Vector Mechanics for Engineers Statics Part 4

Vector Mechanics for Engineers Statics Part 4 PROB LEM 2.58 For the cables of Prooblem 2.46, it is known thaat the maximuum 6 N in cable AC and 7500 N in cable B allowablle tension is 600 BC . Determi ne (a ) the maa ximum force e P that can bbe applied at C , ( b) the c orresponding value of α . SOLU TION T Free-Body D iagram ( a) Law of cosines For ce Triangle P 2 = (600) 2 + (7750)2 − 2(600)(750) cos (25° + 45°) PP = 784.02 N ( b) Law of sines P = 784 N  sin β °+ ° = sin (25 445 ) 784.02 N 600 N β = 46.0° ∴ α = 46.0° + 25° α = 71.0°  P ROBLEM 2.59 For the situatioon described in Figure P2.48, determine ( a ) t he value of α ffor which the tension in rope p BC is as sm mall ass possible, (b ) the corresponn ding value off the tension. SOLU TION T Free-Body D iagram For ce Triangle To be sm mallest, TBC m must be perpenndicular to thee direction of TAC. ( a) ( b) Thus, α = 5.00° TBC = (12 kN)sin 5° α = 5.00 °  TBC = 1.046 kN  PROBL EM E 2.60 Two cable s tied togetherr at C are loaaded as show n. n Determine the range of va lues of Q for which the tennsion will nott exceed 60 N in either cable . SOLU TION T ΣFx = 0: −TBC − Qcoos60° + 75 N = 0 Free-Boody Diagram TBC = 75 N − Q cos60° (1) ΣF y = 0: TAC − Q sin 60° = 0 TAC = Qsinn 60° Requireme ent: TAC = 60 N: From Eq. (2): ( Q s in 60° = 60 N (2) Q = 69.3 N Requireme ent: From Eq. (1): ( TBC = 60 N: 75 N − Q cos 60° = 60 N Q = 300.0 N 30.0 N ≤ Q ≤ 69.3 N  PROBLEM 2.61 A movable bin and its contents have a combined weight of 2.8 kN. Determine the shortest chain sling ACB that can be used to lift the loaded bin if the tension in the chain is not to exceed 5 kN. SOLUTION Free-Body Diagram tan α = h 0.6 m (1) Isosceles Force Triangle Law of sines: sin α = 1 2 (2.8 kN) TAC TAC = 5 kN sin α = 1 2 (2.8 kN) 5 kN α = 16.2602° From Eq. (1): tan16.2602 ° = h 0.6 m ∴ h= 0.175000 m Half-length of chain = AC = (0.6 m) 2 +(0.175 m) 2 = 0.625 m Total length: = 2 × 0.625 m 1.250 m  PROBLEM 2.62 For W = 800 N, P = 200 N, and d = 600 mm, determine the value of h consistent with equilibrium. SOLUTION Free-Body Diagram TAC = TBC = 800 N AC = BC = ΣF y = 0: 2(800 N) (h 2 +d2 h h2 + d 2 ) −P = 0 2 ⎛d ⎞ P 800 = 1+⎜ ⎟ 2 ⎝h ⎠ Data: P = 200 N, d = 600 mm and solving for h 2 800 N = ⎛600 mm ⎞ 200 N 1+ ⎜ ⎟ ⎝ h ⎠ 2 h = 75.6 mm  PROBLEM 2.63 PROBLEM 2.65 Three forces are applied to a bracket as shown. The directions of the two 150-N forces may vary, but the angle between these forces is always 50°. Determine the range of values of α for which the magnitude of the resultant of the forces acting at A is less than 600 N. SOLUTION Combine the two 150-N forces into a resultant force Q: Q = 2(150 N) cos 25 ° = 271.89 N Equivalent loading at A: Using the law of cosines: (600 N)2 = (500 N)2 + (271.89 N)2 + 2(500 N)(271.89 N) cos(55° + α ) cos(55° + α ) = 0.132685 Two values for α : or 55° +α = 82.375 α = 27.4 ° 55° +α = −82.375° 55° +α = 360° − 82.375° α = 222.6 ° For R < 600 lb: 27.4° < α < 222.6D  PROB L LEM 2.66 A 200-kg crate is to bee supported bby the rope-annd-pulley arrangement show wn. Determinee the magnituude and direction of the forrce P that must be exerted on the free ennd of the rope to maintain equilibrium. (H Hint: The tensiion in the ropee is the same oon