Problem: Minimum Pressure
What would be the minimum pressure required to hold the given load on the pneumatic cylinder?
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*Special thanks to Chad Grimmer, who suggested this Air Teaser.
See the Solution
This problem has a clockwise moment (torque) that is equal to the CCW moment. We call this a “state of equilibrium.” The CW moment can be calculated figuring the perpendicular distance to its force. If we made a triangle with the 500 lbs. being the opposite side, and the 144″ is the hypotenuse side of a triangle, we can use the cosine to determine the horizontal perpendicular distance to the load back to the pivot. (See diagram “A”.) Cosine of 45° = adjacent / hypotenuse. Cosine of 45°x 144″ = 101.8″. Our CW moment (torque) is now 101.8″ x 500 lbs. = 50,911.6 lb. inches.
Now for the reaction (cylinder force), we need to find the CCW moment. Looking a diagram “B”, I have drawn a triangle representing a perpendicular distance to the center of the cylinder back to the pivot point. The triangle that we are looking at has the opposite side of a 35°angle. This time we will use the sine function because we have the opposite side and the hypotenuse. Sine of 35° x 72″ = a perpendicular distance of 41.3″. Our CCW moment is 41.3″ x F(unknown force). Now we are ready to write our equilibrium equation where the CCW moment = the CW moment. Note that our cylinder force is our only unknown and I called it “F”. Remember that a moment is a force and a perpendicular distance. Therefore 101.8″ x 500 lbs. = 41.3″ x F (lbs.). F = 101.8 x 500 / 41.3
F = 1232.4 lbs. on our cylinder.
Now we need calculate the cylinder pressure with F = P x A. A = D²x .7854, A = 4²x .7854 or 12.566.
1232.4 / 12.566 = 98 PSI
Deadline past. Not available for submissions.
Harry Pawluk, Jr., Hanco, Ltd, Ardara, PA
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.
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