PHYSICS 1 Forces & Newton s Laws

Advanced Placement PHYSICS 1 Forces & Newton s Laws Presenter 2014-2015

Forces & Newton s Laws What I Absolutel Have to Know to Survive the AP* Exam Force is an push or pull. It is a vector. Newton s Second Law is the workhorse of the AP Phsics 1 exam. It allows ou to write down mathematical relationships that are true. hus, for a single bod, if ou pick an direction and sum up all the positive and negative forces that act on the bod along that line, the sum will equal the product of the bod s mass and its acceleration along that line. A Free Bod Diagram allows ou to identif all of the forces acting on a single bod. Neglect one force or add a fictitious force on our FBD and ou are in trouble. Newton s 1 st Law: in an inertial frame of reference, an object in a state of constant velocit (including zero velocit) will continue in that state unless impinged upon b a net external force. If ΣF=0, then a=0 and the object is at rest or moving at a constant velocit in a straight line. he converse is true also, if an object is in a state of constant velocit (including zero velocit) then a=0 and ΣF=0. Newton s 2 nd Law: A net force acting on a mass causes that mass to accelerate in the direction of the net force. he acceleration (vector) is directl proportional to the net force (vector) acting on the mass and ΣF inversel proportional to the mass of the object being accelerated. a= or Σ F = ma m Newton s 3 rd Law: For ever action force, there exists an equal and opposite reaction force. When one object exerts a force on a second object, the second object alwas responds with a duplicate force in the opposite direction. If ou hit a table with our fist, then the size and direction of the force ou appl must be equal and opposite the force the table applies to ou. Forces are generated in action/reaction pairs that occur on different objects. If ou tr to appl 800 Newtons of force to a table that can onl provide 600 Newtons of reaction force back on ou, ou will never succeed. he table will break as soon as ou exceed 600 Newtons, which is the maximum force it can appl to ou. Σ F= Fnet = ma F = mg W Ke Formulas and Relationships Ffsmax µ sfn ( static) Ffk = µ kfn ( kinetic) Gm1m 2 FG = 2 r kgm ΣF = Sum of the forces is the Net Force Newtons (N) = 2 s a = acceleration m = mass F W = weight g = acceleration due to gravit F = maximum static frictional force fsmax F = kinetic frictional force fk F N = normal force F G = gravitational force r = distance between the centers of two masses AP* is a trademark of the College Entrance Examination Board. he College Entrance Examination Board was not involved in the production of this material. Copright 2013 National Math + Science Initiative, Inc., Dallas, X. All rights reserved.

Forces & Newton s Laws Weight Gravitational F g, W FG Fg = mg Basic Kinds of Forces Alwas directed toward the Earth s center. Force on a free falling bod, if we neglect air friction. mm 1 2 FG = G r 2 A force of attraction between an two massive objects. When the Earth is one of the two bodies involved, then the force felt b the second bod while positioned on the Earth s surface will alwas be directed toward the Earth s center. A force of support, provided to an object b a surface in which the object is in contact. Alwas directed perpendicular to and awa from the surface providing the support. F N Normal F, N N W In the figure above, a box is supported b a table. he figure shows all the forces acting on the box and is called a Free Bod Diagram (FBD). If a box, rests on a level table, then the FN = W = mg. Notice that the normal force sometimes equals the weight but not alwas. F N x mgsinθ θ mgcosθ If the box is placed on an inclined plane, then the FN = mgcosθ, the component of the weight that is equal and opposite the normal force. For the inclined plane above, the normal force and the weight are not equal and not even in the same direction. W Copright 2013 National Math + Science Initiative, Inc., Dallas, X. All rights reserved

