CHAPTER 4 FORCES AND NEWTON'S LAWS OF MOTION

Transcription

1 CHAPTER 4 ORCES AND NEWTON'S LAWS O MOTION CONCEPTUAL QUESTIONS 1. REASONING AND SOLUTION When the car come to a udden halt, the upper part of the bod continue forward (a predicted b Newton' firt law) if the force eerted b the lower back mucle i not great enough to give the upper bod the ame deceleration a the car. The lower portion of the bod i held in place b the force of friction eerted b the car eat and the floor. When the car rapidl accelerate, the upper part of the bod trie to remain at a contant velocit (again a predicted b Newton' firt law). If the force provided b the lower back mucle i not great enough to give the upper bod the ame acceleration a the car, the upper bod appear to be preed backward againt the eat a the car move forward.. REASONING AND SOLUTION When the birdfeeder i hanging freel and no one i pulling on the dangling (lower) cord, there i a tenion in the cord between the birdfeeder and the tree limb (the upper cord), becaue the upper cord upport the weight of the birdfeeder. When the lower cord i pulled down with a low continuou pull, the tenion in both cord increae lowl. Since the upper cord ha a larger tenion to begin with, it alwa ha the greater tenion a the lower cord i pulled. Thu, the upper cord nap firt. On the other hand, when the child give the lower cord a udden, downward pull, the tenion in the lower cord increae uddenl. However, the tenion in the upper cord doe not increae a uddenl. The reaon i that the birdfeeder ha a large ma, o it accelerate ver lowl. Thu, the upper cord i tretched lowl and, conequentl, the tenion in the upper cord rie lowl. Since the tenion rie much fater in the lower cord, it i the firt to nap. 3. SSM REASONING AND SOLUTION If the net eternal force acting on an object i zero, it i poible for the object to be traveling with a nonzero velocit. According to Newton econd law, Σ = ma, if the net eternal force Σ i zero, the acceleration a i alo zero. If the acceleration i zero, the velocit mut be contant, both in magnitude and in direction. Thu, an object can move with a contant nonzero velocit when the net eternal force i zero. 4. REASONING AND SOLUTION According to Newton' econd law, a net force i required to give an object a non-zero acceleration. a. If an object i moving with a contant acceleration of 9.80 m/, we can conclude that there i a net force on the object.

2 156 ORCES AND NEWTON'S LAWS O MOTION b. If an object move with a contant velocit of 9.80 m/, it acceleration i zero; therefore, we can conclude that the net force acting on the object i zero. 5. REASONING AND SOLUTION An object will not necearil accelerate when two or more force are applied to the object imultaneoul. The applied force ma cancel o the net force i zero; in uch a cae, the object will not accelerate. The reultant of all the force that act on the object mut be nonzero in order for the object to accelerate. 6. REASONING AND SOLUTION Since the father and the daughter are tanding on ice kate, there i virtuall no friction between their bodie and the ground. We can aume, therefore, that the onl horizontal force that act on the daughter i due to the father, and imilarl, the onl horizontal force that act on the father i due to the daughter. a. According to Newton' third law, when the puh off againt each other, the force eerted on the father b the daughter mut be equal in magnitude and oppoite in direction to the force eerted on the daughter b the father. In other word, both the father and the daughter eperience puhing force of equal magnitude. b. According to Newton' econd law, = ma. Therefore, a = / m. The magnitude of the net force on the father i the ame a the magnitude of the net force on the daughter, o we can conclude that, ince the daughter ha the maller ma, he will acquire the larger acceleration. 7. REASONING AND SOLUTION a. The force of the gmnat on the trampoline caue the elatic urface of the trampoline to deform. b. The reaction force eerted b the trampoline on the gmnat caue the gmnat to decelerate and come to a momentar top. 8. REASONING AND SOLUTION The two oppoitel directed puhing force predicted b Newton' third law act on different bodie; one force act on the crate, while the other act on the peron who i puhing on the crate. Since the two force act on different object, the cannot cancel each other. Whether or not the crate move depend on the net force that act on the crate. If the crate doe not move under the action of a ingle puhing force, the onl reaonable concluion i that there mut be another force acting on the crate that cancel the puhing force. The other force i the force of tatic friction.

3 Chapter 4 Conceptual Quetion REASONING AND SOLUTION The magnitude of the gravitational force between an two of the particle i given b Newton' law of univeral gravitation: = Gm 1 m / r where m 1 and m are the mae of the particle and r i the ditance between them. Since the particle have equal mae, we can arrange the particle o that each one eperience a net gravitational force that ha the ame magnitude if we arrange the particle o that the ditance between an two of the particle i the ame. Therefore, the particle hould be placed at the corner of an equilateral triangle with all three ide of equal length. 10. REASONING AND SOLUTION The ma of an object i a quantitative meaure of it inertia. The ma of an object i an intrinic propert of the object and i independent of the location of the object. The weight of an object i the gravitational force eerted on the object b the earth. The gravitational force depend on the ditance between the object and the center of the earth. Therefore, when an object i moved from ea level to the top of a mountain, it weight will change, while the ma of the object remain contant. 11. REASONING AND SOLUTION The weight of the ball alwa act downward. The force of air reitance will alwa act in the direction that i oppoite to the direction of motion of the ball. The net force on the ball i the reultant of the weight and the force of air reitance. a. A the ball move upward, the force of air reitance act downward. Since air reitance and the weight of the ball act in the ame direction in thi cae, the net force on the ball will be greater in magnitude than the weight of the ball. b. A the ball fall downward, the force of air reitance i upward. Since air reitance and the weight of the ball act in oppoite direction, the net force that act on the ball will be maller in magnitude than the weight of the ball. Note that in both cae, the net force point downward ince the object accelerate downward in both cae. 1. REASONING AND SOLUTION rom Equation 4.5, W = mg, we know that the weight of an object i directl proportional to it ma. The proportionalit contant on a given planet i g, the magnitude of the acceleration due to gravit on that planet. If object A weigh twice a much a object B at the ame pot on the earth, then the ma of object A i twice a much a the ma of object B. The ma of an object i independent of the object' location in the univere. Since the weight of an object i directl proportional to it ma, object A will weigh twice a much a object B at the ame pot on an planet. While the value of the weight will differ from planet to planet, a the value of g varie from planet to planet, the ratio of the weight (which equal the ratio of their mae) will be the ame on an planet.

