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Chapter 6: Force and Motion II In this chapter, we will take a rigorous look at three particular forces. The first is friction, which is surprisingly difficult to analyze. The second is the drag force, which affects all moving objects other than those in a vacuum. Finally, we will discuss the forces that arise during uniform circular motion. PC131/PC151: Chapter 6, Pg 1 Contents • Friction • Drag • Uniform Circular Motion PC131/PC151: Chapter 6, Pg 2 Friction • Consider the following “thought experiments”… • A book slides across a horizontal surface. The book slows and then stops. This means that the book must have an acceleration parallel to the surface, in the direction opposite to the book’s velocity. • By Newton’s 2nd law, this means that a force exists in the direction opposite to the book’s velocity. This is the frictional force. • Now, push horizontally on the book to make it move at a constant velocity. The force that you are applying can not be the only force on the book, otherwise it would be accelerating in the direction of your push. By Newton’s 2nd law, there must be a second force, directed opposite to your push, that exactly balances it. This is the frictional force. PC131/PC151: Chapter 6, Pg 3 Friction • Here is a second, slightly more complex situation. • In fig (a), a block rests on a tabletop. As we have seen, there are two forces acting on it; the gravitational force Fg and the normal force FN . They must be perfectly balanced, since the block is not accelerating. • In fig. (b), we exert a force F on the block, attempting to pull it to the left. In response, a frictional force f s is directed to the right, balancing F so that the block still does not move. f s is called the static frictional force. • In figs (c-d), the magnitude of the applied force F increases. In response to this, f s also increases so that the block remains at rest. PC131/PC151: Chapter 6, Pg 4 Friction • In fig (e), the applied force reaches a certain magnitude that the block begins to slide; for a brief moment it accelerates to the left. The frictional force that opposes this motion is called the kinetic frictional force, f k . Note that F and f k are not balanced at this moment, which is evident since the block is accelerating. • In response to the onset of motion, you reduce the applied force, as in fig. (f). When F f k , the block moves at constant velocity. Note that f k is constant, regardless of the applied force F (actually, it’s approximately constant, since the microscopic properties of the surface have a slight variation). • The figure to the right is an example of the measured magnitude of the frictional force during steps (a)-(f). PC131/PC151: Chapter 6, Pg 5 Friction – Physical Description • The frictional force is the vector sum of a vast number of forces acting between the surface atoms of the two objects in contact. • If we take two highly-polished (i.e. atomically smooth) and perfectly clean metal surfaces and bring them together in a vacuum (so that no particles are trapped between them), they can cold-weld together instantly, so that no force can cause them to slide apart. This is because the number of atom-to-atom interactions is enormous. • When there are surface irregularities, the proportion of the surface that is actually in contact is much smaller, and thus the frictional force is weaker, as shown in the figure. • As the surfaces slide, there isa continuous breaking and re-forming of welds, so f k varies slightly. PC131/PC151: Chapter 6, Pg 6 Friction - Properties • When an object presses against a surface and a force Fattempts to slide the object in one direction, the resulting frictional force has three properties: • If the object does not move, then the static frictional force f s and the component of F that is parallel to the surface balance each other; they are equal in magnitude, with opposite directions. • The magnitude of f shas a maximum value, f s ,max , that is given by f s ,m ax s FN where μs is the coefficient of static friction and FN is the magnitude of the normal force on the object from the surface. If the magnitude of the component of F that is parallel to the surface exceeds f s ,max , then the object begins to slide along the surface. This property will become more clear when we look at some sample problems. PC131/PC151: Chapter 6, Pg 7 Friction - Properties • If the object begins to slide along the surface, the magnitude of the frictional force rapidly decreases to a value fk, given by f k k FN where μk is the coefficient of kinetic friction. During the sliding, a kinetic frictional force f k with a magnitude given by the above equation opposes the motion. • It should come as no surprise that the frictional force is proportional to the magnitude of the normal force, since the latter describes how firmly the