Preview Section 1 Changes in Motion Section 2 Newton's First Law Section 3 Newton's Second and Third Laws Section 4 Everyday Forces

Forces Forces can change motion. Start movement, stop movement, or change the direction of movement Cause an object in motion to speed up or slow down One common misconception is that “forces cause motion.” Forces actually cause a change in motion, or more specifically, a change in velocity (an acceleration). This will be covered in more detail in the next sections, in the context of Newton’s laws.

Forces Contact forces Field forces Pushes or pulls requiring physical contact between the objects Baseball and bat Field forces Objects create force fields that act on other objects. Gravity, static electricity, magnetism Pictured is a contact force, the bat and the ball, as well as a field force, the static electric field around charged balloon exerting a force on small pieces of paper. Ask students to identify other examples of contact forces.

Units of Force The SI unit of force is the newton (N). Named for Sir Isaac Newton Defined as the force required to accelerate a 1 kg mass at a rate of 1 m/s2 Approximately 1/4 pound Other units are shown below. 1 N = 0.225 pounds (roughly 1/4 pound) Have students determine their approximate weight in newtons to reinforce the size of the unit. When talking about problems, use both units to help them become more comfortable. For example, a 10 000 N car is about a 2500 lb car.

Force Diagrams Forces are vectors (magnitude and direction). Force diagram (a) Shows all forces acting during an interaction On the car and on the wall Free-body diagram (b) Shows only forces acting on the object of interest On the car Students often have trouble isolating the forces acting on an object to draw a free-body diagram for the object. The free-body diagram of the car is analyzed in more detail in the next slide.

Free-Body Diagrams Three forces are shown on the car. Describe each force by explaining the source of the force and where it acts on the car. Is each force a contact force or a field force? For simplicity, all forces are shown acting on the center of the object. Remind students that, when adding vectors, they can be moved parallel without changing the results. Even though the upward force acts on each of the 4 tires, the total is shown acting on the center of the car. Even though the wall strikes the front bumper, that force can be moved to the center of the car without changing the resultant. Gravity (the pull of Earth’s field) acts on every particle in the car but is shown as a single downward force at the center.

Newton’s First Law Experimentation led Galileo to the idea that objects maintain their state of motion or rest. Newton developed the idea further, in what is now known as Newton’s first law of motion: Discuss Galileo’s experiment with balls rolling down and then back up inclines. Each ball returned to its original height even if the angle of incline was changed. He theorized that the ball would roll forever if the track was horizontal because it would never reach the starting height. A short version of this law would be as follows: Fnet = 0 <-----> v = constant If net force is zero, the velocity is constant and, if the velocity is constant, the net force is zero. It sounds simple, but students have a difficult time with this law because they do not “see” the force of friction when they look at moving objects.

Newton’s First Law Called the law of inertia Inertia Tendency of an object not to accelerate Mass is a measure of inertia More mass produces more resistance to a change in velocity Which object in each pair has more inertia? A baseball at rest or a tennis ball at rest Answer: the baseball A tennis ball moving at 125 mi/h or a baseball at rest Students may choose the moving tennis ball if they confuse inertia (mass) with momentum (mass times velocity). Emphasize that inertia depends only on mass, and so the baseball has a greater inertia in both cases.

Net Force - the Sum of the Forces This car is moving with a constant velocity. Fforward = road pushing the tires Fresistance = force caused by friction and air Forces are balanced Velocity is constant because the net force (Fnet) is zero. Ask students how to increase the speed of the car. Answer: Increase the forward force (accelerator) or decrease the resistance force (make the car more aerodynamic). Ask students how to decrease the speed of the car. Answer: Increase the resistance force (the brakes) or decrease the forward force (accelerator). This will provide a nice introduction to Newton’s 2nd Law.

Equilibrium The state in which the net force is zero. All forces are balanced. Object is at rest or travels with constant velocity. In the diagram, the bob on the fishing line is in equilibrium. The forces cancel each other. If either force changes, acceleration will occur. After reviewing this slide, return to the previous slide and ask students if the car is in equilibrium.

Classroom Practice Problem A man is pulling on his dog with a force of 70.0N directed at an angle of +30.0 degrees to the horizontal. Find the x-component and the y-component. Answer: x-component 60.6 N Y-component 35.0 N Be sure students have looked at Sample Problem B in the Student Edition before trying this problem. Give students some time to work on this problem and then go through each step with them. After completing this problem, show the students that any two of the three forces will be cancelled by the third force. These balanced forces produce equilibrium.

