Womack Data Sheet 5: Angle Problems in Fluid Power Applications
Trigonometric functions such as sines, cosines, and tangents are useful in solving angle problems... for example if the cylinder axis is at an angle to the direction of load movement... in beam and crane problems where the cylinder is not at right angles with the beam axis, and where the angle continually changes as the beam rises... applications where a heavy load moves upward or downward at an inclination to the horizontal...
The purpose of this sheet is to show simplified methods of using multipliers and dividers in solving these problems. While these methods are based on trigonometry, a knowledge of the subject is not necessary to use these methods
How to Calculate a Friction Load... Figure 1  Horizontal Friction... Frictional resistance depends on the coefficient of friction between the mating surfaces. Coefficients for various materials are shown in machinery handbooks. For example, steel running on cast iron, lubricated, has a coefficient of 0.2. In this figure, using this coefficient with a load weight of 4500 lbs., friction in a horizontal direction is 4500 x 0.2 = 900 lbs. 
Figure 2  Friction on Inclined Plane... If the load is sliding at an angle with the horizontal, use cosine of angle as a multiplier. Using a coefficient of friction 0. 2, friction of a 4500 lb. load on a 40° incline is 4500 x 0. 2 x 0.766 (cos 40°) = 689 lbs. 
Cylinders Pushing at an Angle to Load Direction...
Figure 3  Additional cylinder force is required if cylinder is not pushing directly into load. Downward acting cylinder increases friction of the load. 
If the cylinder axis is not parallel to direction of load movement, more force is required than if it were pushing straight on. When it is pushing at a downward angle, its force will actually create additional friction. The problem is to calculate the cylinder force required. 
MULTIPLIERS (See Text)  
Coef. of Friction 
Angle of Cylinder Axis to the Horizontal  
10º  15º  20º  25º  30º  35º  40º  45º  50º  
0.10  .103  .107  .111  .116  .123  .132  .143  .158  .178 
0.20  .210  .219  .230  .243  .261  .284  .314  .353  .408 
0.30  .321  .337  .358  .384  .418  .463  .523  .605  .724 
0.40  .436  .463  .498  .542  .600  .677  .784  .942  1.19 
0.50  .557  .598  .650  .719  .811  .939  1.12  1.41  1.93 
0.60  .682  .741  .817  .920  1.06  1.27  1.58  2.13  3.28 
0.70  .810  .891  .998  1.14  1.35  1.68  2.21  3.29  6.59 
0.80  .951  1.12  1.21  1.42  1.74  2.26  3.26  5.95  34.7 
0.90  1.09  1.23  1.43  1.71  2.17  2.97  4.81  12.9     
Example: If the load weighs 10,000 lbs., and the coefficient of friction is 0.4, what cylinder force is required to keep it in motion when the cylinder is at an angle of 15° with the direction of travel?
Solution: On a straight push a force of 10,000 x 0.4 = 4000 lbs. would be required. But it will take more than this because of the 15° angle. For a 15° angle and a coefficient of 0.4, the table shows a multiplier of .463. Cylinder force, then, is 10,000 x .463 = 4630 lbs.
Trig Table

Figure 4  Angle of Repose. As the angle of incline is increased, friction between the load weight and the inclined plane will prevent the weight from sliding downhill until the angle of repose is reached. The coefficient of friction between the two mating surfaces is equal to the tangent of the angle of repose. Figure 5  Load on an Inclined Plane. Cylinder and load axes are in line but load is moving on an upward incline. The cylinder must provide enough force to raise the load to a higher elevation against the force of gravity. If the cylinder were mounted in a vertical position, a force of a little over 5000 lbs. would be needed to lift a 5000 lb. load weight. But with the cylinder inclined as in the illustration, the same 5000 lb. load can be lifted with a smaller cylinder force because of mechanical advantage which is similar to a wedging action. Figure 6  Cylinder Operating a Rotating Lever. Cylinder force, F, is horizontal in this figure. Only that portion, T, which is at right angles to the lever axis is effective for turning the lever. The value of T varies with the acute angle "A" between the cylinder axis and the lever axis. Example: A 4inch bore cylinder working at 750 PSI will develop a 9420 lb. force (12.56 sq. in. area x 750 PSI). Effective force T when working at a 65° angle is: 9420 x .906 (sine of 65°) = 8535 lbs. Figure 7  Cylinder on a Crane. Find the lifting capacity of this crane when its members are at the angles shown (capacity will vary as the beam of the crane raises and lowers). 
© 1988 by Womack Machine Supply Co. This company assumes no liability for errors in data nor in safe and/or satisfactory operation of equipment designed from this information.