4 158 ORCES AND NEWTON'S LAWS O MOTION 13. REASONING AND SOLUTION Equation 4.4 and 4.5 ma be combined to give the acceleration due to gravit at a ditance r from the center of the earth: g = GM E / r. Since thi epreion depend on r, the acceleration of a freel falling object doe depend on it location. The acceleration due to gravit will be greater at Death Valle, California (where r i maller) than it i on the top of Mt. Everet (where r i greater). Since r i meaured from the center of the earth, however, thee value will be ver cloe. 14. REASONING AND SOLUTION Auming that the accelerating mechanim remain attached to the rocket, the acceleration will be greater when the rocket i fired horizontall. The accelerating mechanim provide an acceleration that point in the initial direction of motion of the rocket. The net acceleration i the reultant of the accelerating mechanim and the acceleration due to gravit. When the rocket i fired horizontall, thee acceleration will be at right angle to each other. When the rocket i fired traight up, thee acceleration will be in oppoite direction. The magnitude of the reultant will be greater when thee two acceleration are at right angle rather than when the are in oppoition. Therefore, the acceleration will be greater when the rocket i fired horizontall. If we aume that the accelerating mechanim i not attached to the rocket, then once the rocket i fired, the onl force on the rocket i that due to gravit, and the rocket ha the acceleration due to gravit regardle of it orientation. In thi cae, the acceleration of the rocket i the ame regardle of whether it i fired traight up or fired horizontall. 15. REASONING AND SOLUTION If the elevator were at ret, or moving with a contant velocit, the cale would read the true weight of mg = 98 N. When the elevator i accelerating, the cale reading will differ from 98 N and will dipla the apparent weight,, which i given b Equation 4.6: = mg + ma where a, the acceleration of the elevator, i poitive when the elevator accelerate upward and negative when the elevator accelerate downward. a. When the apparent weight i = 75 N, the apparent weight i le than the true weight (mg = 98 N) o a mut be negative. The elevator i accelerating downward. b. When the apparent weight i = 10 N, the apparent weight i greater than the true weight (mg = 98 N) o a mut be poitive. The elevator i accelerating upward. 16. SSM REASONING AND SOLUTION The apparent weight will differ from the true weight onl in an accelerating elevator. When the cale in an elevator read the true weight, the onl concluion that can be made i that the elevator ha zero acceleration. Therefore, one cannot conclude whether the elevator i moving with an contant velocit upward, an contant velocit downward, or whether the elevator i at ret, ince each of thee condition involve zero acceleration. 17. REASONING AND SOLUTION If the elevator were at ret, or moving with a contant velocit, the cale will read the true weight mg. When the elevator i accelerating, the cale

5 Chapter 4 Conceptual Quetion 159 reading will differ from the true weight and will regiter the apparent weight,, which i given b Equation 4.6: = mg + ma where a, the acceleration of the elevator, i poitive when the elevator accelerate upward and negative when the elevator accelerate downward. Since the cale regiter 600 N when the elevator i moving with contant velocit, we know that the true weight i 600 N. a. The elevator i moving upward. When the elevator low down, it acceleration vector point downward. The term ma will be negative; therefore, the cale reading will be le than 600 N. b. When the elevator i topped, the cale will regiter the true weight; therefore, the cale read 600 N. c. The elevator i moving downward and peeding up. The acceleration vector point downward, and the term ma i negative. Therefore, the cale regiter a value that i le than 600 N. 18. REASONING AND SOLUTION Since the led move with contant velocit, the force of kinetic friction i preent. The magnitude of thi force i given b µ k, where µ k i the coefficient of kinetic friction and i the magnitude of the normal force that act on the led. urthermore, the horizontal component of the applied force mut be equal in magnitude to the force of kinetic friction, ince there i no acceleration. When the peron pull on the led, the vertical component of the pulling force tend to decreae the magnitude of the normal force relative to that when the led i not being pulled or puhed. On the other hand, when the peron puhe on the led, the vertical component of the puhing force tend to increae the normal force relative to that when the led i not being pulled or puhed. Therefore, when the led i pulled, the magnitude of the force of kinetic friction, and therefore the magnitude of the applied force, i le than when the led i puhed. 19. REASONING AND SOLUTION We know that µ = 1.4 µ k for a crate in contact with a cement floor. The maimum force of tatic friction i µ N f = while the force of kinetic friction i fk = µ kn. A long a the crate i on the cement floor, we can conclude that the magnitude of the maimum tatic frictional force acting on the crate will alwa be 1.4 time the magnitude of the kinetic frictional force acting on the moving crate, once the crate ha begun moving. However, the force of tatic friction ma not have it maimum value. Thu, the magnitude of the tatic frictional force i not alwa 1.4 time the magnitude of the kinetic frictional force. 0. REASONING AND SOLUTION A bo ret on the floor of a tationar elevator. Becaue of tatic friction, a force i required to tart the bo liding acro the floor of the