Classroom Practice Problem The man pulls with a force of 25.0N at an angle of 18 degrees to the horizontal. Find the x and y components. Answer: X= 23.8 N and Y=7.73 N Be sure students have looked at Sample Problem B in the Student Edition before trying this problem. Give students some time to work on this problem and then go through each step with them. After completing this problem, show the students that any two of the three forces will be cancelled by the third force. These balanced forces produce equilibrium.

Classroom Practice Problem A crate is pulled to the right with a force of 85N, to the left with a force of 115N, upward with a force of 565N, and downward with a force of 236N. Find the net external force in the x direction. Find the net external force in the y direction. Find the magnitude and direction of he net external force on the crate. Answer: X= -30 N, Y=329 N, 330 N at 94.5 degree angle counter clockwise from the horizontal Be sure students have looked at Sample Problem B in the Student Edition before trying this problem. Give students some time to work on this problem and then go through each step with them. After completing this problem, show the students that any two of the three forces will be cancelled by the third force. These balanced forces produce equilibrium.

Newton’s Second Law Increasing the force will increase the acceleration. Which produces a greater acceleration on a 3-kg model airplane, a force of 5 N or a force of 7 N? Answer: the 7 N force Increasing the mass will decrease the acceleration. A force of 5 N is exerted on two model airplanes, one with a mass of 3 kg and one with a mass of 4 kg. Which has a greater acceleration? Answer: the 3 kg airplane Be sure students understand what is meant by the terms “directly proportional” and “inversely proportional.” A simulation from the Phet web site is available to help students visualize the force and the acceleration. The web address is: http://phet-web.colorado.edu/web-pages/index.html Choose the “Motion” simulations, then select “motion in 2D.” You can turn off the vectors and just allow students to observe the motion. Then ask the students to predict the acceleration vector. Which way will it point? Will it have a constant size? After predicting, show the acceleration vector. Next, have them predict the force vector’s direction and size. After predicting, show the force vector and both vectors. Then you can try the other motions described on the screen and ask them to observe the motion, describe the acceleration, and describe the forces. This exercise allows students to see that accelerations are caused by forces. We see the accelerations, but often do not see the forces.

Newton’s Second Law (Equation Form) F represents the vector sum of all forces acting on an object. F = Fnet Units for force: mass units (kg)  acceleration units (m/s2) The units kg•m/s2 are also called newtons (N). It is often useful to write the equation as a = F/m to show students the relationship between force and acceleration and between mass and acceleration. It is easier to see that forces cause accelerations when the equation is written in this form. Even though students saw these units in section 1, they may not recall the fact that newtons are simply a short name for the SI units of kg•m/s2. When solving problems, they will need to know this equivalence in order to cancel units. Remind students of the other units for force, such as dynes (g•cm/s2) and pounds (slug•ft/s2).

Classroom Practice Problem The net external force on a propeller of a 0.75 kg model airplane is 17 N forward. What is the acceleration of the airplane? Answer: 23 m/s2

Classroom Practice Problem The net external force on a golf cart is 390 N north. If the cart has a total mass of 270 kg, what are the magnitude and direction of its acceleration? Answer: 1.4m/s2 north

Classroom Practice Problem A car has a mass of 1500 kg. What force is required to accelerate the car at 4.5 m/s2 to the east? Answer: 6800 N east

Classroom Practice Problem A 2.0 kg mass starts form rest at the top of an inclined plane 85 cm long and slides down to the bottom in 0.50 s. What net external force acts on the mass along the incline? Answer: 14 N

Classroom Practice Problem A ball pushed with a force of 13.5 N accelerates at 6.5 m/s2 to the right. What is the mass of the ball? Answer: 2.1 kg

What do you think? Two football players, Alex and Jason, collide head-on. They have the same mass and the same speed before the collision. How does the force on Alex compare to the force on Jason? Why do you think so? Sketch each player as a stick figure. Place a velocity vector above each player. Draw the force vector on each and label it (i.e. FJA is the force of Jason on Alex). When asking students to express their ideas, you might try one of the following methods. (1) You could ask them to write their answers in their notebook and then discuss them. (2) You could ask them to first write their ideas and then share them with a small group of 3 or 4 students. At that time you can have each group present their consensus idea. This can be facilitated with the use of whiteboards for the groups. The most important aspect of eliciting student’s ideas is the acceptance of all ideas as valid. Do not correct or judge them. You might want to ask questions to help clarify their answers. You do not want to discourage students from thinking about these questions and just waiting for the correct answer from the teacher. Thank them for sharing their ideas. Misconceptions are common and can be dealt with if they are first expressed in writing and orally. This question will likely produce a wide variety of responses. Some students may believe that the forces are always equal. Many will believe they are equal for the first example but not so for the second and third examples (next slide).

