Chapter 6, Conceptual Questions

Transcription

1 Chapter 6, Conceptual Questions 6.1. (a) Static equilibrium. The barbell is not accelerating and has a velocit of zero. (b) Dnamic equilibrium. The girder is not accelerating but has a nonzero constant velocit. (c) Not in equilibrium. Slowing down means the acceleration is not zero. (d) Dnamic equilibrium. The plane is not accelerating but has a nonzero constant velocit. (e) Not in equilibrium. The bo slows down with the truc, so has a nonzero acceleration. 6.. No. The ball is still changing its speed, and just momentaril has zero velocit No, because the force is not necessaril in the same direction as the motion. For eample, a car using its braes to slow its forward motion has a force opposite its direction of motion Equal. The tension in the cable is equal to the force of gravit, since the force must be zero in order for the elevator to move with constant speed Greater. Since the elevator is slowing down as it moves downward, it has an upward force, so the tension must be greater than the gravitational force The ball filled with lead is more massive. Since the balls are weightless, the astronaut must measure their inertia (mass) directl. One eas option is to move each ball side to side in turn. More force is required to change the more massive lead-filled ball s direction of motion Larger. A free-bod diagram for the boo is shown in the figure. The normal force of the table on the boo is larger than its weight, since the force is zero. 30

2 6.19. Yes, the friction force on a crate dropped on a conveor belt speeds the crate up to the belt s speed North. The friction force on the crate is the onl horizontal force and is responsible for speeding the crate up along with the truc. Therefore the friction force points in the same direction as the motion of the crate. 31

3 Chapter 6, Eercises and Problems 6.1. Model: We can assume that the ring is a single massless particle in static equilibrium. Solve: Written in component form, Newton s first law is F F T1 T T3 0 N = Σ = + + = Evaluating the components of the force vectors from the free-bod diagram: Using Newton s first law: Rearranging: T T1 = T1 T = 0 N T cos30 3 = T3 T 1 = 0 N T = T T sin30 3 = T3 F = Σ F = T1 + T + T3 = 0 N T + T cos30 = 0 N T T sin30 = 0 N = T cos30 = ( 100 N)( ) = 86.7 N T T 1 3 = sin30 = 100 N 0.5 = 50.0 N 3 Assess: Since T 3 acts closer to the -ais than to the -ais, it maes sense that T > T Model: We can assume that the ring is a particle. This is a static equilibrium problem. We will ignore the weight of the ring, because it is ver light, so the onl three forces are the tension forces shown in the free-bod diagram. Note that the diagram defines the angle θ. Solve: Because the ring is in equilibrium it must obe F = 0 N. This is a vector equation, so it has both - and - components: F = T cosθ T = 0 N T3 cosθ = T 3 F = T T sinθ = 0 N T sinθ = T We have two equations in the two unnowns T 3 and θ. Divide the -equation b the -equation: Now we can use the -equation to find T3 sinθ T1 80 N = tanθ = = = 1.6 θ = tan 1 ( 1.6) = 58 T cosθ T 50 N 3 3