6 160 ORCES AND NEWTON'S LAWS O MOTION elevator. The magnitude of thi force i given b µ N f =, where i the magnitude of the normal force eerted on the bo b the floor of the elevator. When the elevator i tationar, the magnitude of the normal force eerted on the bo i mg where m i the ma of the crate and g i the magnitude of the acceleration due to gravit. When the crate accelerate upward, the floor of the elevator will puh againt the bo to accelerate it upward; therefore, when the crate accelerate upward, the magnitude of the normal force will be greater than mg. When the elevator accelerate downward, the normal force between the bo and the floor of the elevator will be le than when the elevator i tationar; therefore, the magnitude of the normal force will be le than mg. Thu, the magnitude of the force required to tart the bo liding acro the floor of the elevator are ranked a follow in acending order: (c) elevator accelerating downward; (a) elevator tationar; (b) elevator accelerating upward. 1. REASONING AND SOLUTION When the rope i tied to a tree and pulled b the ten people, the tenion in the rope i twice a great a it wa when it wa ued in a tug-of-war with five people on each team. Therefore, the rope i more likel to break when it i tied to the tree and pulled b ten people. Note that when the rope i tied to the tree and pulled b ten people, the ituation i equivalent to a tug-of-war with ten people on each team.. REASONING AND SOLUTION An object i in equilibrium when it acceleration i zero. When a tone i thrown from the top of a cliff, it acceleration i the acceleration due to gravit; therefore, the tone i not in equilibrium. 3. SSM REASONING AND SOLUTION An object i in equilibrium when it acceleration i zero. a. If a ingle nonzero force act on an object, the object will accelerate according to Newton' econd law. The object i not in equilibrium. b. If two force that point in mutuall perpendicular direction act on an object, the object will eperience a net force. B Newton' econd law, the object will, therefore, have a nonzero acceleration. The object i not in equilibrium. c. If two force that point in direction that are not perpendicular act on the object, the object ma or ma not be in equilibrium, depending on how the force are oriented. In general, the reultant of two uch force i nonzero, the object will accelerate, and it i not in equilibrium. In the pecial cae where the two force point in oppoite direction and have the ame magnitude, the net force i zero, the object ha zero acceleration, and the object i, therefore, in equilibrium.

7 Chapter 4 Conceptual Quetion REASONING AND SOLUTION A circu performer hang from a tationar rope. Since there i no acceleration, the tenion in the rope mut be equal in magnitude to the weight of the performer. She then begin to climb upward b pulling herelf up, hand-over-hand. Whether the tenion in the rope i greater than or equal to the tenion when he hang tationar depend on whether or not he accelerate a he move upward. When he move upward at contant velocit, the tenion in the rope will be the ame. When he accelerate upward, the rope mut upport the net upward force in addition to her weight; therefore, in thi cae, the tenion in the rope will be greater than when he hang tationar. 5. REASONING AND SOLUTION If a k diver with an open parachute approache the ground with a contant velocit, the acceleration of the k diver i zero, and the k diver i, therefore, in equilibrium. The two force reponible for the equilibrium are the weight of the k diver and the force eerted on the k diver b the parachute. Thee force mut be equal in magnitude and oppoite in direction o that their reultant i zero. 6. REASONING AND SOLUTION There are three force that act on the ring a hown in the figure below. The weight of the block, which act downward, and two force of tenion that act along the rope awa from the ring. Since the ring i at ret, the net force on the ring i zero. The weight of the block i balanced b the vertical component of the tenion in the rope. Clearl, the rope can never be made horizontal, for then there would be no vertical component of the tenion force to balance the weight of the block. T T 7. REASONING AND SOLUTION A freight train i accelerating on a level track. The tenion in the coupling between the engine and the firt car depend on the total ma being pulled b the engine. Therefore, other thing being equal, if ome of the cargo in the lat car were tranferred to an one of the other car, the tenion in the coupling between the engine and the firt car would remain the ame. Thi i becaue the tranfer doe not change the total ma being pulled b the engine. W

8 16 ORCES AND NEWTON'S LAWS O MOTION CHAPTER 4 ORCES AND NEWTON'S LAWS O MOTION PROBLEMS 1. REASONING AND SOLUTION According to Newton econd law, the acceleration i a = Σ/m. Since the pilot and the plane have the ame acceleration, we can write Σ Σ or ( ) m Σ = Σ = PILOT m m m PILOT PILOT PLANE PLANE Therefore, we find N ( Σ ) = PILOT ( 78 kg) 93 N 4 = kg. REASONING Newton econd law of motion give the relationhip between the net force Σ and the acceleration a that it caue for an object of ma m. The net force i the vector um of all the eternal force that act on the object. Here the eternal force are the drive force, the force due to the wind, and the reitive force of the water. SOLUTION We chooe the direction of the drive force (due wet) a the poitive direction. Solving Newton econd law ( Σ = ma) for the acceleration give Σ N 800 N 100 N a = = = m/ m 6800 kg The poitive ign for the acceleration indicate that it direction i due wet. 3. REASONING According to Newton econd law, Equation 4.1, the average net force Σ i equal to the product of the object ma m and the average acceleration a. The average acceleration i equal to the change in velocit divided b the elaped time (Equation.4), where the change in velocit i the final velocit v minu the initial velocit v 0. SOLUTION The average net force eerted on the car and rider i 0 ( ) v v m/ 0 m/ = ma = m = kg = N t t 7.0

9 Chapter 4 Problem REASONING AND SOLUTION The acceleration i obtained from where v 0 = 0 m/. So Newton econd law give = v 0 t + 1 at a = /t 18 ( m) ( 0.95 ) Σ = ma= m = ( 7 kg) = 900 N t 5. SSM REASONING We can ue the appropriate equation of kinematic to find the acceleration of the bullet. Then Newton' econd law can be ued to find the average net force on the bullet. SOLUTION According to Equation.4, the acceleration of the bullet i v v0 715 m/ 0 m/ a = = = m/ t Therefore, the net average force on the bullet i = ma= (15 10 kg)( m/ ) = 490 N 6. REASONING AND SOLUTION The acceleration required i v v0 ( 15.0 m/) a = = =.5 m/ ( 50.0 m) Newton' econd law then give the magnitude of the net force a = ma = (1580 kg)(.5 m/ ) = 3560 N 7. SSM REASONING According to Newton' econd law of motion, the net force applied to the fit i equal to the ma of the fit multiplied b it acceleration. The data in the problem give the final velocit of the fit and the time it take to acquire that velocit. The average acceleration can be obtained directl from thee data uing the definition of average acceleration given in Equation.4. SOLUTION The magnitude of the average net force applied to the fit i, therefore,

