132 ‘The sport of ice hockey depends crucially on having just the right amount of friction between a player's skates and the ice. If there ‘were too much friction, the play- ers would move much more slowly; if there were too litle frie- tion, they could hardly keep from falling over. 8-16 The normal and friction forces arise from interactions between molecules at high points ‘on the surfaces of the block and the floor, CHAPTER 5 APPLICATIONS OF NEWTON'S LAWS 5-4 Frictionat Forces We have seen several problems where a body rests or slides on a surface that exerts forces on the body, and we have used the terms normal force and friction force to describe these forces. Whenever two bodies interact by direct contact (touching) of their surfaces, we call the interaction forces contact forces. Normal and friction forces are bothcontact forces. ‘Our concern in this section is with friction, an important force in many aspects of everyday life. The oil in a car engine minimizes friction between moving parts, but with- out friction between the tires and the road we couldn't drive or turn the car. Air drag—the frictional force exerted by the air on a body moving through it—decreases automotive fuel economy but makes parachutes work. Without friction, nails would pull ut, light bulbs would unscrew effortlessly, and riding a bicycle would be hopeless. KINETIC AND STATIC FRICTION Let’s consider a body sliding across a surface. When you try to slide a heavy box of books across the floor, the box doesn’t move at all unless you push with a certain mi ‘mum force. Then the box starts moving, and you can usually keep it moving with less, force than you needed to get it started. If you take some of the books out, you need less force than before to get it started or keep it moving, What general statements can we make about this behavior? First, when a body rests or slides on a surface, we can always represent the contact force exerted by the surface on the body in terms of components of force perpendicular and parallel to the surface. We call the perpendicular component vector the normal force, denoted by 7. (Recall that normal is a synonym for perpendicular.) The component vec- tor parallel to the surface is the friction force, denoted by f- By definition, 7 and f are always perpendicular to each other. We use script symbols for these quantities to empha- size their special role in representing the contact force. If the surface is frictionless, then the contact force has only a normal component, and f is zero. (Frictionless surfaces are an unattainable idealization, but we can approximate a surface as frictionless if the effects of friction are negligibly small.) The direction of the friction force is alway’s such as to oppose relative motion of the two surfaces. ‘The kind of friction that acts when a body slides over a surface is called a kinetic friction force f,. The adjective “kinetic” and the subscript “k” remind us that the two surfaces are moving relative to each other. The magnitude of the kinetic friction force usually increases when the normal force increases. It takes more force to slide a box full of books across the floor than to slide the same box when it is empty. This principle is also used in automotive braking systems; the harder the brake pads are squeezed against the rotating brake disks, the greater the braking effect. In many cases the magnitude of the kinetic friction force f, is found experimentally to be approximately proportional to the magnitude 7 of the normal force. In such cases we can write Jc= 4% (magnitude of kinetic friction force), 6-5) ‘where i, (pronounced “mu-sub-k”) is a constant called the coefficient of kinetic frie- tion. The more slippery the surface, the smaller the coefficient of friction. Because it is ‘a quotient of two force magnitudes, 1, is a pure number without units. CAUTION » Remember, the friction force and the normal force are always perpen: dicular. Equation (5-5) is not a vector equation, but a scalar relation between the ‘magnitudes of the two perpendicular forces. Equation (5-5) is only an approximate representation of a complex phenomenon. On a microscopic level, friction and normal forces result from the intermolecular forces (fundamentally electrical in nature) between two rough surfaces at points where they come into contact (Fig. 5-15). The actual area of contact is usually much smaller than5-4 FRICTIONAL FORCES the total surface area. As a box slides over the floor, bonds between the two surfaces form and break, and the total number of such bonds varies; hence the kinetic friction force is not perfectly constant. Smoothing the surfaces can actually increase friction, since more molecules are able to interact and bond; bringing two smooth surfaces of the same metal together can cause a “cold weld.” Lubricating oils work