Chapter 11 Friction Brake Theory After studying this chapter, you will be able to: D Define friction. D Explain the relationship of weight and speed to kinetic energy. Define coefficient of friction. D Define static and kinetic friction. D Define brake fade. D Explain the relationship of friction to heat development. o Identify and describe the factors affecting stopping power. Identify outside factors affecting coefficient of friction. o Describe brake lining materials and construction. Important Terms Friction Static friction Weight transfer Metallic Kinetic energy Kinetic friction Inertia Rotm Momentum Brake fade Non-metall ic Drum Coefficient of friction Heat dissipation Semi-metallic 179 180 Auto Brakes Before stu dyi ng the operation of the brake friction possible. This is why engines and drivetrain parts have members, you must have an understanding of the basic elaborate lubrica ti on sys tems. principles of friction. These principles explain what friction In the brake sys tem, friction is put to work to over­ is, how it is used to overcome vehicle motion, and various come the vehicle's momentum; in other words, to stop the factors affecting brake material construction. Thi s chapter vehicle. What momentum is, and how it is overcome is is an introducti on to bl'ake friction principles w hich will explained in th e following paragraphs. prepare you for the disc and drum brake theory and se rvice chapters. Momentum and Kinetic Energy What Is Friction? Kinetic energy is the total amount of energy the brake system must convert to stop the vehicle and is measured in When two things move against each other, there is a foot-pounds (ft-Ibs.) or Newton-meters (N. m). You w ill re sista nce to the movement between th em. This is caused often hear the term momentum used in pl ace of the term by microscopic imperfections (high spots) th at exist on kinetic energy. Although th e two terms are closely related even th e smoothest surfaces, Figure 11-1. Th e imperfec­ mathematica lly and both are functions of the ve hicle's tions on one surface contact the imperfections on the other weight and velocity, kinetic energy is the measurement surface as they move against each other. The resistance you should use when discussing the vehicle's braking sys­ caused by this contact is called friction. tem. As a tec hni cian, you will not be asked to calculate if there was no friction, tires would have no traction kinetic energy. However, for youl' own understa nding of against the road. Without friction, bolts wou ld not tighten, the basic principles of brake operation, you should and doors would not stay closed . However, in many remember the following concepts. instances, it is desirable to minimize friction as much as At any given speed, kinetic energy increases directly with the weight of the vehicle. For example, a 4000 lb. (1812 kg) vehicle traveling at a given speed w ill have tw ice th e kinetic Imperfections in the surfaces energy of a 2000 lb. (906 kg) vehicle trave ling at the same resist movement speed. A 6000 lb. (2722 kg) will have three times the kinetic Two smooth energy of th e 2000 lb. (906 kg) vehicle, Figure 11-2. Speed has much more effect on kinetic energy th an weight does. For example, if a car travels twice as fast as an identical car, it has four times the kinetic energy of the slowel' car. If the faster car is moving at three times the speed of the slower car, it has nine times the kinetic energy. If the fastel' ca r is traveling four times the speed of the slower car, it w ill have sixteen times the kinetic energy. Brakes perform more work stopping a light vehic le going at a high rate of speed th an a hea vier ve hicle going slowly. You would think that a 4000 lb. vehic le moving at 25 mph (40 kph) would have th e same amount of kinet ic energy as a 2000 lb. (906 kg) ve hi c le moving at Figure 11-1. A magnifying glass showing the surface imperfec­ 50 mph (80 kph). However, the 4000 lb. (1812 kg) vehi­ tions which cause friction between the two moving surfaces. cle has only 86,000 ft-Ibs. of kinetic energy at 25 mph, These imperfections act like two pieces of sandpaper being while the 2000 lb. vehicle has over 172,000 ft-Ibs. of rubbed together. kineti c energy at 50 mph. 1x kinetic energy ... 1x kinetic energy ... 20 mph (32 kph) 20 mph (32 kph) 4x kinetic energy ... 40 mph (64 kph) 2x kinetic energy ... 9x kinetic energy ... 