10 164 ORCES AND NEWTON'S LAWS O MOTION v 8.0 m/ 0 m/ = ma= m = ( 0.70 kg) = 37 N t REASONING AND SOLUTION rom Equation.9, 0 v = v + a Since the arrow tart from ret, v 0 = 0 m/. In both cae i the ame o or 1 v a a v a v = a = a v = 1 a Since = ma, it follow that a = /m. The ma of the arrow i unchanged, and or v v = v = v = v = 5.0 m/ = 35.4 m/ ( ) 9. SSM WWW REASONING Let due eat be choen a the poitive direction. Then, when both force point due eat, Newton' econd law give + = ma A B 1 (1) Σ where a 1 = 0.50 m/. When A point due eat and B point due wet, Newton' econd law give = ma A B () Σ where force. a = 0.40 m/. Thee two equation can be ued to find the magnitude of each SOLUTION a. Adding Equation 1 and give

11 Chapter 4 Problem 165 A ( + a ) ( 8.0 kg)( 0.50 m / m / ) m a1 = = = 3.6 N b. Subtracting Equation from Equation 1 give B ( a ) ( 8.0 kg)( 0.50 m / 0.40 m / ) m a1 = = = 0.40 N 10. REASONING AND SOLUTION Newton' econd law, Σ = ma, implie that the acceleration a and the net force are in the ame direction. Thi i 64 N of E. The magnitude of the net force i = ma = (350 kg)(0.6 m/ ) = 0 N 11. REASONING Newton econd law give the acceleration a a = (Σ)/m. Since we eek onl the horizontal acceleration, it i the component of thi equation that we will ue; a = (Σ )/m. or completene, however, the free-bod diagram will include the vertical force alo (the normal force and the weight W). SOLUTION The free bod diagram i hown at the right, where 1 = 59.0 N = 33.0 N θ = 70.0 When 1 i replaced b it and component, we obtain the free bod diagram in the econd drawing. + + θ W 1

12 166 ORCES AND NEWTON'S LAWS O MOTION a Chooing right to be the poitive direction, we have Σ 1coθ a = = m m ( 59.0 N) co 70.0 ( 33.0 N) = = 1.83 m/ 7.00 kg The minu ign indicate that the horizontal acceleration point to the left. + 1 co θ 1 in θ W + 1. REASONING The acceleration of the k diver can be obtained directl from Newton econd law a the net force divided b the k diver ma. The net force i the vector um of the k diver weight and the drag force. SOLUTION rom Newton econd law, Σ = ma (Equation 4.1), the k diver acceleration i f Σ a = + m The free-bod diagram how the two force acting on the k diver, hi weight W and the drag force f. The net force i Σ = f W. Thu, the acceleration can be written a a = f W m The acceleration of the k diver i f W 107 N 915 N a = = = m/ m 93.4 kg W ree-bod diagram Note that the acceleration i poitive, indicating that it point upward. 13. SSM REASONING According to Newton' econd law ( = ma), the acceleration of the object i given b a = / m, where i the net force that act on the object. We mut firt find the net force that act on the object, and then determine the acceleration uing Newton' econd law. SOLUTION The following table give the and component of the two force that act on the object. The third row of that table give the component of the net force.

13 Chapter 4 Problem 167 orce -Component -Component N 0 N (60.0 N) co 45.0 = 4.4 N (60.0 N) in 45.0 = 4.4 N = N 4.4 N The magnitude of i given b the Pthagorean theorem a Σ = (8.4 N) + (4.4) = 9.7 N The angle θ that make with the + ai i θ N = tan = N Σ θ 8.4 N 4.4 N According to Newton' econd law, the magnitude of the acceleration of the object i 9.7 N a = = = m 3.00 kg 30.9 m/ Since Newton' econd law i a vector equation, we know that the direction of the right hand ide mut be equal to the direction of the left hand ide. In other word, the direction of the acceleration a i the ame a the direction of the net force. Therefore, the direction of the acceleration of the object i 7. above the + ai. 14. REASONING Newton econd law, Σ = ma, tate that a net force of Σ mut act on an object of ma m in order to impart an acceleration a to the object. In the impact hock tet the bo i ubjected to a large deceleration and, hence, a correpondingl large net force. To determine the net force we will determine the deceleration in a kinematic calculation and ue it in Newton econd law. SOLUTION According to Newton econd law, the net force i Σ = ma, where the acceleration can be determined with the aid of Equation.4 (v = v 0 + at). According to thi equation v v a = 0 t

14 168 ORCES AND NEWTON'S LAWS O MOTION Subtituting thi reult for the acceleration into the econd law give v v ma m 0 Σ = = t Since the initial velocit (v 0 = +0 m/), final velocit (v = 0 m/), and the duration of the colliion (t = ) are known, we find v v0 0 m/ 0 m/ 6 Σ = m = ( 41 kg) = N t The minu ign indicate that the net force point oppoite to the direction in which the bo i thrown, which ha been aumed to be the poitive direction. The magnitude of the net force i N, which i over three hundred thouand pound. 15. REASONING Equation 3.5a ( = v 1 0 t+ a t ) and 3.5b ( v 1 ) 0 t a t = + give the diplacement of an object under the influence of contant acceleration a and a. We can add thee diplacement a vector to find the magnitude and direction of the reultant diplacement. To ue Equation 3.5a and 3.5b, however, we mut have value for a and a. We can obtain thee value from Newton econd law, provided that we combine the given force to calculate the and component of the net force acting on the duck, and it i here that our olution begin. SOLUTION Let the direction due eat and due north, repectivel, be the + and + direction. Then, the component of the net force are ( ) ( ) Σ = 0.10 N N co 5 = 0.31 N Σ = 0.0 N in 5 = N According to Newton econd law, the component of the acceleration are a a Σ 0.31 N = = = m/ m.5 kg Σ N = = = m/ m.5 kg rom Equation 3.5a and 3.5b, we now obtain the diplacement in the and direction:

