Problem: Minimum Pressure

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