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Short Version : 5. Newton's Laws Applications. Example 5.3. Restraining a Ski Racer A starting gate acts horizontally to restrain a 60 kg ski racer on.

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Presentation on theme: "Short Version : 5. Newton's Laws Applications. Example 5.3. Restraining a Ski Racer A starting gate acts horizontally to restrain a 60 kg ski racer on."— Presentation transcript:

1 Short Version : 5. Newton's Laws Applications

2 Example 5.3. Restraining a Ski Racer A starting gate acts horizontally to restrain a 60 kg ski racer on a frictionless 30  slope. What horizontal force does the gate apply to the skier? since  x y  FgFg n FhFh  x : y :

3 Alternative Approach x y  FgFg n FhFh  Net force along slope (x-direction) :

4 5.2. Multiple Objects Example 5.4. Rescuing a Climber A 70 kg climber dangles over the edge of a frictionless ice cliff. He’s roped to a 940 kg rock 51 m from the edge. (a)What’s his acceleration? (b)How much time does he have before the rock goes over the edge? Neglect mass of the rope. 

5   Tension T = 1N throughout

6 5.3. Circular Motion 2 nd law: Uniform circular motion centripetal

7 Example 5.6. Engineering a Road At what angle should a road with 200 m curve radius be banked for travel at 90 km/h (25 m/s)? x y n  FgFg  a x :y :

8 Example 5.7. Looping the Loop Radius at top is 6.3 m. What’s the minimum speed for a roller-coaster car to stay on track there? Minimum speed  n = 0

9 Conceptual Example 5.1. Bad Hair Day What’s wrong with this cartoon showing riders of a loop-the-loop roller coaster? From Eg. 5.7:  n  m g =  m a =  m v 2 / r ( a  g ) Consider hair as mass point connected to head by massless string. Then T  m g =  m a where T is tension on string. Thus,T = m ( g  a )  0. ( downward ) This means hair points upward ( opposite to that shown in cartoon).

10 Frictional Forces Pushing a trunk: 1.Nothing happens unless force is great enough. 2.Force can be reduced once trunk is going. Static friction  s = coefficient of static friction Kinetic friction  k = coefficient of kinetic friction  k : 1.5 (rough) Rubber on dry concrete :  k = 0.8,  s = 1.0 Waxed ski on dry snow:  k = 0.04 Body-joint fluid:  k = 0.003

11 Example Dragging a Trunk Mass of trunk is m. Rope is massless. Kinetic friction coefficient is  k. What rope tension is required to move trunk at constant speed? x y T  FgFg fsfs n x : y :

12 Rolling wheel:

13 Skidding wheel 滑動的輪子 : kinetic friction 動摩擦  k  0.8 Rolling wheel 滾動的輪子 : static friction 靜摩擦  s  1 Rolling friction 滾動摩擦  r  0.01

14 Dynamics of Wheels F fsfs frfr

15 Example 5.8. Stopping a Car  k &  s of a tire on dry road are 0.61 & 0.89, respectively. If the car is travelling at 90 km/h (25 m/s), (a) determine the minimum stopping distance. (b) the stopping distance with the wheels fully locked (car skidding).  (a)  =  s : (b)  =  k :

16 Steering Bicycle turning to the left. Car turning to the left. More details

17 Example 5.9. Steering A level road makes a 90  turn with radius 73 m. What’s the maximum speed for a car to negotiate this turn when the road is (a) dry (  s = 0.88 ). (b) covered with snow (  s = 0.21 ). (a) (b)

18 5.5. Drag Forces Terminal speed: max speed of free falling object in fluid. Drag force: frictional force on moving objects in fluid. Depends on fluid density, object’s cross section area, & speed. Parachute: v T ~ 5 m/s. Ping-pong ball: v T ~ 10 m/s. Golf ball: v T ~ 50 m/s. Sky-diver varies falling speed by changing his cross-section. Drag & Projectile Motion

19 Simple Machines


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