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Chapter 4

Chapter 4. Forces and Newton’s Laws of Motion. Forces and Newton’s Laws of Motion. Forces Newton’s Three Laws of Motion The Gravitational Force Contact Forces (normal, friction, tension) Application of Newton’s Second Law Apparent Weight. Net Force.

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Chapter 4

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  1. Chapter 4 Forces and Newton’s Laws of Motion PHY 1401- Ch 04b - Revised: 6/9/2010

  2. Forces and Newton’s Laws of Motion • Forces • Newton’s Three Laws of Motion • The Gravitational Force • Contact Forces (normal, friction, tension) • Application of Newton’s Second Law • Apparent Weight PHY 1401- Ch 04b - Revised: 6/9/2010

  3. Net Force The net force is the vector sum of all the forces acting on a body. The net force is the resultant of this vector addition. Bold letters represent vectors. The units of Force are Newtons, or the abbreviation N, which represent the SI units: kg-m/s2 PHY 1401- Ch 04b - Revised: 6/9/2010

  4. Free Body Diagrams The free body diagram (FBD) is a simplified representation of an object, and the forces acting on it. It is called free because the diagram will show the object without its surroundings; i.e. the body is “free” of its environment. We will consider only the forces acting on our object of interest. The object is depicted as not connected to any other object – it is “free”. Label the forces appropriately. Do not include the forces that this body exerts on any other body. The best way to explain the free body diagram is to describe the steps required to construct one. Follow the procedure given below. (1) Isolate the body of interest. Draw a dotted circle around the object that separates our object from its surroundings. (2) Draw all external force vectors acting on that body. (3) You may indicate the body’s assumed direction of motion. This does not represent a separate force acting on the body. (4) Choose a convenient coordinate system. PHY 1401- Ch 04b - Revised: 6/9/2010

  5. y N1 F T x w1 Free Body Diagram The force directions are as indicated in the diagram. The magnitudes should be in proportion if possible. PHY 1401- Ch 04b - Revised: 6/9/2010

  6. Newton’s First Law of Motion:Inertia and Equilibrium Newton’s 1st Law (The Law of Inertia): If no force acts on an object, then the speed and direction of itsmotion do not change. Inertia is a measure of an object’s resistance to changes in its motion. It is represented by the inertial mass. PHY 1401- Ch 04b - Revised: 6/9/2010

  7. Newton’s First Law of Motion If the object is at rest, it remains at rest (velocity = 0). If the object is in motion, it continues to move in a straight line with the same velocity. No force is required to keep a body in straight line motion when effects such as friction are negligible. An object is in translational equilibrium if the net force on it is zero and vice versa. Translational Equilibrium PHY 1401- Ch 04b - Revised: 6/9/2010

  8. Newton’s Second Law of Motion Net Force, Mass, and Acceleration Newton’s 2nd Law: The acceleration of a body is directly proportional to the net force acting on the body and inversely proportional to the body’s mass. Mathematically: This is the workhorse of mechanics PHY 1401- Ch 04b - Revised: 6/9/2010

  9. Newton’s Second Law of Motion An object’s mass is a measure of its inertia. The more mass, the more force is required to obtain a given acceleration. The net force is just the vector sum of all of the forces acting on the body, often written as F. If a = 0, then F = 0. This body can have: Velocity = 0 which is called static equilibrium, or Velocity  0, but constant, which is called dynamic equilibrium. PHY 1401- Ch 04b - Revised: 6/9/2010

  10. Newton’s Third Law of Motion Interaction Pairs Newton’s 3rd Law: When 2 bodies interact, the forces on the bodies, due to each other, are always equal in magnitude and opposite in direction. In other words, forces come in pairs. Mathematically: designates the force on object 2 due to object 1. PHY 1401- Ch 04b - Revised: 6/9/2010

  11. Types of Forces Contact forces: Normal Force & Friction Tension Gravitational Force PHY 1401- Ch 04b - Revised: 6/9/2010

