20. Given: m = 75 kg ; a! = 2.0 m/s 2 [up] Required: F! N Analysis:! F! y. = m a!. Choose up as the positive direction. Solution:! F! y. = m a!!

Transcription

1 Chapter Review, paes Knowlede 1. (b). (a) 3. (c) 4. (a) 5. (d) 6. (a) 7. (d) 8. (c) 9. (b) 10. False. If only three force of equal manitude act on an object, the object may or may not have a non-zero net force. 11. False. When a non-zero net force acts on an object, the object will accelerate in the direction of the net force. 1. True 13. False. When a skater bumps into the boards in an arena, the skater exerts a force on the boards, and the boards exert an equal and opposite reaction force on the skater at the same time. 14. True 15. False. When a person walks on a rouh surface, the frictional force exerted by the surface on the person is in the same direction as the person s motion. 16. False. When you are slidin down a hill on a snowboard, the normal force on you is smaller in manitude than the force of ravity. 17. False. When two objects slip over each other, the force of the friction between them is called kinetic friction. 18. True 19. True Understandin 0. Given: m 75 k ; a.0 m/s [up] Required: F N Analysis: F y m a. Choose up as the positive direction. Solution: F y m a " m m a m( + a) (75 k)(.0 m/s m/s ) 8.8 #10 N Statement: The floor exerts a force of N [up] on the man when the elevator starts movin upward. Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -

2 1. (a) The FBD of the suitcase is shown below. (b) We know that the suitcase is movin at a constant speed, so the net force on the suitcase is zero. Choosin the direction of motion to be the +x-direction, the components of the net force are F x and F y. (c) Refer to the FBD to check the directions of the forces, choosin the +x-direction as the direction in which the handle is pointin. Both the force of ravity and the normal force are vertical, so their x-components are zero. Friction acts in the neative x-direction, so its x-component is neative. The applied force is at an anle, so we can use trionometry to determine its x-component. The x-components of the forces can be expressed as follows: F x ; F Nx ; x F a cos ; and F fx F f. (d) The choice of +x that is more convenient is horizontally forward, because the positive x-direction is the direction of motion. Also, three of the four forces alin with either the horizontal or vertical directions.. (a) A drop of rain fallin with a constant speed has constant velocity and a net force of zero. (b) A cork with a mass of 10 floatin on still water means that the cork is at rest, and that the net force is zero. (c) A stone with a mass of 0.1 k just after it is dropped from the window of a stationary train is acted upon by a net force equal to its weiht. This is expressed as: F m [down] (0.1 k)(9.8 m/s ) [down]. F 1 N [down] (d) The same stone at rest on the floor of a train, which is acceleratin at 1.0 m/s is acted upon by a net force equal to the force of static friction pullin it forward with the train. This is expressed as: F m a [forward] (0.1 k)(1.0 m/s ) [forward] F 0.1 N [forward] Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -3

3 3. (a) The FBD for a saucepan hanin from a hook is shown below. (b) The FBD for a person standin at rest on the floor is shown below. (c) The FBD for a puck slidin in a straiht line on the ice to the riht is shown below. (d) The FBD for a toboan pulled by a rope at an anle above the horizontal to the riht with sinificant friction on the toboan is shown below. 4. The acceleration of the astronaut the instant he is outside the spaceship would be 0 m/s. Once outside the spacecraft, the astronaut continues to move at the velocity the spacecraft had as he exited. By Newton s first law of motion, no net force acts on the astronaut and so he does not accelerate. 5. Action and reaction forces cannot cancel each other, even thouh they are equal and opposite, because one force acts on one object and the other force acts on the second object. These forces do not cancel because they do not act on the same object. Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -4

