Solutions to Phsics: Principles with pplications, 5/E, Giancoli Chapter 9 CHPTER 9 6. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the point from the force diagram for the leg: τ = = 0; (15.0 kg)(9.80 m/s 2 )(0.350 m) (0.805 m), which gives = 63.9 N. ecause there is no acceleration of the hanging mass, we have =, or m = /g = (63.9 N)/(9.80 m/s 2 ) = Hip Joint 6.52 kg. CM 7. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the point from the force diagram for the beam and piano: τ = ( +! F N2 = 0, which gives F N2 = ( +! = ((300 kg)(9.80 m/s 2 ) +!(160 kg)(9.80 m/s 2 ) = 1.52 10 3 N. We write F = ma from the force diagram for the beam and piano: F N1 + F N2 = 0, which gives F N1 = + F N2 = (300 kg)(9.80 m/s 2 ) +!(160 kg)(9.80 m/s 2 ) 1.52 10 3 N = 2.94 10 3 N. The forces on the supports are the reactions to these forces: 2.94 10 3 N down, and 1.52 10 3 N down. F N 1 FN2 8. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the point from the force diagram for the two beams: τ = ( +! F N2 = 0, which gives F N2 = ( +! = ([!(1000 kg)](9.80 m/s 2 ) +!(1000 kg)(9.80 m/s 2 ) = 6.13 10 3 N. We write F = ma from the force diagram for the two beams: F N1 + F N2 = 0, which gives F N1 = + F N2 = (1000 kg)(9.80 m/s 2 ) +!(1000 kg)(9.80 m/s 2 ) 6.13 10 3 N = 8.57 10 3 N. F N 1 F N2 9. We must move the direction of the net force 10 to the right. We choose the coordinate sstem shown, with the -ais in the direction of the original net force. We write F from the force diagram for the tooth: F = cos 20 + cos 20 = F net sin 10 ; (2.0 N) cos 20 + cos 20 = F net sin 10 ; F = + sin 20 + sin 20 = F net cos 10 ; + (2.0 N) sin 20 + sin 20 = F net cos 10. 20 10 F net 20 Page 9 1
Solutions to Phsics: Principles with pplications, 5/E, Giancoli Chapter 9 We have two equations for the two unknowns: F net, and. When we eliminate F net, we get = 2.3 N. Page 9 2
Solutions to Phsics: Principles with pplications, 5/E, Giancoli Chapter 9 10. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the support point from the force diagram for the board and people: τ = m 1 g( d) + m 2 gd = 0; m 1 g d d m 2 g (30 kg)(10 m d) + (70 kg)d = 0, which gives d = 3.0 m from the adult. F N 11. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the support point from the force diagram for the board and people: τ = m 1 g( d) (! d) + m 2 gd = 0; m 1 g d d m 2 g (30 kg)(10 m d) (15 kg)(5.0 m d) + (70 kg)d = 0, which gives d = 3.3 m from the adult. F N 12. From the force diagram for the mass we can write F cos = 0, or = cos 30. F = sin = 0, or sin 30 = = (200 kg)(9.80 m/s 2 ), which gives = 3.9 10 3 N. FT1 Thus we have = cos 30 = (3.9 10 3 N) cos 30 = 3.4 10 3 N. 13. From the force diagram for the hanging light and junction we can write F cos 1 cos 2 = 0; FT1 cos 37 = cos 53 ; F sin 1 + sin 2 = 0; sin 37 + sin 53 = (30 kg)(9.80 m/s 2 ). 2 1 When we solve these two equations for the two unknowns,, and, we get = 1.8 10 2 N, and = 2.4 10 2 N. 14. From the force diagram for the seesaw and children we can write F = F N m 1 g m 2 g = 0, or F N = (m 1 + m 2 + M)g = (30 kg + 25 kg + 2.0 kg)(9.80 m/s 2 ) = 5.6 10 2 N. d F N m 1 g m 2 g Page 9 3
