* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Document
Routhian mechanics wikipedia , lookup
Density of states wikipedia , lookup
Jerk (physics) wikipedia , lookup
Classical mechanics wikipedia , lookup
Eigenstate thermalization hypothesis wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Fictitious force wikipedia , lookup
Center of mass wikipedia , lookup
Photon polarization wikipedia , lookup
Wave packet wikipedia , lookup
Equations of motion wikipedia , lookup
Spinodal decomposition wikipedia , lookup
Mass versus weight wikipedia , lookup
Work (thermodynamics) wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Matter wave wikipedia , lookup
Hunting oscillation wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Chap 12 11. The forces on the ladder are shown in the diagram below. F1 is the force of the window, horizontal because the window is frictionless. F2 and F3 are components of the force of the ground on the ladder. M is the mass of the window cleaner and m is the mass of the ladder. The force of gravity on the man acts at a point 3.0 m up the ladder and the force of gravity on the ladder acts at the center of the ladder. Let be the angle between the ladder and the ground. We use cos d / L or sin L2 d 2 /L to find = 60º. Here L is the length of the ladder (5.0 m) and d is the distance from the wall to the foot of the ladder (2.5 m). (a) Since the ladder is in equilibrium the sum of the torques about its foot (or any other point) vanishes. Let be the distance from the foot of the ladder to the position of the window cleaner. Then, Mg cos mg L / 2 cos F1L sin 0 , and F1 ( M mL / 2) g cos [(75kg) (3.0 m)+(10 kg) (2.5m)](9.8m/s 2 ) cos 60 L sin (5.0 m)sin 60 2.8 102 N. This force is outward, away from the wall. The force of the ladder on the window has the same magnitude but is in the opposite direction: it is approximately 280 N, inward. (b) The sum of the horizontal forces and the sum of the vertical forces also vanish: F1 F3 0 F2 Mg mg 0 The first of these equations gives F3 F1 2.8 102 N and the second gives F2 (M m) g (75kg 10 kg) (9.8m/s 2 ) 8.3 102 N The magnitude of the force of the ground on the ladder is given by the square root of the sum of the squares of its components: F F2 F3 ( 2.8 102 N)2 (8.3 102 N)2 8.8 102 N. 2 2 (c) The angle between the force and the horizontal is given by tan = F3/F2 = 830/280 = 2.94, so = 71º. The force points to the left and upward, 71º above the horizontal. We note that this force is not directed along the ladder. 21. The beam is in equilibrium: the sum of the forces and the sum of the torques acting on it each vanish. As we see in the figure, the beam makes an angle of 60º with the vertical and the wire makes an angle of 30º with the vertical. (a) We calculate the torques around the hinge. Their sum is TL sin 30º – W(L/2) sin 60º = 0. Here W is the force of gravity acting at the center of the beam, and T is the tension force of the wire. We solve for the tension: T= W sin60 222N sin 60 = = 192 N. 2 sin30 2 sin 30 (b) Let Fh be the horizontal component of the force exerted by the hinge and take it to be positive if the force is outward from the wall. Then, the vanishing of the horizontal component of the net force on the beam yields Fh – T sin 30º = 0 or Fh = T sin30 = 192.3N sin 30 = 96.1N. (c) Let Fv be the vertical component of the force exerted by the hinge and take it to be positive if it is upward. Then, the vanishing of the vertical component of the net force on the beam yields Fv + T cos 30º – W = 0 or Fv = W T cos30 = 222 N 192.3N cos30 = 55.5 N. 31. The diagram below shows the forces acting on the plank. Since the roller is frictionless the force it exerts is normal to the plank and makes the angle with the vertical. Its magnitude is designated F. W is the force of gravity; this force acts at the center of the plank, a distance L/2 from the point where the plank touches the floor. FN is the normal force of the floor and f is the force of friction. The distance from the foot of the plank to the wall is denoted by d. This quantity is not given directly but it can be computed using d = h/tan. The equations of equilibrium are: horizontal force components vertical force components torques F sin f 0 F cos W FN 0 FN d fh W d L2 cos 0. The point of contact between the plank and the roller was used as the origin for writing the torque equation. When = 70º the plank just begins to slip and f = sFN, where s is the coefficient of static friction. We want to use the equations of equilibrium to compute FN and f for = 70º, then use s = f/FN to compute the coefficient of friction. The second equation gives F = (W – FN)/cos and this is substituted into the first to obtain f = (W – FN) sin /cos = (W – FN) tan . This is substituted into the third equation and the result is solved for FN: FN = d L/2 cos + h tan d + h tan W h(1 tan 2 ) ( L / 2)sin W, h(1 tan 2 ) where we have use d = h/tan and multiplied both numerator and denominator by tan . We use the trigonometric identity 1+ tan2 = 1/cos2 and multiply both numerator and denominator by cos2 to obtain L FN = W 1 cos 2 sin . 