Tài liệu Physics exercises_solution: Chapter 11 docx

11.1: Take the origin to be at the center of the small ball; then, (1.00 kg)(0)  (2.00 kg )(0.580 m) xcm   0.387 m 3.00 kg from the center of the small ball 11.2: The calculation of Exercise 11.1 becomes (1.00 kg)(0)  (1.50 kg )(0.280 m)  (2.00 kg )(0.580 m) xcm   0.351 m 4.50 kg This result is smaller than the one obtained in Exercise 11.1 11.3: In the notation of Example 11.1, take the origin to be the point S , and let the child’s distance from this point be x Then, M ( D 2)  mx MD scm   0, x   1.125 m, M m 2m which is ( L  D 2) 2, halfway between the point S and the end of the plank 11.4: a) The force is applied at the center of mass, so the applied force must have the same magnitude as the weight of the door, or 300 N In this case, the hinge exerts no force b) With respect to the hinge, the moment arm of the applied force is twice the distance to the center of mass, so the force has half the magnitude of the weight, or 150 N The hinge supplies an upward force of 300 N  150 N  150 N 11.5: F (8.0 m) sin 40  (2800 N)(10.0 m), so F  5.45 kN, keeping an extra figure 11.6: The other person lifts with a force of 160 N  60 N  100 N Taking torques about the point where the 60 - N force is applied,  160 N    2.40 m (100 N) x  (160 N)(1.50 m), or x  (1.50 m)  100 N  11.7: If the board is taken to be massless, the weight of the motor is the sum of the ( 2.00 m )( 600 N ) applied forces, 1000 N The motor is a distance (1000 N )  1.200 m from the end where the 400-N force is applied 11.8: The weight of the motor is 400 N  600 N  200 N  800 N Of the myriad ways to this problem, a sneaky way is to say that the lifters each exert 100 N to the lift the board, leaving 500 N and 300 N to the lift the motor Then, the distance of the motor ( 2.00 m )( 300 N )  0.75 m The center of from the end where the 600-N force is applied is (800 N ) gravity is located at ( 200 N )(1.0 m )  (800 N )( 0.75 m ) (1000 N )  0.80 m from the end where the 600 N force is applied 11.9: cot θ h, and the torque due to Ty is Ty D  Lw The torque due to Tx is  Tx h   Lw D The sum of these torques is Lw(1  Dh cot θ ) From Figure (11.9(b)), h  D tan θ , so the net torque due to the tension in the tendon is zero 11.10: a) Since the wall is frictionless, the only vertical forces are the weights of the man and the ladder, and the normal force For the vertical forces to balance, n2  w1  wm  160 N  740 N  900 N, and the maximum frictional forces is μs n2  (0.40)(900 N)  360 N (see Figure 11.7(b)) b) Note that the ladder makes contact with the wall at a height of 4.0 m above the ground Balancing torques about the point of contact with the ground, (4.0 m)n1  (1.5 m)(160 N)  (1.0 m)(3 5))(740 N)  684 N  m, so n1  171.0 N, keeping extra figures This horizontal force about must be balanced by the frictional force, which must then be 170 N to two figures c) Setting the frictional force, and hence n1 , equal to the maximum of 360 N and solving for the distance x along the ladder, (4.0 m)(360 N)  (1.50 m)(160 N)  x(3 5)(740 N), so x = 2.70 m, or 2.7 m to two figures 11.11: Take torques about the left end of the board in Figure (11.21) a) The force F at the support point is found from F (1.00 m)  (280 N)(1.50 m)  (500 N)(3.00 m), or F  1920 N b) The net force must be zero, so the force at the left end is (1920 N)  (500 N)  (280 N)  1140 N, downward 11.12: a) b) x  6.25 m when FA  0, which is 1.25 m beyond point B c) Take torques about the right end When the beam is just balanced, FA  0, so FB  900 N The distance that point B must be from the right end is then ( 300 N )( 4.50 m ) ( 900 N )  1.50 m 11.13: In both cases, the tension in the vertical cable is the weight ω a) Denote the length of