SECTION 4. Main Ideas. Explain the difference between mass and weight. Find the direction and magnitude of normal forces.

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1 Photo Researchers, Inc. led Everyday Forces Key Terms weight static friction coefficient of friction normal force kinetic friction Weight How do you know that a bowling ball weighs more than a tennis ball? If you imagine holding one ball in each hand, you can imagine the downward forces acting on your hands. Because the bowling ball has more mass than the tennis ball does, gravitational force pulls more strongly on the bowling ball. Thus, the bowling ball pushes your hand down with more force than the tennis ball does. The gravitational force exerted on the ball by Earth,, is a vector quantity, directed toward the center of Earth. The magnitude of this force,, is a scalar quantity called weight. The weight of an object can be calculated using the equation = m a g, where a g is the magnitude of the acceleration due to gravity, or free-fall accel eration. On the surface of Earth, a g = g, and = mg. In this book, g = 9.81 m/s 2 unless otherwise specified. Weight, unlike mass, is not an inherent property of an object. Because it is equal to the magnitude of the force due to gravity, weight depends on location. For example, if the astronaut in Figure 4.1 weighs 800 N (180 lb) on Earth, he would weigh only about 130 N (30 lb) on the moon. The value of a g on the surface of a planet depends on the planet s mass and radius. On the moon, a g is about 1.6 m/s 2 much smaller than 9.81 m/s 2. Even on Earth, an object s weight may vary with location. Objects weigh less at higher altitudes than they do at sea level because the value of a g decreases as distance from the surface of Earth increases. The value of a g also varies slightly with changes in latitude. The Normal Force FIGURE 4.1 Imagine a television set at rest on a table. We know that the gravitational force is acting on the television. How can we use Newton s laws to explain why the television does not continue to fall toward the center of Earth? An analysis of the forces acting on the television will reveal the forces that are in equilibrium. First, we know that the gravitational force of Earth,, is acting downward. Because the television is in equilibrium, we know that another force, equal in magnitude to but in the opposite direction, must be acting on it. This force is the force exerted on the television by the table. This force is called the normal force,. Differentiated Instruction English Learners English learners might have difficulty differentiating between homophones. For example, this page includes terms that have homophones with similar sounds but very different meanings. For example: weigh and way night and knight SECTION 4 Main Ideas Explain the difference between mass and weight. Find the direction and magnitude of normal forces. Describe air resistance as a form of friction. Use coefficients of friction to calculate frictional force. weight a measure of the gravitational force exerted on an object; its value can change with the location of the object in the universe Weight on the Moon On the moon, astronauts weigh much less than they do on Earth. normal force a force that acts on a surface in a direction perpendicular to the surface Forces and the Laws of Motion 133 When discussing concepts as a class, check occasionally to be sure that students know which meaning of a term is used when homophones are discussed. While students may eventually figure it out themselves, they may also miss valuable information in the meantime. 5/4/2011 2:57:41 PM SECTION 4 Plan and Prepare Preview Vocabulary Visual Vocabulary Students may have difficulty understanding the difference between the weight of an object and its mass. Sketch a diagram showing an astronaut with a weight on Earth of 180 lb. Then sketch the same astronaut on different planets with different gravitational forces. Make the astronaut the same size to emphasize that mass is the same everywhere. A person who weighs 180 lb on Earth would have these weights on different bodies in the solar system: Mercury: 68 lb, Venus: 163 lb, moon: 30 lb, Jupiter: 426 lb. Teach TEACH FROM VISUALS FIGURE 4.1 Point out that it is easier to lift a massive object on the moon than on Earth because the object weighs less on the moon, even though its mass remains the same. Also, an object s inertia is the same regardless of the magnitude of free-fall acceleration. Ask Will a dart shot from a dart gun go farther horizontally on Earth or on the moon? Disregard air resistance. Answer: The dart will travel farther on the moon. Because the dart is accelerated downward more slowly on the moon than on Earth, it is in motion for a longer time on the moon. The horizontal velocity will be the same in each case. Teaching Tip For practical purposes, the gravitational field near the surface of Earth is constant. For example, a person who weighs lb at sea level would weigh lb at an altitude of 9.0 km above sea level. This difference in weight is only 0.3%. Forces and the Laws of Motion 133