4 T cosθ 50 N 94 N cos58 3 = = = The tension in the third rope is 94 N directed 58 below the horizontal. T 6.5. Please refer to the Figure EX6.5. Solve: Appling Newton s second law to the diagram on the left, For the diagram on the right: ( F ) 4 N N a = = = 1.0 m/s m g ( F ) 4 N N a = = = 1.0 m/s m g ( F ) 3 N 3 N a = = = 0 m/s m g ( F ) 6.7. Please refer to Figure EX6.7. Solve: (a) Appl Newton s second law in both the and directions. 3 N 1 N N a = = = 0 m/s m g ( F ) = = 5.0 N cos37.0 N 5.0 g a = 0.40 m/s a ( F ) = + =.0 N 5.0 N sin N 5.0 g a = 0.0 m/s a (b) The angle that the 5.0 N force maes with the -ais is 37. Appl Newton s second law for both the and direction. ( F ) = + = 3.0 N 5.0 N sin37.0 N 5.0 g a = 0.80 m/s a ( F ) = = 4.0 N 5.0 N cos g a = 0.0 m/s a Assess: The orientation of the coordinate aes is chosen for convenience, and does not alwas need to conform to the horizontal and vertical Model: We assume the rocet is a particle moving in a vertical straight line under the influence of onl two forces: gravit and its own thrust. Solve: (a) Using Newton s second law and reading the forces from the free-bod diagram, Fthrust FG ma Fthrust ma mgearth = = + = 0.00 g 10 m/s m/s = 3.96 N (b) Liewise, the thrust on the moon is (0.00 g)(10 m/s m/s ) =.3 N. Assess: The thrust required is smaller on the moon, as it should be, given the moon s weaer gravitational pull. The magnitude of a few newtons seems reasonable for a small model rocet Model: We assume that the safe is a particle moving onl in the -direction. Since it is sliding during the entire problem, we can use the model of iic friction. 33

5 Solve: The safe is in equilibrium, since it s not accelerating. Thus we can appl Newton s first law in the vertical and horizontal directions: Then, for iic friction: F = Σ F = F + F f = 0 N f = F + F = 350 N N = 735 N B C B C ( F ) F n FG n FG mg = Σ = = = = = = 3 0 N 300 g 9.80 m/s N f 735 N f = µ n µ = = = n N Assess: The value of µ = 0.50 is hard to evaluate without nowing the material the floor is made of, but it seems reasonable Model: We will represent the crate as a particle. Solve: (a) When the belt runs at constant speed, the crate has an acceleration a = 0 m/s and is in dnamic equilibrium. Thus F = 0. It is tempting to thin that the belt eerts a friction force on the crate. But if it did, there would be a force because there are no other possible horizontal forces to balance a friction force. Because there is no force, there cannot be a friction force. The onl forces are the upward normal force and the gravitational force on the crate. (A friction force would have been needed to get the crate moving initiall, but no horizontal force is needed to eep it moving once it is moving with the same constant speed as the belt.) (b) If the belt accelerates gentl, the crate speeds up without slipping on the belt. Because it is accelerating, the crate must have a horizontal force. So now there is a friction force, and the force points in the direction of the crate s motion. Is it static friction or iic friction? Although the crate is moving, there is no motion of the crate relative to the belt. Thus, it is a static friction force that accelerates the crate so that it moves without slipping on the belt. (c) The static friction force has a maimum possible value ( fs) ma = µ sn. The maimum possible acceleration of the crate is ( f ) s ma µ sn ama = = m m If the belt accelerates more rapidl than this, the crate will not be able to eep up and will slip. It is clear from the freebod diagram that n = FG = mg. Thus, ama = µ s g = (0.5)(9.80 m/s ) = 4.9 m/s 6.0. Model: We assume that the truc is a particle in equilibrium, and use the model of static friction. 34

6 Solve: The truc is not accelerating, so it is in equilibrium, and we can appl Newton s first law. The normal force has no component in the -direction, so we can ignore it here. For the other two forces: ( F ) F f ( F ) f ( F ) mg θ = Σ = = 0 N = = sin = 4000 g 9.80 m/s sin15 = 10,145 N s G s G Assess: The truc s weight (mg) is roughl 40,000 N. A friction force that is 5% of the truc s weight seems reasonable Model: The car is a particle subject to Newton s laws and inematics. Solve: Kiic friction provides a horizontal acceleration which stops the car. From the figure, appling Newton s first and second laws gives Combining these two equations with f = µ n ields Kinematics can be used to determine the initial velocit. a F = f = ma F = n F = 0 n = F = mg G G = µ g = m/s = 4.9 m/s vf = vi + a vi = a v i = 4.9 m/s 65 m 0 m = 5 m/s Thus is a reasonable speed to have initiall for a vehicle to leave 65-meter- Assess: The initial speed of long sid mars. 5 m/s 56 mph 6.5. Model: We assume that the sdiver is shaped lie a bo and is a particle. 35