2h Now we use this expression for FN in f = (W – FN) tan to find the friction: f = WL 2 sin cos . 2h We substitute these expressions for f and FN into s = f/FN and obtain s = L sin 2 cos . 2h L sin cos2 Evaluating this expression for = 70º, we obtain s = 6.1m sin2 70cos70 = 0.34. 2 3.05m 6.1m sin70cos2 70 35. The diagrams below show the forces on the two sides of the ladder, separated. FA and FE are the forces of the floor on the two feet, T is the tension force of the tie rod, W is the force of the man (equal to his weight), Fh is the horizontal component of the force exerted by one side of the ladder on the other, and Fv is the vertical component of that force. Note that the forces exerted by the floor are normal to the floor since the floor is frictionless. Also note that the force of the left side on the right and the force of the right side on the left are equal in magnitude and opposite in direction. Since the ladder is in equilibrium, the vertical components of the forces on the left side of the ladder must sum to zero: Fv + FA – W = 0. The horizontal components must sum to zero: T – Fh = 0. The torques must also sum to zero. We take the origin to be at the hinge and let L be the length of a ladder side. Then FAL cos – W(L/4) cos – T(L/2) sin = 0. Here we recognize that the man is one–fourth the length of the ladder side from the top and the tie rod is at the midpoint of the side. The analogous equations for the right side are FE – Fv = 0, Fh – T = 0, and FEL cos – T(L/2) sin = 0. There are 5 different equations: Fv FA W 0, T Fh 0 FA L cos W ( L / 4) cos T ( L / 2)sin 0 FE Fv 0 FE L cos T ( L / 2)sin 0. The unknown quantities are FA, FE, Fv, Fh, and T. (a) First we solve for T by systematically eliminating the other unknowns. The first equation gives FA = W – Fv and the fourth gives Fv = FE. We use these to substitute into the remaining three equations to obtain T Fh 0 WL cos FE L cos W ( L / 4) cos T ( L / 2)sin 0 FE L cos T ( L / 2)sin 0. The last of these gives FE = Tsin/2cos = (T/2) tan. We substitute this expression into the second equation and solve for T. The result is T 3W . 4 tan To find tan, we consider the right triangle formed by the upper half of one side of the ladder, half the tie rod, and the vertical line from the hinge to the tie rod. The lower side of the triangle has a length of 0.381 m, the hypotenuse has a length of 1.22 m, and the vertical side has a length of . mg b 0.381 mg 116 . m . This means b122 2 2 tan = (1.16m)/(0.381m) = 3.04. Thus, T 3(854 N) 211N. 4(3.04) (b) We now solve for FA. Since Fv = FE and FE = T sin2cos, Fv = 3W/8. We substitute this into Fv + FA – W = 0 and solve for FA. We find FA W Fv W 3W / 8 5W / 8 5(884 N)/8=534 N. (c) We have already obtained an expression for FE: FE = 3W/8. Evaluating it, we get FE = 320 N. 36. (a) The Young’s modulus is given by stress 150 106 N/m2 E slope of the stress-strain curve 7.5 1010 N/m 2 . strain 0.002 (b) Since the linear range of the curve extends to about 2.9 × 108 N/m2, this is approximately the yield strength for the material. 39. (a) Let FA and FB be the forces exerted by the wires on the log and let m be the mass of the log. Since the log is in equilibrium FA + FB – mg = 0. Information given about the stretching of the wires allows us to find a relationship between FA and FB. If wire A originally had a length LA and stretches by L A , then LA FA LA / AE , where A is the cross–sectional area of the wire and E is Young’s modulus for steel (200 × 109 N/m2). Similarly, LB FB LB / AE . If is the amount by which B was originally longer than A then, since they have the same length after the log is attached, LA LB . This means FA LA FB LB . AE AE We solve for FB: FB FA LA AE . LB LB We substitute into FA + FB – mg = 0 and obtain FA mgLB AE . LA LB c h 2 The cross–sectional area of a wire is A r 2 120 . 