Sample from chapter 12 of Solution Manual Halliday Physics 12th Edition

1. Email: smtb98@gmail.com Telegram: https://t.me/solutionmanual Contact me in order to access the whole complete document. WhatsApp: https://wa.me/message/2H3BV2L5TTSUF1 smtb98@gmail.com smtb98@gmail.com Chapter 12 1. (a) The center of mass is given by 0 0 0 ( )(2.00 m) ( )(2.00 m) ( )(2.00 m) m x = + + + + + m m m = 1.00 m. com 6 (b) Similarly, we have 0 ( )(2.00 m) ( )(4.00 m) ( )(4.00 m) ( )(2.00 m) 0 m m y = + + + + + m m = 2.00 m. com 6 m (c) Using Eq. 12.1.14 and noting that the gravitational effects are different at the different locations in this problem, we have 6 x m g x m g x m g x m g x m g m g m g m g = (d) Similarly, we have 6 y m g y m g y m g y m g y m g m g m g m g = + + = + + = 2. Our notation is as follows: M = 1360 kg is the mass of the automobile; L = 3.05 m is the horizontal distance between the axles; horizontal distance from the rear axle to the center of mass; F1 is the force exerted on each front wheel; and F2 is the force exerted on each back wheel. (a) Taking torques about the rear axle, we find (1360 kg)(9.80 m/s )(1.27 m) 2 2(3.05 m) L  + + + + + + i i i x m g m g x m g m g x m g = = = =  0.987 m. 1 6 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 i cog + + + + m g 1 1 2 2 3 3 4 4 5 5 6 6 i i 1 i  + + + + + + i i i y m g m g y m g m g y m g = = =  1 6 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 i cog + + + + m g 1 1 2 2 3 3 4 4 5 5 6 6 i i 1 i + + + + + 0 (2.00)(7.80 ) (4.00)(7.60 ) (4.00)(7.40 ) (2.00)(7.60 ) 0 8.0 7.80 7.60 1.97 m. m m m m m m m + 7.40 7.60 7.80 m m m = (3.05 1.78) m − = 1.27 m is the  2 Mg  = = = 2.77 10 N. × 3 F 1 603 complete document is available on https://solumanu.com/ *** contact me if site not loaded

2. 604 CHAPTER 12 F = 3.89 10 N. × + = 3 (b) Equilibrium of forces leads to 3. THINK Three forces act on the sphere: the tension force T of the wall N F , and the force of gravity EXPRESS The free-body diagram is shown to the right. The tension force T wall N F acts horizontally away from the wall, and the force of gravity mg acts downward. Since the sphere is in static equilibrium, they sum to zero. Let θ be the angle between the rope and the vertical. Then Newton’s second law gives vertical component: T horizontal component: F ANALYZE (a) We solve the first equation for the tension: θ = + to obtain mg L T L from which we obtain 2 2 , F F Mg 2 1 2  of the rope, the force  mg .  acts along the rope, the force of the  cos – θ – N = 0 0. mg θ = sin T = θ /cos . We substitute T mg 2 2 cos / L L r + 2 2 r = + 2 2 2 (0.85 kg)(9.8 m/s ) (0.080 m) (0.042 m) = 0.080 m = 9.4 N. (b) We solve the second equation for the normal force: = θ Using sinθ = sin . F T N + 2 2 we obtain / , r L r + 2 2 2 (0.85 kg)(9.8 m/s )(0.042 m) (0.080 m) Tr L mg L r r mgr L = = = = = 4.4 N. F N L + + 2 2 2 2 r L r LEARN Since the sphere is in static equilibrium, the vector sum of all external forces acting on it must be zero. 4. The situation is somewhat similar to that depicted for problem 10 (see the figure that accompanies that problem in the text). By analyzing the forces at the “kink” where F the angle (taken positive) between each segment of the string and its “relaxed” position (when the two segments are collinear). Setting T = F therefore yields θ = 30º. Since α = 180º – 2θ is the angle between the two segments, then we find α = 120º.  is exerted, we find (since the acceleration is zero) 2 sin θ = where θ is , T F complete document is available on https://solumanu.com/ *** contact me if site not loaded

