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5. Newton's Laws Applications 1. 2. 3. 4. 5. Using Newton’s 2nd Law Multiple Objects Circular Motion Friction Drag Forces 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. 5.1. Using Newton’s 2nd 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. Fnet n Fg m a n 0 , ny nx Fg x m ax ny Fg y m a y a ax , 0 Fg m g sin , cos y x: 2 a ax 9.8 m / s 2 sin 32 5.2 m / s n a y: Fg m g sin m ax x ny m g cos 0 n ny 65 kg 9.8 m / s 2 cos 32 540 N Example 5.2. Bear Precautions Mass of pack in figure is 17 kg. What is the tension on each rope? Fnet T1 T2 Fg m a 0 T1 T1 cos , sin T2 T2 cos , sin x : T1 cos T2 cos 0 y T2 y : T1 sin T2 sin m g 0 T1 x Fg T 17 kg 9.8 m / s 2 2 sin 22 a0 since Fg 0 , m g 220 N T1 T2 T T mg 2 sin 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? Fnet Fh n Fg ma0 Fh Fh , 0 n n sin , cos y n x: Fh n sin 0 y: n cos m g 0 x Fh Fh Fg a0 since Fg 0 , m g Fh n sin n mg cos mg 2 sin 60 kg 9.8 m / s tan 30 340 N cos Alternative Approach Net force along slope (x-direction) : Fh cos Fg sin 0 Fh Fg tan y n 60 kg 9.8 m / s 2 tan 30 340 N Fh x Fg m g sin GOT IT? 5.1. 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? n x: T cos mg sin 0 T mg tan T Smaller smaller T Fg x 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. Frock Tr Fg r n mr a r ac ar a Fclimber Tc Fg c mc ac Tc Tr T Tr Tr , 0 Fg r 0 , mr g Tc 0 , Tc Fg c 0 , mc g Tr mr ar mr g n 0 Tc mc g mc ac n 0 , n ar ar , 0 ac 0 , ac T mr a mr g n 0 T mc g mc a T mr a mr g n 0 T mc g mc a a mc g mr mc 70 kg 9.8 m / s 2 940 kg 70 kg 0.679 m / s 2 x x0 v0 t x x0 51 m 1 2 at 2 t 2 x x0 a v0 0 Tension T = 1N throughout 2 51 m 0.679 m / s 2 12 s GOT IT? 5.1. What are 1N (a)the rope tension and 1N (b)the force exerted by the hook on the rope? 5.3. Circular Motion Uniform circular motion 2nd law: v2 Fnet m a m r centripetal 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. T Fg m a T T cos , sin a a , 0 Fg 0 , m g x: T cos m a y : T sin m g 0 y T T T a cos g cot m a v2 r x v Fg mg sin ar g L cot cos 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)? n Fg m a n n sin , cos Fg 0 , m g y v2 x : n sin m r n y : n cos m g 0 25 m / s v2 tan r g 200 m 9.8 m / s 2 2 a Fg v2 a , 0 r x 0.318877... 0.32 17.74... 18 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? n Fg m a n 0 , n Fg 0 , m g v2 a 0 , r v2 n m g m r Minimum speed n = 0 v gr 9.8 m / s 6.3 m 2 7.9 m / s 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 v2 / r Consider hair as mass point connected to head by massless string. Then T+mg=ma 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 ). 5.4. Friction Some 20% of fuel is used to overcome friction inside an engine. The Nature of Friction Frictional Forces Pushing a trunk: 1.Nothing happens unless force is great enough. 2.Force can be reduced once trunk is going. Static friction f s s n v0 s = coefficient of static friction Kinetic friction f k k n v0 k = coefficient of kinetic friction k s k : < 0.01 (smooth), > 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 Application of Friction Walking & driving require static friction. foot pushes ground ground pushes you No slippage: Contact point is momentarily at rest static friction at work 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 a , 0 n Fg f f m a n 0 , n Fg 0 , m g n m a a v v 2 a x x0 2 v0 2 0 v02 x 2a n m (a) = s : (b) = k : f f n , 0 nm g 0 g x x v02 25 m / s 2 0.89 9.8 m / s 2 36 m v02 25 m / s 2 0.61 9.8 m / s 2 52 m 2s g 2k g 2 2 Application: Antilock Braking Systems (ABS) Skidding wheel: kinetic friction Rolling wheel: static friction 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 ). v2 a ,0 r n Fg f f m a n 0 , n Fg 0 , m g v2 s n m r v s r n m f f s n , 0 nm g 0 s r g (a) v 0.88 73 m 9.8 m / s 2 25 m / s 90 km / h (b) v 0.21 73 m 9.8 m / s2 12 m / s 44 km / h Example 5.10. Avalanche! Storm dumps new snow on ski slope. s between new & old snow is 0.46. What’s the maximum slope angle to which the new snow can adhere? n Fg f f m a n 0 , n a0 f f s n , 0 Fg m g sin , cos y n x: fs m g sin s n 0 y: n m g cos 0 tan s Fg x tan 1 s tan 1 0.46 25 Example 5.11. 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? n Fg f f T m a a0 n 0 , n f f k n , 0 Fg 0 , m g y x: n T fs k n T cos 0 n x Fg T T cos , sin T T k cos k T cos mg sin y: k n m g T sin 0 cos m g T sin 0 m g k cos k sin GOT IT? 5.4 Is the frictional force (a) less than, (b) equal to , or (c) greater than the weight multiplied by the coefficient of friction? Reason: Chain is pulling downward, thus increasing n. 5.5. Drag Forces Drag force: frictional force on moving objects in fluid. Depends on fluid density, object’s cross section area, & speed. Terminal speed: max speed of free falling object in fluid. Parachute: vT ~ 5 m/s. Ping-pong ball: vT ~ 10 m/s. Golf ball: vT ~ 50 m/s. Drag & Projectile Motion Sky-diver varies falling speed by changing his cross-section.