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Chapter 4

Chapter 4. Forces in One Dimension. Section 4.1 Force and Motion. Essential Questions : What is a force? What is the relationship between force and acceleration? How does motion change when the net force is zero?. Section 4-1 Force and Motion.

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Chapter 4

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  1. Chapter 4 Forces in One Dimension

  2. Section 4.1 Force and Motion • Essential Questions: • What is a force? • What is the relationship between force and acceleration? • How does motion change when the net force is zero?

  3. Section 4-1 Force and Motion • Force (F) - a push or pull. All forces are vectors. • Force is measured in Newtons (N). 1 N = kgm/s2 • Force vectors are normally blue in color. • When considering how a force affects motion, it is important to identify the object or objects of interest, called the system. • Everything around the system with which the system can interact is called the external world.

  4. Types of Forces • Contact forces – directly touching the object with a force. • Field forces – these forces are exerted without contact. • Gravity • Magnetism • Electrical fields

  5. Types of Field Forces • Gravitational force - an attractive force that exists between all objects. This is the weakest of the four field forces. • Electromagnetic force - the forces that give materials their strength. • Strong nuclear force - holds particles in the nucleus together. • Weak force - a form of electromagnetic force.

  6. Agents • Forces result from interactions; every contact and field force has a specific and identifiable cause, called the agent. • I.e. When you push your textbook, your hand (agent) exerts a force on the textbook (the system).

  7. Free-body Diagrams (FBD) • A free-body diagram is a physical representation that shows the forces acting on a system.

  8. Free-body Diagram Guidelines • The FBD is drawn separately from the sketch of the problem. • Apply the particle model, and represent the object with a dot. • Represent each force with an arrow that points in the direction the force is applied. Always draw the force vectors pointing away from the particle, even when pushed.

  9. Free-body Diagram Guidelines • Make the length of each arrow proportional to the size of the force. Make your best estimate. • Label each force. Use the symbol F with a subscript label to identify both the agent and the object on which the force is exerted. • Choose a direction to be positive, and indicate this on the diagram. • Practice FN Ftable on box +y Fbox on table Fw

  10. Combining Forces • Multiple forces on an object can be defined. • Net force (Fnet) – the vector sum of all the forces on an object. F1 = 20 N F2 = 18 N 2N

  11. Ex: A skydiver falls downward through the air at a constant velocity. (The air exerts an upward force on the person.) Draw a free body diagram. Fdrag Fg

  12. Ex: Two horizontal forces, 225 N and 165 N, are exerted on a canoe. If these forces are applied in opposite direction, find the net horizontal force on the canoe. Be sure to draw a free-body diagram. 225 N 225 N 165 N 165 N 60 N toward the larger force

  13. Ex: If the same two forces as in the previous problem are exerted on the canoe in the same direction, what is the net horizontal force on the canoe? Be sure to indicate the direction of the net force. 225 N 165 N 225 N 165 N 390 N

  14. Acceleration and Force • The slope on an acceleration-force graph is as follows: m/s2= m/s2 = 1 N kgm/s2 kg Slope = 1/mass

  15. Newton’s Second Law • Acceleration is found to be directly proportional to force. • Force = mass times acceleration • F = ma • Unit for force. • 1 N = kgm/s2

  16. Ex: What force is required to accelerate a 1500 kg race car at +3.00 m/s2? F = ma F = (1500 kg)(+3.00 m/s2) F = 4500 kgm/s2 F = + 4,500 N

  17. Ex: A spring scale is used to exert a net force of 2.7 N on a cart. If the cart’s mass is 0.64 kg, what is the cart’s acceleration? Fnet = 2.7 N m = 0.64 kg F = ma 2.7 N = (0.64 kg)a a = 4.21875 m/s2 a = +4.2 m/s2

  18. Ex: Timmy is holding a pillow with a mass of 0.30 kg when John decides that he wants it and tries to pull it away from Timmy. If John pulls horizontally on the pillow with a force of 10.0 N and Timmy pulls with a horizontal force of 11.0 N, what is the horizontal acceleration of the pillow? John Timmy F = ma 1.0 N = (0.30 kg)a a = 3.3 m/s2 10.0 N 11.0 N 11.0 N 1.0 N in the direction of Timmy 10.0 N

  19. Newton’s First Law of Motion • An object with no force acting on it moves with constant velocity. • An object with no force acting on it remains at rest or moves with constant velocity in a straight line.

  20. Inertia • Newton’s first law is sometimes called the law of inertia. • Inertia is the tendency of an object to resist changes in velocity. • Inertia is not a force! It is not an interaction between two objects.