7 The sdiver falls straight down toward the earth s surface, that is, the direction of fall is vertical. Since the sdiver falls feet first, the surface perpendicular to the drag has the cross-sectional area A = 0 cm 40 cm. The phsical conditions needed to use Equation 6.16 for the drag force are satisfied. The terminal speed corresponds to the situation when the force acting on the sdiver becomes zero. Solve: The epression for the magnitude of the drag with v in m/s is 1 D Av = 0.5( ) v N = 0.00 v N 4 The gravitational force on the sdiver is F mg G = = 75 g 9.8 m/s = 735 N. The mathematical form of the condition defining dnamical equilibrium for the sdiver and the terminal speed is F = FG + D = 0 N vterm N 735 N = 0 N vterm = 19 m/s 0.0 Assess: The result of the above simplified phsical modeling approach and subsequent calculation, even if approimate, shows that the terminal velocit is ver high. This result implies that the sdiver will be ver badl hurt at landing if the parachute does not open in time We used the force-versus-time graph to draw the acceleration-versus-time graph. The pea acceleration was calculated as follows: a F m 10 N m/s 5 g ma ma = = = Solve: The acceleration is not constant, so we cannot use constant acceleration inematics. Instead, we use the more general result that v( t) = v + area under the acceleration curve from 0 s to t 0 The object starts from rest, so v 0 = 0 m/s. The area under the acceleration curve between 0 s and 6 s is 1 (4 s) ( m/s ) = 4.0 m/s. We ve used the fact that the area between 4 s and 6 s is zero. Thus, at t = 6 s, v = 4.0 m/s. 36

8 6.9. Model: You can model the beam as a particle in static equilibrium. Solve: Using Newton s first law, the equilibrium equations in vector and component form are: F = T + T + F = 0 N Using the free-bod diagram ields: 1 G F = T + T + F = 0 N 1 G F = T + T + F = 0 N 1 G T sinθ + T sinθ = 0 N T cosθ + T cosθ F = 0 N G The mathematical model is reduced to a simple algebraic sstem of two equations with two unnowns, T 1 and T. Substituting θ1 = 0, θ = 30, and FG = mg = 9800 N, the simultaneous equations become T sin 0 + T sin 30 = 0 N T cos 0 + T cos30 = 9800 N 1 1 You can solve this sstem of equations b simple substitution. The result is T1 = 6397 N and T = 4376 N. Assess: The above approach and result seem reasonable. Intuition indicates there is more tension in the left rope than in the right rope Model: Represent the rocet as a particle that follows Newton s second law. Solve: (a) The -component of Newton s second law is 5 F Fthrust mg N a = a = = = 9.80 m/s = 5. m/s m m 0,000 g (b) At 5000 m the acceleration has increased because the rocet mass has decreased. Solving the equation of part (a) for m gives m 5 Fthrust N 5000 m a5000 m + g 6.0 m/s m/s = = = g 37