because an oil film between two surfaces (such as the pistons and cylinder walls in a car engine) prevents, - them from coming into actual contact. Table 51 shows a few representitive values of ,. Although these values are given with two significant figures, they-are only approximate, since friction forces can also depend on the speed of the body relative to the surface. We'll ignore this effect and assume that 11, and f, are independent of speed so that we can concentrate on the sim- plest cases. Table 5—I also lists coefficients of static friction; we'll define these shortly. TABLE 5-1 APPROXIMATE COEFFICIENTS OF FRICTION. MATERIALS STATIC, KINETIC. i Steel on steel 074 037 ‘Aluminum on steel, ost 07 Copper on steel 033 036 Brass on steel 031 os Zine on cast con oss 21 Copper on cast ion 1.05 029 Giasson glass 094 040 Copper on glass 0.68 053 ‘Teflon on Teflon 0.04 0.08 Teflon on steel 0.04 008 Rubber on concrete (dry) LO os Rubber on concrete (wet) 0.30 2s Friction forces may also act when there is no relative motion. If you try to slide that “box of books across the floor, the box may not move at all because the floor exerts an ‘equal and opposite friction force on the box. This is called a static friction force f., In Fig. 5-16a the box is at rest in equilibrium under the action of its weight and the ‘upward normal force 7%, which is equal in magnitude to thé weight and exerted on the box by the floor. Now we tie a rope to the box (Fig. 5—16b) and gradually increase the tension T in the rope. At first the box remains at rest because, as T increases, the force ‘of static friction f, also increases (staying equal in magnitude to 7). At some point, T becomes greater than the maximum static friction force f, the sur- face can exert. Then the box “breaks loose” (the tension 7 is able to break the bonds ‘between molecules in the surfaces of the box and floor) and start to slide. Figure 5~16e is the force diagram when T's at this critical value. If T exceeds this value, the box is no mnger in equilibrium. For a given pair of surfaces the maximum value of f, depends on normal force. Experiment shows that in many cases this maximum value, called na iS approximately proportional to 7; we call the proportionality factor 4. (pro- junced “mu-sub-s”) the coefficient of static frietion. Some representative values of 1, shown in Table 5-1. In a particular situation, the actual force of static friction can ve any magnitude between zero (when there is no other force parallel to the surface) \d a maximum value given by 4.7. In symbols, FS 4.N (magnitude of static friction force). 6-6) ike Eq, (5-5), this is a relation between magnitudes, not a vector relation, The 133 im @ (No sliding) n 4 Ksmit © (Gust about to side) in ie faun © (Now sliding) 5-16 (a) (b), (6) When there is no relative motion of the surfaces, the ‘magnitude of the static frition force fis less than or equal to 4,7. (@) When there is relative motion, the magnitude of the kinetic fric- tion force f, equals 1472.134 ‘Truck Pulls Crate 26 Pushing a Crate 27 Pushing a Crate Up a Wall 28 ‘Skier Goes Down a Slope 29 Skior and Rope Tow 2.10 Pushing a Crate Up an Incline 2a2 ‘Truck Pulls Two Crates 5-17 In response to an externally applied force, the friction force increases 10 (f,)yae Then the sur- faces begin to slide across one another, and the frictional force drops back to a nearly constant value f. The kinetic fiction force varies somewhat as intermolecular bonds form and break. a CHAPTER 5 APPLICATIONS OF NEWTON'S LAWS equality sign holds only when the applied force 7; parallel to the surface, has reached the critical value at which motion is about to start (Fig. 5—16c). When T is less than 1 value (Fig. 5~16b), the inequality sign holds. In that case we have to use the equilibrium conditions (2 F'= 0) to find f. If there is no applied force (T'= 0) as in Fig. 5—16a, then there is no static friction force either (f,= 0). ‘As soon as sliding starts (Fig. 5~16d), the friction force usually decreases; it's eas- ier to keep the box moving than to start it moving. Hence the coefficient of kinetic friction is usually less than the coefficient of static friction for any given pair of surfaces, as shown in Table 5~1. If we start with no applied force (T=0) at time t= 0 and gradu- ally increase the force, the friction force varies somewhat, as shown in Fig. S-17. In some situations the surfaces will alternately stick (static friction) and slip (kinetic friction). This is what causes the horrible squeak made by chalk held at the wrong angle while writing on the blackboard. Another stick-slip phenomenon is the squeaky noise your windshield-wiper blades make when the glass is nearly dry; still another is the out- raged shriek of tires sliding on asphalt pavement. A more positive example is the motion of a violin bow