Here is what you will be able to do when you complete each objective: 1. State the general laws of static and kinetic friction.

Transcription

1 Friction I Learning Outcome When you complete this module you will be able to: Describe and solve problems involving friction. Learning Objectives Here is what you will be able to do when you complete each objective: 1. State the general laws of static and kinetic friction. 2. Define the coefficient of friction. 3. Solve problems involving friction forces on a horizontal plane. 1

2 FRICTION The force of friction can be defined as a force that opposes motion of one surface over another. This opposition to motion is due to the irregularities of the two surfaces. The direction of this force of friction is opposite to the direction of any force that is trying to move or is moving an object. The control of friction is necessary in many situations. Friction may be reduced by the use of lubricants that in effect separate the two surfaces and results in only the resistance of the lubricant itself causing friction. An example of this is a rotating shaft, which is supported by an oil-supplied bearing. The oil, which is under pressure, causes the shaft to ride on a cushion of oil instead of allowing metal to metal contact. Friction may, in some cases, be necessary and an increase in friction may be achieved by the proper selection of materials, as well as the use of forces in the proper direction. An example of this could be two pulleys joined by a belt. The selection of the belt material and belt tension may increase the friction. In other words, friction can be an asset or a liability depending upon the application. TYPES OF FRICTION Friction occurs in several different forms: 1. Standing or static friction is the resistance that opposes the initial movement of a body at rest. In other words, to start an object moving, a certain amount of frictional resistance must be overcome; this resistance is called static friction. 2. Sliding or kinetic friction is the resistance that opposes the continued movement of an object. To keep an object moving at a constant speed requires a constant force to overcome the kinetic friction. 3. Rolling friction is the resistance that opposes the motion of a wheel or roller as it rolls along a surface. An example of this is a tire rolling over concrete. 4. Fluid friction is the resistance to movement within the layers of a fluid. Friction always exists to some extent and a percentage of the energy input to any machine is consumed in overcoming frictional forces. The two main types of friction that will be examined are static friction and kinetic friction. 2

3 LAWS GOVERNING FRICTION The laws that govern friction are not precise; however, these laws have been experimentally determined to hold true for any situation encountered. The laws can be stated as follows: 1. The force of friction is proportional to the force that presses the two surfaces together. For static or kinetic friction this will depend upon the mass of an object on a horizontal surface, or more precisely the total force normal (perpendicular) to the surface. If the downward force is doubled, the frictional force between the two surfaces is doubled. 2. Static friction (standing friction) is always greater than kinetic friction (moving friction). The amount of force required to start an object moving is greater than the amount of force required to keep the object moving at constant speed. 3. The force of friction, whether static or kinetic, is not affected by the area of the two surfaces in contact. If the same downward force is distributed over twice as much area, there is no change in the frictional forces between the two surfaces. 4. Kinetic friction is not affected by the speed of the body. Within reasonable limits, the forces of kinetic friction between two surfaces will remain unchanged as the speed of the object is increased or decreased. 5. The force of friction (either static or kinetic) is affected by the relative roughness of the two surfaces in contact. The rougher the surfaces, the greater the forces required to overcome friction. If a greater force of friction is desired, it is only necessary to increase the roughness on one of the surfaces. 6. Kinetic friction (sliding friction) is greater than rolling friction. This fact explains why roller type bearings are used extensively. 3

4 COEFFICIENT OF FRICTION The coefficient of friction (µ) is a ratio of the force required to move a body to the normal reaction force. Since the coefficient of friction is a ratio, there are no units. Stated mathematically, µ = F F / R N Where: µ is the coefficient of friction. F F is the force (parallel to the surface) required to start an object moving or to keep an object moving at constant speed. At the instant the object moves or is kept moving at a constant speed, F A = F F (Friction Force) when dealing with horizontal forces and surfaces. (F A is show in Fig. 1) R N is the reaction normal (perpendicular) to the surface. R N is opposite in direction and numerically equal to the downward force of the object due to gravity when dealing with horizontal surfaces. NOTE: F F and R N must be at right angles. Table 1 shows some examples of the coefficient of friction between two surfaces. The values have been determined experimentally and should be considered as typical ranges. Surfaces Coefficient of Friction metal on metal 0.15 to 0.65 greased metal on metal 0.02 to 0.06 metal on wood 0.20 to 0.60 wood on wood - dry 0.25 to 0.55 wood on wood - wet 0.10 to 0.45 rubber on concrete 0.60 to 0.95 Table 1 Coefficients of Friction 4