What do you think? Suppose Alex has twice the mass of Jason. How would the forces compare? Why do you think so? Sketch as before. Suppose Alex has twice the mass and Jason is at rest. How would the forces compare?

Forces always exist in pairs. Newton’s Third Law Forces always exist in pairs. You push down on the chair, the chair pushes up on you Called the action force and reaction force Occur simultaneously so either force is the action force Emphasize that the action and reaction forces occur at the same time.

Newton’s Third Law For every action force there is an equal and opposite reaction force. The forces act on different objects. Therefore, they do not balance or cancel each other. The motion of each object depends on the net force on that object.

Hammer Striking a Nail What are the action/reaction pairs for a hammer striking a nail into wood? Force of hammer on nail = force of nail on hammer Force of wood on nail = force of nail on wood Which of the action/reaction forces above act on the nail? Force of hammer on nail (downward) Force of wood on nail (upward) Does the nail move? If so, how? Fhammer-on-nail > Fwood-on-nail so the nail accelerates downward This example is continued on the next slide.

Hammer Striking a Nail What forces act on the hammer? Force of nail on hammer (upward) Force of hand on hammer (downward) Does the hammer move? If so, how? Fnail-on-hammer > Fhand-on-hammer so the hammer accelerates upward or slows down The hammer and nail accelerate in opposite directions. Use this example to stress the fact that the action and reaction forces do not cancel each other because they act on different objects. The best way to handle this is by drawing free body diagrams of each object next to each other. The free-body diagram for the nail is show on the previous slide. Ask students to draw the free-body diagram for the hammer. Then students can visualize the action-reaction forces and see that they do not balance each other. Each object accelerates or maintains constant motion based on the forces acting on that object.

Action-Reaction: A Book on a Desk Action Force The desk pushes up on the book. Reaction Force The book pushes down on the desk. Earth pulls down on the book (force of gravity). The book pulls up on Earth. Have students observe a book sitting on a desk for this slide. After students see the action force on the slide, they should be able to state the reaction force before you show it to them. Often students think the reaction force for the desk pushing up on the book is Earth pulling down on the book. Remind them that these forces act on the same object, the book, so they are not an action-reaction pair.

Action-Reaction: A Falling Book The book pulls up on Earth. What is the result of the reaction force? Unbalanced force produces a very small upward acceleration (because the mass of Earth is so large). Action Earth pulls down on the book (force of gravity). What is the result of the action force (if this is the only force on the book)? Unbalanced force produces an acceleration of -9.81 m/s2. Now, remove the book from the desk and allow it to fall to the floor. Ask students if the forces on the book are still balanced. What is the result of this unbalanced force? Acceleration. Have students calculate the acceleration of Earth. Assume the book’s mass is 2.0 kg, so the force on the book is (2.0 kg)(-9.8 m/s2) or 19.6 N downward. Therefore, the upward force on Earth is also 19.6 N. The mass of Earth is about 6 x 1024 kg, so students can calculate the upward acceleration and see how small it will be. You could also choose a falling distance and have students calculate the time required to fall the distance Earth would move upward during that time (using the equations from Chapter 2).

Weight and Mass Mass is the amount of matter in an object. Kilograms, slugs Weight is a measure of the gravitational force on an object. Newtons, pounds Depends on the acceleration of gravity Weight = mass  acceleration of gravity W = mag where ag = 9.81 m/s2 on Earth Depends on location ag varies slightly with location on Earth. ag is different on other planets. Mention that weight is less on the moon because ag on the moon is 1.6 m/s2 . Reinforce that converting between mass and weight is simple, just multiply or divide by 9.81 m/s2 . Point out that each kg has a weight of 9.81 N on Earth.