4 6. Given: m A 4.0 k; m B 6.0 k ; F a 0.0 N [E] Required: a Analysis: F m a ". Choose east as positive. Solution: F m T a m T a " a m 1 + m 0.0 N 10.0 k a.0 m/s Statement: The blocks accelerate at.0 m/s [E]. 7. The readin of the balance while it is in the air is 0 N, because there is no tension between its ends. The force of ravity acts on the sprin balance, and it accelerates at the acceleration of ravity, with no other force actin upon it. 8. No, it is not always necessary for the coefficient of friction to be less than one. There are some materials with a coefficient of friction reater than one. Examples include sticky or tacky surfaces, wet packed snow on skis, or materials like Velcro. 9. Answers may vary. Sample answer: Static friction may be useful on inclined planes for keepin objects in place; for example, a parked car does not slide away. Static friction is a problem on an inclined surface when you want to put an object in motion. For example, if I do not have the riht wax on my cross-country skis, I will not slide downhill even if I lean forward. Kinetic friction is useful on an incline to keep objects movin at a constant speed; for example, with the riht wax and the riht incline, I can lide on my cross-country skis without acceleratin. Kinetic friction is a problem on an incline if a reat amount of acceleration is desired. For example, downhill ski racers want to accelerate quickly with their skis only rippin by the edes on the turns. 30. A linear actuator is a device that uses enery to apply a force. This force may lower the counter for a cashier, open or close power windows in cars, raise a workstation for an extremely tall worker, or lift a patient into a harness and onto a stretcher for transportation, or tihten the screws fastenin the dashboard to a vehicle in an automobile assembly line. All of these tasks ease the effort a worker must ive and so reduce the possibility of strain and injury. Analysis and Application 31. A trebuchet applies the principles of linear motion by usin the force of ravity to accelerate a lare mass downward. The machine is constructed so that this forces the less massive arm to swin upwards with a hih acceleration. The bucket holdin the ammunition is even less massive and is swun with even hiher acceleration. The bucket is anled to release the ammunition at a pre-calculated anle for the required anle of the projectile. 3. The force of reaction exerted by the block on the rope when the block is restin on a smooth horizontal surface is equal and opposite to the force of the rope on the block. The reaction force is 31.5 N. Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -5

5 33. Given: m H k ; m k; a 1 m/s [up] Required: Analysis: F y m a. Choose up as positive. Solution: F y m a " m P m P a m P ( + a) P (40 k)(9.8 m/s +1 m/s ) 9. #10 3 N Statement: The normal force exerted by the crew and passeners on the floor is N [downward]. 34. Given: F A 45 N [E] ; F B 5 N [N] ; m 3 k Required: a Analysis: F $ # F (F x ) + (F y ) ; tan "1 y # F ' & ). Choose east and north as positive. % x ( Solution: For the x-components of the force: F x F Ax + F Bx 5 N + 0 N F x 5 N For the y-components of the force: F y F Ay + F By + 9 N F y 9 N Construct F : F (F x ) + (F y ) (45 N) + (9 N) F N (two extra diits carried) $ # F tan "1 y # F ' & ) % x ( $ 9 N ' tan "1 % & 45 N ( ) 33 Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -6

6 Solve for the acceleration, a : F m a a F m N [E 33 N] 3 k a.3 m/s [E 33 N] Statement: The object s acceleration is.3 m/s [E 33 N]. 35. Given: F A 47 N [E 31 N]; F B 58 N [E 46 N] ; F Required: F ; θ Analysis: F A + F B + F. Choose north and east as positive. Solution: For the x-components of the force: F Ax + F Bx + F x 0 F x F Ax F Bx (47 N)cos31 (58 N)cos46 F x N (two extra diits carried) For the y-components of the force: F Ay + F By + F y 0 F y F Ay F By Construct F : F (F x ) + (F y ) (47 N)sin31 (58 N)sin46 F y N (two extra diits carried) (80.58 N) + (65.93 N) 100 N F N $ # F tan "1 y # F ' & ) % x ( $ N ' tan "1 % & N ( ) 39 Statement: F is N [W 39 S]. Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -7

7 36. (a) Given: m A.3 k; m B 3.5 k; F KA 5.4 N Required: a Analysis: F m a Solution: For block A (mass m A ): F x m A a F T " F K m A a For block B (mass m B ): F y m B a m B " F T m B a Add the final equations to eliminate the strin tension. Solvin for a : F T F KA ( ) + m B F ( T ) m A a + mb a m B F KA (m A + m B ) a a m B F KA m A + m B (3.5 k)(9.8 m/s ) 5.4 N 5.8 k m/s (two extra fiures carried) a 5.0 m/s Statement: The manitude of acceleration of the blocks is 5.0 m/s. (b) Given: m A.3 k; F KA 5.4 N Required: F T Analysis: F T F K m A a Solution: F T F K m A a F T m A a + FK (.3 k)(4.983 m/s ) N F T 17 N Statement: The manitude of tension in the strin is 17 N. Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -8

8 37. (a) The FBD for the man is shown below. The FBD for the mattress is shown below. (b) The reaction to the force of ravity on the man is an upward force of the man on Earth. The reaction to the force of ravity on the mattress is an upward force of the mattress on Earth. The normal force of the mattress on the man and the force of the man on the mattress are an action reaction pair. The reaction force to the normal force of the water on the mattress is the mattress pushin down on the water. (c) Given: m k ; m mattress 7.0 k; F mattress man Required: normal force of the water on the mattress, F" Nw Analysis: F Solution: w m m mattress w m + m mattress (110 k)(9.8 m/s ) + (7.0 k)(9.8 m/s ) 1078 N N w 1.1"10 3 N Statement: The normal force of the water on the mattress is N. (d) Given: m man k ; F man Required: normal force of the mattress on the man, F Nm Analysis: F Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -9

