1 / 85

Forces and Uniform Circular Motion

Forces and Uniform Circular Motion. Physics Unit 2. 4.1-4.4 Newton’s Laws of Motion. Kinematics How things move Dynamics Why things move. Force A push or a pull Is a vector Unit: Newton (N) Measured by a spring scale. 4.1-4.4 Newton’s Laws of Motion. Newton’s First Law of Motion

omana
Download Presentation

Forces and Uniform Circular Motion

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Forces and Uniform Circular Motion Physics Unit 2

  2. 4.1-4.4 Newton’s Laws of Motion • Kinematics • How things move • Dynamics • Why things move • Force • A push or a pull • Is a vector • Unit: Newton (N) • Measured by a spring scale

  3. 4.1-4.4 Newton’s Laws of Motion Newton’s First Law of Motion • A body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force. • Inertia • Property of objects to remain in constant motion or rest. • Mass is a measure of inertia

  4. 4.1-4.4 Newton’s Laws of Motion Newton’s Second Law of Motion • Acceleration of a system is directly proportional to and in the same direction as the net force acting on the system, and inversely proportional to its mass. • Net force is the vector sum of all the forces.

  5. 4.1-4.4 Newton’s Laws of Motion • Free-body diagram • Draw only forces acting on the object • Represent the forces are vector arrows

  6. 4.1-4.4 Newton’s Laws of Motion • Weight • Measure of force of gravity • Unit: N • Depends on local gravity • Mass • Not a force • Measure of inertia or amount of matter • Unit: kg • Constant

  7. F R T W 4.1-4.4 Newton’s Laws of Motion • A 1000 kg sailboat’s engine pushes forward with 5000 N thrust. The wind blows the boat at an angle of 30° starboard with a force of W. Another boat is pulling it at 30° port with force of F. There is a resistance force of 2000 N. There is an acceleration of 5.0 m/s2 forward. What are W and F? • W = F = 1160 N

  8. 4.1-4.4 Newton’s Laws of Motion Newton’s Third Law of Motion • Whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that it exerts. • Every force has an equal and opposite reaction force. • You push down on your chair, so the chair pushed back up on you.

  9. 4.1-4.4 Newton’s Laws of Motion • A 70-kg diver jumps off a boat on a smooth lake. He applies 75 N horizontally against the 500-kg boat. What are the accelerations of the diver and the boat? • ad = 1.07 m/s2 • ab = -0.15 m/s2

  10. Day 14 Homework • Force yourself to do these problems • 4P1-4a, 6, 8-9, 13, 15-16 • Read 4.5-4.6 • 4CQ21-22 • Answers • 1) 265 N • 2) 9.26 s • 3) 13.3 m/s2 • 4a) 56.0 kg • 6) 4.12×105 N • 8) 253 m/s2 • 9) 0.130 m/s2, 0 m/s2 • 13) 1.5×103 N, 150 kg • 15) 2.64×103 N • 16) 692 N, 932 N

  11. FN 4.5 Normal, Tension, and Other Forces4.6 Problem-Solving Strategies • When two objects touch there is often a force • Normal Force • Perpendicular component of the contact force between two objects

  12. 4.5 Normal, Tension, and Other Forces4.6 Problem-Solving Strategies • Weight pushes down • So the table pushes up • Called Normal force • Newton’s 3rd Law • Normal force doesn’t always = weight • Draw a freebody diagram to find equation

  13. 4.5 Normal, Tension, and Other Forces4.6 Problem-Solving Strategies • A box is sitting on a ramp angled at 20°. If the box weighs 50 N, what is the normal force on the box? • 47 N FN 20° 20° w

  14. 4.5 Normal, Tension, and Other Forces4.6 Problem-Solving Strategies • Tension • Pulling force from rope, chain, etc. • Everywhere the rope connects to something, there is an identical tension

  15. 4.5 Normal, Tension, and Other Forces4.6 Problem-Solving Strategies • Problems-Solving Strategy • Identify the principles involved and draw a picture • List your knows and Draw a free-body diagram • Apply • Check your answer for reasonableness

  16. 4.5 Normal, Tension, and Other Forces4.6 Problem-Solving Strategies • A lady is weighing some bananas in a grocery store when the floor collapses. If the bananas mass is 2 kg and the floor is accelerating at -2.25 m/s2, what is the apparent weight (normal force) of the bananas? • FN = 15.1 N

