# Newton s Second Law. First of only two important equations in this chapter: r =

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1 Newton s First Law Unless they are acted upon by an external force, objects at rest will stay at rest, and object in motion will stay in motion with a constant velocity. Only applies in inertial reference frames: e.g. a soda can will tend to roll backward in a T-train which is accelerating forward In a sense, it is really a statement about different inertial reference frames being equivalent: every object which moves with constant velocity has an inertial reference frame where it is at rest. If you are in an inertial reference frame, you can declare that you stand still, while the whole universe moves around you. Newton s First law is really just a special case of Newton s Second Law (for a=0) 1

2 Newton s Second Law First of only two important equations in this chapter: F r = r ma Sum of external forces = (mass)(acceleration) Defines mass (the quantity which resists acceleration) Unit of force = Newton = kg m / s 2 Note that it is a vector equation (really three equations) x = max, Fy = ma y, F F = z ma z Best understood by doing lots of examples Example: To stop a 2000 kg car moving at 14m/s in 2.0s requires an average breaking force of (2000kg)(-14m/s)/(2s)= N. 2

3 Other important equation Newton s Third Law r r F AB = F BA The force exerted by A on B is equal and opposite the force exerted by B on A. Always a pair: action force and reaction force (which is which is arbitrary) Always act on different bodies; do not cancel each other Example: A tennis racket pushing on a tennis ball with 50 N of force experience 50 N of force from the ball to the tennis racket. Both objects have net forces on them. 3

4 Weight Special Forces r W = mg r = mgj ˆ Weight = (mass)(acceleration of gravity) gravitational force on a body always points down depends on the planet you are on (don t confuse it with mass) a 50 kg woman weights (50kg)(9.8m/s 2 )=490 N on the surface of the Earth W r m How do astronauts measure their mass in space? 4

5 Special Forces Tension ropes, cables, string, etc. are used to transmit force for the limit of a massless string, both ends pull with the same tension T: T Tension is always pulling along the direction of the rope (for a rigid rod, tension can be positive or negative) Pulleys (ideal) change the direction of the tension, without changing its magnitude: T T T 5

6 Special Forces Example with tension: a r (frictionless surface) = 2.0m/s T1 T2 m 1 m 2 F r 8.0kg 2.0kg What is the tension in each (massless) rope? 6

7 Special Forces Example with tension and weight: F r T 2 a r = 2.0m/s T 1 m 2 2.0kg m 1 8.0kg What is the tension in each (massless) rope? 7

8 Special Forces Normal Force Always comes from a surface (table, wall, ramp, etc.) Direction is directly outward from the surface, perpendicular to the surface N r Normal force cannot be negative (pushes, but never pulls) 8

9 Special Forces Example with tension, weight, and normal force T m 1 =5.0kg m 2 =2.0kg θ o = 20 Find N,a, T 9

10 Newton s Laws problem solving strategy Eight basic steps 1. Draw a sketch (if not already provided) with each body labeled: T m 1 =5.0kg m 2 =2.0kg θ o = 20 10

11 Newton s Laws problem solving strategy 2. Draw a free body diagram for each body (not the same thing as a sketch) 3. Draw and label all of the forces on each body m 2 =2.0kg T g m 2 T N g m 1 m 1 =5.0kg 11

12 Newton s Laws problem solving strategy 4. Draw the direction of acceleration: Not on the bodies, because the acceleration is not a force 5. Assign convenient coordinate axes (try to make as many forces and accelerations align with one axis or the other) y m 2 =2.0kg T m 2 g a x a T N m 1 =5.0kg m 1 g θ o = 20 12

13 Newton s Laws problem solving strategy 6. Break the off-axis forces into components (it is helpful to put a squiggly line through the force that has been broken up so that you don t count it twice) m 2 =2.0kg T m 2 g a m 1 =5.0kg x a T y N m 1 g sinθ m 1 g cosθ m 1 g θ o = 20 13

14 Newton s Laws problem solving strategy 7. Write down F=ma equations for each body for each coordinate direction m 2 =2.0kg T m a x : T W1 sinθ = m1a y : N W cos 0 1 θ = a T a m 1 =5.0kg T W 2 m2g x W2 = 2 W 1 y N sinθ 8. Algebra W 1 cosθ W = 1 m1 g θ o = 20 14

15 Newton s Law Problems Example with tension and weight: F r T 2 a r = 2.0m/s T 1 m 2 2.0kg m 1 8.0kg What is the tension in each (massless) rope? 15

16 Newton s Law Problems Example with tension, weight, and angles: Tarzan crossing a gorge on a vine. θ θ θ o = lb What is the tension in the vine? 16

