Chapter 5 & 6

1. Chapter 5 & 6 Force and Motion-I & II

2. 5.2 Newtonian Mechanics • Study of relation between force and acceleration of a body: • Newtonian Mechanics. • Newtonian Mechanics does not hold good for all situations. • Examples: • Relativistic or near-relativistic motion • Motion of atomic-scale particles

3. 5.3 Newton’s First Law Newton’s First Law: If no force acts on a body, the body’s velocity cannot change; that is, the body cannot accelerate. If the body is at rest, it stays at rest. If it is moving, it continues to move with the same velocity (same magnitude and same direction).

4. 5.6 Newton’s Second Law The net force on a body is equal to the product of the body’s mass and its acceleration. In component form, The acceleration component along a given axis is caused only by the sum of the force components along that same axis, and not by force components along any other axis.

5. 5.8 Newton’s Third Law • The minus sign means that these two forces are in opposite directions • The forces between two interacting bodies are called a third-law force pair. When two bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction.

6. Newton’s laws are valid only in an inertial reference frame.

7. In a non-inertial reference frame the laws of physics vary depending on the acceleration of that frame with respect to an inertial frame, and the usual physical forces must be supplemented by fictitious forces , e.g. coriolis force and centrifugal force.

8. 滑輪無摩擦力 無質量 的滑輪與繩子 人與座椅的質量 = 100 kg (1)若人與座椅系統以等速度往上移動, 他人要拉繩子的力量為何 ? (1)若人與座椅系統以等速度往上移動, 人要拉繩子的力量為何 ? (2)若人與座椅系統以2.0 m/s2加速度往上移動, 他人要拉繩子的力量又為何 ? (2)若人與座椅系統以2.0 m/s2加速度往上移動, 人要拉繩子的力量又為何 ? (3)座椅對人的作用力量為何 ? (4)滑輪施於天花板的力量為何?

10. In a non-inertial reference frame the laws of physics vary depending on the acceleration of that frame with respect to an inertial frame, and the usual physical forces must be supplemented by fictitious forces , e.g. coriolis force and centrifugal force.

11. Inertial Reference Frames If the puck is sent sliding along a long ice strip extending from the north pole, and if it is viewed from a point on the Earth’s surface, the puck’s path is not a simple straight line. The apparent deflection is not caused by a force, but by the fact that we see the puck from a rotating frame. In this situation, the ground is a noninertial frame. An inertial reference frame is one in which Newton’s laws hold. If a puck is sent sliding along a short strip of frictionless ice—the puck’s motion obeys Newton’s laws as observed from the Earth’s surface. (a) The path of a puck sliding from the north pole as seen from a stationary point in space. Earth rotates to the east. (b) The path of the puck as seen from the ground.

12. Coriolis Force(科里奧利) (fictitious force) Force that appears if the particle has a velocity relative to a rotating frame. Consider a platform ( frame S’ ) of radius R which is rotating with constant angular velocity ω with respect to an inertial frame S. For the person ininertial frame S , the ball moves in a straight line.  S’    慣性座標上的觀察者的觀點 At time t =0, a person at the center O ( ininertial frame S ) throws a ball at speed v’toward a person P( innoninertial frame S’ ). By the time t when the ball reaches the rim, P has moved a distance (ωR) t.

13. Coriolis Force(科里奧利) 非慣性座標上的觀察者的觀點 ( For the personPin a noninertial frame [rotating frame] ) 

15. Motion of an object relative to some frame of reference( Inertial Reference Frame ) o' o P.3

16. Special case : S' S u t x' x P.3

17. Special case : S' S u t x' x Galilean Transformation P.3

18. Accelerated Noninertial Frame Galilean Transformation  Ax u u(t)

19. A m Accelerated Noninertial Frame Suppose that no horizontal forces acting on mass m Fx= 0 A

20. 5.9 Applying Newton’s Laws • The reading is equal to the magnitude of the normal force on the passenger from the scale. • We can use Newton’s Second Law only in an inertial frame. If the cab accelerates, then it is not an inertial frame. So we choose the ground to be our inertial frame and make any measure of the passenger’s acceleration relative to it. Sample Problem, Part a

22. The physiological effects of acceleration on the human body.

23. Atwood's machineis a device where two blocks with masses m1 and m2 are connected by a cord ( of negligible mass) passing over a frictionless pulley ( also of negligible mass). Assume that m2 > m1 . Atwood's machine is attached to the ceiling of an elevator, as shown in Figure. When the elevator accelerates downword with an accerelation “ a ” ( relative to an inertial frame ). Find the magnitude of blocks’ accelerationrelative to the pulley. Find the tension in the cord.