9 Solution: m m man m m man (110 k)(9.8 m/s ) m 1.1"10 3 N Statement: The normal force of the mattress on the man is N. 38. Answers may vary. Sample answer: One version of Newton s first law of motion says that objects at rest tend to stay at rest. I can demonstrate this by settin a textbook on top of a piece of paper on a desk with a bit of the paper showin over the ede of the desk. If I suddenly pull horizontally on the paper, the book remains restin on the desk. The force of static friction between the paper and the book breaks so quickly, and the force of kinetic friction is so low, that the only force exerted on the textbook is a tiny external force, and so it hardly moves at all. 39. (a) Given: F a N [E] N [E] ; m 4 t k Required: a Analysis: F m a. Choose east as the positive direction. Solution: F m a m a F a a m 1. "105 N 4. "10 4 k.857 m/s (two extra diits carried) a.9 m/s Statement: The acceleration produced by the enines is.9 m/s [E]. (b) Given: a.9 m/s ; v i 0 m/s; v f 71 m/s [E] Required: Δ d Analysis: d v " v f i a Solution: d v " v f i a (71 m/s) " (0 m/s) (.857 m/s ) d 8.8 #10 m Statement: The minimum lenth of the runway needed is m. 40. (a) Given: m k ; m.4 k ; F a 8.6 N [W] Required: a ; m T m 1 + m Analysis: F m a. Choose west as positive. Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -10

10 Solution: m T m 1 + m 1.3 k +.4 k F m T a m T a a m T m T 3.7 k 8.6 N [W] 3.7 k.34 m/s [W] (two extra diits carried) a.3 m/s [W] Statement: The acceleration of the masses is.3 m/s [W]. (b) Given: a.34 m/s [W]; m k Required: F Analysis: F m a Solution: F m a (1.3 k)(.34 m/s [W]) F 3.0 N [W] Statement: The net force actin on the mass is 3.0 N [W]. 41. (a) Given: m 1 11 k; m 19 k ; " N Required: " F T Analysis: F m a Solution: F m T a m T a a m T 530 N 30 k a m/s (two extra diits carried) For mass m : F m a F T (19 k)(1.767 m/s ) F T 34 N Statement: The tension on the strin when the applied force pulls directly on the 11 k mass is 34 N. Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -11

11 (b) Given: m 1 11 k; a m/s Required: F T Analysis: F m a Solution: F m 1 a F T (11 k)(1.767 m/s ) F T 19 N Statement: The tension on the strin when the applied force pulls directly on the 19 k mass is 19 N. 4. (a) Given: m.5 k; ir 1 N [riht]; F Required: F T Analysis: F ; F T ( F Tx ) + ( F Ty ). Choose riht and up as positive. Solution: For the x-components of the force: F x F Tx + F air F Tx " F air F Tx "1 N For the y-components of the force: F y F Ty " m F Ty m (.5 k)(9.8 m/s ) F Ty 4.5 N (one extra diit carried) Construct F T from its components: F T (F Tx ) + (F Ty ) F T (1 N) + (4.5 N) 7.8 N 7 N Statement: The tension in the rope is 7 N. (b) Given: F Tx 1 N ; F Ty 4.5 N Required: θ # FTy & Analysis: tan "1 % (. Choose riht and up as positive. $ F Tx ' Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -1

12 FTy # & Solution: tan "1 % ( $ F Tx ' # 4.5 N & tan "1 $ % 1 N ' ( 64 Statement: The rope makes an anle of 64 with the horizontal. 43. (a) Given: m 54 k ; 35.0 Required: tension in horizontal rope, F T1 ; tension in vertical rope, F T Analysis: F Solution: For the y-components of the force: F Ty m F Ty m F T m sin" (54 k)(9.8 m/s ) sin N (two extra diits carried) F T 9. #10 N For the x-components of the force: F T1x + F Tx F T1 + F T cos" F T1 F T cos" (9.6 N)cos35.0 F T1 7.6 #10 N Statement: The tension in the horizontal rope is N, and the tension in the vertical rope is N. (b) If the horizontal rope were slihtly loner and attached to the wall at a hiher point, its tension would have a vertical component supportin some of the weiht of the performer. As a result the tension in the second rope would be reduced. This means the horizontal component of the first tension would also be reduced. As lon as the anle of the first rope with the horizontal is small, the tension in the first rope would be reduced as well. 44. Given: m 65 k ; v 5 m/s [S] ; F a N [S] Required: F w Analysis: F ; F w F N + F f. Choose south and up as positive. Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -13

