2. SECOND LAW OF MOTION If there is a net force acting on an object, the object will have an acceleration and the object’s velocity will change. Newton's Second Law of Motion states that for a particular force, the acceleration of an object is directly proportional to the net force and inversely proportional to the mass of the object. The direction of the force is the same as that of the acceleration. In equation form:

3. SECOND LAW OF MOTION In the SI system, the unit for force is the newton (N): A newton is that net force which, when applied to a 1-kg mass, gives it an acceleration of 1 m/s 2.

5. Acceleration and Force With Zero Friction Forces Pushing the cart with twice the force produces twice the acceleration. Three times the force triples the acceleration.

6. Acceleration and Mass Again With Zero Friction F F a a/2 Pushing two carts with same force F produces one-half the acceleration. The acceleration varies inversely with the mass.

7. THIRD LAW OF MOTION According to Newton's third law of motion, when one body exerts a force on another body, the second body exerts on the first an equal force in opposite direction. For every action force, there must be an equal and opposite reaction force. The Third Law of Motion applies to two different forces on two different objects: " For every action force, there must be an equal and opposite reaction force. " Action and reaction forces never balance out because they act on different objects.

8. Newton’s Third Law Third Law: For every action force, there must be an equal and opposite reaction force. Forces occur in pairs. Action Reaction Action Reaction

9. Acting and Reacting Forces Use the words by and on to study action/reaction forces below as they relate to the hand and the bar: The action force is exerted by the _____ on the _____. The reaction force is exerted by the _____ on the _____. bar handsbar hands Action Reaction

10. MASS The property that a body has of resisting any change in its state of rest or of uniform motion is called inertia. The inertia of a body is related to the amount of matter it contains. A quantitative measure of inertia is mass. The unit of mass is the kilogram (kg).

11. WEIGHT (F G ) Weight (a vector quantity) is different from mass (a scalar quantity). The weight of a body varies with its location near the Earth (or other astronomical body), whereas its mass is the same everywhere in the universe. The weight of a body is the force that causes it to be accelerated downward with the acceleration of gravity g. F G = mg Units: Newtons (N) g = acceleration due to gravity = 9.8 m/s 2

13. THE NORMAL FORCE A normal force is a force exerted by one surface on another in a direction perpendicular to the surface of contact. Note: The gravitational force and the normal force are not an action-reaction pair.

14. Applying Newton’s Second Law Read the problem and draw a sketch. Write down the data in correct units: m = kgF G = N Draw a free-body diagram for each object. Choose x or y-axis along motion and choose direction of motion as positive. Write Newton’s law for both axes:  F x = m a x  F y = m a y Solve for unknown quantities. Read the problem and draw a sketch. Write down the data in correct units: m = kgF G = N Draw a free-body diagram for each object. Choose x or y-axis along motion and choose direction of motion as positive. Write Newton’s law for both axes:  F x = m a x  F y = m a y Solve for unknown quantities.

15. 4.1 A 5.0-kg object is to be given an upward acceleration of 0.3 m/s 2 by a rope pulling straight upward on it. What must be the tension in the rope? FTFT FGFG a (+) m = 5 kg a = 0.3 m/s 2 ΣF y = F T - F G = ma F T = m(a + g) = 5(0.3 + 9.8) = 50.5 N 5 kg N2L

16. 4.2 A cord passing over a frictionless pulley has a 7.0 kg mass hanging from one end and a 9.0-kg mass hanging from the other. (This arrangement is called Atwood's machine). a. Find the acceleration of the masses. FTFT FTFT F G1 F G2 a (+) m 1 = 7 kg m 2 = 9 kg 7kg 9 kg N2L

17. FTFT FTFT F G1 F G2 a (+) ΣF = F T - F G1 - F T + F G2 = m TOTAL a = 1.22 m/s 2

18. FTFT FTFT F G1 F G2 a (+) b. Find the tension of the cord Using either side of the pulley yields the same answer! F T – F G1 = m 1 a F T = m 1 a + F G1 = m 1 (a + g) = 7(1.22 + 9.8) = 77.1 N