Forces & Newton s Laws Friction is produced b the atomic interaction between two bodies as the either slide over one another (kinetic friction) or sit motionless in contact with one another (static friction). Friction F f, f F µ F (static) fsmax s N Static Friction opposes the intended direction of relative sliding. he static frictional force will onl be as high as it needs to be to keep the sstem in equilibrium. If successivel greater and greater forces are applied, the static frictional force will counter each push with a force of equal and opposite magnitude until the applied force is great enough to shear the bonding between the two surfaces. When ou calculate f s or Ffsusing the equation above, ou are finding the maximum static frictional force, one of an infinite number of possible frictional forces that could be exerted between the two bodies. µ s is a proportionalit constant called the coefficient of static friction. It is the ratio of the static frictional force between the surfaces divided b the normal force acting on the surface. F = µ F fk k N (kinetic) Kinetic Friction or Dnamic Friction or Sliding Friction is alwas opposite the direction of motion. he statement that kinetic friction is a function of the normal force onl (surface area is independent) is true onl when dealing with rigid bodies that are sliding relative to each other. When ou calculate f k or Ff using the equation above, ou are finding the single, constant kinetic frictional force that exists between the two bodies sliding relative to one another. No matter their velocit (assuming heating does not alter the coefficient of kinetic friction) the kinetic frictional force will alwas be the same. µ is a proportionalit k constant called the coefficient of kinetic friction. It is the ratio of the kinetic frictional force between the surfaces divided b the normal force acting on the surface WARNING! he two quantities f s and f k ma look the same, but the tell us different things. Kinetic friction is tpicall less than static friction for the same two surfaces in contact. Note that the normal force sometimes equals the weight but not alwas. When ou draw a free bod diagram of forces acting on an object or sstem of objects, be sure to include the frictional force as opposing the relative motion (or potential for relative motion) of the two surfaces in contact. Copright 2013 National Math + Science Initiative, Inc., Dallas, X. All rights reserved

Forces & Newton s Laws ension Applied F, FSubscript F is a force that is applied to a bod b a rope, string, or cable. F is applied along the line of the string and awa from the bod in question. Push me, pull ou force that does not fall into one of the above categories, for example, a friend shoves ou. he magnitude of the applied force is characterized b an F, with an subscript that makes sense to solve the problem. Later in the ear, ou will encounter additional forces, like the electric force, F E, and the magnetic force, F B. Strateg on Force Problems 1. ake one bod in the sstem and draw a Free Bod Diagram (FBD) for it. 2. Choose x and axes and place them beside our FBD. One axis must be in the direction of the acceleration ou are tring to find. If there is no acceleration, then Σ F = 0. 3. If there are forces on the FBD that are not along the x and directions, find their respective x and components. 4. Using Newton s 2 nd Law, sum the forces in the x direction and set them equal to ma. If a x second equation is needed, sum the forces in the direction and set them equal to ma. 5. Repeat the above process for all the bodies in the sstem or until ou have the same number of equations as unknowns and solve the problem. Effective Problem Solving Strategies Free Bod Diagram (FBD) A Free Bod Diagram is normall depicted as a box showing all the forces acting on the bod. hese forces are depicted as arrows. he don t have to be drawn to scale, but the should have a length that is appropriate for their magnitude. Also, the force vectors do need to be directionall accurate and labeled. Do not include components of the force vectors on our FBD. When drawing Free Bod Diagrams show onl the force(s) that act on the bod in question and do not show forces that the bod applies to other bodies. Also, do not include velocit or acceleration vectors on our free bod force diagrams, since ou will lose points for extraneous vectors on our FBD. Copright 2013 National Math + Science Initiative, Inc., Dallas, X. All rights reserved

Forces & Newton s Laws Example 1 A rope supports an empt bucket of mass 3.0 kg. Determine the tension in the rope when the bucket is (a) at rest and (b) the bucket is accelerated upward at 2.0 m/s 2. Solution Step 1 Draw a Free Bod Diagram for each bod in the sstem. F W Step 2 Choose x and axes. In this case we could choose the axis to be in the same dimension as the tension and the weight. Step 3 If there are off axes forces, then find the x and components. Step 4 Using Newton s 2 nd law, sum the forces in the dimension and set equal to ma. (a) When the bucket is at rest, the net force is zero, so that the tension in the rope equals the weight of the bucket. Σ F = ma F mg = 0 m F = mg = ( 3.0kg) 10 = 30N 2 s (b) When the bucket accelerates upward, the net force is ma. Copright 2013 National Math + Science Initiative, Inc., Dallas, X. All rights reserved