13 Solution: For the y-components of the force: F y " m m (65 k)(9.8 m/s ) 637 N (one extra diit carried) For the x-components of the force: F x " F f F f F a F x 100 N (two extra diits carried) Construct F w : F w ( ) + ( F f ) F w (637 N) + (100 N) N # tan "1 FN & % ( $ ' F f # 637 N & tan "1 $ % 100 N ' ( 8 Statement: The force of the water on the skier s ski is N [N 8 up]. 45. Given: m k ; v i 0 m/s ; v f 3 m/s; Δ t s Required: F " a Analysis: F m a. Choose forward and up as positive. Solution: a v v f i "t (3 m/s) (0 m/s) s a 909 m/s (two extra diits carried) Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -14

14 Determine the net force: F x ma ma (0.160 k)(909 m/s ) 4.7 "10 N Statement: The manitude of the force applied to the puck is N. 46. Given: N ; a 6.9 m/s Required: F a Analysis: F m a. Choose up as the positive direction. Solution: In deep space: F m a m a F m a a 1.7 "104 N 6.9 m/s m.464 "10 3 k (two extra diits carried) On Earth: F m a " m m a m( + a) (.464 #10 3 k)(9.8 m/s m/s ) 4.1#10 4 N Statement: A force of N is required for the spacecraft to accelerate at the same rate upward from Earth. 47. Given: m k ; m 4.0 k ; m k Required: a ; F TA ; F TB Analysis: F m a Equation for mass m 1 : F m 1 a F TA " m 1 m 1 a Equation for mass m : F m a " F TA + F TB + m m a Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -15

15 Equation for mass m 3 : F m 3 a " F TB + m 3 m 3 a We can now add the three equations above to solve for a. m 1 + m + m 3 (m 1 + m + m 3 ) a Solution: a (m 1 + m + m 3 ) m 1 + m + m 3 a (m 1 + m + m 3 ) m 1 + m + m 3 (6.0 k k k )9.8 m/s 6.0 k k k m/s (two extra diits carried) a 0.75 m/s Substitute in the equation for mass m 1 : F TA m 1 m 1 a F TA m 1 ( + a) (6.0 k)(9.8 m/s m/s ) F TA 63 N Substitute in the equation for mass m 3 : F TB + m 3 m 3 a F TB m 3 ( a) (3.0 k)(9.8 m/s m/s ) F TB 7 N Statement: The acceleration of the system is 0.75 m/s. The tension in strin A is 63 N, and the tension in strin B is 7 N. 48. Given: m 60.0 k; direction of F T1 is [up 15 left]; direction of F T is [up 15 riht] ( ) Required: F T F T1 F T Analysis: F Solution: For the x-components of the force: F x F x + F 1x + F x 0 N (0 N) " F T1 cos15 + F T cos15 F T1 F T Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -16

16 For the y-components of the force: F y F y + F 1y + F y 0 N "m + F T1 sin15 + F T sin15 0 N "(60.0 k)(9.8 m/s ) + F T sin15 F T (60.0 k)(9.8 m/s ) sin15 F T 1.1#10 3 N Statement: The tension in the rope on both sides of the tihtrope walker is 3 F N. T 49. Given: m k ; µ K 0.35 Required: Analysis: F. The forces on crate (mass m ) include the force of ravity, the normal force of the floor, kinetic friction and the horizontal normal force from crate 1 (mass m 1 ). For crate, it is the horizontal normal force that counters the kinetic friction. Solution: For the y-components of the force (mass m ): F y " m m ( k)(9.8 m/s ) 15.6 N (two extra diits carried) For the x-components of the force (mass m ): F x " F K F K µ K FN 0.35(15.6 N) 75 N Statement: The manitude of the normal force between the two crates is 75 N. 50. (a) Given: F a N [riht 35 down] ; a 0.6 m/s [riht] ; F f N [left] Required: m Analysis: F m a Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -17

17 Solution: For the x-components of the force: F x m a cos" # F f m a F m a cos" # F f a (0 N)cos35 #140 N 0.6 m/s 64.8 k (two extra diits carried) m 65 k Statement: The total mass of the chair and person is 65 k. (b) Given: m 64.8 k ; F a N [riht 35 down] Required: F N Analysis: F m a Solution: Equation for the y-components: F y " sin# " m sin# + m (0 N) sin (64.8 k)(9.8 m/s ) 7.6 $10 N Statement: The normal force actin on the chair is N. 51. Given: µ S 0.5 ; m k Required: F a Analysis: F ; F S max µ FN S Solution: For the y-components of the force: F y " m m (130 k)(9.8 m/s ) 174 N (two extra diits carried) Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -18