  21. Equilibrium • If the net force on an object is zero, then the object is in equilibrium. • An object is in equilibrium if it is moving at a constant velocity. • An object is also in equilibrium if it is not moving (v = 0 m/s).

  22. Section 4.1 Force and Motion • Did We Answer the Essential Questions: • What is a force? • What is the relationship between force and acceleration? • How does motion change when the net force is zero?

  23. Section 4.2 Weight and Drag Force • Essential Questions: • How are the weight and the mass of an object related? • How do actual weight and apparent weight differ? • What effect does air have on falling objects?

  24. Weight • An object’s weight is the gravitational force experienced by that object. Fg = mg = Fw • The gravitational field (g) is a vector quantity that relates the mass of an object to the gravitational force it experiences at a given location. (g = 9.80 m/s2 towards Earth’s center)

  25. Ex: Find the weight of a 2.26 kg bag of sugar. m = 2.26 kg g = -9.80 m/s2 Fw = ? N FW = mg FW = (2.26 kg)(-9.80 m/s2) FW = -22.148 kgm/s2 FW = -22.1 N Fg = 22.1 N

  26. Scales • Inside a scale, springs provide the upward force necessary to make the net force equal zero. You can only use this formula if the net force is zero! (The object is NOT moving!) Fscale = -Fg

  27. Ex: You place a watermelon on a spring scale calibrated to measure in newtons. If the watermelon’s mass is 4.0 kg, what is the scale’s reading? Fscale = 39 N m = 4.0 kg g = -9.80 m/s2 Fscale = ? Fscale = -Fg Fscale = -mg Fscale = -(4.0 kg)(-9.80 m/s2) 39.2 kgm/s2 39 N Fg = 39 N

  28. Ex: You place a 22.50 kg TV on a spring scale. If the scale reads 235.2 N, what is the gravitational field at that location? m = 22.50 kg Fscale = 235.2 N g = ? m/s2 Fscale = -Fg Fscale = -mg 235.2 N = -(22.50 kg)g g = -10.4533 m/s2 g = -10.45 m/s2 Fscale = 235.2 N Fg = 235.2 N

  29. Net Force • Remember….net force is equal to total force. When applying net force on an object in vertical position, we will use the following formula: Fnet = Fa + Fg • If the object is not moving then, Fnet = 0. If the object is moving, then Fnet = ma.

  30. Ex: A grocery sack can withstand a maximum of 230 N before it rips. Will a bag holding 15 kg of groceries that is lifted from the checkout counter at an acceleration of 7.0 m/s2 hold? Fa = ? N m = 15 kg a = 7.0 m/s2 g = -9.80 m/s2 Is Fa greater than or less than 230 N? Fnet = Fg + Fa ma = mg + Fa (15 kg)(7.0 m/s2) = (15 kg)(-9.8 m/s2) + Fa 105 = -147 + Fa 252 N = Fa The bag will rip! Fnet = 105 N Fg = 147 N

  31. Apparent Weight • Apparent weight – the support force exerted on an object. • Standing on a scale in an elevator. • Accelerating upward, you feel heavier • Accelerating downward, you feel lighter

  32. Weightlessness • Weightlessness – there are no contact forces acting to support an object. The object’s apparent weight is zero. • Astronauts experience weightlessness in orbit because they and their spacecraft are in free fall. • Vomit comet

  33. Ex: Your mass is 75.0 kg and you are standing on a bathroom scale in an elevator. Starting from rest, the elevator accelerates upward at 2.00 m/s2 for 2.00 seconds and then continues at a constant speed. Is the scale reading during acceleration greater than, equal to, or less than the scale reading when the elevator is at rest? m = 75.0 kg a = 2.00 m/s2 g = -9.80 m/s2 t = 2.00 s At rest: Fscale = -Fg Fscale = -mg Fscale = -(75.0 kg)(-9.80 m/s2) Fscale = 735 N Fscale = 735 N Fg = 735 N

  34. Ex: Your mass is 75.0 kg and you are standing on a bathroom scale in an elevator. Starting from rest, the elevator accelerates upward at 2.00 m/s2 for 2.00 seconds and then continues at a constant speed. Is the scale reading during acceleration greater than, equal to, or less than the scale reading when the elevator is at rest? m = 75.0 kg a = 2.00 m/s2 g = -9.80 m/s2 t = 2.00 s Accelerating Upward: Fnet = Fscale + Fg ma = Fscale + mg (75.0)(2.00) = Fscale + (75.0)(-9.80) 150 N = Fscale + (-735 N) Fscale = 885 N Scale reading is greater than during acceleration! FScale = ? N Fnet = 150 N Fg = 735 N

  35. Drag Force • A drag force is the force exerted by a fluid on an object opposing motion through the fluid. • Gases and liquids are fluids.