against the string, ‘When a body slides on a layer of gas, friction can be made very small. In the linear air track used in physics laboratories, the gliders are supported on a layer of ait. The fric~ tional force is velocity-dependent, but at typical speeds the effective coefficient of friction is ofthe order of 0.001. A similar device is the air table, where the pucks are sup- ported by an array of small air jets about 2 em apart. Je No elative—s4e— Relative motion Friction in horizontal motion A delivery company has just horizontal force of magnitude 230 N. Once it “breaks loose” and unloaded a SO0-N crate full of home exercise equipment on the _ starts to move, you can keep it moving at constant velocity with sidewalk in front of your house (Fig. 5~18a). You find that fo get only 200 N. What ae the coefficients of static and kinetic it started moving toward your front door, you have to pull with a friction? @ w=500N o © {5-18 (a) Pulling a crate with a horizontal force. (b) Free-body diagram for the crate as it starts to ‘move. (c) Free-body diagram forthe crate moving at constant velocitySOLUTION ‘The state of rest and the state of motion with con- stant velocity are both equilibrium conditions, so we use Eqs. (5-2). An instant before the rate stats to move, the static fric~ tion force has its maximum possible valu, (fj, = A The appropriate force diagram is Fig. S~18b, We find = 230N= (Dau = 0 an = 230.N, -500N =0, n= SOON, SLs In Example 5~13, what is the friction force if the crate is at rest ‘on the surface and a horizontal force of 50.N is applied to it? SOLUTION From the equilibrium conditions we have +f) = 50N~ S.= 50N. 5-4 FRICTIONAL FORCES 135 After the erate starts to move, the forces are as shown in Fig. 5-186, and we have BR=T+CK. f= 200N, ER =n+(-w) f= 4R — (qotion occurs), =f = 200N _ n= = TX = 0.40. It’s easier to keep the crate moving than to start it moving, and so the coefficient of kinetic friction is less than the coefficient of static friction. In this case fis less than the maximum value (f.)g = 47: The frictional force can prevent motion for any horizontal applied force up to 230'N, 13, suppose you try to move your erate Full of ‘exercise equipment by tying a rope around it and pulling upward the rope at an angle of 30° above the horizontal (Fig. 5-19). hhard do you have to pull to keep the crate moving with ity? Is this easier or harder than pulling horizon- 00 N and p= 0.40. 8 onthe erate. The kinetic friction force fis still equal to , bat now the normal force 72s not equal in magnitude tothe gh of te crate. The force exerted by the rope has an add nal vertical component that tends to lift the crate ofthe floor, @ “The crate isin equilibrium, since its velocity is constant, so EF, = Te0s30° + (=ji)= T eos30° - 0.407 = 0, ER = Tsin30° + + (-S00N) = 0. ‘These are two simultaneous equations for the two unknown quantities Tand 2. To solve them, we can eliminate one unknown and solve for the other. There are many ways to do this; here is one way. Rearrange the second equation to the form n= 500 N ~ Tsin 30°. ‘Substitute this expression for 72 back into the first equation: T €0830° ~ 0.40(500 N ~ T'sin 30°) = 0. o136 CHAPTER 5 APPLICATIONS OF NEWTON'S LAWS Finally, solve this equation for 7, then substitute the result back into either of the original equations to obtain 7. The results are T=188N, 2 =406N. Note that the normal force is less than the weight of the box Eee € ‘Toboggan ride with friction! Let's go back tothe toboggan that ‘we studied in Example 5-9 (Section S~3). The wax has worn off, and there is now a coefficient of kinetic friction 1. The slope has just the right angle to make the toboggan slide with constant speed, Derive an expression forthe slope angle in terms of wand SOLUTION Figure 5-20 shows the free-body diagram. The slope angle is The forces on the toboggan are its weight w and the normal and frictional components of the contact force exerted on itby the sloping surface. We take axes perpendicular and parallel to the surface and represent the weight in terms of its compo- nents in these two directions, as shown. The toboggan is in equilibrium because its velocity is constant, and the equilibrium conditions are EF = wsina + (-f) = wsina — y= BE + (-weosa) = 0, Rearranging, we get n= Just as in Example 5-9, the normal force 7is not equal to the weight w, When we divide the first of these equations by the see- ond, we find n= weosa. sin a, = tana. {gan and coefficient of friction as in Example 5-16, but a steeper hill? This time the toboggan accelerates, although not as much, as in Example 5-9, when there was no friction. Derive an expression for the acceleration in terms of gc 4, and ve SOLUTION The free-body diagram (Fig, 521) is almost the ‘same as for Example 516, but the body is no longer in equilib- rium; a is still zero, but a is not. Using w = mg, Newton's second law gives us the two equations EE = mgsinat (-f,) = may, ER =2+ (mgcosa) = 0. From the second equation and Eq. (5-5), = #4 7 We get an expression for fi: = mgcosa, f= m= pymg cos a (w= 500 N) because the vertical component of tension pulls upward on the crate. Despite this, the tension required is a litle less than the 200-N force needed when you pulled horizontally, in Example 513. Try pulling at 22°; you'll find that you need ‘even less force (see Challenge Problem 5-111). 