18 For the x-components of the force: F x " F S F S µ s FN 0.5(174 N) " 3. #10 N Statement: The manitude of the minimum force required is N. 5. (a) Given: m 85 k ; F a N Required: µ S Analysis: F ; F S µ S FN Solution: For the y-components of the force: F y " m m For the x-components of the force: F " x " " " F S " F S " µ S F " N F " a µ S F" a m 330 N (85 k )(9.8 m/s ) µ S 0.40 Statement: The coefficient of static friction between the floor and the trunk is (b) Given: v f.0 m/s; Δ t 5.0 s ; F a N ; m 85 k Required: µ K Analysis: F m a Solution: a v t.0 m/s 5.0 s a 0.40 m/s Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -19

19 Use the x-components of the forces: F x m a " F K m a F K " m a µ FN K F a " m a F µ K a " m a m 330 N " (85 k )(0.40 m/s ) (85 k )(9.8 m/s ) µ K 0.36 Statement: The coefficient of kinetic friction is Given: a 4.0 m/s [forward] Required: µ S Analysis: F x m a Solution: Equation for x-component of motion: F x m a F S max m a µ S m m a a µ S 4.0 m/s 9.8 m/s µ S 0.41 Statement: The coefficient of static friction between the driver s tires and the road is Given: v i 35 m/s; v f 0 m/s ; Δ d 95 m Required: µ K Analysis: v f v i + ad. Choose forward as positive. Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -0

20 Solution: Determine the acceleration, a : v f v i + ad a " v i d (35 m/s) " (95 m) a "6.447 m/s (two extra diits carried) Calculate µ : K F x m a " F K m a "µ K m m a a µ K " m/s 9.8 m/s µ K 0.66 Statement: The coefficient of kinetic friction between the puck and the ice is Given: m.4 k ; F a 450 N [down] ; µ S 0.67 Required: F N Analysis: F S µ FN S ; F y Solution: Equation for y-direction: F y F " S " m " F S m + (.4 k)(9.8 m/s ) N F S 36.8 N (two extra diits carried) Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -1

21 Use the breakin value of static friction to solve for the normal force: F S µ FN S F S µ S 36.8 N N Statement: Each jaw of the vise exerts a horizontal normal force of N on the block. 56. Given: m 1 15 k (Crate 1) ; m 35 k (Crate ) ; a 1.7 m/s ; 9.8 m/s Required: F S1 ; F N1 ; F 1 ; F S ; F N ; F Analysis: F m a Solution: The FBD of Crate 1 is shown below. The FBD of Crate is shown below. For the x-components of the force (Crate 1): F x m 1 a F S1 m 1 a (15 k)(1.7 m/s ) F S1 5 N Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -

22 For the y-components of the force (Crate 1): F y 1 " m 1 1 m 1 (15 k)(9.8 m/s ) 147 N (one extra diit carried) 1 1.5#10 N For the x-components of the force (Crate ): F x m a F S " F S1 m a F S F S1 + m a (5.5 N) + (35 k)(1.7 m/s ) F S 85 N For the y-components of the force (Crate ): F y " 1 " m 1 + m Force of ravity on Crate : F m 147 N + (35 k)(9.8 m/s ) 147 N N 490 N 4.9 #10 N 35 k(9.8 m/s ) F N Statement: The normal force on Crate 1, 1, is N, which is the same as the force of ravity, F 1. The static friction action reaction pair of forces between Crate 1 and Crate, F S1, is 5 N. The force of ravity on Crate, F, is N. The normal force between Crate and the truck,, is N. The static friction between Crate and the truck, F S, is 85 N. 57. (a) Given: µ 0.70 ; m.0 k; θ 3 Required: F f Analysis: The followin forces are actin on the block: force of ravity [down], normal force [up perpendicular to plane], force of friction [up parallel to the plane]. The force of friction must act up the plane because the force of ravity pulls the block down the plane. Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -3

23 If the block does not move, there is no net force and the friction is static. If the block moves at all, it will move down the plane and the friction is kinetic. In either case, the manitude of the force of friction is less than the component of ravity down the plane. Write out the net force in components and then calculate the normal force and the component of ravity alon the plane. Compare the manitude of the force of ravity alon the plane with µ F. N Solution: y-direction of net force: F y " m cos# m cos# (.0 k)(9.8 m/s )cos N (two extra diits carried) x-component of force of ravity: F x " m sin# m sin# m sin# (.0 k)(9.8 m/s )sin3 10 N maximum value of F " f : µ 0.70(16.6 N) N µ F N 1 N Since µ F N is reater than m sinθ, the force of friction must just balance the force of ravity alon the plane, F f 10 N [up the plane]. Statement: The force of friction actin on the block is 10 N [up the plane]. (b) Since F f is less than µ F, the force of friction is static. N 58. Given: µ S 0.45 ; m 66 k Required: F a Analysis: F S µ FN S ; F Solution: For the y-components of the force: F y " m m (66 k)(9.8 m/s ) N (two extra diits carried) Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -4