2 Teach continued TEACH FROM VISUALS FIGURE 4.2 Tell students that the television is in equilibrium, so the normal force from the table must be equal in magnitude and opposite in direction to the gravitational force exerted on the television. Instruct students to draw free-body diagrams for the television, table, and Earth and to identify the third-law pairs. Ask Do the forces in Figure 4.2 constitute an action-reaction pair (Newton s third law)? Answer: No, both forces act on the television and therefore cannot be an action-reaction pair. Teaching Tip Be sure students understand why the two angles labeled θ in Figure 4.3 are equal. Recognizing equal angles in free-body diagrams is an important problem-solving skill for this chapter. FIGURE 4.2 Normal Force In this example, the normal force,, is equal and opposite to the force due to gravity,. FIGURE 4.3 Normal Force When an Object Is on a Ramp The normal force is not always opposite the force due to gravity, as shown by this example of a refrigerator on a loading ramp. static friction the force that resists the initiation of sliding motion between two surfaces that are in contact and at rest FIGURE 4.4 Overcoming the Force of Friction The word normal is used because the direction of the contact force is perpendicular to the table surface and one meaning of the word normal is perpendicular. Figure 4.2 shows the forces acting on the television. The normal force is always perpendicular to the contact surface but is not always opposite in direction to the force due to gravity. Figure 4.3 shows a free-body diagram of a refrigerator on a loading ramp. The normal force is perpendicular to the ramp, not directly opposite the force due to gravity. In the absence of other forces, the normal force,, is equal and opposite to the component of that is perpendicular to the contact surface. The magnitude of the normal force can be calculated as = mg cos θ. The angle θ is the angle between the normal force and a vertical line and is also the angle between the contact surface and a horizontal line. The Force of Friction Consider a jug of juice at rest (in equilibrium) on a table, as in Figure 4.4(a). We know from Newton s first law that the net force acting on the jug is zero. Newton s second law tells us that any additional unbalanced force applied to the jug will cause the jug to accelerate and to remain in motion unless acted on by another force. But experience tells us that the jug will not move at all if we apply a very small horizontal force. Even when we apply a force large enough to move the jug, the jug will stop moving almost as soon as we remove this applied force. Friction opposes the applied force. When the jug is at rest, the only forces acting on it are the force due to gravity and the normal force exerted by the table. These forces are equal and opposite, so the jug is in equilibrium. When you push the jug with a small horizontal force F, as shown in Figure 4.4(b), the table exerts an equal force in the opposite direction. As a result, the jug remains in equilibrium and therefore also remains at rest. The resistive force that keeps the jug from moving is called the force of static friction, abbreviated as F s. F F s F F k (a) Because this jug of juice is in equilibrium, any unbalanced horizontal force applied to it will cause the jug to accelerate. (b) When a small force is applied, the jug remains in equilibrium because the static-friction force is equal but opposite to the applied force. (c) The jug begins to accelerate as soon as the applied force exceeds the opposing static-friction force. 134 Chapter 4 Differentiated Instruction Below Level Hands-on exploration will help students to better understand the difference between static friction and kinetic friction. Encourage them to explore friction by gently pushing objects of varying mass across their desks. Which objects required greater force to overcome static friction? Students may say larger or heavier objects. Remind them that these objects have greater mass. Encourage students to discuss examples of friction in everyday life. How is static friction involved in walking? Static friction holds your foot in place while you move your other foot. Compare the friction of walking on an icy surface compared with a dry one. Less static friction on ice means that the friction is easier to overcome, which makes you slip and fall. When pushing something large, why is it easier to keep it moving than to start it moving? Static friction is greater than kinetic friction. Untitled /4/2011 2:57: Chapter 4