object presses against the surface. • Note however that the proportionality between frictional and normal forces only refers to their magnitudes; the frictional force is always directed along the surface while the normal force is always directed perpendicular to the surface. PC131/PC151: Chapter 6, Pg 8 Friction - Properties • The coefficients μs and μk are dimensionless, and are determined experimentally. Their values depend on the materials that make up both the object and the surface, as well as the particular physical preparation of the materials (are they atomically smooth? weathered? heat-treated?) • The field of study that deals with friction (as well as lubrication and wear) is called tribology. It is generally considered a sub-discipline of mechanical engineering. PC131/PC151: Chapter 6, Pg 9 Problem #1: Sliding a Block PROBLEM (Sample Problem, pp. 120-121) A block of mass m = 3.0 kg is pulled along a floor by a force F of magnitude 12.0 N. The force is directed at an adjustable angle θ above the vertical, where θ can vary from 0 to 90 degrees. The coefficient of kinetic friction between the block and the floor is μk = 0.40. What angle θ results in the maximum value of the block’s acceleration magnitude a? SOLUTION If there was no friction, then the answer would be obvious; the angle should be zero. This way, the component of F that accelerates the block along the floor would be maximized. When friction is accounted for, we must note that having a slight upward component to F acts to reduce the normal force acting on the block by the floor. This in turn reduces the frictional force that opposes the acceleration. PC131/PC151: Chapter 6, Pg 10 Problem #1: Sliding a Block (Sample Problem, pp. 120-121) Because the block is moving, a kinetic frictional force acts on it: fk = μkFN. We need to find FN. The top figure is a free-body diagram showing the forces on the block along the vertical axis. These forces are the gravitational force, the normal force, and the vertical component of the applied force, which is illustrated in the bottom figure and has a magnitude of Fy= F sin θ. We can now write Newton’s 2nd law for the vertical component of motion, keeping in mind that the block does not accelerate vertically: FN F sin mg ma 0 and therefore, FN mg F sin PC131/PC151: Chapter 6, Pg 11 Problem #1: Sliding a Block (Sample Problem, pp. 120-121) Now let’s look at the horizontal motion. The free-body diagram on the right shows the two forces along this axis; the x-component of the applied force, Fx= F cos θ, and the frictional force fk. which opposes the direction of motion and is thus directed to the left. Newton’s 2nd law for the horizontal motion gives us F cos k FN ma The value a – the acceleration of the block – is what we are seeking. Substituting our calculated value of FN and solving for a, we get F F a cos k g sin m m In this expression, F, m, and μk are constant. We now know acceleration a as a function of θ. PC131/PC151: Chapter 6, Pg 12 Problem #1: Sliding a Block (Sample Problem, pp. 120-121) To find the value of θ that maximizes a, we take the derivative of a with respect to θ and set the result equal to zero: da F F sin k cos 0 d m m A bit of rearranging - the F / m term cancels out - results in tan k and, since μk = 0.40, we find that the acceleration will be maximum if tan 1 (0.40 ) 21 .8 It is worth noting that the optimum angle is a function only of μk and does not depend on the block’s mass or the applied force F. PC131/PC151: Chapter 6, Pg 13 Problem #2: Two Blocks and a Pulley (Problem 6.28) PROBLEM Consider the figure on the right, in which two blocks are connected by a frictionless and massless pulley. The mass of block A is mA = 10 kg, and the coefficient of kinetic friction between block A and the incline is μk = 0.20. The angle θ of the incline is 30 degrees. You observe that block A is sliding down the incline at a constant speed. What is the mass of block B? SOLUTION There are a few different forces at work here. Gravity is acting on both blocks, tension in the rope is pulling on both blocks, and block A experiences kinetic friction (opposing its downward slide). There’s also a normal force upon block A, oriented perpendicular to the incline. PC131/PC151: Chapter 6, Pg 14 Problem #2: Two Blocks and a Pulley (Problem 6.28) We can draw free-body diagrams for each of the blocks separately: For a downhill slide, f k must be directed uphill; its value is negative for the orientation shown. For block A we take the +x direction to be uphill and the +y direction to be in the direction of the normal force. For block B, the +x direction is vertically downward. As with the problem in the last chapter, this is done so that the motion of both blocks can be properly coupled (when A moves uphill, B moves downward by the same amount). The +y direction for block B is irrelevant, as no