  17. y 116° T1 T2 x W 4.5 Normal, Tension, and Other Forces4.6 Problem-Solving Strategies • A stoplight is suspended by two cables over a street. Weight of the light is 110 N and the cables make a 116° angle with each other. Find the tension in each cable. • 104 N

  18. Day 15 Homework • The tension is mounting… I can’t wait to see what’s next! • 4P18-20, 23-27a, 28, 34-35 • Read 4.7-4.8 • 4CQ23, 25-27 • Answers • 18) 779 N • 19) 7.84×10-4 N, 1.89×10-3 N, 2.41 time tension in vert strand • 20) 588 N, 678 N • 23) 6.20 m/s2 • 24) 1.03×103 kg • 25) 3.43×103 N • 26) 3.14×103 N • 27a) 4.41×105 N • 28) 9.89×104 kg, 1.70×104 N • 34) 1.42×103 N, 539 N • 35) 73 N

  19. 4.7 Further Applications of Newton’s Laws of Motion4.8 The Four Basic Forces • More fun problems! • The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is 53,800 N. The lift force Lgenerated by the rotating blade makes an angle of 21.0° with respect to the vertical. What is the magnitude of the lift force? • 57600 N

  20. 4.7 Further Applications of Newton’s Laws of Motion4.8 The Four Basic Forces • A 1380-kg car is moving due east with an initial speed of 27.0 m/s. After 8.00 s the car has slowed down to 17.0 m/s. Find the magnitude and direction of the net force that produces the deceleration.

  21. 4.7 Further Applications of Newton’s Laws of Motion4.8 The Four Basic Forces • A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 535 N. Find the tensions in the rope to the left and to the right of the mountain climber.

  22. 4.7 Further Applications of Newton’s Laws of Motion4.8 The Four Basic Forces • Four Basic Forces • All forces are made up of only 4 forces • Gravitational - gravity • Electromagnetic – static electricity, magnetism • Weak Nuclear - radioactivity • Strong Nuclear – keeps nucleus of atoms together

  23. 4.7 Further Applications of Newton’s Laws of Motion4.8 The Four Basic Forces • All occur because particles with that force property play catch with a different particle • Electromagnetic uses photons • Scientists are trying to combine all forces together in Grand Unified Theory • Have combined electric, magnetic, weak nuclear • Gravity is the weakest • We feel it because the electromagnetic cancels out over large areas • Nuclear forces are strong but only over short distance

  24. Day 16 Homework • Electromagnetic forces are responsible for doing homework. • 4P40-44, 46, 52-54 • Read 5.1 • 5CQ1-3 • Answers • 40) 10.23 m/s2 @ 85.33° above horizontal • 41) 376 N • 42) 194 N • 43) -68.5 N • 44) 7.43 m/s, 2.97 m • 46) 4.20 m/s, 29.4 m/s2, 4.31×103 N • 52) 10-13, 10-11 • 53) 10-38, 10-25, 10-36, gravity has small nuclear influence • 54) 102

  25. 5.1 Friction • Normal force – perpendicular to surface • Friction force – parallel to surface, and opposes motion • Comes from rough surface • Not well understood

  26. 5.1 Friction • Static Friction • Keeps things from moving. • Cancels out applied force until the applied force gets too big. • Depends on force pushing down and roughness of surface

  27. 5.1 Friction • Static Friction • Depends on force pushing down and roughness of surface • More pushing down (FN), more friction • is coefficient of static friction (0.01 to 1.5)

  28. 5.1 Friction • Kinetic Friction • Once motion happens • is usually less than

  29. W fs Sofa FN 5.1 Friction • A sofa rests on the horizontal bed of a moving van. The coefficient of static friction between the sofa and the van bed is 0.30. The van starts from rest and accelerates for a time of 5.1 s. What is the maximum distance the van can travel in this time period without having the sofa slide? • 38.2 m

  30. W fk FN 5.1 Friction • A car skids to a stop after initially going 30.0 m/s. k = 0.800. How far does the car go before stopping? • 57.3 m

  31. Day 17 Homework • Don’t let these problems cause friction between us • 5P1-5b, 8-11, 17-19 • Read 5.2 • 5CQ6-8 • Answers • 1) 5.00 N • 2) 1.00×103 N, 30.0 N • 3) 10 N, 97.0 N • 4) 588 N, 1.96 m/s2 • 5a) 4.90 m/s2, 5b) will not slip • 8) work • 9) work • 10) 1.83 m/s2 • 11) 0.737 m/s2, 5.71° • 17) 272 N, 512 N, 0.268 • 18) 51.0 N, 0.720 m/s2 • 19) 46.5 N, 0.655 m/s2

  32. 5.2 Drag Forces • Drag • Resistive force from moving through a fluid • Size depends on area, speed, and properties of the fluid • For large objects • = drag coefficient • = density of fluid • = area of object • = speed of object relative to the fluid

  33. 5.2 Drag Forces • The drag coefficient of a car is measured in a wind tunnel. If the wind is blowing at , the area is , the density of air is , and the drag force is measured at . What is the drag coefficient of the car?