5-20 Free-body diagram for the toboggan with friction, ‘The weight w doesn't appear in this expression. Any toboggan, regardless of its weight, slides down an incline with constant speed if the coefficient of kinetic frietion equals the tangent of, the slope angle of the incline, The steeper the slope, the greater the coefficient of friction has to be for the toboggan to slide with constant velocity. This is just what we should expect. 5-21 Free-body diagram for the toboggan with friction, going down a steeper hil‘We substitute this back into the x-component equation. The result is imgsin a+ (mg cosa) = ma,, @, = g(sin a — 1, cosa). oes this result make sense’ Here are some special cases we ‘can check. First, ifthe hill is vertical, «= 90; then sin c= {cos c= 0, and a, = g. This is free fall, just what we would ‘expect. Second, on a hill at angle « with néfriction, 4, =0. Then sin o The situation is the same asin Example 5-9, and “we get the same result; that’s encouraging! Next, suppose that there is just enough friction to make the toboggan move with constant velocity In that case, a, =0, and our result gives, sina =j4,cosa and ‘This agrees with our result from Example 5~16; good! Finally, rote that there may be so much friction that 4, cos cris actually, greater than sin a In that case, a, is negative; if we give the a, = tan a. LING FRICTION 5-4 FRICTIONAL FORCES 137 toboggan an initial downhill push to start it moving, it will slow down and eventually stop. We have pretty much beaten the toboggan problem to death, but there is an important lesson to be leamed. We started out with a simple problem and then extended it to more and more ‘general situations. Our most general result, found in this exam pie, includes ail the previous ones as special cases, and that’s a nice, neat package! Don't memorize this package; ic is useful only for this one set of problems. But do try to understand how ‘we obtained it and what it means, (ne final variation that you may want to try out is the case in which we give the toboggan an initial push up the hill. The direction ofthe friction force is now reversed, so the accelera- tion is different from the downhill value. It turns out that the expression for a, i the same as for downhill motion except that the minus sign becomes plus. Can you prove this? ‘W's a lot easier to move a loaded filing cabinet across a horizontal floor by using a cart ‘with wheels than to slide it. How much easier? We can define a coefficient of rolling friction 1 which is the horizontal force needed for constant speed on a flat surface divided by the upward normal force exerted by the surface. Transportation engineers call 1u, the tractive resistance. Typical values of u, are 0,002 to 0.003 for steel wheels on steel ils and 0.01 to 0.02 for rubber tires on concrete. These values show one reason why jlroad trains are in general much more fuel-efficient than highway trucks. Eee = Motion with rolling friction A typical car weighs about 12,000 N (about 2700 Ib). I the coefficient of rolling friction is ‘010, what horizontal foree must you apply to push the at constant speed on a level road” Neglect ar resistance. SOLUTION The normal force 7 is equal to the weight w; because the road surface is horizontal and there are no other vertical forces. From the definition of x, the rolling friction force fis f= HM = (0.010)(12,000 N) = 120 N___ (about 27 Ib). From Newton's frst law, a forward force with this magnitude is ‘needed to keep the car moving with constant speed. ‘We invite you to apply this analysis to your erate of exercise ‘equipment (Example 5~13). Ifthe delivery company brings it on arubber-wheeled dolly with ,= 0.02, only a LO-N force is needed to keep it moving at constant velocity. Can you verify this? UID RESISTANCE AND TERMINAL SPEED Sticking your hand out the window of a fast-moving car will convince you of the exis- tence of fluid resistance, the force that a fluid (a gas or liquid) exerts on a body moving. ‘through it. The moving body exerts a force on the fluid to push it out of the way. By Newton's third law, the fluid pushes back on the body with an equal and opposite force. The direction of the fluid resistance force acting on a body is always opposite the direction of the body’s velocity relative to the fluid. The