103 m 4.52 106 m2 . Both LA and LB may be taken to be 2.50 m without loss of significance. Thus FA (103kg) (9.8 m/s2 ) (2.50 m)+(4.52 106 m2 ) (200 109 N/m2 ) (2.0 103 m) 2.50 m+2.50 m 866 N. (b) From the condition FA + FB – mg = 0, we obtain FB mg FA (103kg) (9.8m/s2 ) 866 N=143 N. (c) The net torque must also vanish. We place the origin on the surface of the log at a point directly above the center of mass. The force of gravity does not exert a torque about this point. Then, the torque equation becomes FAdA – FBdB = 0, which leads to dA FB 143 N 0.165. dB FA 866 N 40. (a) Since the brick is now horizontal and the cylinders were initially the same length , then both have been compressed an equal amount . Thus, FA AA E A and F B AB EB which leads to FA AA E A ( 2 AB )( 2 EB ) 4. FB AB EB AB EB When we combine this ratio with the equation FA + FB = W, we find FA/W = 4/5 = 0.80 . (b) This also leads to the result FB/W = 1/5 = 0.20. (c) Computing torques about the center of mass, we find FAdA = FBdB which leads to d A FB 1 0.25. d B FA 4 Chap 13 6. The gravitational forces on m5 from the two 5.00g masses m1 and m4 cancel each other. Contributions to the net force on m5 come from the remaining two masses: Fnet 6.67 10 = 11 N m 2 /kg 2 2.50 103 kg 3.00 10 3 kg 1.00 10 3 kg 2 101 m 2 = 1.67 1014 N. The force is directed along the diagonal between m2 and m3, towards m2. In unit-vector notation, we have Fnet Fnet (cos 45ˆi sin 45ˆj) (1.18 1014 N) ˆi (1.18 1014 N) ˆj 15. The acceleration due to gravity is given by ag = GM/r2, where M is the mass of Earth and r is the distance from Earth’s center. We substitute r = R + h, where R is the radius of Earth and h is the altitude, to obtain ag = GM /(R + h)2. We solve for h and obtain h GM / ag R . According to Appendix C, R = 6.37 106 m and M = 5.98 1024 kg, so h 6.67 10 11 m3 / s 2 kg 5.98 1024 kg 4.9m / s 2 6.37 106 m 2.6 106 m. 21. Using the fact that the volume of a sphere is 4 R3/3, we find the density of the sphere: M total 1.0 104 kg 2.4 103 kg/m3 . 3 3 4 4 3R 3 1.0 m When the particle of mass m (upon which the sphere, or parts of it, are exerting a gravitational force) is at radius r (measured from the center of the sphere), then whatever mass M is at a radius less than r must contribute to the magnitude of that force (GMm/r2). (a) At r = 1.5 m, all of Mtotal is at a smaller radius and thus all contributes to the force: Fon m GmM total m 3.0 107 N/kg . r2 (b) At r = 0.50 m, the portion of the sphere at radius smaller than that is 4 M = r 3 = 1.3 103 kg. 3 Thus, the force on m has magnitude GMm/r2 = m (3.3 107 N/kg). (c) Pursuing the calculation of part (b) algebraically, we find Fon m Gm 43 r 3 r 2 N mr 6.7 107 . kg m 31. (a) The work done by you in moving the sphere of mass mB equals the change in the potential energy of the three-sphere system. The initial potential energy is Ui GmA mB GmA mC GmB mC d L Ld and the final potential energy is Uf GmA mB GmAmC GmB mC . Ld L d The work done is 1 1 1 1 W U f U i GmB mA mC L d d d Ld 1 1 (6.67 1011 m3 / s 2 kg) (0.010 kg) (0.080 kg) 0.040 m 0.080 m 1 1 (0.020 kg) 0.080 m 0.040 m 5.0 1013 J. (b) The work done by the force of gravity is (Uf Ui) = 5.0 1013 J. 32. Energy conservation for this situation may be expressed as follows: K1 U1 K 2 U 2 K1 GmM GmM K2 r1 r2 where M = 5.0 1023 kg, r1 = R = 3.0 106 m and m = 10 kg. (a) If K1 = 5.0 107 J and r2 = 4.0 106 m, then the above equation leads to 1 1 K 2 K1 GmM 2.2 107 J. r2 r1 (b) In this case, we require K2 = 0 and r2 = 8.0 106 m, and solve for K1: 1 1 K1 K2 GmM 6.9 107 J. r1 r2 33. (a) We use the principle of conservation of energy. Initially the particle is at the surface of the asteroid and has potential energy Ui = GMm/R, where M is the mass of the asteroid, R is its radius, and m is the mass of the particle being fired upward. The initial kinetic energy is 1 2 mv 2 . The particle just escapes if its kinetic energy is zero when it is infinitely far from the asteroid. The final potential and kinetic energies are both zero. Conservation of energy yields GMm/R + ½mv2 = 0. We replace GM/R with agR, where ag is the acceleration due to gravity at the surface. Then, the energy equation becomes agR + ½v2 = 0. We solve for v: v 2ag R 2(3.0 m/s2 ) (500 103 m) 1.7 103 m/s. (b) Initially the particle is at the surface; the potential energy is Ui = GMm/R and the kinetic energy is Ki = ½mv2. Suppose the particle is a distance h above the surface when it momentarily comes to rest. The final potential energy is Uf = GMm/(R + h) and the final kinetic energy is Kf = 0. Conservation of energy yields GMm 1 2 GMm mv . R 2 Rh We replace GM with agR2 and cancel m in the energy equation to obtain ag R 2 1 2 ag R v . 2 ( R h) The solution for h is h 2ag R 2 2ag R v 2 R 2(3.0 m/s 2 ) (500 103 m) 2 (500 103 m) 2 3 2 2(3.0 m/s ) (500 10 m) (1000 m/s) 2.5 105 m. (c) Initially the particle is a distance h above the surface and is at rest. Its potential energy is Ui = GMm/(R + h) and its initial kinetic energy is Ki = 0. Just before it hits the asteroid its potential energy is Uf = GMm/R. Write 1 2 mv2f for the final kinetic energy. Conservation of energy yields GMm GMm 1 2 mv . Rh R 2 We substitute agR2 for GM and cancel m, obtaining ag R 2 Rh ag R 1 2 v . 