3. smtb98@gmail.com 605 CHAPTER 12  is exerted, we find (since the acceleration is zero) 2T sinθ = F, where θ is the angle smtb98@gmail.com 5. The object exerts a downward force of magnitude F = 3160 N at the midpoint of the rope, causing a “kink” similar to that shown for problem 10 (see the figure that accompanies that problem in the text). By analyzing the forces at the “kink” where F (taken positive) between each segment of the string and its “relaxed” position (when the two segments are collinear). In this problem, we have 1 0.35 m tan 1.72 m  Therefore, / 2sin 7.92 1 . ( ) 0 N T F θ = = × 6. Let 1 1.5 m =  and 2 (5.0 1.5) m 3.5 m. = − =  closer to the window as F1 and that in the other cable as F2. The force of gravity on the scaffold itself (of magnitude msg) is at its midpoint, (a) Taking torques about the end of the plank farthest from the window washer, we find (80 kg)(9.8 m/s )(3.5 m) (60 kg)(9.8 m/s )(2.5 m) w s m g m g F = = + = × (b) Equilibrium of forces leads to (60 kg 80 kg)(9.8 m/s ) 1.4 10 N, s w F F m g m g + = + = + which (using our result from part (a)) yields 7. The forces on the ladder are shown in the diagram below. −     θ = = ° 11.5 .  3 We denote tension in the cable = from either end. 2.5 m  3 + + 2 2    2 3 1 5.0 m  1 2 2 8.4 10 N. = × 2 3 1 2 F = 5.3 10 N. × 2 2 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.

4. 606 CHAPTER 12 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 θ be the angle between the ladder and the ground. We use cos d L θ = 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, ( cos /2 cos Mg mg L θ +  and ( /2) cos [(75 kg)(3.0 m)+(10 kg)(2.5 m)](9.8 m/s )cos60 sin 2.8 10 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: F F Mg − − The first of these equations gives 3 1 2.8 10 N F F = = ( ) (75 kg 10 kg)(9.8 m/s ) F M m g = + = + 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: (2.8 10 N) F F F = + = × (c) The angle φ between the force and the horizontal is given by tan   / (830 N)/(280 N) F F φ= = so φ = 71º. The force points to the left and upward, 71º above the horizontal. We note that this force is not directed along the ladder. to find θ = 60º. θ = − 2 2 / or sin / L d L ) θ − θ = sin 0, FL 1 + θ ° 2 M mL L g  = = F 1 θ ° (5.0 m)sin60 2 − = = 0 0. F 1 3 mg 2 × 2 and the second gives = 8.3 10 N. × 2 2 2 + (8.3 10 N) × = 8.8 10 N. × 2 2 2 2 2 2 2 2 3 = 2.94, 3 2 Email: smtb98@gmail.com Telegram: https://t.me/solutionmanual Contact me in order to access the whole complete document. WhatsApp: https://wa.me/message/2H3BV2L5TTSUF1

5. smtb98@gmail.com 607 CHAPTER 12 smtb98@gmail.com    τ = × 8. From page (relative to the fulcrum), and persons 5 through 8 exert torques pointing into the page. (a) Among persons 1 through 4, the largest magnitude of torque is (330 N)(3 m) = 990 N·m, due to the weight of person 2. (b) Among persons 5 through 8, the largest magnitude of torque is (330 N)(3 m) = 990 N·m, due to the weight of person 7. 9. THINK In order for the meter stick to remain in equilibrium, the net force acting on it must be zero. In addition, the net torque about any point must also be zero. EXPRESS Let the x axis be along the meter stick, with the origin at the zero position on the scale. The forces acting on the stick are shown to the right. The coins are at x1 = 0.120 m, and 10.0 g m = is their total mass. The knife edge is at x2 = 0.455 m and provides a supporting force . F The mass of the meter stick is M, and the force of gravity acts at the center of the stick, x3 = 0.500 m. Since the meter stick is in equilibrium, the sum of the torques about x2 must vanish: Mg(x3 – x2) – mg(x2 – x1) = 0. ANALYZE Solving the equation for M, we find the mass of the meter stick to be 0.455 m 0.500 m x x −    LEARN Since the net torque about any point is zero, we could have chosen x1 as the location of the rotation axis. Then the balance of torques requires that ( ) F x x Mg x − − The fact that the net force is zero indicates that the supporting force is Substituting this into the equation gives the same result as before: x M x  , we note that persons 1 through 4 exert torques pointing out of the r F      − − −      0.120 m 0.455 m x x = = = (10.0 g) 74.4 g. 2 1 M m 3 2 − = ( ) 0. x 2 1 3 1 = + ( ) . F M m g     − − x x = . 2 1 m 3 2