5 Figure 1 (a) Static Friction Figure 1 (a) shows the forces acting on a block under static friction conditions. If force F A is less than force F F, then the block will not move. When force F A is equal to force F F, then the block will just start to move. If force F A is further increased, then the block will accelerate. Figure 1 (b) Kinetic Friction Figure 1 (b) shows the forces acting on a block under kinetic friction conditions. If the force F A is equal to the force F F, then the block will move at constant velocity. If the force F A is further increased, then the block will accelerate. 5

6 Example 1: Find the coefficient of friction between a 100 kg box and a concrete floor if a horizontal force of 460 N is required to start the box moving. Solution: First, R N must be calculated from the 100 kg mass. The mass (m) multiplied by the acceleration due to gravity (g) will be equal to the downward force (F W ). R N is numerically equal to the downward force when the surface and applied force are both horizontal. Therefore: F W = m g = 100 kg x 9.81 m/s 2 = 981 N Then: R N = 981 N Sketch 2 And: µ s = F F / R N = 460 N / 981 N = = 0.47 (Ans.) The coefficient of static friction is

7 Example 2: Find the coefficient of sliding (kinetic) friction if the box in Example 1 requires a force of 430 N to keep it moving at constant speed. Solution: R N = 981 N as before. µ k = F F / R N = 430 N / 981 N = = 0.44 (Ans.) The coefficient of kinetic friction is NOTE: The coefficient of static friction is greater than the coefficient of kinetic friction. µ s > µ k The subscripts s and k are not always included with µ since the wording of a problem often indicates whether the frictional forces are static or kinetic. 7

8 Example 3: Find the force required to move a pump and the skid on which it rests. The total mass is 2000 kg. The coefficient of static friction between the floor and the skid is Solution: First find the value of R N : F W = 2000 kg x 9.81 m/s 2 Thus: R N = N = N Now substitute in the equation: µ k = F F / R N 0.26 = F F / N 0.26 x N = F F Therefore: F F = 0.26 x N = 5101 N (Ans.) A force of 5101 N will start the pump and skid moving. 8

9 Example 4: Find the mass of an object when a horizontal force of 2100 N just starts to move the object along a horizontal surface. The coefficient of static friction is Solution: µ s = 0.40 F F = 2100 N µ s = F F / R N 0.40 = 2100 N / R N 0.40 R N = 2100 N R N = 2100 N / 0.40 = 5250 N The downward force (F W ) is numerically equal to R N, since the surface and applied force are horizontal. Therefore: And: F W F W = 5250 N = m g m = F W / g = 5250 N / 9.81 m/s 2 = 535 kg (Ans.) 9

10 Self Test After completion of the Self-Test, check your answers against the answer guide that follows. 1. (a) Find the coefficient of static friction of a crate on a wooden floor if the crate has a mass of 400 kg and the horizontal force being applied is 1600 N. (b) (c) Find the coefficient of kinetic friction if a force of 1400 is required to keep the same crate moving at constant speed. Find the coefficient of kinetic friction between the same crate and the floor if water is used as a lubricant and the force required to move the crate at a constant speed is 1100 N. 2. Find the force required to keep a wooden block moving along a horizontal surface if the block has a mass of 25 kg and the frictional resistance between the two surfaces is

11 Self Test Answers 1. (a) 0.41 (b) 0.36 (c) N 11

12 References and Reference Material 1. Levinson, Irving J. Introduction to Mechanics. Englewood Cliffs, NJ: Prentice-Hall Inc. 2. Spiegel, Murray R. Theory and Problems of Theoretical Mechanics. New York, NY: Schaum Publishing Co. 12