  34. 5.2 Drag Forces • Terminal Velocity • Falling objects will accelerate until the downward force of gravity = drag force

  35. 5.2 Drag Forces • Find the terminal velocity of a falling mouse in air () and a falling human falling flat (). The density of air is . • Mouse: 12.7 m/s • Human: 44.4 m/s

  36. 5.2 Drag Forces • Terminal Velocity of very small objects (like pollen) • Stoke’s Law • = radius of object • = viscosity of fluid (kg/m·s) Table 12.1 • = velocity of object • What is terminal velocity for a Ragweed pollen grain? , density of pollen is , and viscosity of air is .

  37. Day 18 Homework • Even though homework can be a drag, it’s good for you • 5P20-23, 25-28 • Read 5.3 • 5CQ9, 11, 14-16 • Answers • 20) 115 m/s, 414 km/h • 21) Assume A=0.14 m2 49.2 s • 22) 25.1 m/s, 9.9 m/s • 23) 44.8 N, 91.5 N; 357 N, 729 N • 25) 313 m/s, 9.8 m/s • 26) work • 27) for water , 2.38×10-6 m/s • 28) 0.76 kg/m·s

  38. 5.3 Elasticity: Stress and Strain • Forces that deform • Change shape • For small deformations • Object returns to original shape • Deformation proportional to force • Hooke’s Law

  39. 5.3 Elasticity: Stress and Strain • Hooke’s Law • The amount of stretching () depends on several things • Tension and Compression • depends • applied force • cross-sectional area • property called elastic modulus or Young’s modulus

  40. 5.3 Elasticity: Stress and Strain • For Tension and Compression Hooke’s Law becomes • Where • is change in length • is applied force • is Young’s modulus (see Table 5.3) • is cross-sectional area • is the original length

  41. 5.3 Elasticity: Stress and Strain • Rearranging produces

  42. 5.3 Elasticity: Stress and Strain • What is the Young’s Modulus for nylon rope? It’s radius is 0.005 m and a 1 meter length stretches 0.0016 m with 50 kg is hung on it’s end.

  43. 5.3 Elasticity: Stress and Strain • Sideways Stress: Shear Modulus • Force perpendicular to length of material • is shear modulus

  44. 5.3 Elasticity: Stress and Strain • Changes in Volume: Bulk Modulus • Deforming material in all directions • Gases very easy • Solids and liquids more difficult • is bulk modulus • Depends on molecular arrangement within material and overcome electromagnetic forces

  45. 5.3 Elasticity: Stress and Strain • If it didn’t break, how much would the end of a 2 m long wooden pole, 3 cm in diameter, bend if a 80 kg person hung from the end of it?

  46. Day 19 Homework • Let these problems stress and strain your mind. • 5P30-39, 41-42 • Read 6.1-6.2 • 6CQ1-2 • Answers • 30) • 31) • 32) • 33) • 34) • 35) • 36) • 37) • 38) • 39) , • 41) • 42)

  47. 6.1 Rotation Angle and Angular Velocity6.2 Centripetal Acceleration • Newton’s Laws of motion primarily relate to straight-line motion. • Uniform Circular Motion • Motion in circle with constant speed • Rotation Angle () • Angle through which an object rotates

  48. 6.1 Rotation Angle and Angular Velocity6.2 Centripetal Acceleration • Arc Length is the distance around part of circle • Angle Units: • Revolutions: 1 circle = 1 rev • Degrees: 1 circle = 360° • Radians: 1 circle = • Arc Length formula must use radians and angle unit

  49. 6.1 Rotation Angle and Angular Velocity6.2 Centripetal Acceleration • Convert 60° to radians • Angular Velocity • How fast an object rotates

  50. 6.1 Rotation Angle and Angular Velocity6.2 Centripetal Acceleration • Angular Velocity () • Unit: rad/s • CCW +, CW – • A CD rotates 320 times in 2.4 s. What is its angular velocity in rad/s? What is the linear velocity of a point 5 cm from the center?

More Related