the horizontal part of the cable by L Taking torques about the pivot point, TL tan 30.0  wL  w( L 2), from which T  2.60w The pivot exerts an upward vertical force of w and a horizontal force of 2.60 w , so the magnitude of this force is 3.28w , directed 37.6 from the horizontal b) Denote the length of the strut by L , and note that the angle between the diagonal part of the cable and the strut is 15.0 Taking torques about the pivot point, TL sin 15.0  wL sin 45.0  ( w 2) L sin 45, so T  4.10w The horizontal force exerted by the pivot on the strut is then T cos 30.0  3.55ω and the vertical force is (2w)  T sin 30  4.05w, for a magnitude of 5.38w, directed 48.8 11.14: a) Taking torques about the pivot, and using the 3-4-5 geometry, (4.00 m) (3 5)T  (4.00 m)(300 N)  (2.00 m)(150 N), so T  625 N b) The horizontal force must balance the horizontal component of the force exerted by the rope, or T (4 5)  500 N The vertical force is 300 N  150 N  T (3 5)  75 N, upwards 11.15: To find the horizontal force that one hinge exerts, take the torques about the other hinge; then, the vertical forces that the hinges exert have no torque The horizontal force is found from FH (1.00 m)  (280 N)(0.50 m), from which FH  140 N The top hinge exerts a force away from the door, and the bottom hinge exerts a force toward the door Note that the magnitudes of the forces must be the same, since they are the only horizontal forces 11.16: (a) Free body diagram of wheelbarrow:  wheel   (450 N)(2.0 m)  (80 N)(0.70 m)  WL (0.70 m)  WL  1200 N (b) From the ground 11.17: Consider the forces on Clea nr  89 N, nf  157 N nr  nf  w so w  246 N 11.18: a) Denote the length of the boom by L, and take torques about the pivot point The tension in the guy wire is found from TL sin 60  (5000 N) L cos 60.0  (2600 N)(0.35 L) cos 60.0, so T  3.14 kN The vertical force exerted on the boom by the pivot is the sum of the F  weights, 7.06 kN and the horizontal force is the tension, 3.14 kN b) No; tan  v    FH  11.19: To find the tension TL in the left rope, take torques about the point where the rope at the right is connected to the bar Then, TL (3.00 m) sin 150  (240 N)(1.50 m)  (90 N)(0.50 m), so TL  270 N The vertical component of the force that the rope at the end exerts must be (330 N)  (270 N) sin 150  195 N, and the horizontal component of the force is  (270 N) cos 150, so the tension is the rope at the right is TR  304 N and θ  39.9 11.20: The cable is given as perpendicular to the beam, so the tension is found by taking torques about the pivot point; T (3.00 m)  (1.00 kN)(2.00 m) cos 25.0  (5.00 kN)(4.50 m) cos 25.0, or T  7.40 kN The vertical component of the force exerted on the beam by the pivot is the net weight minus the upward component of T , 6.00 kN  T cos 25.0  0.17 kN The horizontal force is T sin 25.0  3.13 kN 11.21: a) F1 (3.00 m)  F2 (3.00 m  l )  (8.00 N)(l ) This is given to have a magnitude of 6.40 N.m, so l  0.80m b) The net torque is clockwise, either by considering the figure or noting the torque found in part (a) was negative c) About the point of contact of F , the torque due to F is  F1l , and setting the magnitude of this torque to 6.40 N  m gives l  0.80 m, and the direction is again clockwise 11.22: From Eq (11.10), l (0.200 m) Y F F  F (1333 m  ) 2 lΑ (3.0  10 m)(50.0  10  m ) Then, F  25.0 N corresponds to a Young’s modulus of 3.3  10 Pa, and F  500 N corresponds to a Young’s modulus of 6.7  10 Pa A 11.23: Fl (400 N)(2.00 m)   1.60  10 6 m , 10 2 Yl (20  10 Pa )(0.25  10 m) and so d  A   1.43  10 3 m, or 1.4 mm to two figures 11.24: a) The strain, from Eq (11.12), is and A   d2 l l0  F YA For steel, using Y from Table (11.1)  1.77  10 4 m , (4000 N) l   1.1  10  11 4 l (2.0  10 Pa )(1.77  10 m ) Similarly, the