  12. Contact Forces Contact forces: these are forces that arise due to of an interaction between the atoms in the surfaces of the bodies in contact. PHY 1401- Ch 04b - Revised: 6/9/2010

  13. Normal force of the ground on the box N Normal force of the ramp on the box N w w Normal Forces Normal force: this force acts in the direction perpendicular to the contact surface. PHY 1401- Ch 04b - Revised: 6/9/2010

  14. y N x w Normal Forces Example: Consider a box on a table. FBD for box Apply Newton’s 2nd law This just says the magnitude of the normal force equals the magnitude of the weight; they are not Newton’s third law interaction partners. PHY 1401- Ch 04b - Revised: 6/9/2010

  15. Frictional Forces Friction: a contact force parallel to the contact surfaces. Static friction acts to prevent objects from sliding. Kinetic friction acts to make sliding objects slow down. Sometimes called Dynamic friction. PHY 1401- Ch 04b - Revised: 6/9/2010

  16. Frictional Forces PHY 1401- Ch 04b - Revised: 6/9/2010

  17. Tension This is the force transmitted through a “rope” from one end to the other. An ideal cord has zero mass, does not stretch, and the tension is the same throughout the cord. PHY 1401- Ch 04b - Revised: 6/9/2010

  18. y T x w Example (text problem 4.77): A pulley is hung from the ceiling by a rope. A block of mass M is suspended by another rope that passes over the pulley and is attached to the wall. The rope fastened to the wall makes a right angle with the wall. Neglect the masses of the rope and the pulley. Find the tension in the rope from which the pulley hangs and the angle . FBD for the mass M Apply Newton’s 2nd Law to the mass M. PHY 1401- Ch 04b - Revised: 6/9/2010

  19. y F T  x T Example continued: Apply Newton’s 2nd Law: FBD for the pulley: This statement is true only when  = 45 and PHY 1401- Ch 04b - Revised: 6/9/2010

  20. F21 F12 M2 M1 r Gravitational Forces Gravity is the force between two masses. Gravity is a long-range force. No contact is needed between the bodies. The force of gravity is always attractive! r is the distance between the two masses M1 and M2 and G = 6.671011 Nm2/kg2. PHY 1401- Ch 04b - Revised: 6/9/2010

  21. Gravitational Forces Let M1 = ME = mass of the Earth. Here F = the force the Earth exerts on mass M2. This is the force known as weight, w. Near the surface of the Earth PHY 1401- Ch 04b - Revised: 6/9/2010

  22. Gravitational Forces Note that is the gravitational force per unit mass. This is called the gravitational field strength. It is also referred to as the acceleration due to gravity. What is the direction of g? What is the direction of w? PHY 1401- Ch 04b - Revised: 6/9/2010

  23. Gravitational Forces Example: What is the weight of a 100 kg astronaut on the surface of the Earth (force of the Earth on the astronaut)? How about in low Earth orbit? This is an orbit about 300 km above the surface of the Earth. On Earth: In low Earth orbit: The weight is reduced by about 10%. The astronaut is NOT weightless! PHY 1401- Ch 04b - Revised: 6/9/2010

  24. Applying Newton’s Second Law The one equation everyone remembers! Sumof the forces acting on the objects in the system “m” is the System Mass “a” is the System Response This equation is just the tip of the “iceberg” of the mechanics problem. The student will need to anlyze the forces in the problem and sum the force vector components to build the left hand side of the equation. PHY 1401- Ch 04b - Revised: 6/9/2010

  25. block 2 block 1 F Applying Newton’s Second Law Example: A force of 10.0 N is applied to the right on block 1. Assume a frictionless surface. The masses are m1 = 3.00 kg and m2 = 1.00 kg. Find the tension in the cord connecting the two blocks as shown. Assume that the rope stays taut so that both blocks have the same acceleration. PHY 1401- Ch 04b - Revised: 6/9/2010

  26. y y N1 N2 F T T x x w2 w1 FBD for block 2: FBD for block 1: Apply Newton’s 2nd Law to each block: PHY 1401- Ch 04b - Revised: 6/9/2010