19. APPARENT WEIGHT The actual weight of a body is the gravitational force that acts on it. The body's apparent weight is the force the body exerts on whatever it rests on. Apparent weight can be thought of as the reading on a scale a body is placed on. scale

20. 4.3 What will a spring scale read for the weight of a 75 kg man in an elevator that moves: m = 75 kg F G = 75(9.8) = 735 N

21. ΣF y = F N - F G = 0 F N = F G = 735 N FNFN FGFG a. With constant upward speed of 5 m/s m = 75 kg F G = 735 N b. With constant downward speed of 5 m/s F N = 735 N N1L

22. ΣF = F N - F G = ma F N = ma + F G = 75 (2.45) + 735 = 919 N FNFN FGFG b. Going up with an acceleration of 0.25 g a = 2.45 m/s 2 N2L

23. ΣF = F G - F N = ma F N = F G - ma = 735 - 75 (2.45) = 551 N FNFN FGFG c. Going down with an acceleration of 0.25 g a = 2.45 m/s 2 N2L

24. For free fall the only force acting is F G so F N = 0 N FGFG e. In free fall? N2L

25. “Weightlessness” More properly, this effect is called apparent weightlessness, because the gravitational force still exists. It can be experienced on Earth, but only briefly:

27. FRICTION: STATIC AND KINETIC FRICTION Frictional forces act to oppose relative motion between surfaces that are in contact. Such forces act parallel to the surfaces. Static friction occurs between surfaces at rest relative to each other. When an increasing force is applied to a book resting on a table, for instance, the force of static friction at first increases as well to prevent motion. In a given situation, static friction has a certain maximum value called starting friction. When the force applied to the book is greater than the starting friction, the book begins to move across the table.

28. The origin of friction: on a microscopic scale, most surfaces are rough.

29. FRICTION: STATIC AND KINETIC FRICTION The kinetic friction (or sliding friction) that occurs afterward is usually less than the starting friction, so less force is needed to keep the book moving than to start it moving.

30. COEFFICIENT OF FRICTION The frictional force between two surfaces depends on the normal force F N pressing them together and on the natures of the surfaces. The latter factor is expressed quantitatively in the coefficient of friction  (mu) whose value depends on the materials in contact. The frictional force is experimentally found to be: Static friction Kinetic friction

32. 4.4 A horizontal force of 140 N is needed to pull a 60.0 kg box across the horizontal floor at constant speed. What is the coefficient of friction between floor and box? F a = 140 N m = 60 kg ΣF y = F N – F G = 0 F N = F G =mg = 60(9.8) = 588 N FNFN FGFG FfFf FaFa ΣF x = F a – F f = 0 F a = F f = μF N = 0.24 N1L

33. 4.5 A 70-kg box is slid along the floor by a horizontal 400-N force. Find the acceleration of the box if the value of the coefficient of friction between the box and the floor is 0.50. m = 70 kg F a = 400 N μ = 0.5 ΣF y = F N – F G = 0 F N = F G =mg = 70(9.8) = 686 N ΣF x = F a – F f = ma F f = μF N = (0.5)(686) = 343 N FNFN FaFa FGFG FfFf N2L

34. m = 70 kg F a = 400 N μ = 0.5 ΣF x = F a – F f = ma = 0.81 m/s 2 FNFN FaFa FGFG FfFf

35. 4.6 A 70-kg box is pulled by a rope with a 400-N force at an angle of 30  to the horizontal. Find the acceleration of the box if the coefficient of friction is 0.50 m = 70 kg F a = 400 N, 30  μ = 0.5 FGFG FaFa FfFf FNFN F ax = 400 cos 30  = 346.4 N F ay = 400 sin 30  = 200 N ΣF y = F N + F ay - F g = 0 F N = F G - F ay = 70(9.8) - 200 = 486 N F f = μF N = (0.5)(486) = 243 N N2L

36. FGFG FaFa FfFf FNFN ΣF x = F ax – F f = ma = 1.47 m/s 2 m = 70 kg F a = 400 N, 30  μ = 0.5

37. FNFN FfFf FGFG F Gy F Gx θ 4.7 A box sits on an incline that makes an angle of 30˚ with the horizontal. Find the acceleration of the box down the incline if the coefficient of friction is 0.30. θ = 30  μ = 0.3 N2L ΣF y = F N - F Gy = 0 F N = F Gy ΣF x = F Gx – F f = ma F f = μF N ΣF x = F Gx – μ F Gy = ma