3 led As long as the jug does not move, the force of static friction is always equal to and opposite in direction to the component of the applied force that is parallel to the surface ( F s = - F applied ). As the applied force increases, the force of static friction also increases; if the applied force decreases, the force of static friction also decreases. When the applied force is as great as it can be without causing the jug to move, the force of static friction reaches its maximum value, F s,max. Kinetic friction is less than static friction. When the applied force on the jug exceeds F s,max, the jug begins to move with an acceleration to the left, as shown in Figure 4.4(c). A frictional force is still acting on the jug as the jug moves, but that force is actually less than F s,max. The retarding frictional force on an object in motion is called the force of kinetic friction (F k ). The magnitude of the net force acting on the object is equal to the difference between the applied force and the force of kinetic friction ( F applied - F k ). At the microscopic level, frictional forces arise from complex interactions between contacting surfaces. Most surfaces, even those that seem very smooth to the touch, are actually quite rough at the microscopic level, as illustrated in Figure 4.5. Notice that the surfaces are in contact at only a few points. When two surfaces are stationary with respect to each other, the surfaces stick together somewhat at the contact points. This adhesion is caused by electrostatic forces between molecules of the two surfaces. Tips and Tricks In free-body diagrams, the force of friction is always parallel to the surface of contact. The force of kinetic friction is always opposite the direction of motion. To determine the direction of the force of static friction, use the principle of equilibrium. For an object in equilibrium, the frictional force must point in the direction that results in a net force of zero. The force of friction is proportional to the normal force. It is easier to push a chair across the floor at a constant speed than to push a heavy desk across the floor at the same speed. Experimental observations show that the magnitude of the force of friction is approximately proportional to the magnitude of the normal force that a surface exerts on an object. Because the desk is heavier than the chair, the desk also experiences a greater normal force and therefore greater friction. Friction can be calculated approximately. Keep in mind that the force of friction is really a macroscopic effect caused by a complex combination of forces at a microscopic level. However, we can approximately calculate the force of friction with certain assumptions. The relationship between normal force and the force of friction is one factor that affects friction. For instance, it is easier to slide a light textbook across a desk than it is to slide a heavier textbook. The relationship between the normal force and the force of friction provides a good approximation for the friction between dry, flat surfaces that are at rest or sliding past one another. kinetic friction the force that opposes the movement of two surfaces that are in contact and are sliding over each other FIGURE 4.5 Microscopic View of Surfaces in Contact On the microscopic level, even very smooth surfaces make contact at only a few points. Forces and the Laws of Motion 135 5/4/2011 2:57:43 PM Demonstration Static vs. Kinetic Friction Purpose Show that kinetic friction is less than static friction. Materials rectangular block, hook, spring scale Procedure Use the spring scale to measure the force required to start the rectangular block moving. Then, use the spring scale to measure the frictional force for constant velocity. Perform several trials. Have students record all data and find the average for each. Point out that the normal force and the surfaces remain the same, so the only difference in the two average values is due to motion. Demonstration Friction of Different Surfaces Purpose Show students that the force of friction depends on the surface. Materials large cube with different materials (such as glass, carpeting, and sandpaper) covering each of four sides, with two sides left uncovered; hook; spring scale Procedure Attach the hook to one of the two uncovered sides of the block. Pull the block across the table with the spring scale. Repeat the demonstration with a new surface of the cube exposed to the table. Repeat the demonstration for the two remaining covered sides. Have students summarize the results and reach a conclusion concerning the nature of the surfaces in contact and the frictional force. Forces and the Laws of Motion 135