strain for copper (Y  1.10  1011 Pa ) is 2.1  104 b) Steel: (1.1  104 )  (0.750 m)  8.3  105 m Copper: (2.1  104 )(0.750 m)  1.6  104 m  11.25: From Eq (11.10), Y (5000 N)(4.00 m)  2.0  1011 Pa 4 2 (0.50  10 m )(0.20  10 m) 11.26: From Eq (11.10), (65.0 kg)(9.80 m s )(45.0 m) Y  6.8  10 Pa 3 ( (3.5  10 m) )(1.10 m) 11.27: a) The top wire is subject to a tension of (16.0 kg)(9.80 m s )  157 N and hence a tensile strain of (157 N ) ( 201010 Pa )( 2.510  m )  3.14  10 3 , or 3.1  10 3 to two figures The bottom wire is subject to a tension of 98.0 N, and a tensile strain of 1.96  103 , or 2.0  103 to two figures b) (3.14  103 )(0.500 m)  1.57 mm, (1.96  103 )(0.500 m)  0.98 mm 11.28: a) ( 8000 kg )( 9.80 m ) s  (12.510  m )  1.6  106 Pa b) 1.610 Pa 2.01010 Pa  0.8  10 5 c) (0.8  105 )  (2.50 m)   105 m 11.29: (2.8  1)(1.013  10 Pa )(50.0 m )  9.1  10 N 11.30: a) The volume would increase slightly b) The volume change would be twice as great c) The volume is inversely proportional to the bulk modulus for a given pressure change, so the volume change of the lead ingot would be four times that of the gold 11.31: a) 11.32: 250 N 0.7510 -4 m  3.33  106 Pa b) (3.33  106 Pa)(2)(200  10 4 m )  133 kN a) Solving Eq (11.14) for the volume change, ΔV   kVP  (45.8  1011 Pa 1 )(1.00 m )(1.16  108 Pa  1.0  105 Pa )  0.0531 m b) The mass of this amount of water not changed, but its volume has decreased to 103 kg  1.09 10 kg m 1.000 m  0.053 m  0.947 m , and the density is now 1.003.947 m3 B 11.33: 11.34: (600 cm3 )(3.6  106 Pa )  4.8  109 Pa , k   2.1  1010 Pa 1 (0.45 cm ) B a) Using Equation (11.17), Shear strain  F|| (9  105 N)   2.4  10 AS [(.10 m)(.005 m)][7.5  1010 Pa ] b) Using Equation (11.16), x  Shear stain  h  (.024)(.1 m)  2.4  10 3 m 11.35: The area A in Eq (11.17) has increased by a factor of 9, so the shear strain for the larger object would be that of the smaller 11.36: Each rivet bears one-quarter of the force, so Shear stress  F|| (1.20  104 N)   6.11  108 Pa A (.125  10 m) 11.37: 11.38: F A ( 90.8 N )   ( 0.9210 3 m )  3.41  107 Pa , or 3.4  107 Pa to two figures a) (1.6  103 )(20  1010 Pa )(5  106 m )  1.60  103 N b) If this were the case, the wire would stretch 6.4 mm c) (6.5  103 )(20  1010 Pa )(5  106 m )  6.5  103 N 11.39: Ftot (2.40  108 Pa ) (3.00  104 m )  a  9.80 m s  10.2 m s (1200 kg) m 11.40: A 350 N 4.7 10 Pa  7.45  107 m , so d  A   0.97 mm 11.41: a) Take torques about the rear wheel, so that fωd  ωxcm , or xcm  fd b) (0.53)(2.46 m)  1.30 m to three figures 11.42: If Lancelot were at the end of the bridge, the tension in the cable would be (from taking torques about the hinge of the bridge) obtained from T (12.0 N)  (600 kg)(9.80 m s )(12.0 m)  (200 kg)(9.80 m s )(6.0 m), so T  6860 N This exceeds the maximum tension that the cable can have, so Lancelot is going into the drink To find the distance x Lancelot can ride, replace the 12.0 m multiplying Lancelot’s weight by x and the tension T by Tmax  5.80  10 N and solve for x; x (5.80  103 N)(12.0 m)  (200 kg)(9.80 m s )(6.0 m)  9.84 m (600 kg)(9.80 m s ) 11.43: For the airplane to remain in level flight, both  F  and    Taking the clockwise direction as positive, and taking torques about the center of mass, Forces:  Ftail  W  Fwing  Torques:  (3.66 m) Ftail  (.3 m) Fwing  A shortcut method is to write a second torque equation for torques about the tail, and solve for the Fwing : (3.66 m)(6700 N)  (3.36 m) Fwing  This gives Fwing  7300 N(up), and Ftail  6700 N  7300 N  600 N(down) Note that the rear stabilizer provides a downward force, does not hold up the tail of the aircraft, but serves to counter the torque produced by the wing Thus balance, along with weight, is a crucial factor in airplane loading 11.44: The