20 mph (32 kph) 4000 Ibs. (1812 kg) A 8 60 mph (92 kph) 2000 Ibs. (906 kg) Figure 11-2. A vehicle's kinetic energy is a combination of its mass (weight) and velocity (speed). A-Kinetic energy increases in direct proportion to the weight. B-A vehicle 's kinetic energy increases exponentially as the speed increases. Chapter 11 Friction Brake Theory 181 shoes or pads become glazed, they do not create as much Putting Friction to Work friction against the drum or rotor surface. This is similar to making the box out of a slicker material. The job of the brake system is to overcome momen­ tum and stop the vehicle. To do this, it uses hydraulics, pneumatics, mechanical leverage, and friction. Note: The coefficient of friction will always Hydraulics, pneumatics, and mechanical leverage were be less than one. discussed in earlier chapters. The following sections explain how friction is put to work to stop the vehicle. Static and Kinetic Friction Coefficient of Friction The two basic types of friction are stationary or static As stated earlier, friction is always present between friction and kinetic friction, sometimes called sliding or two matel'ials that slide against each other. The coefficient dynamic friction. Keep in mind that static friction is a of friction is the amount of friction that can be produced holding action that keeps a stationary object in place, as two materials sl ide across each other. The coefficient of while kinetic friction slows a moving object by converting friction is determined by a simple calculation. In the exam­ momentum to heat. Note that static friction is always ple shown in Figure 11-3, the coefficient of friction is cal­ higher than ki netic friction. culated by measuring th e force required to slide a block The most obvious use of static friction is the parking over a surface and then dividing it by the weight of the brake, Figure 11-5. When the parking brake is appl ied, block. static friction between the applied brake components If it takes 10 Ibs. (22 kg) of force to sl ide a 10 lb. resists movement. To move the vehicle, static friction must (16 kg) block over a flat surface, the coefficient of friction be eliminated by releasing the brakes. Since the vehicle for the block is 1. Another 10 Ib, (22 kg) object, made from has no movement, there is no momentum to overcome, a different material, may only require Sibs. (8 kg) of force and no heat is generated. to slide it across the same surface. In this case, the coeffi­ As discussed earlier, vehicle weight times vehicle cient of friction for that block is .5. See Figure 11-4. speed equals momentum. Applying the brakes on a mov­ In the previous example, the way the box slides ing vehicle causes the stationary friction members (pads or across the floor was altered by changing the material that shoes) to be forced into contact with the rotating friction the box was made of. When the box was made out of a members (rotors or drums). This contact, Figure 11-6, slicker material, it became easier to slide. However, if you causes friction and heat, which results in the rotating parts look at the basic relationships between the box, the floor, slowing and eventually stopping. Since the momentum of and the pulling force, you will see th at the same effect the rotating parts is called kinetic energy, the friction used could have been achieved by decreasing the box weight or to stop the rotating parts is called kinetic friction. by using a different material on the floor. Now that you understand the basic relationships measured by the coefficient of friction, you can apply these relationships to the operation of a vehicle's brake system. Imagine the box is the bl'ake pad or shoe, the Bilk 1 weight of the box is the hydraulic pressure used to apply the brakes. The shop floor is drum or rotor surface, and the If pull required pulling force is the momentum in the turning drum or rotor. weight is 10 pounds, 10 Ibs. By applying more pressure to the brakes, you can slow the C.O.F. =100% or 1.0 momentum of the spinning drum or rotor. This is the same as increasing the weight of the box, making it more diffi­ A ~ Shopfloor cult to pull across the shop floor. Similarly, when brake Block 2 1 If pull required If pull required Weight is 5 pounds, I Weight is 100 pounds, 101bs. C.O.F. =.5 t 100 Ibs. C.O.F. = 100% or 1.0 B Shopfloor Shop floor ~ Figure 11-3. A 100 pound (45,359 kg) weight being pulled Figure 11-4. Two examples illustrating the coefficient of friction. across the shop floor. The coefficient of friction (C.OF) is 1.0.