4 Teach continued Demonstration Friction and Surface Area Purpose Show the relation between surface area and frictional forces. Materials rectangular block, hook, spring scale Procedure Attach the hook to the block. Pull the block across the table with the spring scale. Have students note the force required to pull the block at a constant velocity. Repeat the demonstration for another surface area in contact with the table and have students note the force. Ask students to summarize the results and to reach a conclusion concerning the areas in contact and frictional forces. TEACH FROM VISUALS FIGURE 4.6 Remind students that frictional force depends on the coefficient of friction and the normal force. Ask What changes in environment might cause a change in the frictional force experienced by the snowboarder on the way down the hill? Answer: Answers will vary but could include the following: surface conditions (such as wet or dry snow, ice, and dirt), whether the snowboarder is moving or not moving, and the angle of the hill. coefficient of friction the ratio of the magnitude of the force of friction between two objects in contact to the magnitude of the normal force with which the objects press against each other FIGURE 4.6 Minimizing Friction Snowboarders wax their boards to minimize the coefficient of friction between the boards and the snow. FIGURE 4.7 The force of friction also depends on the composition and qualities of the surfaces in contact. For example, it is easier to push a desk across a tile floor than across a floor covered with carpet. Although the normal force on the desk is the same in both cases, the force of friction between the desk and the carpet is higher than the force of friction between the desk and the tile. The quantity that expresses the dependence of frictional forces on the particular surfaces in contact is called the coefficient of friction. The coefficient of friction between a waxed snowboard and the snow will affect the acceleration of the snowboarder shown in Figure 4.6. The coefficient of friction is represented by the symbol μ, the lowercase Greek letter mu. The coefficient of friction is a ratio of forces. The coefficient of friction is defined as the ratio of the force of friction to the normal force between two surfaces. The coefficient of kinetic friction is the ratio of the force of kinetic friction to the normal force. μ k = _ F k The coefficient of static friction is the ratio of the maximum value of the force of static friction to the normal force. μ s = _ F s,max If the value of μ and the normal force on the object are known, then the magnitude of the force of friction can be calculated directly. F f = μ Figure 4.7 shows some experimental values of μ s and μ k for different materials. Because kinetic friction is less than or equal to the maximum static friction, the coefficient of kinetic friction is always less than or equal to the coefficient of static friction. COEFFICIENTS OF FRICTION (APPROXIMATE VALUES) μ s μ k μ s μ k steel on steel waxed wood on wet snow aluminum on steel waxed wood on dry snow 0.04 rubber on dry concrete metal on metal (lubricated) rubber on wet concrete 0.5 ice on ice wood on wood Teflon on Teflon glass on glass synovial joints in humans Chapter 4 Problem Solving Mark Gallup/Pictor/Image State Reality Check Encourage students to rely on their common sense and experience with real objects when checking answers. For example, if they are asked to calculate and compare the coefficient of static friction of the same object on a rough surface and a smooth surface, and their answer gives a greater value for the smooth surface, then they should recognize that there is an error in the calculation. They know from experience that friction is less on smoother surfaces compared with rougher surfaces, and they should apply this real-life experience to the mathematical model represented by the equation. If they are calculating the coefficient of kinetic friction and the coefficient of static friction for an object being dragged across a paved surface, they should expect the coefficient of kinetic friction to be the lesser value, because they have experienced that it takes less force to keep an object moving than to start it moving. Untitled /4/2011 2:57: Chapter 4

5 Coefficients of Friction Sample Problem D A 24 kg crate initially at rest on a horizontal floor requires a 75 N horizontal force to set it in motion. Find the coefficient of static friction between the crate and the floor. ANALYZE Given: F s,max = F applied = 75 N SOLVE Tips and Tricks Because the crate is on a horizontal surface, the magnitude of the normal force ( ) equals the crate s weight (mg). m = 24 kg Unknown: μ s =? Use the equation for the coefficient of static friction. _ F s,max = _ F s,max μ s = mg μ 75 N s = 24 kg 9.81m/s 2 μ s = 0.32 PREMIUM CONTENT 1. Once the crate in Sample Problem D is in motion, a horizontal force of 53 N keeps the crate moving with a constant velocity. Find μ k, the coefficient of kinetic friction, between the crate and the floor. 