simplest way to this is to consider the changes in the forces due to the extra weight of the box Taking torques about the rear axle, the force on the front wheels m  1200 N, so the net force on the front wheels is decreased by 3600 N 1.00 3.00 m is10,780 N  1200 N  9.58  10 N to three figures The weight added to the rear wheels is then 3600 N  1200 N  4800 N, so the net force on the rear wheels is 8820 N  4800 N  1.36  10 N, again to three figures b) Now we want a shift of 10,780 N away from the front axle Therefore, W 1.00 m 3.00 m  10,780 N and so w  32,340 N 11.45: Take torques about the pivot point, which is 2.20 m from Karen and 1.65 m from Elwood Then wElwood (1.65 m)  (420 N)(2.20 m)  (240 N)(0.20 m), so Elwood weighs 589 N b) Equilibrium is neutral 11.46: a) Denote the weight per unit length as α, so w1  α (10.0 cm), w2  α (8.0 cm), and w3  αl The center of gravity is a distance xcm to the right of point O where xcm   w1 (5.0 cm)  w2 (9.5 cm)  w3 (10.0 cm  l 2) w1  w2  w3 (10.0 cm)(5.0 cm)  (8.0 cm)(9.5 cm)  l (10.0 cm  l 2) (10.0 cm)  (8.0 cm)  l Setting xcm  gives a quadratic in l , which has as its positive root l  28.8 cm b) Changing the material from steel to copper would have no effect on the length l since the weight of each piece would change by the same amount     Let ri   ri  R ,where R is the vector from the point O to the point P    The torque for each force with respect to point P is then i  ri  Fi , and so the net torque is 11.47:    τ  r  R  F     r  F   R  F      r  F  R  F   i i i i i i i i i In the last expression, the first term is the sum of the torques about point O, and the second term is given to be zero, so the net torques are the same  11.48: From the figure (and from common sense), the force F1 is directed along the length of the nail, and so has a moment arm of (0.0800 m) sin 60 The moment arm of  F2 is 0.300 m, so F2  F1 (0.0800 m) sin 60  (500 N)(0.231)  116 N (0.300 m) 11.67: Consider torques around the point where the person on the bottom is lifting The center of mass is displaced horizontally by a distance (0.625 m  0.25 m) sin 45 and the horizontal distance to the point where the upper person is lifting is (1.25 m) sin 45 , and sin 45 so the upper lifts with a force of w 01 375  (0.300) w  588 N The person on the 25 sin 45 bottom lifts with a force that is the difference between this force and the weight, 1.37 kN The person above is lifting less 11.68: (a) Elbow  FB (3.80 cm)  (15.0 N)(15.0 cm) FB  59.2 N (b)  E  FB (3.80 cm)  (15.0 N)(15.0 cm)  (80.0 N)(33.0 cm) FB  754 N 11.69: a) The force diagram is given in Fig 11.9   0, axis at elbow wL  T sin θ D  h hD sin θ  so w  T 2 h D L h2  D hD wmax  Tmax L h2  D dwmax Tmax h  D2  1  ; the derivative is positive   dD L h2  D  h  D  c) The result of part (b) shows that wmax increases when D increases b) 11.70: By symmetry, A=B and C=D Redraw the table as viewed from the AC side τ (about right end)  : A(3.6 m)  90.0 N (1.8 m)  1500 N (0.50 m) A  130 N  B F  : A  B  C  D  1590 N Use A  B  130 N and C  D C  D  670 N By Newton’s third law of motion, the forces A, B, C, and D on the table are the same as the forces the table exerts on the floor ... (2.1  104 )(0.750 m)  1.6  104 m  11. 25: From Eq (11. 10), Y (5000 N)(4.00 m)  2.0  1 011 Pa 4 2 (0.50  10 m )(0.20  10 m) 11. 26: From Eq (11. 10), (65.0 kg)(9.80 m s )(45.0 m) Y ... of the gold 11. 31: a) 11. 32: 250 N 0.7510 -4 m  3.33  106 Pa b) (3.33  106 Pa)(2)(200  10 4 m )  133 kN a) Solving Eq (11. 14) for the volume change, ΔV   kVP  (45.8  10? ?11 Pa 1 )(1.00... object would be that of the smaller 11. 36: Each rivet bears one-quarter of the force, so Shear stress  F|| (1.20  104 N)   6 .11  108 Pa A (.125  10 m) 11. 37: 11. 38: F A ( 90.8 N )   ( 0.9210