magnitude of the fluid resis- tance force usually increases with the speed of the body through the fluid, Contrast this, behavior with that of the kinetic friction force between two surfaces in contact, which ‘we can usually regard as independent of speed. For low speeds, the magnitude f of the resisting force of the fluid is approximately proportional to the body's speed vs f = kv (fluid resistance at low speed), 6)138 y 5-22 Free-body diagram for a body falling through a fluid, ol 5-23 Graphs of acceleration, velocity, and position versus time for a body falling with fluid resis- tance proportional to v, shown as dark color curves. The light color curves show the corresponding. relations if there is no fluid resis- tance. CHAPTER 5 APPLICATIONS OF NEWTON'S LAWS where k isa proportionality constant that depends on the shape and size of the body and _ the properties of the fluid. In motion through air at the speed of a tossed tennis ball or faster, the resisting force is approximately proportional to v” rather than to v. It is then called air drag or simply drag. Airplanes, falling raindrops, and cars moving at high speed all experience air drag. In this case we replace Eq. (5—7) by f= Dv® (fluid resistance at high speed), 6-8) Beeiiuse of the u* dependence, air drag increases rapidly with increasing speed. The air drag on a typical car is negligible at low speeds but comparable to or greater than the force of rolling resistance at highway speeds. The value of D depends on the shape and size of the body and on the density of the air. We invite you to show that the units of the constant k in Eg. (5—7) are N -s/m or kg/s and that the units of the constant D in Eq. (5-8) are N + s"/m* or kg/m. Because of the effects of fluid resistance, an object falling in a fluid will nor have a constant acceleration. To describe its motion, we can’t use the constant-acceleration relationships from Chapter 2; instead, we have to start over, using Newton’s second law Let’s consider the following situation. You release a rock at the surface of a deep pond, and it falls to the bottom, The fluid resistance force in this situation is given by Eq. (S—1). What are the acceleration, velocity, and position of the rock as functions of time? ‘The free-body diagram is shown in Fig 5-22. We take the positive direction to be downward and neglect any force associated with buoyancy in the water. There are no components, and Newton's second law gives EE, = mg + (kv) ‘When the rock first starts to move, v = 0, the resisting force is zero, and the initial accel- eration is a= g. As its speed increases, the resisting force also increases until finally itis equal in magnitude to the weight. At this time, mg — kv = 0, the acceleration becomes zero, and there is no further increase in speed. The final speed v,, called the terminal speed, is given by mg kv,=0, or y= fe (terminal speed, fluid resistance f = kv), (6-9) Figure 5-23 shows how the acceleration, velocity, and position vary with time. As time ‘goes by, the acceleration approaches zero, and the velocity approaches v, (remember that wwe chose the positive y-direction to be down). The slope of the graph of y versus 1 becomes constant as the velocity becomes constant.Skydiver In deriving the terminal speed in Eq. (5-9), we assumed that the fluid resistance force was proportional to the speed. For an object falling through the air at high speeds, ‘so that the fluid resistance is proportional to u” as in Eq. (5-8), we invite you to show that the terminal speed , is given by [me “VD “This expression for terminal speed explains the observation that heavy objects in air tend ‘to fall faster than light objects. Two objects with the same physical size but different mass (say, a table-tennis ball and a lead ball with the same radius) have the same value ‘of D but different values of m. The more massive object has a larger terminal speed and % (terminal speed, fluid resistance f = Dv*). (5-13) By changing the positions of their arms and legs while falling, sky- divers can change the value of the falls faster. The same idea explains why a sheet of paper falls faster if you first crumple constant D in Eq, (5-8) and hence it into a ball; the mass m is the same, but the smaller size makes D smaller (less air drag adjust the terminal speed of their for a given speed) and v, larger. fall (Eq. (5-13) Terminal speed of a sky diver For a human body falling through When the sky diver deploys the parachute, the value of D air in a spread-eagle position, the numerical value of the cor- inereases greatly, and the terminal speed of the sky diver and stant D in Eq. (5-8) is about 0.25 kg/m. For an 80-kg sky diver parachute is (thankfully) much less than 56 m/s. terminal velocity i, from Eq. (5-13), [mg _ [G0 kg)0-8 mis?) YD" 025kgim = 56 m/s (about 200 km/h, or 125 mi).