2. A 25 kg chair initially at rest on a horizontal floor requires a 165 N horizontal force to set it in motion. Once the chair is in motion, a 127 N horizontal force keeps it moving at a constant velocity. a. Find the coefficient of static friction between the chair and the floor. b. Find the coefficient of kinetic friction between the chair and the floor. 3. A museum curator moves artifacts into place on various different display surfaces. Use the values in Figure 4.7 to find F s,max and F k for the following situations: a. moving a 145 kg aluminum sculpture across a horizontal steel platform b. pulling a 15 kg steel sword across a horizontal steel shield c. pushing a 250 kg wood bed on a horizontal wood floor d. sliding a 0.55 kg glass amulet on a horizontal glass display case Interactive Demo HMDScience.com Forces and the Laws of Motion 137 Classroom Practice Coefficients of Friction A refrigerator is placed on a ramp. The refrigerator begins to slide when the ramp is raised to an angle of 34. What is the coefficient of static friction? Answer: 0.67 PROBLEM guide D Use this guide to assign problems. SE = Student Edition Textbook PW = Sample Problem Set I (online) PB = Sample Problem Set II (online) Solving for: µ SE Sample, 1 2; Ch. Rvw. 35, 36*, 37*, 49 PW 4 7, 10* PB 8 10 F f SE 3 Pw Sample, 1 3, 7, 10* PB 5 7, m PW 8 9 PB Sample, 1 4 *Challenging Problem Answers Practice D a b a N, N b N, 84 N c N, N d. 5 N, 2 N titled /4/2011 2:57:44 PM Forces and the Laws of Motion 137

6 Teach continued Classroom Practice Overcoming Friction Two students are sliding a 225-kg sofa at constant speed across a wood floor. One student pulls with a force of 225 N at an angle of 13 above the horizontal. The other student pushes with a force of 250 N at an angle of 23 below the horizontal. What is the coefficient of kinetic friction between the sofa and the floor? Answer: 0.22 How could the students make moving the sofa easier? Answer: They could change the angles, put the sofa on rollers, or wax the floors. Overcoming Friction Sample Problem E A student attaches a rope to a 20.0 kg box of books. He pulls with a force of 90.0 N at an angle of 30.0 with the horizontal. The coefficient of kinetic friction between the box and the sidewalk is Find the acceleration of the box. F k ANALYZE Given: m = 20.0 kg μ k = PLAN F applied 30 Unknown: a =? Diagram: F applied = 90.0 N at θ = 30.0 F k F applied Choose a convenient coordinate system, and find the x and y components of all forces. The diagram at left shows the most convenient coordinate system, because the only force to resolve into components is F applied. F applied,y = (90.0 N)(sin 30.0 ) = 45.0 N (upward) F applied,x = (90.0 N)(cos 30.0 ) = 77.9 N (to the right) Choose an equation or situation: A. Find the normal force,, by applying the condition of equilibrium in the vertical direction: ΣF y = 0. B. Calculate the force of kinetic friction on the box: F k = μ k. C. Apply Newton s second law along the horizontal direction to find the acceleration of the box: ΣF x = ma x. PREMIUM CONTENT Interactive Demo HMDScience.com SOLVE 138 Chapter 4 Problem Solving Substitute the values into the equations and solve: A. To apply the condition of equilibrium in the vertical direction, you need to account for all of the forces in the y direction:,, and F applied,y. You know F applied,y and can use the box s mass to find. F applied,y = 45.0 N = (20.0 kg)(9.81 m/s 2 ) = 196 N Continued Houghton Mifflin Harcourt Alternative Approaches When evaluating the measures of F applied, x and F applied, y, remind students that they may use the Pythagorean theorem. This is the equation that compares the three sides of a right triangle. The theorem can be written as: a 2 + b 2 = c 2 When solving for F applied, x and F applied, y, use the equation as so: (F applied ) 2 = (F applied, x ) 2 + (F applied, y ) 2 = (77.9) 2 + (45) 2 F applied = 90 N Untitled /4/2011 2:57: Chapter 4

7 Overcoming Friction (continued) Tips and Tricks Remember to pay attention to the direction of forces. Here, is subtracted from and F applied,y because is directed downward. Tips and Tricks F k is directed toward the left, opposite the direction of F applied,x. As a result, when you find the sum of the forces in the x direction, you need to subtract F k from F applied, x. CHECK YOUR WORK Next, apply the equilibrium condition, ΣF y = 0, and solve for. ΣF y = + F applied,y - = N N = 0 = N N = 151 N B. Use the normal force to find the force of kinetic friction. F k = μ k = (0.500)(151 N) = 75.5 N C. Use Newton s second law to determine the horizontal acceleration. ΣF x = F applied, x - F k = ma x a x = F applied, x - F k m = 77.9 N N = _ 2.4 N 20.0 kg 20.0 kg 2.4 kg m/s2 = 20.0 kg a = 0.12 m/s 2 to the right The normal force is not equal in magnitude to the weight because the y