Normal Force Force on an object perpendicular to the surface (Fn) It may equal the weight (Fg), as it does here. It does not always equal the weight (Fg), as in the second example. Fn = mg cos  Point out that the equation for normal force applies to the first example also. Because cos(0)=1, the equation reduces to Fn = mg when the forces are directly opposite one another.

Static Friction Force that prevents motion Abbreviated Fs How does the applied force (F) compare to the frictional force (Fs)? Would Fs change if F was reduced? If so, how? If F is increased significantly, will Fs change? If so, how? Are there any limits on the value for Fs? These questions should help students understand that static friction balances the external force (F), so it increases and decreases as F increases and decreases. Eventually, F will be so large that the static frictional force (Fs) will no longer be able to balance it, and the net force will cause the object to slide. At this point, frictional forces become kinetic (see next slide).

Kinetic Friction Force between surfaces that opposes movement Abbreviated Fk Does not depend on the speed Using the picture, describe the motion you would observe. The jug will accelerate. How could the person push the jug at a constant speed? Reduce F so it equals Fk. Ask students if it requires more force to get an object moving when it is at rest or to keep it moving once it is already in motion. When pushing an object, we exert enough force to overcome static friction. At that point the object moves. The opposing force is now kinetic friction, which is less than static friction. Therefore, in order to maintain a constant speed and not accelerate, the force pushing the object is reduced.

Friction Click below to watch the Visual Concept. Visual Concept This Visual Concept discusses the nature of friction and the factors that affect the amount of friction. Before running the video, ask students to explain what causes the force of friction. Lead this discussion by asking students how the force of friction is affected by changing the types of surfaces or by adding a lubricant (such as water or glycerin on glass tubing). They should see these things clearly when you play the video. Make sure the students are focused on the magnified view of the surfaces. This will help them understand the effect of increased normal force and the effect of different surface types. During the comments on swimming, ask them how swimmers reduce the force of friction. Have them draw a free-body diagram of a swimmer showing the force propelling him forward (water pushing against his hand) and the frictional force in the opposite direction. Ask students how the swimmer can accelerate. They should respond that he can reduce friction or increase the force of the water on his hand.

Calculating the Force of Friction (Ff) Ff is directly proportional to Fn (normal force). Coefficient of friction (): Determined by the nature of the two surfaces s is for static friction. k is for kinetic friction. s > k Point out to students that Ff is the general term for both static friction (Fs) and kinetic friction (Fk).

Typical Coefficients of Friction Values for  have no units and are approximate. Point out that static is greater than kinetic for each example. Also explain that the coefficient is generally less than 1 but there could be sticky surfaces where the frictional force was greater than the normal force. This would lead to coefficients greater than 1.

Everyday Forces Click below to watch the Visual Concept.

Classroom Practice Problem 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. Draw a free-body diagram and use it to find: the weight the normal force (Fn) the force of friction (Ff) Find the coefficient of friction. Answer: s = 0.32 This is a relatively simple example from the book (Sample Problem D). Ask students to follow the steps. It is easy to get the answer by skipping the free-body diagram, but they need this diagram to understand that normal force = weight, and the 75 N horizontal push is equal to the force of friction. More complicated problems (next slide) can’t be solved without a free- body diagram.

Classroom Practice Problem Once the crate is in motion, a horizontal fore of 53 N keeps the crate moving with a constant velocity. Find the coefficient of kinetic friction between the crate and the floor. Find the coefficient of friction. Answer: k = 0.23 This is a relatively simple example from the book (Sample Problem D). Ask students to follow the steps. It is easy to get the answer by skipping the free-body diagram, but they need this diagram to understand that normal force = weight, and the 75 N horizontal push is equal to the force of friction. More complicated problems (next slide) can’t be solved without a free- body diagram.

Classroom Practice Problem A museum curator moves artifacts into place on many different display surfaces. Use the Table in 4-2 to find Fs,max and Fk for the following situations: Moving a 145kg aluminum sculpture across a horizontal steel platform Pulling a 15 kg steel sword across a horizontal steel shield Pushing a 227 kg wood bed on a wood floor Sliding a 0.55kg glass amulet on a glass display case. This is a relatively simple example from the book (Sample Problem D). Ask students to follow the steps. It is easy to get the answer by skipping the free-body diagram, but they need this diagram to understand that normal force = weight, and the 75 N horizontal push is equal to the force of friction. More complicated problems (next slide) can’t be solved without a free- body diagram.