17 Newton s Law Problems Example with tension, weight, and (trivial) normal force m 1 =5.0kg T m 2 =2.0kg Find N,a, T 17

18 Newton s Law problems Example with tension, weight, and (non-trivial) normal force T m 1 =5.0kg m 2 =2.0kg θ o = 20 Find N,a, T 18

19 Friction Friction is actually a complicated (and not entirely understood) interaction between two surfaces, but it can be described by some fairly simple approximate laws involving coefficients of friction. So far, we have neglected effect of friction in all calculations, but it is common and important A few cases where you want low friction: wheel bearings skis levitated trains frying pans A few cases where you want high friction: automobile tires rock climbing shoes You will test/measure the properties of friction in the lab this week. 19

20 Kinetic Friction Start with kinetic friction (sliding friction), which is a bit easier to deal with: Direction is: tangent to surface opposing the relative motion of the object on the surface f k f = µ k k N Magnitude of kinetic frictional force = (coefficient of kinetic friction)(normal force) N v v N W W f k 20

21 Kinetic Friction Problem Example 1: If m 2 is moving down with constant velocity, what is the coeficient of friction between m 1 and the table? m 1 =5.0kg T m 2 =2.0kg Example 2: If m k =0.10 and the m 1 is moving to the left, what is the acceleration? 21

22 More kinetic friction problems Careful: the normal force can depend on the force applied: What is the magnitude of F needed to keep the box moving at constant speed? o 30 F m = 10kg µ k = 0.50 F N f k W = mg N f f k k + F sin 30 = µ k N o F cos 30 o mg = 0 = 0 What if, instead the force is applied 30 degrees below horizontal? 22

23 Static Friction static is for stationary. Not sliding (subscript s can cause confusion) Direction is: tangent to surface f = µ s (max) s N maximum magnitude of static frictional force = (coefficient of static friction)(normal force) f s (max) f s 0 opposing the sum of all other forces on the object ask what direction would object slide if friction vanished? : the direction of the static frictional force is directly opposite this direction 23

24 Friction Note that for a given pair of surfaces, µ s µ k in other words, for where friction dominates, it takes a larger force to get something moving from rest than to keep it moving at constant speed. v=0 (just breaking free) v=const f s(max) f k 24

25 Static Friction Example 1with static friction: At what angle will the block just begin to slide? m 1 =5.0kg µ s = 0.10 f s(max) N θ =? θ W = mg 25

26 More static friction problems Example 2 with static friction: For what values of m 2 will this system remain stationary? m 1 =5.0kg T 10 = 0. µ s m 2 =? θ o = 30 26

27 NL1. A constant force is exerted on a cart that is initially at rest on an air track. Friction between the cart and the track is negligible. The force acts for a short time interval and gives the cart a certain final speed. To reach the same final speed with a force that is only half as big, the force must be exerted on the cart for a time interval 1. four times as long as 2. twice as long as 3. equal to 4. half as long as 5. a quarter of that for the stronger force. 27

28 NL2. A constant force is exerted for a short time interval on a cart that is initially at rest on an air track. This force gives the cart a certain final speed. The same force is exerted for the same length of time on another cart, also initially at rest, that has twice the mass of the first one. The final speed of the heavier cart is 1. one fourth 2. four times 3. half 4. double 5. the same as that for the stronger force. 28

29 NL3. A constant force is exerted for a short time interval on a cart that is initially at rest on an air track. This force gives the cart a certain final speed. Now we repeat the experiment but, instead of starting from rest, the cart is already moving with constant speed in the direction of the force at the moment we begin to apply the force. After we exert the same constant force for the same time interval, the increase in the cart's speed 1. is equal to two times its initial speed. 2. is equal to the square of its initial speed. 3. is equal to four times its initial speed. 4. is the same as when it started from rest. 5. Cannot be determined from the information provided. 29

30 NL4. Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is 1. larger than 2. identical to 3. smaller than the downward weight W of the person. 30

31 NL5. A locomotive pulls a series of wagons. Which is the correct analysis of the situation? 1. The train moves forward because the locomotive pulls forward slightly harder on the wagons than the wagons pull backward on the locomotive. 2. Because action always equals reaction, the locomotive cannot pull the wagons--the wagons pull backward just as hard as the locomotive pulls forward, so there is no motion. 3. The locomotive gets the wagons to move by giving them a tug during which the force on the wagons is momentarily greater than the force exerted by the wagons on the locomotive. 4. The locomotive's force on the wagons is as strong as the force of the wagons on the locomotive, but the frictional force on the locomotive is forward and large while the backward frictional force on the wagons is small. 5. The locomotive can pull the wagons forward only if it weighs more than the wagons. 31