15 Chapter 4 Problem 169 ( )( ) ( )( ) = v t+ a t = 0.11 m/ m/ 3.0 = m ( )( ) ( )( ) = v t+ a t = 0 m/ m/ 3.0 = m The magnitude of the reultant diplacement i ( ) ( ) r = + = m m = 0.78 m The direction of the reultant diplacement i m θ = tan = 1 outh of eat m 16. REASONING or both the tug and the ateroid, Equation.8 ( vt 1 at ) = + applie with 0 v 0 = 0 m/, ince both are initiall at ret. In appling thi equation, we mut be careful and ue the proper acceleration for each object. Newton econd law indicate that the acceleration i given b a = Σ/m. In thi epreion, we note that the magnitude of the net force acting on the tug and the ateroid are the ame, according to Newton actionreaction law. The mae of the tug and the ateroid are different, however. Thu, the ditance traveled for either object i given b, where we ue for Σ onl the magnitude of the pulling force 1 1 Σ = vt 0 + at = t m SOLUTION Let L be the initial ditance between the tug and the ateroid. When the two object meet, the ditance that each ha traveled mut add up to equal L. Therefore, Solving for the time t give 1 1 T A T A L= + = a t + a t 1 Σ 1 Σ L= t t m + = Σ + T m A mt m t A ( ) L 450 m t = = = Σ ( 490 N) + + m T m A 3500 kg 600 kg

16 170 ORCES AND NEWTON'S LAWS O MOTION 17. SSM WWW REASONING We firt determine the acceleration of the boat. Then, uing Newton' econd law, we can find the net force that act on the boat. Since two of the three force are known, we can olve for the unknown force once the net force i known. SOLUTION Let the direction due eat be the poitive direction and the direction due north be the poitive direction. The and component of the initial velocit of the boat are then v 0 = (.00 m/) co 15.0 = 1.93 m/ v 0 = (.00 m/) in 15.0 = m/ Thirt econd later, the and velocit component of the boat are v = (4.00 m/) co 35.0 = 3.8 m/ v = (4.00 m/) in 35.0 =.9 m/ Therefore, according to Equation 3.3a and 3.3b, the and component of the acceleration of the boat are v v0 3.8 m/ 1.93 m/ a = = = m/ t 30.0 W a v v.9 m/ m/ = = = m/ t Thu, the and component of the net force that act on the boat are = ma = (35 kg) ( m/ ) = 14.6 N = ma = (35 kg) ( m/ ) = 19. N The following table give the and component of the net force and the two known force that act on the boat. The fourth row of that table give the component of the unknown force. W

17 Chapter 4 Problem 171 orce -Component -Component N (31.0 N) co 15.0 = 9.9 N 19. N (31.0 N) in 15.0 = 8.0 N W = 1 (3.0 N ) co 15.0 =. N (3.0 N) in 15.0 = 5.95 N 14.6 N 9.9 N +. N = 6.9 N 19. N 8.0 N N = 17.1 N The magnitude of W i given b the Pthagorean theorem a The angle θ that W W = (6.9 N) + (17.1N ) = 18.4 N make with the ai i θ N = tan = N θ 17.1 N Therefore, the direction of W i 68, north of eat. 6.9 N 18. REASONING The magnitude of the gravitational force that each part eert on the other i given b Newton law of gravitation a = Gm m r. To ue thi epreion, we need 1 / the mae m 1 and m of the part, wherea the problem tatement give the weight W 1 and W. However, the weight i related to the ma b W = mg, o that for each part we know that m = W/g. SOLUTION The gravitational force that each part eert on the other i ( / )( / ) Gm1m GW 1 g W g = = r r 11 ( N m / kg )( N)( 3400 N) = = ( 9.80 m/ ) ( 1 m) N 19. REASONING Newton law of univeral gravitation indicate that the gravitational force that each uniform phere eert on the other ha a magnitude that i inverel proportional

18 17 ORCES AND NEWTON'S LAWS O MOTION to the quare of the ditance between the center of the phere. Therefore, the maimum gravitational force between two uniform phere occur when the center of the phere are a cloe together a poible, and thi occur when the urface of the phere are touching. Then, the ditance between the center of the phere i the um of the two radii. SOLUTION When the bowling ball and the billiard ball are touching, the ditance between their center i r = r Bowling + r Billiard. Uing thi epreion in Newton law of univeral gravitation give Gm m Gm m = = r r r Bowling Billiard Bowling Billiard ( Bowling + Billiard ) 11 ( N m / kg )( 7. kg)( 0.38 kg) ( 0.11 m m) = = N 0. REASONING AND SOLUTION The force that act on the rock are hown at the right. Newton' econd law (with the direction of motion a poitive) i Σ = mg R= ma R Solving for the acceleration a give ( 45 kg)( 9.80 m/ ) ( 18 N) mg R mg a = = = 9.4 m/ m 45 kg 1. SSM REASONING AND SOLUTION a. Combining Equation 4.4 and 4.5, we ee that the acceleration due to gravit on the urface of Saturn can be calculated a follow: g M ( ) ( kg ) Saturn 11 Saturn = G = N m /kg = 10.5 m/ 7 r Saturn ( m) b. The ratio of the peron weight on Saturn to that on earth i W Saturn mgsaturn gsaturn 10.5 m/ = = = = 1.07 W earth mgearth gearth 9.80 m/

19 Chapter 4 Problem 173. REASONING A dicued in Conceptual Eample 7, the ame net force i required on the moon a on the earth. Thi net force i given b Newton econd law a Σ = ma, where the ma m i the ame in both place. Thu, from the given ma and acceleration, we can calculate the net force. On the moon, the net force come about due to the drive force and the oppoing frictional force. Since the drive force i given, we can find the frictional force. SOLUTION Newton econd law, with the direction of motion taken a poitive, give 3 ( ) ( )( ) Σ = ma or 1430 N f = kg 0.0 m/ Solving for the frictional force f, we find 3 ( ) ( )( ) f = 1430 N kg 0.0 m/ = 130N 3. SSM REASONING AND SOLUTION According to Equation 4.4 and 4.5, the weight of an object of ma m at a ditance r from the center of the earth i 7 GM Em mg = r In a circular orbit that i m above the urface of the earth ( radiu = m, 4 ma = kg ), the total ditance from the center of the earth i 7 r = m m. Thu the acceleration g due to gravit i 11 4 E 7 6 GM ( N m /kg )( kg) g = = = r ( m m) 0.3 m/ 4. REASONING Each particle eperience two gravitational force, one due to each of the remaining particle. To get the net gravitational force, we mut add the two contribution, taking into account the direction. The magnitude of the gravitational force that an one particle eert on another i given b Newton law of gravitation a = Gm m r. Thu, for particle A, we need to appl thi law to it interaction with particle B and with particle C. or particle B, we need to appl the law to it interaction with particle A and with particle C. Latl, for particle C, we mut appl the law to it interaction with particle A and with particle B. In conidering the direction, we remember that the gravitational force between two particle i alwa a force of attraction. 1 / SOLUTION We begin b calculating the magnitude of the gravitational force for each pair of particle: 6