6. Email: smtb98@gmail.com Telegram: https://t.me/solutionmanual Contact me in order to access the whole complete document. WhatsApp: https://wa.me/message/2H3BV2L5TTSUF1 608 CHAPTER 12 10. (a) Analyzing vertical forces where string 1 and string 2 meet, we find w T φ (b) Looking at the horizontal forces at that point leads to 1sin35 T T = ° = (c) We denote the components of T3 as Tx (rightward) and Ty (upward). Analyzing horizontal forces where string 2 and string 3 meet, we find Tx = T2 = 28 N. From the vertical forces there, we conclude Ty = wB = 50 N. Therefore, T T T = + (d) The angle of string 3 (measured from vertical) is T T     11. THINK The diving board is in equilibrium, so the net force and net torque acting on it must be zero. EXPRESS We take the force of the left pedestal to be F1 at x = 0, where the x axis is along the diving board. We take the force of the right pedestal to be F2 at position x = d. W is the weight of the diver, located at x = L. The following two equations result from setting the sum of forces equal to zero (with upward being the positive direction), and the sum of torques about x2 equal to zero: F F Fd W L d + ANALYZE (a) The second equation gives 3.0 m 1.5 m d    which should be rounded off to 1 1.2 10 N. F = − × (b) Since F1 is negative, this force is downward. 40 N cos35 = = = 49 N. A 1 ° cos ° = (49 N)sin35 28 N. 2 = 2 x 2 y 57 N. 3          1 28 50  − − θ = = = ° 1 tan tan 29 . x y + − − = = 0 0. W 1 2 ( ) 1 L d −         = − = − = − (580 N) 1160 N, F W  1 | 1.2 10 N. F = × 3 3 Thus, | 1 complete document is available on https://solumanu.com/ *** contact me if site not loaded

7. smtb98@gmail.com 609 CHAPTER 12 smtb98@gmail.com = − = 580 N 1160 N + = (c) The first equation gives which should be rounded off to (d) The result is positive, indicating that this force is upward. (e) The force of the diving board on the left pedestal is upward (opposite to the force of the pedestal on the diving board), so this pedestal is being stretched. (f) The force of the diving board on the right pedestal is downward, so this pedestal is being compressed. 1740 N, F W F 2 1 F = 1.7 10 N. × | 1.7 10 N. = × 3 3 Thus, | F 2 2 −      L d L = − LEARN We may relate F1 and F2 via 2, which indicates that the two F F 1 forces are in opposite directions, one acting downward and the other upward. 12. The angle of each half of the rope, measured from the dashed line, is 1 0.30 m tan θ =   Analyzing forces at the “kink” (where F 550 N 2sin 2sin1.9 θ 13. The (vertical) forces at points A, B, and P are FA, FB, and FP, respectively. We note that FP = W and is upward. Equilibrium of forces and torques (about point B) lead to F F bW aF − (a) From the second equation, we find / (15/5) 3 A F bW a W W = = = (b) The direction is upward since FA > 0. (c) Using this result in the first equation above, we obtain 4 4(900 N) B A F W F W = − = − = − −     = ° 1.9 . 9.0 m  is exerted), we find F = = = 8.3 10 N. × 3 T ° + + = = 0 0. W A B A = = 2.7 10 N. × 3 3(900 N) 3.6 10 N, = − × 3 complete document is available on https://solumanu.com/ *** contact me if site not loaded

8. 610 CHAPTER 12 = 3.6 10 N. × 3 or (d) FB points downward, as indicated by the negative sign. 14. With pivot at the left end, Eq. 12.1.9 leads to – ms g2 where ms is the scaffold’s mass (50 kg) and M is the total mass of the paint cans (75 kg). The variable x indicates the center of mass of the paint can collection (as measured from the left end), and TR is the tension in the right cable (722 N). Thus we obtain x = 0.702 m. 15. (a) Analyzing the horizontal forces (which add to zero), we find Fh = F3 = 5.0 N. (b) Equilibrium of vertical forces leads to Fv = F1 + F2 = 30 N. (c) Computing torques about point O, we obtain ( 2 3 F b F a d = +  = | | F B L – Mgx + TRL = 0, )( ) ( 30 N )( ) 10 N 3.0 m + 5.0 N 2.0 m = 1.3 m. v F d 16. The forces exerted horizontally by the obstruction and vertically (upward) by the floor are applied at the bottom front corner C of the crate, as it verges on tipping. The center of the crate, which is where we locate the gravity force of magnitude 500 N, mg = is a horizontal distance =  magnitude F = 350 N is a vertical distance h from C. Taking torques about C, we obtain (500 N)(0.375 m) 350 N F 17. (a) With the pivot at the hinge, Eq. 12.1.9 gives from C. The applied force of 0.375 m mg  = = = 0.536 m. h L θ = co s – 0. TL mg 2 This leads to (b) A higher (steeper) slope for the cable results in a smaller tension. Thus, making D greater than the value of part (a) should prevent rupture. 18. With pivot at the left end of the lower scaffold, Eq. 12.1.9 leads to L – mgd + TRL2 = 0, ° Then the geometric relation tanθ= L/D gives D = 0.64 m. θ = 78 . – m2g 2 2 complete document is available on https://solumanu.com/ *** contact me if site not loaded