24 F x " F S F S µ s FN 0.45(646.8 N).9 #10 N Statement: The minimum horizontal force that a person must apply is.9 10 N. 59. Given: θ 6 ; a 1.96 m/s 5 Required: µ K Analysis: F m a. Solution: For the y-components of the force: F y " m cos# m cos# For the x-components of the force: F x m a m sin" # F K m a F K m sin" # m a $ µ K m cos" m sin" # m ' % & 5 ( ) µ K $ m sin6 # 1 ' % & 5( ) m cos6 µ K 0.7 Statement: The coefficient of kinetic friction is (a) The FBD of the person is shown below. (b) Given: µ S 0.43 Required: a Analysis: FS µ SF ; F m a N Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -5

25 Solution: For the y-components of the force: F y " m m For the x-components of the force: F x m a F S m a µ S FN m a µ S m m a a µ S 0.43(9.8 m/s ) a 4. m/s Statement: The maximum acceleration of the train before the person starts to slip is 4. m/s. (c) Answers may vary. Sample answer: To keep from slippin, the passener can push down on the round to increase the normal force. 61. Given: µ S 0.16 Required: a Analysis: F S µ FN S Solution: For the y-components of the force: F y " m m For the x-components of the force: F x m a F S m a µ S FN m a µ S m m a a µ S 0.16(9.8 m/s ) a 1.6 m/s Statement: The maximum acceleration of the train that will allow the box to remain stationary is 1.6 m/s. Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -6

26 6. Given: a.5 m/s ; θ 6 Required: µ K Analysis: F m a Solution: For the y-components of the force: F y " m cos# m cos# For the x-components of the force: F x m a m sin" # F K m a m sin" # µ K FN m a m sin" # µ K m cos" m a µ K sin" # a cos" ( 9.8 m/s )sin6 #.5 m/s ( 9.8 m/s )cos6 µ K 0.0 Statement: The coefficient of kinetic friction between the block and the incline is Given: m. k ; µ K 0.41; F a 18 N [E] Required: a Analysis: F m a Solution: For the y-components of the force: F y " m m For the x-components of the force: F x m a " F K m a " µ FN K m a " µ K m m a F a a " µ K m m 18 N " 0.41(. k )(9.8 m/s ). k a 4. m/s Statement: The acceleration of the block is 4. m/s [E]. Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -7

27 64. Given: a 6.6 m/s [backward] Required: µ K Analysis: F m a ; F K µ K FN. Choose east and up as positive. Solution: y-component of net force: F y " m m x-component of net force: F x m a " F K m a "µ K m m a a µ K " # "6.6 m/s & "% ( $ 9.8 m/s ' µ K 0.67 Statement: The coefficient of kinetic friction between the object and the plane is Given: m 1. k (Box 1) ; m 3.8 k (Box ) ; µ S 0.5; µ K 0.3 Required: Analysis: F m a Solution: For the y-components of the force (Box 1): F y 1 " m 1 1 m 1 For the x-components of the force (Box 1): F x m 1 a F S m 1 a µ S m 1 m 1 a a µ S For the y-components of the force (Box ): F y F " 1 " m F m + 1 F (m 1 + m ) Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -8

28 For the x-components of the force (Box ): F x m a " F K " F S m a F K + F S + m µ S µ K (m 1 + m ) + µ S m 1 + m µ S (µ K + µ S )(m 1 + m ) (0.57)(6.0)(9.8 m/s ) 34 N Statement: The maximum horizontal force that can be applied to the larer box is 34 N. Evaluation 66. Action and reaction forces cannot cancel each other, even thouh they are equal and opposite, because one force acts on one object and the other force acts on the second object. These forces do not cancel because they do not act on the same object. Therefore, the student s statement is not valid. 67. Answers may vary. Sample answer: I could try pullin forward and up on the rope attached to the crate. This reduces the normal force and reduces the upper limit of static friction. I miht try alterin the surface between the crate and the round to reduce the coefficient of static friction. By tippin the crate a bit I could slide a plastic tarp partway underneath or toss under some ice cubes. I miht also tie the rope around a fence post and back on the crate, and then lean aainst the middle of the rope to et the crate movin. 68. (a) This setup provides an advantae to climbin the rope because the forces actin on the person are the force of ravity (down) and the two rope tensions (up). The two rope tensions share the person s weiht. The person is pullin on the left rope creatin its tension. As a result, her pullin force is equal to only part of her weiht. If she were climbin the rope directly, she would be pullin aainst her whole weiht. (b) Given: m 59. k; F Required: F T1 Analysis: F y Solution: F y F T " m F T m (59. k)(9.8 m/s ) F T.9 #10 N Statement: The manitude of force that the person exerts on the rope is.9 10 N. (c) Answers may vary. Sample answer: We assumed that the weiht of the swin and the rope are neliible and that the pulley is frictionless. Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -9