Forces & Newton s Laws Σ F = ma F mg = ma ( ) F = mg + ma = m g + a m m F = 3.0kg 10 + 2.0 36 N upward 2 2 = s s Example 2 In the diagram below, two bodies of different masses (M 1 and M 2 ) are connected b a string which passes over a pulle of negligible mass and friction. What is the magnitude of the acceleration of the sstem in terms of the given quantities and fundamental constants? M 1 M 2 Solution Step 1: Draw a FBD for each bod in the sstem. here are two forces acting on each of the bodies: weight downward and the tension in the string upward. he tension is distributed throughout the string. he pulle (negligible mass and friction) changes onl the direction of motion, not the tension, so the tension is the same on each side of the pulle. Our FBDs should look like this: Copright 2013 National Math + Science Initiative, Inc., Dallas, X. All rights reserved

Forces & Newton s Laws W 1 W 2 Step 2 Choose x and axes. In this case we could choose the axis to be in the same dimension as the tension and the weight for both masses. Step 3 If there are off axes forces, then find the x and components. Step 4 Using Newton s 2 nd law, for each bod sum the forces in the dimension and set equal to ma and/or the x dimension. For m 1 the tension is positive and the weight is negative since the acceleration is upward. Σ F = ma 1 F W = ma 1 1 For m 2 the tension is negative and the weight is positive since the acceleration is downward. Σ F = m a F + W = m a 2 2 2 Copright 2013 National Math + Science Initiative, Inc., Dallas, X. All rights reserved

Forces & Newton s Laws F W = ma 1 1 F + W = m a 2 2 ( ) W W = m + m a a 2 1 1 2 2 1 2 1 = = = ( ) ( ) ( 2 1) ( ) W W m g m g m m g m + m m + m m + m 1 2 1 2 1 2 Example 3 A block of mass m rests on a horizontal table. A string is tied to the block, passed over a pulle, and another block of mass M is hung on the other end of the string, as shown in the figure below. he coefficient of kinetic friction between block m and the table is μ k. Find the magnitude of the acceleration of the sstem in terms of the given quantities and fundamental constants. µ k m M Copright 2013 National Math + Science Initiative, Inc., Dallas, X. All rights reserved

Forces & Newton s Laws Solution Step 1: Draw a FBD for each bod. Note that according to the FBD, the vertical acceleration of the block on the table is zero, since the normal force is directed upward and the weight force is directed downward. here is a horizontal acceleration for the block since the tension is greater than the frictional force. F N f k W 1 W 2 Step 2 Choose x and axes. In this case we could choose the axis to be in the same dimension as the tension and the weight for both masses. We could choose the x axis to be in the same dimension as the frictional force and the tension pulling the mass to the right. Step 3 If there are off axes forces, then find the x and components. Step 4 Using Newton s 2 nd law, for each bod sum the forces in the dimension and set equal to ma and/or the x dimension. For block m the net force verticall is 0 since the block is accelerating to the right and not upward nor downward. Summing the forces in the dimension we find that the normal force equals the weight of the block. Σ F = ma FN W1 = 0 F = W = mg N 1 Summing the forces in the horizontal dimension, the direction of the acceleration for block m, we find that Copright 2013 National Math + Science Initiative, Inc., Dallas, X. All rights reserved

