at end of cylinder rod)
Step 4: Load / sine of 60 degrees = 2462.9 lbs. on the cylinder
Step 5: F = PA 2462.9 / 100 psi = 24.63 sq. in. needed
Step 6: Because we are finding the rod end area, we need to add the area of the rod to the required area. Rod area is 2.5² x 0.7854 or 4.9 sq. in. Add 24.63 and 4.9 = 29.53 sq. in.
Step 7: Find diameter of cylinder: A = D² x 0.7854. Working backwards, 29.53 / 0.7854 = and square root that number 6.13”. The question asks for standard size cylinder, and then we would up the size to a 7-inch diameter.
Deadline past. Not available for submissions.
Robert Justin Allison, CFPS, Butler, PA, Rotork Hiller
George Fling, CFPS, Dallas, TX, Southwestern Controls, Inc. (retired)
Karl Kersker, CFPS, CFPE, Pleasant View, UT, ATK Launch Systems
Richard Throop, CFPAI, CFPAJPP , CFPIHM, CFPIHT , CFPMHM , CFPMHT, CFPMM , CFPMT, CFPPM , CFPS , CFPPT, Fenton, MI, Neff Engineering Co., Inc.
Andrew Van Beusekom, CFPS, Loretto, MN
By Ernie Parker, AI, AJPP, AJPPCC, S, MT, MM, MIH, MIP, MMH,
Fluid Power Instructor, Hennepin Technical College, EParker@Hennepintech.edu
This teaser is printed in the Fluid Power Journal. Those who submit the correct answer before the deadline will have their names printed in the Society Page newsletter and in Fluid Power Journal. The winners will also be entered into a drawing for a special gift.