Solutions to Phsics: Principles with pplications, 5/E, Giancoli Chapter 9 15. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the support point from the force diagram for the cantilever: τ = F 2 d +! = 0; F 2 (20.0 m) + (1200 kg)(9.80 m/s 2 )(25.0 m) = 0, which gives F 2 = 1.47 10 4 N. For the forces in the -direction we have F = F 1 + F 2 = 0, or F 1 = F 2 = (1200 kg)(9.80 m/s 2 ) (1.47 10 4 N) = F 1 F 2 d 2.94 10 3 N (down). 16. From the force diagram for the sheet we can write F = cos cos = 0, which gives. F sin + sin = 0; 2 sin = ; 2 sin 3.5 = (0.60 kg)(9.80 m/s 2 ), which gives = 48 N; The tension is so much greater than the weight because onl the vertical components balance the weight. 17. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the lower hinge from the force diagram for the door: τ = F (H 2)!w = 0; w F F F [2.30 m 2(0.40 m)] (13.0 kg)(9.80 m/s 2 )!(1.30 m), H F which gives F = 55.2 N. We write F = ma from the force diagram for the door: F = F + F = 0; 55 N + F = 0, which gives F = 55.2 N. The top hinge pulls awa from the door, and the bottom hinge pushes on the door. F = F + F = 0. ecause each hinge supports half the weight, we have F = F =!(3.0 kg)(9.80 m/s 2 ) = 63.7 N. F Thus we have top hinge: F = 55.2 N, F = 63.7 N; bottom hinge: F = 55.2 N, F = 63.7 N. 18. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the support point from the force diagram for the seesaw and bos: τ = + m 2 g! + m 3 g m 1 g! = 0; + (35 kg)!(3.6 m) + (25 kg) (50 kg)!(3.6 m) = 0, which gives = 1.1 m. The third bo should be 1.1 m from pivot on side of lighter bo. m 1 g F N m 3 g m 2 g Page 9 4
Solutions to Phsics: Principles with pplications, 5/E, Giancoli Chapter 9 19. We choose the coordinate sstem shown, with positive torques clockwise. For the torques about the point we have τ = F 1 d + = 0; F 1 (1.0 m) + (60 kg)(9.80 m/s 2 )(3.0 m) = 0, which gives F 1 = 1.8 10 3 N (down). For the torques about the point we have τ = F 2 d + ( + d)= 0; F 2 (1.0 m) = (60 kg)(9.80 m/s 2 )(4.0 m + 1.0 m), which gives F 2 = 2.4 10 3 N (up). F 1 F2 d 20. We choose the coordinate sstem shown, with positive torques clockwise. For the torques about the point we have τ = F 1 d + + [!( + d) d] = 0; F 1 (1.0 m) = (60 kg)(9.80 m/s 2 )(3.0 m) (35 kg)(9.80 m/s 2 )(1.0 m), which gives F 1 = 2.1 10 3 N (down). For the torques about the point we have τ = F 2 d + +!( + d) = 0; F 2 (1.0 m) = (60 kg)(9.80 m/s 2 )(3.0 m) + (35 kg)(9.80 m/s 2 )(2.0 m), which gives F 2 = 3.0 10 3 N (up). F 1 d F2 21. From Eample 7 16 we have = (20.4/100)(1.60 m) = 0.326 m. From Table 7 1, we have = [(52.1 4.0)/100](1.60 m) = 0.770 m; M = = 10.35 kg. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the hip joint from the force diagram for the leg: τ = = 0; (10.35 kg)(9.80 m/s 2 )(0.326 m) (0.770 m) = 0. which gives = 42.9 N. ecause there is no acceleration of the hanging mass, we have =, or m = /g = (42.9 N)/(9.80 m/s 2 ) = 4.38 kg. Hip Joint CM 22. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the support point from the force diagram for the beam: τ = + F 1 (d 1 + d 2 + d 3 + d 4 ) F 3 (d 2 + d 3 + d 4 ) F 4 (d 3 + d 4 ) F 5 d 4!(d 1 + d 2 + d 3 + d 4 ) = 0; F 1 (10.0 m) (4000 N)(8.0 m) (3000 N)(4.0 m) (2000 N)(1.0 m) (250 kg)(9.80 m/s 2 )(5.0 m) = 0, which gives F 1 = 5.8 10 3 N. We write τ = Iα about the support point from the force diagram for the beam: τ = F 2 (d 1 + d 2 + d 3 + d 4 ) + F 3 d 1 F 4 (d 1 + d 2 ) F 5 (d 1 + d 2 + d 3 )!(d 1 + d 2 + d 3 + d 4 ) = 0; F 2 (10.0 m) (4000 N)(2.0 m) (3000 N)(6.0 m) F 1 F 3 F4 F5 d 1 d 2 d 3 d 4 F 2 Page 9 5
Solutions to Phsics: Principles with pplications, 5/E, Giancoli Chapter 9 (2000 N)(9.0 m) (250 kg)(9.80 m/s 2 )(5.0 m) = 0, which gives F 2 = 5.6 10 3 N. Page 9 6