Forces & Newton s Laws Σ F = ma x f = ma k x For mass two, summing the forces in the vertical dimension, the vertical acceleration for block M, we find that Σ F = M a + W = Ma 2 Adding the two equations together we determine the acceleration of the sstem f = ma k + W = Ma 2 2 k ( ) W f = m+ M a a W f Mg µ mg 2 k = = a = ( m+ M) ( m+ M) ( M µ km) g ( m+ M) Since f = µ F = µ mg k k N k k Copright 2013 National Math + Science Initiative, Inc., Dallas, X. All rights reserved

AP Phsics Multiple Choice Practice Dnamics SECION A Linear Dnamics 1. A ball of mass m is suspended from two strings of unequal length as shown above. he magnitudes of the tensions 1 and 2 in the strings must satisf which of the following relations? (A) l = 2 (B) 1 > 2 (C) 1 < 2 (D) l + 2 = mg Questions 2 3 A 2-kg block slides down a 30 incline as shown above with an acceleration of 2 m/s2. 2. Which of the following diagrams best represents the gravitational force W. the frictional force f, and the normal force N that act on the block? (D) 3. Which of the following correctl indicates the magnitudes of the forces acting up and down the incline? (A) 20 N down the plane, 16 N up the plane (B) 4 N down the plane, 4 N up the plane (C) 0 N down the plane, 4 N up the plane (D) 10 N down the plane, 6 N up the plane 4. When the frictionless sstem shown above is accelerated b an applied force of magnitude the tension in the string between the blocks is (A) F (B) 2/3 F (C) ½ F (D) 1/3 F 5. A ball falls straight down through the air under the influence of gravit. here is a retarding force F on the ball with magnitude given b F = bv, where v is the speed of the ball and b is a positive constant. he ball reaches a terminal velocit after a time t. he magnitude of the acceleration at time t/2 is (A) Increasing (B) Decreasing (C) 10 m/s/s (D) Zero 39

6. A block of weight W is pulled along a horizontal surface at constant speed v b a force F, which acts at an angle of with the horizontal, as shown above. he normal force exerted on the block b the surface has magnitude (A) greater than W (B) greater than zero but less than W (C) equal to W (D) zero 7. A uniform rope of weight 50 N hangs from a hook as shown above. A box of weight 100 N hangs from the rope. What is the tension in the rope? (A) 75 N throughout the rope (B) 100 N throughout the rope (C) 150 N throughout the rope (D) It varies from 100 N at the bottom of the rope to 150 N at the top. 8. When an object of weight W is suspended from the center of a massless string as shown above, the tension at an point in the string is (A) 2Wcos (B) ½Wcos (D) W/(2cos ) (E) W/(cos ) 9. A block of mass 3m can move without friction on a horizontal table. his block is attached to another block of mass m b a cord that passes over a frictionless pulle, as shown above. If the masses of the cord and the pulle are negligible, what is the magnitude of the acceleration of the descending block? (A) g/4 (B) g/3 (C) 2g/3 (D) g 40

A plane 5 meters in length is inclined at an angle of 37, as shown above. A block of weight 20 N is placed at the top of the plane and allowed to slide down. 10. he magnitude of the normal force exerted on the block b the plane is (A) greater than 20 N (B) greater than zero but less than 20 N (C) equal to 20 N (D) zero 11. Multiple correct: hree forces act on an object. If the object is moving to the right in translational equilibrium, which of the following must be true? Select two answers. (A) he vector sum of the three forces must equal zero. (B) All three forces must be parallel. (C) he magnitudes of the three forces must be equal. (D) he object must be moving at a constant speed. 12. For which of the following motions of an object must the acceleration alwas be zero? (A) An motion in a straight line (B) Simple harmonic motion (C) An motion at constant speed (D) An single object in motion with constant momentum 13. A rope of negligible mass supports a block that weighs 30 N, as shown above. he breaking strength of the rope is 50 N. he largest acceleration that can be given to the block b pulling up on it with the rope without breaking the rope is most nearl (A) 6.7 m/s2 (B) 10 m/s2 (C) 16.7 m/s2 (D) 26.7 m/s2 Questions 14-15 A horizontal, uniform board of weight 125 N and length 4 m is supported b vertical chains at each end. A person weighing 500 N is hanging from the board. he tension in the right chain is 250 N. 14. What is the tension in the left chain? (A) 125 N (B) 250 N (C) 375 N (D) 625 N 41