component of the student s pull on the rope helps support the box. 1. A student pulls on a rope attached to a box of books and moves the box down the hall. The student pulls with a force of 185 N at an angle of 25.0 above the horizontal. The box has a mass of 35.0 kg, and μ k between the box and the floor is Find the acceleration of the box. 2. The student in item 1 moves the box up a ramp inclined at 12 with the horizontal. If the box starts from rest at the bottom of the ramp and is pulled at an angle of 25.0 with respect to the incline and with the same 185 N force, what is the acceleration up the ramp? Assume that μ k = A 75 kg box slides down a 25.0 ramp with an acceleration of 3.60 m/s 2. a. Find μ k between the box and the ramp. b. What acceleration would a 175 kg box have on this ramp? 4. A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exerted downward at an angle of 35.2 below the horizontal. Find μ k between the box and the floor. PROBLEM guide E Use this guide to assign problems. SE = Student Edition Textbook PW = Sample Problem Set I (online) PB = Sample Problem Set II (online) Solving for: F f, a SE Sample, 1 3; Ch. Rvw. 38*, 39*, 47a b*, 48c PW 5 7 PB 4 7, m SE Ch. Rvw. 21, 29, 41, 50, 52* PW Sample, 1 3 PB 8 10 µ SE 3, 4; Ch. Rvw , 48b* PW 4 PB Sample, 1 3 *Challenging Problem Answers Practice E m/s 2 in the positive x direction m/s 2 up the ramp 3. a b m/s 2 down the ramp (Note that m cancels in the solution, and a is the same in both cases; the slight difference is due to rounding.) Forces and the Laws of Motion 139 titled /4/2011 2:57:51 PM Forces and the Laws of Motion 139

8 Teach continued Why It Matters Driving and Friction The coefficient of friction between the ground and the tires of a car is less when rain or snow is on the ground. Snow and rain tires are excellent examples of ways that tires are adapted to regain some of the necessary frictional forces. Point out to students that the friction between a tire and pavement is more complex than the simple sliding friction between dry surfaces, which they have been studying. The force of friction on a car tire is not necessarily simply proportional to the normal force. The fact that there is not a simple proportion between the frictional and normal forces is due in part to the fact that the tires are rolling, so they peel vertically away from the surface rather than continuously slide across it. Also, when the road is covered with water or snow, other factors such as viscosity come into play. A ccelerating a car seems simple to the driver. It is just a matter of pressing on a pedal or turning a wheel. But what are the forces involved? A car moves because as its wheels turn, they push back against the road. It is actually the reaction force of the road pushing on the car that causes the car to accelerate. Without the friction between the tires and the road, the wheels would not be able to exert this force and the car would not experience a reaction force. Thus, acceleration requires this friction. Water and snow provide less friction and therefore reduce the amount of control the driver has over the direction and speed of the car. As a car moves slowly over an area of water on the road, the water is squeezed out from under the tires. If the car moves too quickly, there is not enough time for the weight of the car to squeeze the water out from under the tires. The water trapped between the tires and the road will lift the tires and car off the road, a phenomenon called hydroplaning. When this situation Air resistance is a form of friction. Driving and Friction Another type of friction, the retarding force produced by air resistance, is important in the analysis of motion. Whenever an object moves through a fluid medium, such as air or water, the fluid provides a resistance to the object s motion. For example, the force of air resistance, F R, on a moving car acts in the direction opposite the direction of the car s motion. At low speeds, the magnitude of F R is roughly proportional to the car s speed. At higher speeds, F R is roughly proportional to the square of the car s speed. When the magnitude of F R equals the magnitude of the force moving the car forward, the net force is zero and the car moves at a constant speed. A similar situation occurs when an object falls through air. As a free-falling body accelerates, its velocity increases. As the velocity increases, the resistance of the air to the object s motion also constantly increases. When the upward force of air resistance balances the downward gravitational force, the net force on the object is zero and the object continues to move downward with a constant maximum speed, called the terminal speed. occurs, there is very little friction between the tires and the water, and the car becomes difficult to control. To prevent hydroplaning, rain tires, such