38. FNFN FfFf FGFG F Gy F Gx θ ΣFx = F Gx – μ F Gy = ma mg sin θ - μ mg cos θ = ma = 2.35 m/s 2

39. 4.8 Two blocks m 1 (300 g) and m 2 (500 g), are pushed by a force F. If the coefficient of friction is 0.40. a. What must be the value of F if the blocks are to have an acceleration of 200 cm/s 2 ? m 1 = 0.3 kg m 2 = 0.5 kg μ = 0.4 a = 2 m/s 2 F G1 = F N1 F G2 = F N2 F G1 F G2 F N1 F N2 F F f2 F f1 ΣF x = F - F f1 - F f2 = m T a F = m T a + F f1 + F f2 = m T a + μ F NT = m T a + μ m T g = 0.8(2) + (0.4) 0.8 (9.8) = 4.7 N N2L

40. b. How large a force does m 1 then exert on m 2 ? m 2 alone ΣF x = m 2 a F 12 - F f2 = m 2 a F 12 = F f2 + m 2 a = μ m 2 g + m 2 a = 0.4(0.5) (9.8) + 0.5 (2) = 2.96 N F f2 F 12 F N2 F G2

41. 4.9 An object m A = 25 kg rests on a tabletop. A rope attached to it passes over a light frictionless pulley and is attached to a mass m B = 15 kg. The coefficient of friction between the table and block A is 0.20. a. What is the acceleration of the 25 kg block? m A = 25 kg m B = 15 kg μ = 0.2 F N = F GA = 25(9.8) = 245 N F fA = μF N = 0.2 (245) = 49 N FNFN F GA FfFf FTFT FTFT F GB 25 kg 15 kg N2L

42. FNFN F GA FfFf FTFT FTFT F GB ΣF = F T - F fA + F GB - F T = m T a = 2.45 m/s 2 m A = 25 kg m B = 15 kg μ = 0.2

43. FNFN F GA FfFf FTFT FTFT F GB ΣF = F T - F f = m A a F T = m A a + F f = 25 (2.45) + 49 = 110.25 N b. What is the tension on the string?

44. 4.10 Blocks 1 and 2 of masses m l and m 2, respectively, are connected by a light string, as shown above. These blocks are further connected to a block of mass M by another light string that passes over a pulley of negligible mass and friction. Blocks l and 2 move with a constant velocity v down the inclined plane, which makes an angle  with the horizontal. The kinetic frictional force on block 1 is f and that on block 2 is 2f.

45. a. On the figure below, draw and label all the forces on block m l. FGFG FNFN FTFT FfFf

46. Express your answers to each of the following in terms of m l, m 2, g, , and f. b. Determine the coefficient of kinetic friction between the inclined plane and block 1.

47. c. Determine the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity down the plane. Mg m 1 g m 2 g f 2f FTFT FTFT FTFT F N1 F N2 N1L FTFT

48. m 1 g f F N1 d. The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1 while it is on the inclined plane. N2L

49. Summary Newton’s Second Law: A net force produces an acceleration in the direction of the force that is directly proportional to the force and inversely proportional to the mass.

50. Summary: Procedure Read and write data with units.Read and write data with units. Draw free-body diagram for each object.Draw free-body diagram for each object. Choose x or y-axis along motion and choose direction of motion as positive.Choose x or y-axis along motion and choose direction of motion as positive. Write Newton’s law for both axes:Write Newton’s law for both axes:  F x = m x  F y = m y  F x = m a x  F y = m a y Solve for unknown quantities.Solve for unknown quantities. Read and write data with units.Read and write data with units. Draw free-body diagram for each object.Draw free-body diagram for each object. Choose x or y-axis along motion and choose direction of motion as positive.Choose x or y-axis along motion and choose direction of motion as positive. Write Newton’s law for both axes:Write Newton’s law for both axes:  F x = m x  F y = m y  F x = m a x  F y = m a y Solve for unknown quantities.Solve for unknown quantities. N = (kg)(m/s 2 )