2 The solution for v is v 2a g R 2a g R 2 Rh 2(3.0 m/s 2 ) (500 103 m) 2(3.0 m/s 2 )(500 103 m) 2 (500 103 m) + (1000 103 m) 1.4 103 m/s. 45. (a) The greatest distance between the satellite and Earth’s center (the apogee distance) is Ra = (6.37 106 m + 360 103 m) = 6.73 106 m. The least distance (perigee distance) is Rp = (6.37 106 m + 180 103 m) = 6.55 106 m. Here 6.37 106 m is the radius of Earth. From Fig. 13-14, we see that the semi-major axis is a Ra Rp 2 6.73 106 m + 6.55 106 m 6.64 106 m. 2 (b) The apogee and perigee distances are related to the eccentricity e by Ra = a(1 + e) and Rp = a(1 e). Add to obtain Ra + Rp = 2a and a = (Ra + Rp)/2. Subtract to obtain Ra Rp = 2ae. Thus, e Ra R p 2a Ra R p Ra R p 6.73 106 m 6.55 106 m 0.0136. 6.73 106 m 6.55 106 m 51. In our system, we have m1 = m2 = M (the mass of our Sun, 1.99 1030 kg). With r = 2r1 in this system (so r1 is one-half the Earth-to-Sun distance r), and v = r/T for the speed, we have r T T Gm1m2 m1 2 r r 2 2 2 2 r 3 . GM With r = 1.5 1011 m, we obtain T = 2.2 107 s. We can express this in terms of Earth-years, by setting up a ratio: 2.2 107 s T T (1y) = 1 y 0.71 y. 7 1y 3.156 10 s 56. Although altitudes are given, it is the orbital radii which enter the equations. Thus, rA = (6370 + 6370) km = 12740 km, and rB = (19110 + 6370) km = 25480 km (a) The ratio of potential energies is GmM UB r 1 rB A . U A GmM rB 2 rA (b) Using Eq. 13-38, the ratio of kinetic energies is GmM KB r 1 2rB A . K A GmM rB 2 2rA (c) From Eq. 13-40, it is clear that the satellite with the largest value of r has the smallest value of |E| (since r is in the denominator). And since the values of E are negative, then the smallest value of |E| corresponds to the largest energy E. Thus, satellite B has the largest energy. (d) The difference is E EB EA GmM 1 1 . 2 rB rA Being careful to convert the r values to meters, we obtain E = 1.1 108 J. The mass M of Earth is found in Appendix C. 60. (a) The pellets will have the same speed v but opposite direction of motion, so the relative speed between the pellets and satellite is 2v. Replacing v with 2v in Eq. 13-38 is equivalent to multiplying it by a factor of 4. Thus, 11 3 2 24 GM E m 2(6.67 10 m / kg s ) 5.98 10 kg 0.0040 kg K rel 4 4.6 105 J. 3 (6370 500) 10 m 2r (b) We set up the ratio of kinetic energies: K rel K bullet Chap 14 4.6 105 J 1 2 0.0040 kg 950 m/s 2 2.6 102. 4. The magnitude F of the force required to pull the lid off is F = (po – pi)A, where po is the pressure outside the box, pi is the pressure inside, and A is the area of the lid. Recalling that 1N/m2 = 1 Pa, we obtain pi po F 480 N 1.0 105 Pa 3.8 104 Pa. A 77 104 m 2 18. (a) The force on face A of area AA due to the water pressure alone is FA p A AA w ghA AA w g (2d )d 2 2 1.0 103 kg m3 9.8 m s 2 5.0 m 3 2.5 106 N. Adding the contribution from the atmospheric pressure, F0= (1.0 105 Pa)(5.0 m)2 = 2.5 106 N, we have FA ' F0 FA 2.5 106 N 2.5 106 N 5.0 106 N. (b) The force on face B due to water pressure alone is 5 5 3 5d FB pavgB AB g d 2 w gd 3 1.0 103 kg m3 9.8 m s 2 5.0 m 2 2 2 3.1106 N. Adding the contribution from the atmospheric pressure, F0= (1.0 105 Pa)(5.0 m)2 = 2.5 106 N, we have FB ' F0 FB 2.5 106 N 3.1106 N 5.6 106 N. 19. (a) At depth y the gauge pressure of the water is p = gy, where is the density of the water. We consider a horizontal strip of width W at depth y, with (vertical) thickness dy, across the dam. Its area is dA = W dy and the force it exerts on the dam is dF = p dA = gyW dy. The total force of the water on the dam is F D 0 gyW dy 1 gWD 2 2 1 2 1.00 103 kg m3 9.80 m s 2 314 m 35.0 m 1.88 109 N. 2 (b) Again we consider the strip of water at depth y. Its moment arm for the torque it exerts about O is D – y so the torque it exerts is d = dF(D – y) = gyW (D – y)dy and the total torque of the water is D 0 1 1 1 gyW D y dy gW D 3 D 3 gWD3 2 3 6 1 3 1.00 103 kg m3 9.80 m s 2 314 m 35.0 m 2.20 1010 N m. 6 (c) We write = rF, where r is the effective moment arm. Then, r F 1 6 1 2 gWD3 D 35.0 m 11.7 m. gWD 2 3 3 23. Eq. 14-13 combined with Eq. 5-8 and Eq. 7-21 (in absolute value) gives mg = kx A1 . A2 With A2 = 18A1 (and the other values given in the problem) we find m = 8.50 kg. 25. (a) The anchor is completely submerged in water of density w. Its effective weight is Weff = W – w gV, where W is its actual weight (mg). Thus, V W Weff 200 N 2.04 102 m3 . 3 2 w g 1000 kg/m 9.8 m/s (b) The mass of the anchor is m = V, where is the density of iron (found in Table 14-1). Its weight in air is W mg Vg 7870 kg/m3 (2.04 102 m3 ) 9.80 m/s2 1.57 103 N . 32. (a) An object of the same density as the surrounding liquid (in which case the “object” could just be a packet of the liquid itself) is not going to accelerate up or down (and thus won’t gain any kinetic energy). Thus, the point corresponding to zero K in the graph must correspond to the case where the density of the object equals liquid. Therefore, ball = 1.5 g/cm3 (or 1500 kg/m3). (b) Consider the liquid = 0 point (where Kgained = 1.6 J). In this case, the ball is falling through perfect vacuum, so that v2 = 2gh (see Eq. 2-16) which means that K = 1 2 mv2 = 1.6 J can be used to solve for the mass. We obtain mball = 4.082 kg. The volume of the ball is then given by mball/ball = 2.72 103 m3. 