Solutions to Phsics: Principles with pplications, 5/E, Giancoli Chapter 9 23. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the point from the force diagram for the beam: τ = ( sin α) +! = 0; F W sin 50 + (30 kg)(9.80 m/s 2 )! = 0, which gives = 1.9 10 2 N. Note that we find the torque produced b the tension b finding the torques produced b the components. We write F = ma from the force diagram for the beam: F = F W cos α = 0; F W (1.9 10 2 N) cos 50 = 0, which gives F W = 123 N. F = F W + sin α = 0; F W + (1.9 10 2 N) sin 50 (30 kg)(9.80 m/s 2 ) = 0, which gives F W = 147 N. For the magnitude of F W we have F W = (F W 2 + F W 2 ) 1/2 = [(123 N) 2 + (147 N) 2 ] 1/2 = 1.9 10 2 N. We find the direction from tan = F W /F W = (147 N)/(123 N) = 1.19, which gives = 50. Thus the force at the wall is F W = 1.9 10 2 N, 50 above the horizontal. α 24. ecause the backpack is at the midpoint of the rope, the angles are equal. The force eerted b the backpacker is the tension in the rope. From the force diagram for the backpack and junction we can write F cos cos = 0, or = = F; F sin + sin = 0, or 2F sin =. (a) We find the angle from tan = h/! = (1.5 m)/!(7.6 m) = 0.395, or = 21.5. When we put this in the force equation, we get 2F sin 21.5 = (16 kg)(9.80 m/s2 ), which gives F = 2.1 10 2 N. (b) We find the angle from tan = h/! = (0.15 m)/!(7.6 m) = 0.0395, or = 2.26. When we put this in the force equation, we get 2F sin 2.26 = (16 kg)(9.80 m/s2 ), which gives F = 2.0 10 3 N. 25. We choose the coordinate sstem shown, with positive torques clockwise. For the torques about the CG we have τ CG = F N1 ( ) F N2 = 0; (35.1 kg)g(170 cm ) (31.6 kg)g = 0, which gives = 89.5 cm from the feet. F N1 CG F N2 Page 9 7
Solutions to Phsics: Principles with pplications, 5/E, Giancoli Chapter 9 26. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the point from the force diagram for the beam and sign: τ = ( sin ) + +! = 0; (sin 41.0 )(1.35 m) + (215 N)(1.70 m) + (135 N)!(1.70 m) = 0, which gives = 542 N. Note that we find the torque produced b the tension b finding the torques produced b the components. We write F = ma from the force diagram for the beam and sign: F = F hinge cos = 0; F hinge (542 N) cos 41.0 = 0, which gives F hinge = 409 N. F = F hinge + sin = 0; F hinge + (542 N) cos 41.0 215 N 135 N = 0, which gives F hinge = 6 N (down). F hingev F hingeh 27. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the point from the force diagram for the pole and light: τ = H + cos +! cos = 0; (3.80 m) + (12.0 kg)(9.80 m/s 2 )(7.5 m) cos 37 + (8.0 kg)(9.80 m/s 2 )!(7.5 m) cos 37 = 0, which gives = 2.5 10 2 N. We write F = ma from the force diagram for the pole and light: F = F H = 0; F H 2.5 10 2 N = 0, which gives F H = 2.5 10 2 N. F = F V = 0; F V (12.0 kg)(9.80 m/s 2 ) (8.0 kg)(9.80 m/s 2 ) = 0, which gives F V = 2.0 10 2 N. H F V F H 28. We choose the coordinate sstem shown, with positive torques counterclockwise. We write τ = Iα about the point from the force diagram for the ladder: τ = (!) cos F N2 sin = 0, which gives F N2 = /2 tan. We write F = ma from the force diagram for the ladder: F fr F N2 = 0, which gives F fr = F N2 = /2 tan. We write F = ma from the force diagram for the ladder: F N1 = 0, which gives F N1 =. For the bottom not to slip, we must have F fr ² µf N1, or /2 tan ² µ, from which we get tan ³ 1/2µ. The minimum angle is min = tan 1 (1/2µ). F fr F N1 O F N2 Page 9 8