9 The mass of fuel burned is m m m 3 fuel = initial 5000 = g. m Model: We will model the bo as a particle, and use the models of iic and static friction. The pushing force is along the +-ais, but the force of friction acts along the -ais. A component of the gravitational force on the bo acts along the -ais as well. The bo will move up if the pushing force is at least equal to the sum of the friction force and the component of the gravitational force in the -direction. Solve: Let s determine how much pushing force ou would need to eep the bo moving up the ramp at stead speed. Newton s second law for the bo in dnamic equilibrium is ( F ) = Σ F = n + F + ( f ) + ( F ) = 0 N mg sinθ f + F = 0 N G push push ( F ) = Σ F = n + F + ( f ) + ( F ) = n mg cosθ + 0 N + 0 N = 0 N G push The -component equation and the model of iic friction ield: F = mg sin θ + f = mg sinθ + µ n push Let us obtain n from the -component equation as n = mg cos θ, and substitute it in the above equation to get F = mg sin θ + µ mg cos θ = mg(sin θ + µ cos θ ) push = (100 g)(9.80 m/s )(sin cos0 ) = 888 N The force is less than our maimum pushing force of 1000 N. That is, once in motion, the bo could be ept moving up the ramp. However, if ou stop on the ramp and want to start the bo from rest, the model of static friction applies. The analsis is the same ecept that the coefficient of static friction is used and we use the maimum value of the force of static friction. Therefore, we have F = mg θ + µ θ = + = push (sin s cos ) (100 g)(9.80 m/s )(sin cos 0 ) 1160 N Since ou can push with a force of onl 1000 N, ou can t get the bo started. The big static friction force and the weight are too much to overcome Model: We will model the sled and friend as a particle, and use the model of iic friction because the sled is in motion. The force on the sled is zero (note the constant speed of the sled). That means the component of the pulling force along the +-direction is equal to the magnitude of the iic force of friction in the -direction. Also note that ( F ) = 0 N, since the sled is not moving along the -ais. 38

10 Solve: Newton s second law is ( F ) = Σ F = n + F + ( f ) + ( F ) = 0 N + 0 N f + F cosθ = 0 N G pull pull ( F ) = Σ F = n + F + ( f ) + ( F ) = n mg + 0 N + F sinθ = 0 N G pull pull The -component equation using the iic friction model f = µ n reduces to The -component equation gives µ n = F θ pull cos n = mg F sinθ pull We see that the normal force is smaller than the gravitational force because opposite to the direction of the gravitational force. In other words, equation, µ can now be obtained as µ F cosθ pull = = mg Fpull sinθ F pull has a component in a direction F pull is partl lifting the sled. From the -component ( 75 N)( cos30 ) ( 60 g)( 9.80 m/s ) ( 75 N)( sin30 ) = 0.1 Assess: A quic glance at the various µ values in Table 6.1 suggests that a value of 0.1 for µ is reasonable Model: We will model the shuttle as a particle and assume the elastic cord to be massless. We will also use the model of iic friction for the motion of the shuttle along the square steel rail. Solve: The upward tension component T = T sin 45 = 14.1 N is larger than the gravitational force on the shuttle. Consequentl, the elastic cord pulls the shuttle up against the rail and the rail s normal force pushes downward. Newton s second law in component form is ( F ) = Σ F = T + ( f ) + ( n) + F = T cos45 f + 0 N + 0 N = ma = ma G ( F ) = Σ F = T + ( f ) + ( n) + F = T sin N n mg = ma = 0 N G The model of iic friction is f = µ n. We use the -component equation to get an epression for n and hence f. Substituting into the -component equation and using the value of µ in Table 6.1 gives us T cos45 µ T sin 45 mg a = m ( 0 N) cos45 ( 0.60) + ( 0 N) sin 45 ( g)( 9.80 m/s ) = = 13.0 m/s g Assess: The -component of the tension force is 14.1 N. On the other hand, the force on the shuttle in the - direction is ma = (0.800 g)(13.0 m/s ) = 10.4 N. This value for ma is reasonable since a part of the 14.1 N tension force is used up to overcome the force of iic friction Model: Assume the ball is a particle on a slope, and that the slope increases as the -displacement increases. Assume that there is no friction and that the ball is being accelerated to the right so that it remains at rest on the slope. 39

11 Although the ball is on a slope, it is accelerating to the right. Thus we ll use a coordinate sstem with horizontal and vertical aes. Solve: Newton s second law is Combining the two equations, we get The curve is described b (b) The acceleration at Σ F = nsin θ = ma Σ F = ncosθ F = ma = 0 N G F ma G = sinθ = mg tanθ a = g tanθ cosθ =. Its slope a position is tanθ, which is also the derivative of the curve. Hence, = 0.0 m is d = tanθ = a = ( ) g d a = = ()(0.0)(9.8 m/s ) 3.9 m/s. 40