35. The volume Vcav of the cavities is the difference between the volume Vcast of the casting as a whole and the volume Viron contained: Vcav = Vcast – Viron. The volume of the iron is given by Viron = W/giron, where W is the weight of the casting and iron is the density of iron. The effective weight in water (of density w) is Weff = W – gw Vcast. Thus, Vcast = (W – Weff)/gw and Vcav W Weff W 6000 N 4000 N 6000 N 2 3 2 g w g iron (9.8 m/s ) (1000 kg/m ) (9.8 m/s ) (7.87 103 kg/m3 ) 0.126 m3 . 38. (a) Since the lead is not displacing any water (of density w), the lead’s volume is not contributing to the buoyant force Fb. If the immersed volume of wood is Vi, then m Fb wVi g 0.900 wVwood g 0.900 w g wood , wood which, when floating, equals the weights of the wood and lead: m Fb 0.900 w g wood (mwood mlead ) g . wood Thus, (0.900) (1000 kg/m3 ) (3.67 kg) m mlead 0.900 w wood mwood 3.67 kg 1.84 kg . 600 kg/m3 wood (b) In this case, the volume Vlead = mlead/lead also contributes to Fb. Consequently, m Fb 0.900 w g wood w mlead g (mwood mlead ) g , wood lead which leads to mlead 0.900 ( w / wood )mwood mwood 1 w / lead 1.84 kg 1 (1.00 10 kg/m3 /1.13 10 4 kg/m 3 ) 3 2.01 kg. 47. (a) The equation of continuity leads to v2 A2 v1 A1 r12 v2 v1 2 r2 which gives v2 = 3.9 m/s. (b) With h = 7.6 m and p1 = 1.7 105 Pa, Bernoulli’s equation reduces to p2 p1 gh 1 v12 v22 8.8 104 Pa. 2 55. (a) Since Sample Problem 14-8 deals with a similar situation, we use the final equation (labeled “Answer”) from it: v 2gh v v0 for the projectile motion. The stream of water emerges horizontally (0 = 0° in the notation of Chapter 4), and setting y – y0 = –(H – h) in Eq. 4-22, we obtain the “time-of-flight” t 2( H h) g 2 ( H h). g Using this in Eq. 4-21, where x0 = 0 by choice of coordinate origin, we find x v0t 2 gh 2( H h) 2 h( H h) 2 (10 cm)(40 cm 10 cm) 35 cm. g (b) The result of part (a) (which, when squared, reads x2 = 4h(H – h)) is a quadratic equation for h once x and H are specified. Two solutions for h are therefore mathematically possible, but are they both physically possible? For instance, are both solutions positive and less than H? We employ the quadratic formula: h2 Hh x2 H H 2 x2 0h 4 2 which permits us to see that both roots are physically possible, so long as x H. Labeling the larger root h1 (where the plus sign is chosen) and the smaller root as h2 (where the minus sign is chosen), then we note that their sum is simply h1 h2 H H 2 x2 H H 2 x2 H. 2 2 Thus, one root is related to the other (generically labeled h' and h) by h' = H – h. Its numerical value is h ' 40cm 10 cm 30 cm. (c) We wish to maximize the function f = x2 = 4h(H – h). We differentiate with respect to h and set equal to zero to obtain df H 4 H 8h 0 h dh 2 or h = (40 cm)/2 = 20 cm, as the depth from which an emerging stream of water will travel the maximum horizontal distance. 59. (a) The continuity equation yields Av = aV, and Bernoulli’s equation yields p 12 v 2 12 V 2 , where p = p1 – p2. The first equation gives V = (A/a)v. We use this to substitute for V in the second equation, and obtain p 12 v 2 12 A a v 2 . 2 We solve for v. The result is 2p A2 2 1 a v 2a 2 p . A2 a 2 (b) We substitute values to obtain 2(32 104 m 2 ) 2 (55 103 Pa 41 103 Pa) v 3.06 m/s. (1000 kg / m3 ) (64 104 m 2 ) 2 (32 10 4 m 2 ) 2 Consequently, the flow rate is Av (64 104 m2 ) (3.06 m/s) 2.0 102 m3 / s. Chap 15 5. (a) The motion repeats every 0.500 s so the period must be T = 0.500 s. (b) The frequency is the reciprocal of the period: f = 1/T = 1/(0.500 s) = 2.00 Hz. (c) The angular frequency is = 2f = 2(2.00 Hz) = 12.6 rad/s. (d) The angular frequency is related to the spring constant k and the mass m by k m . We solve for k: k = m2 = (0.500 kg)(12.6 rad/s)2 = 79.0 N/m. (e) Let xm be the amplitude. The maximum speed is vm = xm = (12.6 rad/s)(0.350 m) = 4.40 m/s. (f) The maximum force is exerted when the displacement is a maximum and its magnitude is given by Fm = kxm = (79.0 N/m)(0.350 m) = 27.6 N. 15. The maximum force that can be exerted by the surface must be less than sFN or else the block will not follow the surface in its motion. Here, µs is the coefficient of static friction and FN is the normal force exerted by the surface on the block. Since the block does not accelerate vertically, we know that FN = mg, where m is the mass of the block. If the block follows the table and moves in simple harmonic motion, the magnitude of the maximum force exerted on it is given by F = mam = m2xm = m(2f)2xm, where am is the magnitude of the maximum acceleration, is the angular frequency, and f is the frequency. The relationship = 2f was used to obtain the last form. We substitute F = m(2f)2xm and FN = mg into F µsFN to obtain m(2f)2xm µsmg. The largest amplitude for which the block does not slip is xm b0.50gc9.8 m / s h 0.031 m. = = b2f g b2 2.0 Hzg sg 2 2 2 A larger amplitude requires a larger force at the end points of the motion. The surface cannot supply the larger force and the block slips. 