15. Which of the following describes where the person is hanging? (A) between the chains, but closer to the left-hand chain (B) between the chains, but closer to the right-hand chain (C) Right in the middle of the board (D) directl below one of the chains 16. Multiple correct: he cart of mass 10 kg shown above moves without frictional loss on a level table. A 10 N force pulls on the cart horizontall to the right. At the same time, a 30 N force at an angle of 60 above the horizontal pulls on the cart to the left. Which of the following describes a manner in which this cart could be moving? Select two answers. (A) moving left and speeding up (B) moving left and slowing down (C) moving right and speeding up (D) moving right and slowing down 17. wo people are pulling on the ends of a rope. Each person pulls with a force of 100 N. he tension in the rope is: (A) 0 N (B) 50 N (C) 100 N (D) 200 N 18. he parabola above is a graph of speed v as a function of time t for an object. Which of the following graphs best represents the magnitude F of the net force exerted on the object as a function of time t? (A) (B) (C) (D) 42

1976 B1 (modified) An elevator car that has a mass of 400 kg is in contact with two rails that each exert a frictional force of 200 N on the car. he elevator is attached to a cable that passes over a pulle with negligible friction and is attached to a counterweight M. Under these conditions, the elevator has an upward acceleration of 2 m s 2. a. On the diagrams below draw all the forces acting on the elevator car and the counterweight. Identif the source of all the forces.

b. Write two equations that relate our free bod diagrams to the mass of the elevator car m, the mass of the counterweight M, and the acceleration a, of the sstem. Elevator (forces directed up positive) ΣF = ma F 2F f mg = ma Counterweight (forces directed down positive) ΣF = Ma Mg F = Ma c. Calculate the tension in the cable supporting the elevator. F 2F f mg = ma F = 2F f + mg + ma ( ) + 400kg 10 m s 2 F = 2 200N F = 5200 N + 400kg 2 m s 2 d. Calculate the mass M, of the counterweight. Mg F = Ma Mg Ma = F ( ) = F M g a M = F g a 10 m s 2 m = 650 kg 2 s 2 ( ) = 5200 N

2007B (modified) A child pulls a sled along the ground with a force of 200 N which makes an angle θ = 37 with the horizontal. A dog rides on top of the sled so that the dog-sled sstem has a total mass of 40 kg. he coefficient of friction between the sled and the ground is 0.5. Use sin(37 ) = 0.6 and cos(37 ) = 0.8. a. On the diagram below, draw the forces acting on the dog-sled sstem. Identif the source of all the forces. b. Write two equations that relate our free bod diagram to the mass of the dog-sled sstem m, and the acceleration a, of the dog-sled sstem.

Acceleration of dog-sled sstem is horizontal Vertical Forces: ΣF = 0 F N + F sinθ mg = 0 F N + F sinθ = mg Horizontal Forces: ΣF = ma F cosθ F f = ma c. Qualitativel discuss what happens to the normal force as the angle, θ is increased or decreased. Calculate the normal force acting on the dog-sled sstem. As the angle, θ increases the vertical component of the tension force increases. Hence, the normal force decreases. As the angle, θ decreases the vertical component of the tension force decreases. Hence, the normal force increases. F N + F sinθ = mg F N = mg F sinθ ( ) 10 m s 2 F N = 40kg 200N ( )( sin37 ) = 280 N d. Calculate the friction force acting on the dog-sled sstem. F f = µf N = 0.5( 280N ) = 140 N e. Calculate the acceleration of the dog-sled sstem. a = F cosθ F f m = 200N cos37 140N 40kg = 0.5 m s 2