forces upon or motion of B take place in any direction other than vertically. PC131/PC151: Chapter 6, Pg 15 Problem #2: Two Blocks and a Pulley (Problem 6.28) The problem stipulates that block A slides downhill at a constant speed. This implies that block B moves upward at a constant speed. Both blocks have zero acceleration. We can now write Newton’s 2nd law for three different acceleration components of motion, the x- and y-motion of block A and the x-motion of block B. With all acceleration equal to zero, these equations respectively read T f k m A g sin 0 FN m A g cos 0 mB g T 0 PC131/PC151: Chapter 6, Pg 16 Problem #2: Two Blocks and a Pulley (Problem 6.28) What we have then is three equations in three unknowns (T, FN, and mB). Note that since fk = μkFN, it is not considered as an independent unknown quantity. Those of you who have taken linear algebra have learned a slick way to solve such a problem (it involves a 3 x 3 matrix). However, because at least one of the equations (all three, in fact) involve only two of the three unknowns, this particular problem can be solved by simple back- substitution. I will go through the details in class. The result is mB mA sin k cos 3.3 kg PC131/PC151: Chapter 6, Pg 17 The Drag Force • When an object moves through a fluid, it experiences a drag force D that points in the direction in which the fluid flows relative to the object. • A fluid is anything that can flow; that is, a gas or a liquid. • The magnitude of D depends on several factors, most importantly: • the density of the fluid (air produces less drag than water) • the velocity of the object (a faster-moving object will be “hit” by more of the fluid particles in a given time interval) • the shape of the object (a blunt object like a sphere is affected more than a slender object like a javelin…this is the basis of aerodynamics) PC131/PC151: Chapter 6, Pg 18 The Drag Force • In PC131/PC151, we will only examine blunt objects (approximated as spheres), traveling through still air. • In this case, the magnitude of D can be expressed as 1 D CAv2 2 Here, C is the drag coefficient, which is determined experimentally. ρ is the density of the air (mass per unit volume), v is the object’s velocity, and A is the object’s effective cross-sectional area; that is, the area of a cross-section of the object, taken perpendicular to the direction of its velocity. • We consider C to be a constant for a given object. This is not strictly true; if the velocity varies over a wide range, then C will vary as well. We will ignore this fact. • Typically, C is in the range from 0.4 to 1.0. It is dimensionless (you should be able to determine this by the above equation). PC131/PC151: Chapter 6, Pg 19 The Drag Force • Athletes are well aware of the drag force. To minimize D, they try to minimize their effective cross-sectional area. • Sky divers, on the other hand, want to enjoy the ride as long as possible, so they attempt to increase A. PC131/PC151: Chapter 6, Pg 20 The Drag Force – Terminal Speed • Our previous analysis of free-fall indicates that an object falling in a vacuum will accelerate indefinitely; acceleration is constant, velocity increases linearly with time, and position increases quadratically with time. • When we account for drag, this is no longer the case. We can show this using Newton’s 2nd law. • Imagine that you drop an object from the top of a tall building. Because v is directed downward, D is directed upward. Its magnitude D is initially zero, since v is zero. As v increases, so does D. In this case, Newton’s 2nd law (in 1D, the vertical axis) tells us that D Fg ma that is, the net force on the object equals its mass multiplied by its acceleration. PC131/PC151: Chapter 6, Pg 21 The Drag Force – Terminal Speed • As the object accelerates, its velocity will eventually reach a point where the drag force and the gravitational force balance each other perfectly. As a result, the acceleration will be zero, and the object will fall with a constant speed. We call this the terminal speed, vt. • To find vt, we need only to set a = 0 in Newton’s second law, and insert the formula for D: 1 2 Fg CAvt Fg 0 vt 2 2 CA • In order to use this formula, we must be able to find an expression for Fg. If we are given the mass m of the object, then we can simply use Fg = mg. Many problems only state the physical properties of the object, in which case a little more work is necessary. • If we throw an object downward with an initial speed that is greater than vt, it will slow down. PC131/PC151: Chapter 6, Pg 22 The Drag Force – Terminal Speed • Another measure of the drag force is the 95% distance. This is the distance through which the object must fall from rest to reach 95% of its terminal speed. We won’t be concerned with this measure, but those of you going on to PC235 will eventually understand its meaning. The table