20. Both parts of this problem deal with the critical case when the maximum acceleration becomes equal to that of free fall. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is am = 2xm, where is the angular frequency; this is the expression we set equal to g = 9.8 m/s2. (a) Using Eq. 15-5 and T = 1.0 s, we have 2 I F G HT J Kx 2 gT 2 = 0.25 m. m = g xm = 4 2 (b) Since = 2f, and xm = 0.050 m is given, we find 2 f 2 xm = g f= 1 2 g 2.2 Hz. xm 22. They pass each other at time t, at x1 x2 12 xm where x1 xm cos(t 1 ) and x2 xm cos(t 2 ). From this, we conclude that cos(t 1 ) cos(t 2 ) 12 , and therefore that the phases (the arguments of the cosines) are either both equal to /3 or one is /3 while the other is –/3. Also at this instant, we have v1 = –v2≠0where v1 xm sin(t 1 ) and v2 xm sin(t 2 ). This leads to sin(t + 1) = – sin(t + 2). This leads us to conclude that the phases have opposite sign. Thus, one phase is /3 and the other phase is – /3; the wt term cancels if we take the phase difference, which is seen to be /3 – (– /3) = 2 /3. 25. To be on the verge of slipping means that the force exerted on the smaller block (at the point of maximum acceleration) is fmax = µs mg. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is am =2xm, where k / (m M) is the angular frequency (from Eq. 15-12). Therefore, using Newton’s second law, we have mam = s mg which leads to xm = 0.22 m. k xm = s g m+ M 26. We wish to find the effective spring constant for the combination of springs shown in Fig. 15-35. We do this by finding the magnitude F of the force exerted on the mass when the total elongation of the springs is x. Then keff = F/x. Suppose the left-hand spring is elongated by x and the right-hand spring is elongated by xr. The left-hand spring exerts a force of magnitude kx on the right-hand spring and the right-hand spring exerts a force of magnitude kxr on the left-hand spring. By Newton’s third law these must be equal, so x xr . The two elongations must be the same and the total elongation is twice the elongation of either spring: x 2 x . The left-hand spring exerts a force on the block and its magnitude is F kx . Thus keff kx / 2xr k / 2 . The block behaves as if it were subject to the force of a single spring, with spring constant k/2. To find the frequency of its motion replace keff in f 1 / 2 keff / m with k/2 to obtain a f f= 1 k . 2 2m With m = 0.245 kg and k = 6430 N/m, the frequency is f = 18.2 Hz. 33. (a) Eq. 15-12 (divided by 2) yields f= 1 2 k 1 1000 N / m 2.25 Hz. m 2 5.00 kg (b) With xo = 0.500 m, we have U0 12 kx02 125 J . (c) With vo = 10.0 m/s, the initial kinetic energy is K0 12 mv02 250 J . (d) Since the total energy E = Ko + Uo = 375 J is conserved, then consideration of the energy at the turning point leads to 1 2E E = kxm2 xm = 0.866 m. 2 k 39. (a) We take the angular displacement of the wheel to be = m cos(2t/T), where m is the amplitude and T is the period. We differentiate with respect to time to find the angular velocity: = –(2/T)msin(2t/T). The symbol is used for the angular velocity of the wheel so it is not confused with the angular frequency. The maximum angular velocity is m = a fa f 2 m 2 rad = = 39.5 rad / s. T 0.500 s (b) When = /2, then /m = 1/2, cos(2t/T) = 1/2, and sin 2 t T 1 cos2 2 t T 1 1 2 3 2 2 where the trigonometric identity cos2+sin2= 1 is used. Thus, = 3I Ia fF G K H2 J K= 34.2 rad / s. F I F HKH 2 2 t 2 msin = rad T T 0.500 s During another portion of the cycle its angular speed is +34.2 rad/s when its angular displacement is /2 rad. (c) The angular acceleration is d 2 2 2 m cos 2 t / T . 2 dt T T 2 2 When = /4, 2 2 = 124 rad/s , 0.500 s 4 2 or | | 124 rad/s 2 . 51. This is similar to the situation treated in Sample Problem 15-5, except that O is no longer at the end of the stick. Referring to the center of mass as C (assumed to be the geometric center of the stick), we see that the distance between O and C is h = x. The parallel axis theorem (see Eq. 15-30) leads to I= F IJ G H K 1 L2 mL2 + mh 2 = m + x2 . 