20 174 ORCES AND NEWTON'S LAWS O MOTION AB BC AC 11 ( N m / kg )( 363 kg)( 517 kg) ( m) GmAmB 5 = = = N r 11 ( N m / kg )( 517 kg)( 154 kg) ( m) GmBmC 5 = = = N r 11 ( N m / kg )( 363 kg)( 154 kg) ( m) GmAmC 6 = = = N r In uing thee magnitude we take the direction to the right a poitive. a. Both particle B and C attract particle A to the right, the net force being A = AB + AC = N N = N, right b. Particle C attract particle B to the right, while particle A attract particle B to the left, the net force being = = N N = N, right B BC AB c. Both particle A and B attract particle C to the left, the net force being C = AC + BC = N N = N, left 5. REASONING The magnitude of the gravitational force eerted on the atellite b the earth i given b Equation 4.3 a = Gm m r, where r i the ditance between atellite earth / the atellite and the center of the earth. Thi epreion alo give the magnitude of the gravitational force eerted on the earth b the atellite. According to Newton econd law, the magnitude of the earth acceleration i equal to the magnitude of the gravitational force eerted on it divided b it ma. Similarl, the magnitude of the atellite acceleration i equal to the magnitude of the gravitational force eerted on it divided b it ma. SOLUTION a. The magnitude of the gravitational force eerted on the atellite when it i a ditance of two earth radii from the center of the earth i 11 4 ( N m / kg )( 45 kg)( kg) 6 ( )( m) Gmatellitemearth 3 = = = N r

21 Chapter 4 Problem 175 b. The magnitude of the gravitational force eerted on the earth when it i a ditance of two earth radii from the center of the atellite i 11 4 ( N m / kg )( 45 kg)( kg) 6 ( )( m) Gmatellitemearth 3 = = = N r c. The acceleration of the atellite can be obtained from Newton econd law. a atellite N = = = m 45 kg atellite 3.45 m / d. The acceleration of the earth can alo be obtained from Newton econd law N aearth = = = m / m 4 earth kg 6. REASONING The weight of a peron on the earth i the gravitational force earth that it eert on the peron. The magnitude of thi force i given b Equation 4.3 a m m earth peron earth = G rearth where r earth i the ditance from the center of the earth to the peron. In a imilar fahion, the weight of the peron on another planet i m m planet peron planet = G rplanet We will ue thee two epreion to obtain the weight of the traveler on the planet. SOLUTION Dividing planet b earth we have or m m planet peron G planet planet planet earth r m r = = m m m r earth earth peron earth G planet rearth

22 176 ORCES AND NEWTON'S LAWS O MOTION planet mplanet r earth earth mearth r planet = Since we are given that planet i m m planet earth rearth = and r planet 1 =, the weight of the pace traveler on the 3 ( )( ) 1 planet = 540 N = 10 N 3 7. SSM REASONING AND SOLUTION a. According to Equation 4.4, the weight of an object of ma m on the urface of Mar would be given b GM Mm W = R where M M i the ma of Mar and R M i the radiu of Mar. On the urface of Mar, the weight of the object can be given a W = mg (ee Equation 4.5), o M Subtituting value, we have mg = GM m = GM R M or g M M RM 11 3 ( N m /kg )( kg) g = = 6 ( m) 3.75 m/ b. According to Equation 4.5, W = mg = (65 kg)(3.75 m/ ) =.4 10 N

23 Chapter 4 Problem REASONING AND SOLUTION The figure at the right how the three phere with phere 3 being the phere of unknown ma. Sphere 3 feel a force 31 due to the preence of phere 1, and a force 3 due to the preence of phere. The net force on phere 3 i the reultant of 31 and 3. Note that ince the phere form an equilateral triangle, each interior angle i 60. Therefore, both 31 and 3 make a 30 angle with the vertical line a hown m 30 1 urthermore, 31 and 3 have the ame magnitude given b GMm3 = r where M i the ma of either phere 1 or and m 3 i the ma of phere 3. The component of the two force are hown in the following drawing: co θ co θ in θ in θ Clearl, the horizontal component of the two force add to zero. Thu, the net force on phere 3 i the reultant of the vertical component of 31 and 3 : GMm3 = co θ = co θ r 3 The acceleration of phere 3 i given b Newton' econd law: a 11 3 GM 3 = = θ = m 3 r 10 ( 1.0 m) ( ) ( N m /kg ).80 kg co co 30.0 =.5 10 m/

24 178 ORCES AND NEWTON'S LAWS O MOTION 9. SSM WWW REASONING AND SOLUTION There are two force that act on the balloon; the are, the combined weight of the balloon and it load, Mg, and the upward buoant force. If we take upward a the poitive direction, then, initiall when the B balloon i motionle, Newton' econd law give B Mg 0 =. If an amount of ma m i dropped overboard o that the balloon ha an upward acceleration, Newton' econd law for thi ituation i B ( M m) g = ( M m) a But B = mg, o that ( ) ( ) Mg M m g = mg = M m a Solving for the ma m that hould be dropped overboard, we obtain Ma (310 kg )(0.15 m/ ) m = = = 4.7 kg g + a 9.80 m/ m/ 30. REASONING The drawing how the point between the earth and the moon where the gravitational force eerted on the pacecraft b the earth balance that eerted b the moon. The magnitude of the gravitational force eerted on the pacecraft b the earth i m m earth pacecraft earth = G while that eerted on the pacecraft b the moon i m moon pacecraft moon = G ( rearth-moon r) B etting thee two epreion equal to each other (ince the gravitational force balance), we will be able to find the ditance r. r m Point where the gravitational force balance Moon Earth r r earth-moon