below lists some typical terminal speeds in air. Object Term. Speed (m/s) 95% Distance (m) Shot (from shot put) 145 2500 Skydiver 60 430 Baseball 42 210 Tennis ball 31 115 Basketball 20 47 Ping-pong ball 9 10 Raindrop (1.5 mm rad.) 7 6 Parachutist 5 3 PC131/PC151: Chapter 6, Pg 23 Problem #3: Falling Raindrop (Sample Problem, p. 123) PROBLEM A raindrop with radius R = 1.5 mm falls from a cloud that is at height h = 1200 m above the ground. The drag coefficient C for the raindrop is 0.60, and we assume that the drop remains spherical. The density of water is ρw = 1000 kg/m3 and the density of air is ρa = 1.2 kg/m3. What is the terminal speed of the raindrop? SOLUTION First of all, note that there are two different densities specified in this problem, for the raindrop and for the air. The latter is what we simply called ρ in the previous slides; this describes the medium that produces the drag. The density of the water is necessary to calculate the mass of the raindrop (and hence the gravitational force). PC131/PC151: Chapter 6, Pg 24 Problem #3: Falling Raindrop (Sample Problem, p. 123) We previously showed that (with ρ = ρa ) 2 Fg vt C aA Because the raindrop is assumed spherical, its effective cross-sectional area A is πR2. As for the gravitational force, we know that Fg = mg, where m is the mass of the raindrop. This mass is the density ρw times the volume, which is (4/3)πR3 for a sphere. Thus, we have 4 3 Fg R w g All together, 3 2 Fg 8R 3 wg 8R wg vt 7.4 m/s C aA 3C aR 2 3C a PC131/PC151: Chapter 6, Pg 25 Problem #3: Falling Raindrop (Sample Problem, p. 123) Note that the height of the cloud did not enter into the equation; the table shown a few slides ago indicates that the raindrop reaches terminal speed long before it hits the ground. What would be the drop’s speed just before impact if there were no drag force? In the absence of drag, we are back to the case of 1D constant acceleration (with magnitude g). Recall from chapter 2 that in this case, the speed reached by an object that starts from rest and falls through a distance h is v 2gh For h = 1200 m, this speed is 153 m/s, or 550 km/h (!) PC131/PC151: Chapter 6, Pg 26 Uniform Circular Motion • In chapter 4, it was mentioned that when an object moves in a circle (or a circular arc) of radius R at a constant speed v, it is said to be in uniform circular motion. We also showed that such an object has a centripetal acceleration, directed toward the center of the circle, with a magnitude of v2 a R • Newton’s 2nd law dictates that a centripetal force must be responsible for this acceleration. We can investigate the nature of this force by way of some examples. PC131/PC151: Chapter 6, Pg 27 Uniform Circular Motion • As a first example, consider the figure on the right. A hockey puck moves with constant speed v along a circle of radius R. It is tied to a string which is looped around a central peg. • Here, the centripetal force is the tension in the string, which pulls radially inward on the puck. • For a second example, imagine that you are on board a space shuttle that orbits the earth. In this case, the centripetal force is gravity, which from your point of view is always pointing radially inward (toward the center of the earth). PC131/PC151: Chapter 6, Pg 28 Uniform Circular Motion • As a third example, imagine sitting in the back seat of a car. The car turns left, and you slide across the seat toward the right, jamming up against the inside of the passenger-side door. • In this example, the turn of the car is only possible because the tendency of the car to continue in a straight line is opposed by the frictional force between the tires and the road. This frictional force is directed radially inward; it is centripetal. The driver hopes that the maximum static friction is sufficient to keep the car from sliding. • As for your motion in the back seat, your tendency to move in a straight line is initially opposed by friction between the car seat and your @$$. But apparently the maximum static friction could not provide enough centripetal force; this is why you slid to the right. Now, the centripetal force is provided by the car door, which is forcing you to the left (radially inward). PC131/PC151: Chapter 6, Pg 29 Uniform Circular Motion • These examples indicate that a centripetal force is not a specific type of force such as friction, tension, etc. For any situation, A centripetal force accelerates an object by changing the direction of its velocity without changing its speed. • From Newton’s 2nd law, we can write the magnitude F of a centripetal force as mv 2 F R • An object undergoing uniform circular motion has constant values of v and R, and therefore the magnitude of F is constant. However, its direction is continuously changing so that it is directed inward. • Later on in the course, we will re-visit uniform circular motion in the context