12 12 And Eq. 15-29 gives I T 2 2 mgh c x h 2 cL 12 x h. L2 12 2 2 gx 2 12 gx (a) Minimizing T by graphing (or special calculator functions) is straightforward, but the standard calculus method (setting the derivative equal to zero and solving) is somewhat awkward. We pursue the calculus method but choose to work with 12gT2/2 instead of T (it should be clear that 12gT2/2 is a minimum whenever T is a minimum). d e j 0 d d 12 xi L 12 12 gT 2 2 L2 x dx 2 dx x2 which yields x L / 12 (1.85 m)/ 12 0.53 m as the value of x which should produce the smallest possible value of T. (b) With L = 1.85 m and x = 0.53 m, we obtain T = 2.1 s from the expression derived in part (a). 60. (a) From Hooke’s law, we have k= 500 kg 9.8 m/s2 10cm 4.9 102 N/cm. (b) The amplitude decreasing by 50% during one period of the motion implies e bT 2 m 1 2 . where T 2 Since the problem asks us to estimate, we let k / m . That is, we let 49000 N / m 9.9 rad / s, 500 kg so that T 0.63 s. Taking the (natural) log of both sides of the above equation, and rearranging, we find b= b gb g 2 500 2m ln2 0.69 = 1.1 103 kg / s. T 0.63 Note: if one worries about the ´ approximation, it is quite possible (though messy) to use Eq. 15-43 in its full form and solve for b. The result would be (quoting more figures than are significant) b= 2 ln 2 mk (ln 2)2 + 4 2 = 1086 kg / s which is in good agreement with the value gotten “the easy way” above. 63. With M = 1000 kg and m = 82 kg, we adapt Eq. 15-12 to this situation by writing 2 k . T M 4m If d = 4.0 m is the distance traveled (at constant car speed v) between impulses, then we may write T = v/d, in which case the above equation may be solved for the spring constant: 2 v k 2 v = k M 4m . d M 4m d 2 Before the people got out, the equilibrium compression is xi = (M + 4m)g/k, and afterward it is xf = Mg/k. Therefore, with v = 16000/3600 = 4.44 m/s, we find the rise of the car body on its suspension is F I = 0.050 m. HK 4mg 4mg d xi x f = = k M + 4m 2v 2 Chap 16 6. (a) The amplitude is ym = 6.0 cm. (b) We find from 2/ = 0.020: = 1.0×102 cm. (c) Solving 2f = = 4.0, we obtain f = 2.0 Hz. (d) The wave speed is v = f = (100 cm) (2.0 Hz) = 2.0×102 cm/s. (e) The wave propagates in the –x direction, since the argument of the trig function is kx + t instead of kx – t (as in Eq. 16-2). (f) The maximum transverse speed (found from the time derivative of y) is umax 2 fym 4.0 s 1 6.0 cm 75cm s. (g) y(3.5 cm, 0.26 s) = (6.0 cm) sin[0.020(3.5) + 4.0(0.26)] = –2.0 cm. 8. With length in centimeters and time in seconds, we have u = du dt = 225 sin (x 15t) . Squaring this and adding it to the square of 15y, we have u2 + (15y)2 so that = (225 )2 [sin2 (x 15 t) + cos2 (x 15 t)] u = (225)2 - (15y)2 = 15 152 - y2 . Therefore, where y = 12, u must be 135. Consequently, the speed there is 424 cm/s = 4.24 m/s. 19. (a) We read the amplitude from the graph. It is about 5.0 cm. (b) We read the wavelength from the graph. The curve crosses y = 0 at about x = 15 cm and again with the same slope at about x = 55 cm, so = (55 cm – 15 cm) = 40 cm = 0.40 m. (c) The wave speed is v / , where is the tension in the string and is the linear mass density of the string. Thus, v 3.6 N 12 m/s. 25 103 kg/m (d) The frequency is f = v/ = (12 m/s)/(0.40 m) = 30 Hz and the period is T = 1/f = 1/(30 Hz) = 0.033 s. (e) The maximum string speed is um = ym = 2fym = 2(30 Hz) (5.0 cm) = 940 cm/s = 9.4 m/s. (f) The angular wave number is k = 2/ = 2/(0.40 m) = 16 m–1. (g) The angular frequency is = 2f = 2(30 Hz) = 1.9×102 rad/s (h) According to the graph, the displacement at x = 0 and t = 0 is 4.0 10–2 m. The formula for the displacement gives y(0, 0) = ym sin . We wish to select so that 5.0 10–2 sin = 4.0 10–2. The solution is either 0.93 rad or 2.21 rad. In the first case the function has a positive slope at x = 0 and matches the graph. In the second case it has negative slope and does not match the graph. We select = 0.93 rad. (i) The string displacement has the form y (x, t) = ym sin(kx + t + ). A plus sign appears in the argument of the trigonometric function because the wave is moving in the negative x direction. Using the results obtained above, the expression for the displacement is y ( x, t ) 5.0 102 m sin (16 m 1 ) x (190s 1 )t 0.93 . 21. The pulses have the same speed v. Suppose one pulse starts from the left end of the wire at time t = 0. Its coordinate at time t is x1 = vt. The other pulse starts from the right end, at x = L, where L is the length of the wire, at time t = 30 ms. If this time is denoted by t0 then the coordinate of this wave at time t is x2 = L – v(t – t0). They meet when x1 = x2, or, what is the same, when vt = L – v(t – t0). We solve for the time they meet: t = (L + vt0)/2v and the coordinate of the meeting point is x = vt = (L + vt0)/2. Now, we calculate the wave speed: v L m (250 N) (10.0 m) 158m/s. 0.100 kg Here is the tension in the wire and L/m is the linear mass density of the wire. The coordinate of the meeting point is 10.0 m (158m/s) (30.0 103 s) x 7.37 m. 2 This is the distance from the left end of the wire. The distance from the right end is L – x = (10.0 m – 7.37 m ) = 2.63 m. 22. (a) The tension in each string is given by = Mg/2. Thus, the wave speed in string 1 is v1 Mg (500 g) (9.80 m/s 2 ) 28.6 m/s. 1 21 2(3.00 g/m) (b) And the wave speed in string 2 is v2 Mg (500g) (9.80 m/s 2 ) 22.1m/s. 2 2 2(5.00g/m) (c) Let v1 M1 g /(21 ) v2 M 2 g /(22 ) and M1 + M2 = M. We solve for M1 and obtain M1 M 1 2 / 1 500 g 187.5g 188g. 1 5.00 / 3.00 (d) And we solve for the second mass: M2 = M – M1 = (500 g – 187.5 g) 313 g. 29. The displacement of the string is given by y ym sin(kx t ) ym sin(kx t ) 2 ym cos 12 sin kx t 12 , where = /2. The amplitude is A 2 ym cos 12 2 ym cos( / 4) 1.41ym . 