25 Chapter 4 Problem 179 SOLUTION Setting earth equal to moon, we have G mearth mpacecraft Solving thi epreion for r give r = G m moon m pacecraft ( r r) earth-moon ( r ) m earth earth-moon m ( 8 moon m) r = = = m mearth m moon 31. REASONING The gravitational force that the un eert on a peron tanding on the earth i given b Equation 4.3 a = GM m/ r, where M un i the ma of the un, m un un un-earth i the ma of the peron, and r un-earth i the ditance from the un to the earth. Likewie, the gravitational force that the moon eert on a peron tanding on the earth i given b = GM m/ r, where M moon i the ma of the moon and r moon-earth i the moon moon moon-earth ditance from the moon to the earth. Thee relation will allow u to determine whether the un or the moon eert the greater gravitational force on the peron. SOLUTION Taking the ratio of un to moon, and uing the ma and ditance data from the inide of the tet front cover, we find GM m un un un-earth un moon-earth r M r = = GM m M r moon moon moon un-earth rmoon-earth kg m 11 = = kg m 178 Therefore, the un eert the greater gravitational force. 3. REASONING AND SOLUTION According to Equation 4.4 and 4.5, the acceleration due to gravit at the urface of the neutron tar i Gm ( N m / kg )(.0 10 kg) a = = = m/ 3 r ( m) 1

26 180 ORCES AND NEWTON'S LAWS O MOTION Since the gravitational force i aumed to be contant, the acceleration will be contant and the peed of the object can be calculated from v = v + a, with v 0 = 0m/ ince the object fall from ret. Solving for v ield ( )( ) v= a = m/ m = m/ 33. REASONING We place the third particle (ma = m 3 ) a hown in the following drawing: L D m m 3 m The magnitude of the gravitational force that one particle eert on another i given b Newton law of gravitation a = Gm 1 m /r. Before the third particle i in place, thi law indicate that the force on each particle ha a magnitude before = Gmm/L. After the third particle i in place, each of the firt two particle eperience a greater net force, becaue the third particle alo eert a gravitational force on them. SOLUTION or the particle of ma m, we have or the particle of ma m, we have Gmm Gmm 3 + after D L L m3 Gmm before md = = + 1 L Gmm Gmm 3 + after ( ) L 3 L D L m = = + 1 Gm m before m( L D) L Since after / before = for both particle, we have 3 Lm3 Lm + 1 = + 1 or D = L D md m L D ( ) ( ) Epanding and rearranging thi reult give uing the quadratic formula: D + LD L = 0, which can be olved for D

27 Chapter 4 Problem 181 ( ) ( )( ) L± L 4 1 L D= = 0.414L or.414l 1 () The negative olution i dicarded becaue the third particle lie on the + ai between m and m. Thu, D = L. 34. REASONING In each cae the object i in equilibrium. According to Equation 4.9b, Σ = 0, the net force acting in the (vertical) direction mut be zero. The net force i compoed of the weight of the object() and the normal force eerted on them. SOLUTION a. There are three vertical force acting on the crate: an upward normal force + that the floor eert, the weight m 1 g of the crate, and the weight m g of the peron tanding on the crate. Since the weight act downward, the are aigned negative number. Setting the um of thee force equal to zero give The magnitude of the normal force i N + ( m1g) + ( mg) =0 Σ = m 1 g + m g = (35 kg + 65 kg)(9.80 m/ ) = 980 N b. There are onl two vertical force acting on the peron: an upward normal force + that the crate eert and the weight m g of the peron. Setting the um of thee force equal to zero give N + ( mg) = 0 Σ The magnitude of the normal force i = m g = (65 kg)(9.80 m/ ) = 640 N 35. SSM REASONING The book i kept from falling a long a the total tatic frictional force balance the weight of the book. The force that act on the book are hown in the following free-bod diagram, where P i the preing force applied b each hand.

28 18 ORCES AND NEWTON'S LAWS O MOTION f f P P W In thi diagram, note that there are two preing force, one from each hand. Each hand alo applie a tatic frictional force, and, therefore, two tatic frictional force are hown. The maimum tatic frictional force i related in the uual wa to a normal force, but in thi problem the normal force i provided b the preing force, o that = P. SOLUTION Since the frictional force balance the weight, we have Solving for P, we find that ( ) ( ) N f = µ = µ P = W W 31 N P = = = µ 0.40 ( ) 39 N 36. REASONING AND SOLUTION a. The apparent weight of the peron i given b Equation 4.6 a = mg + ma = (95.0 kg)(9.80 m/ m/ ) = N b. = (95.0 kg)(9.80 m/ ) = 931 N c. = (95.0 kg)(9.80 m/ 1.30 m/ ) = 808 N 37. REASONING AND SOLUTION The block will move onl if the applied force i greater than the maimum tatic frictional force acting on the block. That i, if > µ = µ mg = (0.650)(45.0 N) = 9. N The applied force i given to be = 36.0 N which i greater than the maimum tatic frictional force, o the block will move.