  27. Example continued: (1) These two equations contain the unknowns: a and T. (2) To solve for T, a must be eliminated. Solve for a in (2) and substitute in (1). PHY 1401- Ch 04b - Revised: 6/9/2010

  28. y y N1 N2 F T T x x w2 w1 Pick Your System Carefully Include both objects in the system. Now when you sum the x-components of the forces the tensions cancel. In addition, since there is no friction, y-components do not contribute to the motion. PHY 1401- Ch 04b - Revised: 6/9/2010

  29. y N x w Apparent Weight Stand on a bathroom scale. FBD for the person: Apply Newton’s 2nd Law: PHY 1401- Ch 04b - Revised: 6/9/2010

  30. Apparent Weight The normal force is the force the scale exerts on you. By Newton’s 3rd Law this is also the force (magnitude only) you exert on the scale. A scale will read the normal force. is what the scale reads. When ay = 0, N = mg. The scale reads your true weight. When ay  0, N > mg or N < mg. In free fall ay = -g and N = 0. The person is weightless. PHY 1401- Ch 04b - Revised: 6/9/2010

  31. y N x w Apparent Weight Example (text problem 4.128): A woman of mass 51 kg is standing in an elevator. The elevator pushes up on her feet with 408 newtons of force. What is the acceleration of the elevator? FBD for woman: Apply Newton’s 2nd Law: (1) PHY 1401- Ch 04b - Revised: 6/9/2010

  32. Apparent Weight Example continued: Given: N = 408 newtons, m = 51 kg, g = 9.8 m/s2 Unknown: ay Solving (1) for ay: The elevator could be (1) traveling upward with decreasing speed, or (2) traveling downward with increasing speed. The change in velocity is DOWNWARD. PHY 1401- Ch 04b - Revised: 6/9/2010

  33. Free Body Diagram PHY 1401- Ch 04b - Revised: 6/9/2010

  34. This is not a methodolgy to solve for the acceleration. It is just graphically demonstrating that the net force is ma PHY 1401- Ch 04b - Revised: 6/9/2010

  35. Same problem but the applied force is angled up PHY 1401- Ch 04b - Revised: 6/9/2010

  36. The normal force, N, is smaller in this case because the upward angled applied force reduces the effective weight of the sled. PHY 1401- Ch 04b - Revised: 6/9/2010

  37. Equilibrium Problem PHY 1401- Ch 04b - Revised: 6/9/2010

  38. Equilibrium Problem PHY 1401- Ch 04b - Revised: 6/9/2010

  39. Equilibrium Problem This is an example of three-vector equilibrium problem. It lends itself to a simple solution because the vector sum of the three vectors closes on itself (equilibrium) and forms a triangle PHY 1401- Ch 04b - Revised: 6/9/2010

  40. Milk Carton Max static friction force Non-slip limit on applied force PHY 1401- Ch 04b - Revised: 6/9/2010

  41. PHY 1401- Ch 04b - Revised: 6/9/2010

  42. Hanging Problems PHY 1401- Ch 04b - Revised: 6/9/2010

  43. Hanging Picture PHY 1401- Ch 04b - Revised: 6/9/2010

  44. Hanging Picture - Free Body Diagram T2 mg T1 PHY 1401- Ch 04b - Revised: 6/9/2010

  45. Hanging Picture • Since this turned out to be a right triangle the simple trig functions are that is needed to find a solution. • If the triangle was not a right triangle then the Law of Sines would have been needed. PHY 1401- Ch 04b - Revised: 6/9/2010

  46. PHY 1401- Ch 04b - Revised: 6/9/2010

  47. PHY 1401- Ch 04b - Revised: 6/9/2010

  48. Atwood Machine and Variations PHY 1401- Ch 04b - Revised: 6/9/2010

  49. Atwood’s Machine PHY 1401- Ch 04b - Revised: 6/9/2010

  50. A 2-Pulley Atwood Machine PHY 1401- Ch 04b - Revised: 6/9/2010

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