of angular momentum. PC131/PC151: Chapter 6, Pg 30 Problem #4: The Gravitron Ride • You may be familiar with the Gravitron carnival ride (the text refers to it as the Rotor). This is essentially a large, hollow cylinder that spins around a central axis (we will ignore the fact that the walls are actually sloped). The riders stand against the inner wall of the cylinder, and are “pinned” to the wall even after the floor drops out from beneath them. • This ride relies on two particular effects – centripetal force and friction. Let’s take a close look at what exactly goes on inside the cylinder… PC131/PC151: Chapter 6, Pg 31 Problem #4: The Gravitron Ride • Let’s assume that the cylinder’s radius R is 5.1 m, and that the coefficient of static friction between the rider and the wall is μs = 0.40. Generally the walls have either a thick carpeting or a vinyl padding. This isn’t just for comfort, it’s to provide a suitably large μs. PROBLEM What minimum speed v must the cylinder (and the rider) have if the rider is not to fall when the floor drops? SOLUTION There are three forces to consider here. Fg downward, f s attempts to accelerate the rider opposes this acceleration, and FN is the centripetal (normal) force. PC131/PC151: Chapter 6, Pg 32 Problem #4: The Gravitron Ride • We start by analyzing the forces along a vertical axis at the position of the rider. Because we are interested in the case where the rider remains pinned to the wall without slipping, the vertical acceleration is zero, and Newton’s second law reads f s mg 0 where m is the mass of the rider. • Then, because fs = μsFN, we can write s FN mg 0 FN mg / s • Next, we will analyze the forces along a radial axis that extends from the center of the cylinder through the rider. Newton’s 2nd law reads v2 FN ma m R PC131/PC151: Chapter 6, Pg 33 Problem #4: The Gravitron Ride • Combining the previous two equations, we find that gR v 7.2 m/s s That is, for speeds of 7.2 m/s or greater, the static frictional force is sufficiently large to prevent slippage due to the gravitational force. For speeds slower than this, the rider is only weakly pressed into the wall, and static friction is overwhelmed by gravity. • Notice that the result is independent of the rider’s mass. A big fat guy has no more difficulty remaining pinned to the wall than a small child. The increase in the gravitational force for a large mass is exactly counteracted by the increase in the centripetal (normal) force (and hence in the static frictional force, which is proportional to FN). PC131/PC151: Chapter 6, Pg 34 Problem #5: Angle of a Banked Road • In the example of the car turning around a corner, it was mentioned that a suitable magnitude of static friction must exist between the wheels and the road so that the car does not slide. • In many cases, roads are banked at an angle in order to increase the maximum possible slip-free speed of the cars. It is possible to bank the road at an angle such that no friction is required to keep the car from slipping. PROBLEM For a car traveling with speed v around a curve of radius R, determine a formula for the angle θ at which the road should be banked so that no friction is required. PC131/PC151: Chapter 6, Pg 35 Problem #5: Angle of a Banked Road SOLUTION Even though the road is banked, the car still moves along a horizontal circle, so the centripetal acceleration remains horizontal. A free-body diagram for the car is shown to the right. There are two forces present, the gravitational force (directed downward, as always) and the normal force (directed perpendicular to the surface). As we have assumed in the problem statement, there are no frictional forces present. Since there is no vertical motion, Newton’s 2nd law tells us that FN cos mg 0 PC131/PC151: Chapter 6, Pg 36 Problem #5: Angle of a Banked Road mg Therefore, FN cos As for the horizontal motion, Newton’s 2nd law reveals that v2 FN sin ma m R Combining these two equations, we get mg v2 v2 sin m or tan cos R gR Thus, for an angle θ = tan-1 (v2/gR), a car can make the turn at a speed of v. PC131/PC151: Chapter 6, Pg 37 Video: Physics of Ski Jumping The YouTube video below (which contains audio) describes how ski jumpers handle concepts of drag and lift. We don’t really talk about lift in PC131/PC151; when air flows underneath an angled surface, it produces a net upward force on that surface. This is what allows aircraft to fly. Videos are not embedded in this document - internet access is required to play them PC131/PC151: Chapter 6, Pg 38 Video: Physics of Bobsleigh The YouTube video below (which contains audio) describes how drag, resistance, and centripetal acceleration all play a role in driving a bobsleigh. Videos are not embedded in this document - internet access is required to play them PC131/PC151: Chapter 6, Pg 39

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