# Problem: Calculate This

**Given:**

Coefficient of Friction: .2

Cylinder: 4″ x 20″ x 2″

**1.** Calculate the pressure to retract the load.

**2.** Calculate the distance that the load can move.

**3.** Calculate the pressure to hold the load.

##### Find Out The Solution

Answers **107+** PSI, **60** inches of travel, **52+** PSI

**Solution Part A:**

Step 1: Sin of the angle and the load:

Sin of 30 = 0.5 x 500 lbs. = 250 lbs. to overcome gravity.

Step 2: Cosine of the angle x the load x the Cf:

Cos of 30 = 0.866 x 500 x 0.2 = 86.6 lbs. to overcome friction.

Step 3: If going uphill, add the two answers. If going downhill, subtract the two answers.

250 lbs. + 86.6 lbs. = 336.6 lbs. to pull up the incline.

*Note: If you are holding a load, subtract the two answers. If step 2 is > than step 1, you will have to pull the load downhill by the difference between the two forces.*

The pulley on the end of cylinder has three (3) strands on the moveable pulley, and pulling on the cylinder will be 3x harder to pull than a straight pull up the incline. Pulling the cylinder a foot also means we have to add 3 ft. of cable and that can only come from the load.

336.6 x 3 = 1009 lbs. of pull on the cylinder Rod end area = 12.566 – 3.14 or 9.42 sq. in.

F = PA 1009 / 9.42 = 107.2 PSI

**Part B:** 20-inch stroke x 3:1 mechanical disadvantage = 60 inches of travel

**Part C: **To hold the load, we can subtract the friction (step 2) from gravity (step 1)

250 – 86.6 = 163.4 lbs.

163.4 x 3:1 mech. disadvantage = 490 lbs. to hold the load on the cylinder

F = PA 490 / 9.42 sq. in. = 52.15 PSI

*Note: With air, one would generally give an extra 30% air pressure to give enough force and speed.*

### Deadline past. Not available for submissions.

*Winner*

**Mark Sterrett,** CFPHS, RH Sheppard Co., Inc.

*Answered correctly*

**Ehren Polheber,** CFPS, Scot Forge Company, Salem, WI

**Dan Jensen (student),** CFPS, Hennepin Tech, Excelsior, MN

**Kirk Miller,** Orbital ATK

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.

Tagged air, cylinder, teaser

1. Calculate the pressure to retract the load = 88,76

2. Calculate the distance that the load can move 3X20=60

3. Calculate the pressure to hold the load = 79,5 PSI

my mistake:

i calculated with the friction force only once. But with the 3:1 mechanical disadvantage i have to ad 3 times in the calc.

my revised results:

107 PSI uphill

52 PSI holding

60 distance

revise my calc:

i calculated with the friction force only once. But with the 3:1 mechanical disadvantage i have to multiply 3 times.

my revised results are :

107 PSI for uphill

52 PSI for holding

60 travel

1. Pressure to retract the load = 50.67 psi

2. Calculate the distance that the load can move = 19.5/3 =6.5 inches (deduct 1/2′ for pulley)

3. Pressure to hold the load (Cf = 1) = 60.93 psi