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5. Newton's Laws Applications 1. Using Newton’s 2 nd Law 2. Multiple Objects 3. Circular Motion 4. Friction 5. Drag Forces.

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Presentation on theme: "5. Newton's Laws Applications 1. Using Newton’s 2 nd Law 2. Multiple Objects 3. Circular Motion 4. Friction 5. Drag Forces."— Presentation transcript:

1 5. Newton's Laws Applications 1. Using Newton’s 2 nd Law 2. Multiple Objects 3. Circular Motion 4. Friction 5. Drag Forces

2 Why doesn’t the roller coaster fall its loop-the loop track? Ans. The downward net force is just enough to make it move in a circular path.

3  5.1. Using Newton’s 2 nd Law Example 5.1. Skiing A skier of mass m = 65 kg glides down a frictionless slope of angle  = 32 . Find (a)The skier’s acceleration (b) The force the snow exerts on him.  x :  n FgFg y x a y :

4 Example 5.2. Bear Precautions Mass of pack in figure is 17 kg. What is the tension on each rope? since  x y  T1T1 T2T2  FgFg x : y :

5 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 :

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

7 A roofer’s toolbox rests on a frictionless 45 ° roof, secured by a horizontal rope. Is the rope tension (a)greater than, (b)less than, or (c)equal to the box’s weight? GOT IT? 5.1. x FgFg T  n  Smaller   smaller T x :

8 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. 

9   Tension T = 1N throughout

10 What are (a)the rope tension and (b)the force exerted by the hook on the rope? 1N GOT IT? 5.1.

11 5.3. Circular Motion 2 nd law: Uniform circular motion centripetal

12 Example 5.5. Whirling a Ball on a String Mass of ball is m. String is massless. Find the ball’s speed & the string tension.  x y T  FgFg a x :y :

13 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 :

14 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

15 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 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 = n. Since n is downward, so is T. This means hair points upward ( opposite to that shown in cartoon ).

16 5.4. Friction Some 20% of fuel is used to overcome friction inside an engine. The Nature of Friction

17 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

18 Application of Friction Walking & driving require static friction. No slippage: Contact point is momentarily at rest  static friction at work foot pushes ground ground pushes you

19 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 :

20 Application: Antilock Braking Systems (ABS) Skidding wheel: kinetic friction Rolling wheel: static friction

21 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)

22 Example Avalanche! Storm dumps new snow on ski slope.  s between new & old snow is What’s the maximum slope angle to which the new snow can adhere? x y n FgFg  fsfs  x : y :

23 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 :

24 Is the frictional force (a)less than, (b) equal to, or (c) greater than the weight multiplied by the coefficient of friction? GOT IT? 5.4 Reason: Chain is pulling downward, thus increasing n.

25 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


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