32 NL6. Consider a horse pulling a buggy. Is the following statement true? The weight of the horse and the normal force exerted by the ground on the horse constitute an interaction pair that are always equal and opposite according to Newton's third law. 1. Yes. 2. No. 32

33 NL7. Consider a car at rest. We can conclude that the downward gravitational pull of Earth on the car and the upward contact force of Earth on the car are equal and opposite because 1. the two forces form an interaction pair. 2. the net force on the car is zero. 3. neither of the above. 33

34 NL8. A capped jar contains a bunch of flies. The scale will register the most weight when the flies are 1. sitting on the bottom of the jar 2. flying around inside the jar 3. the weight of the jar is the same in both cases. 34

35 NL9. When Steroid Stan suspends the 48N telephone book with the rope held vertically the tension in each strand of the rope is 24N. If Stan could suspend the book from the strands pulled horizontally as shown, the tension in each strand would be 1. about 24N 2. about 48N 3. about 96N 24N 4. more than a million Newtons. 48N 24N 48N 35

36 NL10. What does the scale read? 1. 0N 2. 10N 3. 20N N N 36

37 NL11. Carefully compare the following two situations: In case a, you cause block A to accelerate by hanging a 10N weight on the end of the string. In case b, you cause block A to accelerate by applying a 10N force directly to the end of the string. Disregard all frictional forces. The acceleration of block A is N N N N 1. greater in case a 2. greater in case b 3. the same in both cases 37

38 NL12. Will hanging a magnet in front of an iron car, as shown, make the car go? 1. yes, it will go 2. it will move if there is no friction 3. it will not go 38

39 NL13. For every action force there is always an equal and opposite reaction force. Consider the action and reaction forces in the case of a body falling under the influence of the Earth's gravity. If the action force is the force with which the Earth pulls down on the falling body, then the reaction force is the force 1. due to air resistance 2. that acts on the inertia of the falling body 3. of the eventual impact of the ground 4. of the falling body pulling up on the Earth 5. There is no reaction force in this situation. 39

40 NL14. If a stretch limousine and a vintage Volkswagen have a head-on collision, which car will experience the greatest impact force? 1. The limousine 2. The Volkswagen 3. Both experience the same force 40

41 NL15. Compare the consequences of driving a car into a head-on collision with and identical car travelling toward you at the same speed--as opposed to colliding at the same speed against a massive concrete wall. Which of these two situations would result in the greatest impact force? 1. Colliding with the approaching car 2. Colliding with the massive stationary wall 3. Both would experience the same impact force 41

42 NL16. If a can of compressed air is punctured and the escaping air blows to the right, the can will move to the left in a rocket-like fashion. Now consider a vacuum can that is punctured. The air blows in the left as it enters the can. After the vacuum is filled the can will 1. be moving to the left 2. be moving to the right 3. not be moving 42

43 NL17. In the 17th century, Otto von Güricke, a physicist in Magdeburg, fitted two hollow bronze hemispheres together and removed the air from the resulting sphere with a pump. Two eight-horse teams could not pull the halves apart even though the hemispheres fell apart when air was re- admitted. Suppose von Güricke had tied both teams of horses to one side and bolted the other side to a heavy tree trunk. In this case, the tension on the hemispheres would be 1. twice 2. exactly the same as 3. half what it was before. 43

44 NL1. Suppose that both an elephant and a feather fall from a high tree. Which encounters the greatest force of air resistance in falling to the ground? 1. The elephant 2. The feather 3. Both the same 44

45 NL2. You are pushing a crate across the floor at constant speed. You decide to turn the crate on end, reducing by half the surface area in contact with the floor. In the new orientation, to push the same crate across the same floor with the same speed, the force that you apply must be about 1. four times as great 2. two times as great 3. equally great 4. half as great 5. one-fourth as great as the force required before you changed the crate's orientation. 45

46 NL4. Suppose you are sailing a sailboat directly downwind with your sails full in a 10m/s wind. The maximum speed you could hope to attain would be 1. nearly 10m/s 2. between about 10m/s and 20m/s 3. There would be no theoretical speed limit in this case. 46

47 NL5. You are again sailing downwind and you pull your sail in so that it no longer makes a 90 o angle with the keel of the boat. This tactic will 1. decrease the speed of the boat. 2. increase the speed of the boat. 3. not affect the speed of the boat. 47

48 NL6. Keeping the angle of sail relative to the boat the same as in the last question, suppose you now direct your boat so that it sails directly across the wind, rather than directly with the wind. Will you sail faster or slower than before? 1. Faster. 2. Slower. 3. The same. 48

49 NL7. For the four boats shown below, note the orientations of the sails with respect to both wind and keel directions. Which of these boats moves in a forward direction with the greatest speed? 49

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