as the ones shown above, keep water from accumulating between the tire and the road. Deep channels down the center of the tire provide a place for the water to accumulate, and curved grooves in the tread channel the water outward. Because snow moves even less easily than water, snow tires have several deep grooves in their tread, enabling the tire to cut through the snow and make contact with the pavement. These deep grooves push against the snow and, like the paddle blades of a riverboat, use the snow s inertia to provide resistance. The Goodyear Tire & Rubber Company 140 Chapter 4 Differentiated Instruction Pre-AP Objects moving in outer space do not experience air resistance. Thus, Earth continually orbits the sun without slowing down. (Earth s speed is actually decreasing because of frequent collisions with small masses such as meteoroids, but this effect is minor.) Earth s velocity does, however, change slightly as it orbits the sun. Have students do research to find out what factor affects Earth s velocity in its orbit. Earth moves slightly faster in its orbit when it is closer to the sun in January than when it is farthest from the sun in July. This difference is a corollary to Kepler s second law of planetary motion, the equal-area law, which describes the movement of planets in elliptical orbits. Untitled /4/2011 2:57: Chapter 4

9 There are four fundamental forces. At the microscopic level, friction results from interactions between the protons and electrons in atoms and molecules. Magnetic force also results from atomic phenomena. These forces are classified as electromagnetic forces. The electromagnetic force is one of four fundamental forces in nature. The other three fundamental forces are gravitational force, the strong nuclear force, and the weak nuclear force. All four fundamental forces are field forces. The strong and weak nuclear forces have very small ranges, so their effects are not directly observable. The electromagnetic and gravitational forces act over long ranges. Thus, any force you can observe at the macroscopic level is either due to gravitational or electromagnetic forces. The strong nuclear force is the strongest of all four fundamental forces. Gravity is the weakest. Although the force due to gravity holds the planets, stars, and galaxies together, its effect on subatomic particles is negli gible. This explains why electric and magnetic effects can easily overcome gravity. For example, a bar magnet has the ability to lift another magnet off a desk. Assess and Reteach Assess Use the Formative Assessment on this page to evaluate student mastery of the section. Reteach For students who need additional instruction, download the Section Study Guide. Response to Intervention To reassess students mastery, use the Section Quiz, available to print or to take directly online at HMDScience.com. SECTION 4 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. Draw a free-body diagram for each of the following objects: a. a projectile accelerating downward in the presence of air resistance b. a crate being pushed across a flat surface at a constant speed 2. A bag of sugar has a mass of 2.26 kg. a. What is its weight in newtons on the moon, where the acceleration due to gravity is one-sixth that on Earth? b. What is its weight on Jupiter, where the acceleration due to gravity is 2.64 times that on Earth? 3. A 2.0 kg block on an incline at a 60.0 angle is held in equilibrium by a horizontal force. a. Determine the magnitude of this horizontal force. (Disregard friction.) b. Determine the magnitude of the normal force on the block. 4. A 55 kg ice skater is at rest on a flat skating rink. A 198 N horizontal force is needed to set the skater in motion. However, after the skater is in motion, a horizontal force of 175 N keeps the skater moving at a constant velocity. Find the coefficients of static and kinetic friction between the skates and the ice. Critical Thinking 5. The force of air resistance acting on a certain falling object is roughly proportional to the square of the object s velocity and is directed upward. If the object falls fast enough, will the force of air resistance eventually exceed the weight of the object and cause the object to move upward? Explain. Answers to Section Assessment Forces and the Laws of Motion 141 titled a. An arrow labeled should point down, and an arrow labeled F air should point opposite the direction of motion. The arrow should be longer than the arrow F air. b. points down, points up, F applied is horizontal, and F friction points in the opposite direction. The two vertical arrows are equal in length, as are the two horizontal arrows. 2. a N b N 3. a. 34 N b. 39 N , no; Once at equilibrium, the velocity will not increase, so the force of air resistance will not increase. 5/4/2011 2:57:53 PM Forces and the Laws of Motion 141