34. The phasor diagram is shown below. We use the cosine theorem: ym2 ym2 1 ym2 2 2 ym1 ym 2 cos ym2 1 ym2 2 2 ym1 ym 2 cos . We solve for cos : cos ym2 ym2 1 ym2 2 (9.0 mm) 2 (5.0 mm) 2 (7.0 mm) 2 0.10. 2 ym1 ym 2 2(5.0 mm) (7.0 mm) The phase constant is therefore = 84°. 43. (a) Eq. 16–26 gives the speed of the wave: v 150 N 144.34 m/s 1.44 102 m/s. 3 7.20 10 kg/m (b) From the Figure, we find the wavelength of the standing wave to be = (2/3)(90.0 cm) = 60.0 cm. (c) The frequency is f v 1.44 102 m/s 241Hz. 0.600 m 47. (a) The amplitude of each of the traveling waves is half the maximum displacement of the string when the standing wave is present, or 0.25 cm. (b) Each traveling wave has an angular frequency of = 40 rad/s and an angular wave number of k = /3 cm–1. The wave speed is v = /k = (40 rad/s)/(/3 cm–1) = 1.2×102 cm/s. (c) The distance between nodes is half a wavelength: d = /2 = /k = /(/3 cm–1) = 3.0 cm. Here 2/k was substituted for . (d) The string speed is given by u(x, t) = y/t = –ymsin(kx)sin(t). For the given coordinate and time, 9 u (40 rad/s) (0.50cm) sin cm 1 (1.5cm) sin 40 s 1 8 3 s 0. 53. (a) The angular frequency is = 8.00/2 = 4.00 rad/s, so the frequency is f = /2 = (4.00 rad/s)/2 = 2.00 Hz. (b) The angular wave number is k = 2.00/2 = 1.00 m–1, so the wavelength is = 2/k = 2/(1.00 m–1) = 2.00 m. (c) The wave speed is v f (2.00 m) (2.00 Hz) = 4.00 m/s. (d) We need to add two cosine functions. First convert them to sine functions using cos = sin ( + /2), then apply cos cos sin sin 2sin cos 2 2 2 2 2 cos cos 2 2 Letting = kx and = t, we find ym cos(kx t ) ym cos(kx t ) 2 ym cos(kx) cos(t ). Nodes occur where cos(kx) = 0 or kx = n + /2, where n is an integer (including zero). Since k = 1.0 m–1, this means x n 12 (1.00m) . Thus, the smallest value of x which corresponds to a node is x = 0.500 m (n=0). (e) The second smallest value of x which corresponds to a node is x = 1.50 m (n=1). (f) The third smallest value of x which corresponds to a node is x = 2.50 m (n=2). (g) The displacement is a maximum where cos(kx) = 1. This means kx = n, where n is an integer. Thus, x = n(1.00 m). The smallest value of x which corresponds to an anti-node (maximum) is x = 0 (n=0). (h) The second smallest value of x which corresponds to an anti-node (maximum) is x 1.00 m (n=1). (i) The third smallest value of x which corresponds to an anti-node (maximum) is x 2.00 m (n=2). 55. (a) The frequency of the wave is the same for both sections of the wire. The wave speed and wavelength, however, are both different in different sections. Suppose there are n1 loops in the aluminum section of the wire. Then, L1 = n11/2 = n1v1/2f, where 1 is the wavelength and v1 is the wave speed in that section. In this consideration, we have substituted 1 = v1/f, where f is the frequency. Thus f = n1v1/2L1. A similar expression holds for the steel section: f = n2v2/2L2. Since the frequency is the same for the two sections, n1v1/L1 = n2v2/L2. Now the wave speed in the aluminum section is given by 1 / 1 , where 1 is the linear mass density of the aluminum wire. The mass of aluminum in the wire is given by m1 = 1AL1, where 1 is the mass density (mass per unit volume) for aluminum and A is the cross-sectional area of the wire. Thus 1 = 1AL1/L1 = 1A and 1 / 1 A. A similar expression holds for the wave speed in the steel section: v2 / 2 A. We note that the cross-sectional area and the tension are the same for the two sections. The equality of the frequencies for the two sections now leads to n1 / L1 1 n2 / L2 2 , where A has been canceled from both sides. The ratio of the integers is 3 3 n2 L2 2 0.866 m 7.80 10 kg/m 2.50. n1 L1 1 0.600 m 2.60 103 kg/m3 The smallest integers that have this ratio are n1 = 2 and n2 = 5. The frequency is f n1v1 / 2L1 n1 / 2L1 / 1 A. The tension is provided by the hanging block and is = mg, where m is the mass of the block. Thus n f 1 2 L1 mg 2 1 A 2 0.600 m 10.0 kg 9.80 m/s2 2.60 10 kg/m 1.00 10 3 3 6 m2 324 Hz. (b) The standing wave pattern has two loops in the aluminum section and five loops in the steel section, or seven loops in all. There are eight nodes, counting the end points.