29 Chapter 4 Problem 183 The block' acceleration i found from Newton' econd law. Σ f k µ k mg a = = = = 3.7 m/ m m m 38. REASONING A hown in the free-bod diagram below, three force act on the car: the tatic frictional force f (directed up the hill), the normal force (directed perpendicular to the road), and it weight mg. A it it on the hill, the acceleration of the Mercede-Benz i zero (a = a = 0 m/ ). According to the dicuion in Section 4.4, thi mean that the net force in the direction mut be zero ( 0) Σ = and the net force in the direction mut be zero ( Σ = 0). Thee two relation can be ued to determine the normal force and the tatic frictional force. SOLUTION + a. Appling Newton econd law to the direction ( Σ = 0) ield f Σ =+ N mgco15 = 0 (4.b) where the term mg co 15 i the component of the car weight (negative, becaue thi component point along the negative ai). Solving for the magnitude of the normal force, we obtain 15º mg + 15º ( )( ) 4 N = mgco15 = 1700 kg 9.80 m/ co15 = N b. Appling Newton econd law to the direction ( 0) Σ = give Σ =+ mgin15 f = 0 (4.a) where the term mg in 15 i the component of the car weight. Solving thi epreion for the tatic frictional force give ( )( ) 3 f = mgin15 = 1700 kg 9.80 m/ in15 = N 39. SSM REASONING In order to tart the crate moving, an eternal agent mut uppl a force that i at leat a large a the maimum value µ N f =, where µ i the

30 184 ORCES AND NEWTON'S LAWS O MOTION coefficient of tatic friction (ee Equation 4.7). Once the crate i moving, the magnitude of the frictional force i ver nearl contant at the value f k = µ k, where µ k i the coefficient of kinetic friction (ee Equation 4.8). In both cae decribed in the problem tatement, there are onl two vertical force that act on the crate; the are the upward normal force, and the downward pull of gravit (the weight) mg. urthermore, the crate ha no vertical acceleration in either cae. Therefore, if we take upward a the poitive direction, Newton' econd law in the vertical direction give N mg = 0, and we ee that, in both cae, the magnitude of the normal force i N = mg. SOLUTION a. Therefore, the applied force needed to tart the crate moving i f = µ mg = (0.760)(60.0 kg)(9.80 m/ ) = 447 N b. When the crate move in a traight line at contant peed, it velocit doe not change, and it ha zero acceleration. Thu, Newton' econd law in the horizontal direction become P - f k = 0, where P i the required puhing force. Thu, the applied force required to keep the crate liding acro the dock at a contant peed i P = fk = µ k mg = (0.410)(60.0 kg)(9.80 m/ ) = 41 N 40. REASONING In each of the three cae under conideration the kinetic frictional force i given b f k = µ k. However, the normal force varie from cae to cae. To determine the normal force, we ue Equation 4.6 ( = mg + ma) and thereb take into account the acceleration of the elevator. The normal force i greatet when the elevator accelerate upward (a poitive) and mallet when the elevator accelerate downward (a negative). SOLUTION a. When the elevator i tationar, it acceleration i a = 0 m/. Uing Equation 4.6, we can epre the kinetic frictional force a ( ) ( ) f = µ = µ mg+ ma = µ m g+ a k k N k k ( )( ) ( ) ( ) = kg 9.80 m/ + 0 m/ 1. N = b. When the elevator accelerate upward, a = +1.0 m/. Then, ( ) ( ) f = µ = µ mg + ma = µ m g + a k k N k k ( )( ) ( ) ( ) = kg 9.80 m/ m/ 3.8 N =

31 Chapter 4 Problem 185 c. When the elevator accelerate downward, a = 1.0 m/. Then, ( ) ( ) f = µ = µ mg + ma = µ m g + a k k N k k ( )( ) ( ) ( ) = kg 9.80 m/ m/ 18.6 N = 41. REASONING The magnitude of the kinetic frictional force i given b Equation 4.8 a the coefficient of kinetic friction time the magnitude of the normal force. Since the lide into econd bae i horizontal, the normal force i vertical. It can be evaluated b noting that there i no acceleration in the vertical direction and, therefore, the normal force mut balance the weight. To find the plaer initial velocit v 0, we will ue kinematic. The time interval for the lide into econd bae i given a t = 1.6. Since the plaer come to ret at the end of the lide, hi final velocit i v = 0 m/. The plaer acceleration a can be obtained from Newton econd law, ince the net force i the kinetic frictional force, which i known from part (a), and the ma i given. Since t, v, and a are known and we eek v 0, the appropriate kinematic equation i Equation.4 (v = v 0 + at). SOLUTION a. Since the normal force balance the weight mg, we know that = mg. Uing thi fact and Equation 4.8, we find that the magnitude of the kinetic frictional force i ( )( )( ) fk = µ kn = µ kmg = kg 9.8 m/ = 390 N b. Solving Equation.4 (v = v 0 + at) for v 0 give v 0 = v - at. Taking the direction of the plaer lide to be the poitive direction, we ue Newton econd law and Equation 4.8 for the kinetic frictional force to write the acceleration a a follow: Σ µ kmg a = = = µ k g m m The acceleration i negative, becaue it point oppoite to the plaer velocit, ince the plaer low down during the lide. Thu, we find for the initial velocit that ( µ ) ( )( ) ( ) v0 = v kg t = 0 m/ m/ 1.6 = m/

32 186 ORCES AND NEWTON'S LAWS O MOTION 4. REASONING AND SOLUTION The deceleration produced b the frictional force i fk µ kmg a = = = µ k g m m The peed of the automobile after 1.30 have elaped i given b Equation.4 a ( ) ( )( µ )( ) v= v0 + at = v0 + kg t = 16.1 m/ m/ 1.30 = 6.9 m/ 43. SSM REASONING The free-bod diagram for the large cube (ma = M) and the mall cube (ma = m) are hown in the following drawing. In the cae of the large cube, we have omitted the weight and the normal force from the urface, ince the pla no role in the olution (although the do balance). N f P mg N In thee diagram, note that the two block eert a normal force on each other; the large block eert the force on the maller block, while the maller block eert the force on the larger block. In accord with Newton third law thee force have oppoite direction and equal magnitude. Under the influence of the force hown, the two block have the ame acceleration a. We begin our olution b appling Newton econd law to each one. SOLUTION According to Newton econd law, we have Σ = P N = Ma N = ma Large block Small block Subtituting = ma into the large-block epreion and olving for P give P = (M + m) a or the maller block to remain in place againt the larger block, the tatic frictional force mut balance the weight of the maller block, o that f = mg. But f i given b f = µ, where, from the Newton econd law, we know that = ma. Thu, we have µ ma = mg or a = g/µ. Uing thi reult in the epreion for P give