29 69. The correct order of steps for solvin two-dimensional force problems that require Newton s second law of motion is: (f) Read the problem before tryin to solve it. (c) Identify the iven variables and the required variables. (e) Identify the object on which the forces act. (d) Choose a coordinate system, and draw an FBD. Include a label for each force. (b) Determine the x- and y-components of each force, and write the necessary equations. (a) Solve the problem usin Newton s second law of motion. 70. The student miht say this because, if the cross-country skier had no friction slidin down the small hills, his acceleration and speed would be reater. If there were no friction, the skier would not be able to et up the hill because the skier needs traction between the skis and the snow. Without friction, the force of ravity would keep the skier from oin up the hill. 71. (a) Given: coefficient of static friction on dry pavment, µ S1 0.81; coefficient of static friction on wet pavment, µ S 0.58 Required: acceleration on dry pavement, a 1 ; acceleration on wet pavement, a Analysis: FS µ SF ; F m a N Solution: For the y-components of the force: F y " m m For the x-components of the force: F x m a F S m a a 1 µ S1 µ S FN m a µ S m m a a µ S a µ S 0.81(9.8 m/s ) 0.58(9.8 m/s ) m/s (two extra diits carried) m/s (two extra diits carried) a m/s a 5.7 m/s Statement: The maximum acceleration on dry pavement is 7.9 m/s and 5.7 m/s on wet pavement. (b) Given: a m/s ; a m/s ; Δ d 100 m ; v i 0 m/s Required: time runnin 100 m on dry pavement, t 1 ; time runnin 100 m on wet pavement, t Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -30

30 Analysis: d v i t + 1 a ( t ) Solution: d v i t a 1 t 1 d v i t a 1 t 1 t 1 d a 1 ( ) ( ) d v i t + 1 a t d v i t + 1 a t t d a ( ) ( ) t s 00 m m/s t 5.9 s 00 m m/s Statement: The lenth of time you could run 100 m with sustained acceleration when the pavement is dry is 5.0 s, and when wet is 5.9 s. Both of these times are unreasonably short, iven that the Olympic record for 100 m is between 9 s and 10 s. 7. Given: m 5 k ; F N [forward 5 up]; F N [forward 5 down] ; µ K 0.5 Required: a Analysis: F y ; F x m a. Choose forward and up as positive. Solution: For the y-components of the force: F y F 1y + F y + F N " m (5 k)(9.8 m/s ) " (340 N)sin5 " ("170 N)sin N (two extra diits carried) For the x-components of the force: F x m a F 1x + F x " F K m a (340 N)cos5 + (170 N)cos5 " (0.5)(437.8 N) a 5 k a 4.5 m/s Statement: The crate accelerates at 4.5 m/s. 73. Given: µ dry 0.85; µ wet 0.45 Required: d dry v " v f i a ; dry Analysis: F m a ; F K µ FN K ; d v " v f i a positive. ; d wet v " v f i a. Choose up and forward as wet Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -31

31 Solution: For the y-components of the force: F y " m m For the x-components of the force: F x m a " F K m a "µ K FN m a "µ K m m a a "µ K Stoppin distance for dry concrete: d dry v " v f i a dry d dry d dry "v i ("µ dry ) v i µ dry Stoppin distance for dry concrete: d wet v " v f i a wet d wet "v i ("µ wet ) v i d wet µ wet The required comparison of distances is: Δd Δd Δd Δd wet dry wet dry µ µ v i µ wet dry wet µ v i dry Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -3