9. smtb98@gmail.com 611 CHAPTER 12 smtb98@gmail.com where m2 is the lower scaffold’s mass (30 kg) and L2 is the lower scaffold’s length (2.00 m). The mass of the package (m = 20 kg) is a distance d = 0.50 m from the pivot, and TR is the tension in the rope connecting the right end of the lower scaffold to the larger scaffold above it. This equation yields TR = 196 N. Then Eq. 12.1.8 determines TL (the tension in the cable connecting the right end of the lower scaffold to the larger scaffold above it): TL = 294 N. Next, we analyze the larger scaffold (of length L1 = L2 + 2d and mass m1, given in the problem statement) placing our pivot at its left end and using Eq. 12.1.9: L – TL d – TR (L1 – d) + TL1= 0. – m1g 1 2 This yields T = 457 N. 19. Setting up equilibrium of torques leads to a simple “level principle” ratio: d F L 20. Our system consists of the lower arm holding a bowling ball. As shown in the free-body diagram, the forces on the lower arm consist of T the bone of the upper arm, and the gravitational forces, mg and Mg. Since the system is in static equilibrium, the net force acting on the system is zero: 0 ( y F T F m M g = = − − +  In addition, the net torque about O must also vanish: 0 ( )( ) (0) O (a) From the torque equation, we find the force on the lower arms by the biceps muscle to be ( ) [(1.8 kg)(0.15 m) (7.2 kg)(0.33 m)](9.8 m/s ) mD ML g T d = ≈ × (b) Substituting the above result into the force equation, we find F to be ( ) 648 N (7.2 kg 1.8 kg)(9.8 m/s ) F T M m g = − + = − + 2.6 cm 12 cm ⊥= = = (40 N) (40 N) 8.7 N.  from the biceps muscle, F  from ) . net,  = τ = + − − ( )( D mg ) ( ). d T F L Mg net + + 2 = = 0.040 m 2 648 N 6.5 10 N. = = 5.6 10 N. × 2 2 560 N

10. 612 CHAPTER 12 21. (a) We note that the angle between the cable and the strut is α =θ – φ= 45º – 30º = 15º. The angle between the strut and any vertical force (like the weights in the problem) is β = 90º – 45º = 45º. Denoting M = 225 kg and m = 45.0 kg, and  as the length of the boom, we compute torques about the hinge and find ( ) 2 sin sin sin α  The unknown length  cancels out and we obtain T = 6.63 × 103 N. (b) Since the cable is at 30º from horizontal, then horizontal equilibrium of forces requires that the horizontal hinge force be = cos30 =5.74 10 N. x F T ° (c) And vertical equilibrium of forces gives the vertical hinge force component: sin30 y F Mg mg T = + + 22. (a) The problem asks for the person’s pull (his force exerted on the rock) but since we are examining forces and torques on the person, we solve for the reaction force 1 N F (exerted leftward on the hands by the rock). At that point, there is also an upward force of static friction on his hands, f1, which we will take to be at its maximum value 1 N F μ We note that equilibrium of horizontal forces requires F F = (the force exerted leftward on his feet); on his feet there is also an upward static friction force of magnitude μ2FN2. Equilibrium of vertical forces gives β + β β + β Mg mg   sin sin /2. Mg mg α = = T sin × 3 ° = 5.96 10 N. × 3 1. 1 2 N N mg − =3.4 10 N. ×  2 + =0 = f f mg F 1 2 1 N μ μ + 1 2 (b) Computing torques about the point where his feet come in contact with the rock, we find ( 1 1 + =0 = N mg d w f w F h h − −  ( ) − F w μ + mg d w F ) 1 1 N =0.88 m. 1 N (c) Both intuitively and mathematically (since both coefficients are in the denominator), we see from part (a) that N F would increase in such a case. 1