Classroom Practice Problem A 25 kg chair initially at rest on a horizontal floor requires a 365 N horizontal force to set it in motion. Once the chair is in motion, a 327 N horizontal force keeps it moving at a constant velocity Draw a free-body diagram and use it to find: the weight the normal force (Fn) the force of friction (Ff) Find the coefficients of friction. Answer: s = 1.5 and k= 1.3 This is a relatively simple example from the book (Sample Problem D). Ask students to follow the steps. It is easy to get the answer by skipping the free-body diagram, but they need this diagram to understand that normal force = weight, and the 75 N horizontal push is equal to the force of friction. More complicated problems (next slide) can’t be solved without a free- body diagram.

Classroom Practice Problem 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 0.500. Find the magnitude of the acceleration of the box. Start with a free-body diagram. Determine the net force. Find the acceleration. Answer: a = 0.12 m/s2 to the right This is Sample Problem E from the book. The free-body diagram is essential to solving this problem. Students often make the mistake of assuming the normal force equals the weight. These two forces are not equal because the student is pulling upward on the box, thus reducing the normal force. So, Fn = weight - (90.0 N)(sin 30)°. Students can then determine the value for Fk and subtract it from (90.0 N)(cos 30°) to get the net force. At this point, they can use Newton’s second law to find the acceleration.

Classroom Practice Problem A box of books weighing 325 N moves with a constant velocity across the floor when it is pushed with a force of 425 N extereed downward at an angle of 35.2 °below the horizontal. Find µk between the box and the floor. Start with a free-body diagram. Answer: 0.609 This is Sample Problem E from the book. The free-body diagram is essential to solving this problem. Students often make the mistake of assuming the normal force equals the weight. These two forces are not equal because the student is pulling upward on the box, thus reducing the normal force. So, Fn = weight - (90.0 N)(sin 30)°. Students can then determine the value for Fk and subtract it from (90.0 N)(cos 30°) to get the net force. At this point, they can use Newton’s second law to find the acceleration.

Classroom Practice Problem A student moves a box of books down the hall by pulling on a rope attached to the box. The student pulls with a force of 185 N at an angle of 25° above the horizontal. The box has a mass of 35 kg, and μk between box and floor is 0.27. Find the magnitude of acceleration of the box . Answer: a = 2.9 m/s2 This is Sample Problem E from the book. The free-body diagram is essential to solving this problem. Students often make the mistake of assuming the normal force equals the weight. These two forces are not equal because the student is pulling upward on the box, thus reducing the normal force. So, Fn = weight - (90.0 N)(sin 30)°. Students can then determine the value for Fk and subtract it from (90.0 N)(cos 30°) to get the net force. At this point, they can use Newton’s second law to find the acceleration.

Classroom Practice Problem The student in item 2 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° with respect to the incline and with the same 185 N force. Find the magnitude of the acceleration of the box. Assume the μk = 0.27 Answer: a = 0.9 m/s2 This is Sample Problem E from the book. The free-body diagram is essential to solving this problem. Students often make the mistake of assuming the normal force equals the weight. These two forces are not equal because the student is pulling upward on the box, thus reducing the normal force. So, Fn = weight - (90.0 N)(sin 30)°. Students can then determine the value for Fk and subtract it from (90.0 N)(cos 30°) to get the net force. At this point, they can use Newton’s second law to find the acceleration.

Classroom Practice Problem A 75 kg box slides down a 25° ramp with an acceleration of 3.6 m/s2. Find μk between the box and the ramp What acceleration would a 175 kg box have on the ramp? Answer: a) 0.06 b) a = 3.0 m/s2 This is Sample Problem E from the book. The free-body diagram is essential to solving this problem. Students often make the mistake of assuming the normal force equals the weight. These two forces are not equal because the student is pulling upward on the box, thus reducing the normal force. So, Fn = weight - (90.0 N)(sin 30)°. Students can then determine the value for Fk and subtract it from (90.0 N)(cos 30°) to get the net force. At this point, they can use Newton’s second law to find the acceleration.

The Four Fundamental Forces Electromagnetic Caused by interactions between protons and electrons Produces friction Gravitational The weakest force Strong nuclear force The strongest force Short range Weak nuclear force