32 Statement: The car would skid about 90 % farther on wet concrete than on dry concrete, assumin the same initial speed in both cases. This calculation suests that in wet conditions it would be wise to drive more slowly and leave more space between you and other cars to accommodate the much loner skid distance in wet conditions. 74. (a) Given: F air ; m 1 59 k ; m 73 k; θ 3 ; µ K 0.10 Required: a 1 ; a Analysis: a sin " µ K cos " Solution: a sin " µ K cos ir m (9.8 m/s )sin3 " 0.10(9.8 m/s )cos3.97 m/s (two extra diits carried) a.9 m/s Statement: With no air resistance, both skiers would have an acceleration of.9 m/s down the slope. (b) Given: F air 8 N ; a.97 m/s Required: a 1 ; a Analysis: a F sin " µ K cos " air m Solution: a F 1 sin " µ K cos " air F a sin " µ K cos " air.97 m/s " 8 N 59 k a m/s m 1.97 m/s " 8 N 73 k a 1.8 m/s Statement: Assumin the same air resistance acts on both skiers, skier 1 accelerates at 1.5 m/s while skier accelerates at 1.8 m/s. There is a smaller effect from air resistance on the heavier skier compared to the lihter skier; with the iven data, the increase in acceleration is 0.3 m/s or 0 %. (c) There is a difference of 0.3 m/s between the skiers in part (b). This difference represents a 0 % increase in acceleration for the heavier skier. This difference could affect the race because the heavier skier would have the faster time, and therefore, win the race. Reflect on Your Learnin 75. Answers may vary. Sample answer: To explain the common forces I would ask students to describe a situation where they knew that some forces were actin. I would have them brainstorm what forces they thouht were in play. Someone could write the suestions on the board and with the help of the other students oranize them into cateories. To explain FBDs, I would work from a system diaram for one of the student examples to an FBD. I would make sure to do an example with action reaction forces and two bodies. This would address the issue of which body to put a iven force on. m Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -33

33 76. Answers may vary. Sample answer: I was surprised to realize that the physics way of analyzin forces and motion applies everywhere in my life. This approach is especially useful in lookin at sports, such as how to use friction to jump effectively, and how to lean when skiin. 77. Answers will vary. Sample answer: In this chapter I learned about the difference between static and kinetic friction. I suppose I was aware of the two kinds of friction but never really made a distinction. It was interestin to realize that static friction can work in a forward direction to make it possible to walk or drive. 78. Answers may vary. Sample answer: I have noticed that it is often easier to pull a heavy box than to push it. Now I realize that when I pull my applied force has an upward component while when I push my applied force has a downward component. This affects the normal force and the friction I am workin aainst. A question for my fellow students: Give an example of a situation where the normal force is larer than the weiht of the object. 79. Answers may vary. Sample answer: I was most surprised to see how the force of ravity works on an inclined plane. 80. Answers may vary. Sample answer: My car battery is often dead and I am always ettin a push or a tow. I really can see how action reaction force pairs work because of this experience. Research 81. Answers may vary. Sample answer: Students answers should mention the followin points. Students should describe biomechanics as an application of physics and mechanics principles to bioloical systems, includin the motion of athletes. Students should discuss that a force platform is a device for measurin the forces exerted by a body as it moves, and how this information can help athletes learn to move efficiently. 8. Answers may vary. A student presentation may include a short history of seat belts from 1959, when Volvo first made them standard equipment, to the present. It should also explain some of the features of modern seatbelts. One is the lockin retractor, that is, a sprin-loaded reel equipped with an inertial lockin mechanism that stops the belt from extendin off the reel durin sudden slow downs. Another is the pretensioner that preemptively tihtens a seatbelt to prevent an occupant from jerkin forward in a crash. Web clamps clamp the webbin in the event of an accident and limit the distance the webbin can spool out. Automatic seatbelts automatically move into position around an occupant once the car door is closed or the enine is started. Students can also describe current developments such as inflatable seatbelts and improved safety in the professional car racin sector. Explanations should reference the concepts of force and friction and also the principles of mechanics. 83. Answers may vary. A summary of research should include examples of how belts and ropes are used in climbin by professional arborists as well as recreational climbers in rock climbin or mountaineerin. There should be a discussion of how friction keeps knots tied and hitches secure as well as how friction can be a problem. Sometimes friction can be too reat and interfere with the climbin operation. A number of devices have been invented to reduce and/or control friction. These include eneral devices such as carabiners, and pulleys and specific devices such as the Cambrium Saver, the Friction Saver, the Rope Guide, and Buck Blocks that are used by arborists. The summary may Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -34

34 also describe some of the factors that affect friction in the physical world such as softness, rouhness, temperature, and moisture. 84. Student s answers may vary. Students answers should include an explanation of how the principles of motion are applied in archery. Factors to consider should include the bow, the arrow, and atmospheric conditions. Factors affectin the arrow are the initial anle the arrow is aimed, its initial velocity, its weiht, its lenth, the lenth of the arrow's feathers, as well as the heiht of the arrow's feathers. Bow factors can include draw lenth and draw force. Atmospheric factors can include ravity, friction, and wind. Students should use Newton s laws of motion in their explanations. Copyriht 01 Nelson Education Ltd. Chapter : Dynamics -35