24. The pulley in an Atwood's machine is given an upward acceleration a, as shown in Figure. Find the acceleration of each mass and the tension in the string that connectsthem. The pulley in an Atwood's machine is given an upward acceleration a, as shown in Figure. Find the acceleration of each mass and the tension in the string that connects them.

25. The Centrifugal Force Inertial Frame Noninertial Frame

26. The Centrifugal Force

27. Motion in a Noninertial Reference Frame P.3

28. A block of mass m is placed on a wedge of mass M that is on a horizontal table . All surface are frictionless. Find the acceleration of the wedge. y x m M θ Hints:

29. Chapter 6 Force and Motion-II

30. 6.2 Friction Friction • Frictional forces are very common in our everyday lives. • Examples: • If you send a book sliding down a horizontal surface, the book will finally slow down and stop. • If you push a heavy crate and the crate does not move, then the applied force must be counteracted by frictional forces.

31. Friction If we either slide or attempt to slide a body over a surface, the motion is resisted by a bonding between the body and the surface. The resistance is considered to be single force called the frictional force, f. This force is directed along the surface, opposite the direction of the intended motion.

32. 6.2 Frictional Force: motion of a crate with applied forces Finally, the applied force has overwhelmed the static frictional force. Block slides and accelerates. WEAK KINETIC FRICTION There is no attemptat sliding. Thus, no friction and no motion. NO FRICTION Force F attempts sliding but is balanced by the frictional force. No motion. STATIC FRICTION To maintain the speed, weaken force F to match the weak frictional force. SAME WEAK KINETIC FRICTION Force F is now stronger but is still balanced by the frictional force. No motion. LARGER STATIC FRICTION Static frictional force can only match growing applied force. Force F is now even stronger but is still balanced by the frictional force. No motion. EVEN LARGER STATIC FRICTION Kinetic frictional force has only one value (no matching). fs is the static frictional force fkis the kinetic frictional force

37. Fapplied Stick-slip Smooth sliding Static region Kinetic region

38. 6.3 Properties of Friction Property 1. If the body does not move, then the static frictional force and the component of F that is parallel to the surface balance each other. They are equal in magnitude, and is fsdirected opposite that component of F. Property 2. The magnitude of has a maximum value fs,max that is given by where ms is the coefficient of static friction and FN is the magnitude of the normal forceon the body from the surface. If the magnitude of the component of F that is parallel to the surface exceeds fs,max, then the body begins to slide along the surface. Property 3. If the body begins to slide along the surface, the magnitude of the frictional force rapidly decreases to a value fk given by where mk is the coefficient of kinetic friction. Thereafter, during the sliding, a kinetic frictional force fk opposes the motion.

39. Some Coefficients of Friction

40. As the direction of the velocity of the particle changes, there is an acceleration!!! Uniform Circular Motion Here v is the speed of the particle and r is the radius of the circle. CENTRIPETAL (center-seeking) ACCELERATION

41. Uniform circular motion θ

42. Uniform circular motion θ θ θ

43. 6.5 Uniform Circular Motion Uniform circular motion: • Examples: • When a car moves in the circular arc, it has an acceleration that is directed toward the center of the circle. The frictional force on the tires from the road provide the centripetal force responsible for that. • In a space shuttle around the earth, both the rider and the shuttle are in uniform circular motion and have accelerations directed toward the center of the circle. Centripetal forces, causing these accelerations, are gravitational pulls exerted by Earth and directed radially inward, toward the center of Earth. A body moving with speed v in uniform circular motion feels a centripetal acceleration directed towards the center of the circle of radius R.

44. Car in flat circular turn

45. Car in bankedcircular turn (1) Frictionless road

46. Car in bankedcircular turn (2) With Friction ( s ) R

47. Homework • Chapter 5 ( page 110 ) 33 , 43, 51, 55, 58, 59, 61, 65, 76 • Chapter 6 ( page 131 ) 12, 29, 33, 34, 51, 54, 58, 59, 63, 68

48. 6.4 The Drag Force and Terminal Speed When there is a relative velocity between a fluid and a body (either because the body moves through the fluid or because the fluid moves past the body), the body experiences a drag force, D, that opposes the relative motion and points in the direction in which the fluid flows relative to the body.

49. Motion with Resistive Forces Motion can be through a medium Either a liquid or a gas The medium exerts a resistive force, , on an object moving through the medium The magnitude of depends on the medium The direction of is opposite the direction of motion of the object relative to the medium nearly always increases with increasing speed Slide 49

50. Motion with Resistive Forces The magnitude of can depend on the speed in complex ways We will discuss only two cases is proportional to v Good approximation for slow motions or small objects is proportional to v2 Good approximation for large objects ; b is a constant • Cis the drag coefficient ; ρ is the density of medium (air) • A is the cross-sectional area of the object • v is the speed of the object Slide 50