Problem: Bucket Cylinders
This is a sketch of a sandbox loader and bucket. Calculate the vertical distance that the bucket cylinders (2) should be mounted above the pivot points to match the required pressure of the boom cylinders. The bucket cylinders are mounted horizontally.
See the Solution
1. Find area of boom cylinder: 1.25 x 1.25 x .7854 = 1.227 sq. in.
2. Find rod end area of bucket cylinder: .44 – .0768 (rod area) = .365 sq. in.
3. Find perpendicular distance for center of cylinder to pivot:
Arc Tangent of 24/12 = 63.4 degrees. Sin of 63.4 x 12″ = 10.73 in.
4. Write moment equations: 10.7″ x force = 36″ x 50 lbs. Force = 167.7 lbs.
167.7 / 2 cylinders = 83.85 lbs. / cylinder
5. F = PA; 83.85 = P x 1.227; P = 68.3 PSI for boom cylinders
6. Cylinder force for retracting the bucket is F = PA; F = 68.3 x .365 sq. in.
F = 24.9 lbs. x 2 cylinders = 49.8 lbs.
7. Write moment equation: 50 lbs. x 5.3″ = 49.8 x (D) Inches; D = 5.32 in.
Correct Answer is 5.32″
Alternate solution to boom force: 50 lbs. x 36″ / 24″ = 75 lbs. at rod end.
75 / sin 26.565 degrees = 167.7 lbs.
Deadline past. Not available for submissions.
By Ernie Parker, AI, AJPP, AJPPCC, S, MT, MM, MIH, MIP, MMH
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.
Tagged air teaser, bucket cylinder, ernie parker