  36. Terminal Velocity • The constant velocity that is reached when the drag force equals the force of gravity is called terminal velocity. Terminal velocity: Fg = Fdrag

  37. Section 4.2 Weight and Drag Force • Did We Answer The Essential Questions: • How are the weight and the mass of an object related? • How do actual weight and apparent weight differ? • What effect does air have on falling objects?

  38. Sec 4.3 - Newton’s Third Law • Essential Questions: • What is Newton’s Third Law? • What is the Normal Force? Forces come in “Pears”

  39. Interaction Pairs • Forces always come in pairs. A set of two forces that are in opposite directions, have equal magnitudes, and act on different objects are called an interaction pair. • Also known as action-reaction pair.

  40. Newton’s Third Law of Motion • When one object exerts a force on a second object, the second object exerts a force on the first that is equal in magnitude but opposite in direction. • Action-reaction forces FA on B = - FB on A

  41. Ex: A softball has a mass of 0.18 kg. What is the gravitational force on Earth due to the ball, and what is Earth’s resulting acceleration? Earth’s mass is 6.0 x 1024 kg. mball = 0.18 kg mEarth = 6.0 x 1024kg gearth on ball = -9.80 m/s2 Fball on earth = ? N Fball on Earth = - FEarth on ball Fball on Earth = - mballgEarth on ball Fball on E = -(0.18 kg)(-9.8 m/s2) Fball on E = 1.764 N Fball on E = 1.8 N

  42. Ex: A softball has a mass of 0.18 kg. What is the gravitational force on Earth due to the ball, and what is Earth’s resulting acceleration? Earth’s mass is 6.0 x 1024 kg. Fball on E = 1.8 N gball on Earth = ? m/s2 Fball on E = mEgball on Earth 1.8 N = (6.0 x 1024 kg)(gball on Earth) gball on Earth = 3.0 x 10-25 m/s2

  43. Tension • Tension (FT) is simply a specific name for the force that a string or rope exerts. • We will assume that all strings and ropes are massless.

  44. Net Force and Tension • If an object is hanging by a rope or string, then use the following formula: Fnet = FT + Fg

  45. Ex: A 50.0 kg bucket is being lifted by a rope. The rope will not break if the tension is 525 N or less. The bucket started at rest, and after being lifted 3.0 m, it moves at 3.0 m/s. If the acceleration is constant, is the rope in danger of breaking? m = 50.0 kg Is FT equal/less 525 N? vi = 0.0 m/s d = 3.0 m vf = 3.0 m/s Since the bucket is accelerating, we need to find a, vf2 = vi2 + 2ad. (3.0)2 = 0 + 2a(3.0) a = 1.5 m/s2 and now we can find FT. Fnet = FT + Fg ma = FT + mg ma = FT + mg (50.0)(1.5) = FT + (50.0)(-9.80) FT = 565 N (The rope will Break!)

  46. Ex: A block hangs from the ceiling by a massless rope. A 3.0 kg block is attached to the first block and hangs below it on another piece of massless rope. The tension in the top rope is 63.0 N. Find the mass of the top block and the tension in the bottom rope . m2 = 3.0 kg m1 = ? kg T1 = 63.0 N T2 = ? N Fnet = FT1 + Fg 0.0 N = 63.0N + (m1 + 3.0 kg)(-9.80 m/s2) -63.0N = (m1 + 3.0 kg)(-9.80 m/s2) m1 = 3.4 kg Fnet = FT2 + Fg 0.0 N = FT2 + (3.0 kg)(-9.80 m/s2) FT2 = 29.4 N

  47. The Normal Force • The normal force (FN) is the perpendicular contact force that a surface exerts on another surface. This is similar to finding the Fscale when on a horizontal surface. • FN = -Fg

  48. Ex: Three blocks are stacked on top of one another. The top block has a mass of 4.6 kg, the middle one has a mass of 1.2 kg, and the bottom one has a mass of 3.7 kg. Identify and calculate any normal forces between the objects. m1 = 4.6 kg m2 = 1.2 kg m3 = 3.7 kg FN1/2 = ? N FN2/3 = ? N FN3/table = ? N FN = -Fg FN1/2 = -(4.6)(-9.80 m/s2) FN1/2 = 45 N FN2/3 = -(4.6 + 1.2)(-9.80 m/s2) FN2/3 = 57 N FN3/table = -(4.6 + 1.2 + 3.7)(-9.80 m/s2) FN3/table = 93 N

  49. Sec 4.3 - Newton’s Third Law • Did We Answer The Essential Questions: • What is Newton’s Third Law? • What is the Normal Force? Forces come in “Pears”

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