each side of o a simple puulley. This cann be proved by y the methodss of Ch. 4.) SOLU TION T Free-Boody Diagram : Pulley A ⎛ 5 ⎞ ΣF x = 0: − 2 P ⎜ ⎟ +P c os α = 0 ⎝ 281 ⎠ cos α = 0.59655 α = ±53.377° For α = +53.377°: ⎛ 16 6 ⎞ ΣF y = 0: 2 P ⎜ ⎟ +P sin 533.377 ° − 1962 N = 0 ⎝ 281 ⎠ P = 7244 N 53.4°  For α = −53.377°: ⎛ 166 ⎞ ΣF y = 0: 2 P ⎜ ° ) − 19662 N = 0 5 ⎟ +P sin( −53.377 ⎝ 281 ⎠ P = 17773 53.4°  PROBLEM 2.67 PROBLEM 2.68 PR O BLEM 2.669 A lo ad Q is appl i ed to the pullley C, which can roll on the p is heldd in the position shown byy a cablee ACB. The pulley seconnd cable CAD, which pa sses over thee pulley A and a 50 N, determ ine supp orts a load P. Knowing that P = 75 ) the magnitudde of load Q . (a) thhe tension in ccable ACB, (b) SOLU TION T Free-Boody Diagram : Pulley C ΣFx =0: T ACBB (cos 25° − coss55°) − (750 N)cos55° = 0 (a) Henc e: TACB = 129 9 2.88 N = 1293 N  TACB A ΣF y = 0: TACB( sin 25 ° + sin 55°) + (750 N) sin 55° − Q = 0 (1292.88 N)(sin 25 ° + sin 55°) + (750 N) sin 55° − Q = 0 (b) or Q = 2219.8 N Q = 2220 N  PR OBLEM O 2. 770 C which can roll An 1800-N load Q is applied to o the pulley C, r on thhe cable ACB. The pulley is held in the poosition shown by a seccond cable CAD, which ppasses over thhe pulley A and a CB, supports a load P. Determine ( a) the tensionn in cable AC (b) thhe magnitude of load P. SOLU TION T Free-Boody Diagram : Pulley C ΣFx = 0: 0 TACB(cos225° − cos55°) − P cos55° = 0 P = 00.58010TACB (1) or ΣFy = 0: TACB(sin22 5° + sin 55°) + P sin 55 ° −1800 N = 0 TACB + 0.819155 P = 1800 N (2) 1.24177T or (a) Substitute Equation ( 1) into Equat ion (2): 1.244177 TACB + 0.81915(0.580110T ACB) = 18000 N Hence: TACB = 10488.37 N = 1048 N  TACB A (b) Using (1), P = 0.58010(1048.337 N) = 608.166 N P = 608 N  PROBLEM 2.71 Determine (a) the x, y, and z components of the 600-N force, (b) the angles θ x, θ y, and θ z that the force forms with the coordinate axes. SOLUTION (a) F x = (600 N)sin 25 °cos30 D F x = 219.60 N Fx = 220 N  F y = (600 N)cos 25° Fy = 544 N  F y = 543.78 N D F z = (380.36 N)sin 25° sin 30 F z = 126.785 N (b) cos θx = cos θ y = cos θz = Fx 219.60 N = F 600 N Fy Fz = 126.8 N  θ x = 68.5°  543.78 N 600 N θ y = 25.0°  Fz 126.785 N = 600 N F θ z = 77.8°  F = PROBLEM 2.72 Determine (a) the x, y, and z components of the 450-N force, (b) the angles θ x, θ y, and θ z that the force forms with the coordinate axes. SOLUTION (a) Fx = −(450 N)cos 35 °sin 40 D Fx = −236.94 N F x = −237 N  F y = (450 N) sin 35° (b) F y = 258.11 N Fy = 258 N  D Fz = (450 N) cos35 ° cos40 = Fz 282.38 N Fz =282 N  cos θx = cos θy = cos θz = Fx -236.94 N = F 450 N Fy θx = 121.8°  258.11 N 450 N θ y = 55.0°  Fz 282.38 N = 450 N F θ z = 51.1°  F = Note: From the given data, we could have computed directly θ y = 90D − 35 D= 55 D, which checks with the answer obtained. PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.) SOLUTION Recoil force F = 400 N ∴ (a) F H= (400 N)cos40° = 306.42 N Fx = − F H sin 35° = − (306.42 N)sin 35° = −175.755 N Fx = −175.8 N  F y = − F sin 40° = −(400 N)sin 40 ° = −257.12 N Fy = −257 N  Fz = + F H cos35° = +(306.42 N) cos 35 ° = +251.00 N (b) cosθ x = Fx −175.755 N = F 400 N cosθ y = cosθ z = Fy F = −257.12 N 400 N Fz 251.00 N = F 400 N Fz = + 251 N  θ x = 116.1 °  θ y = 130.0°  θ z = 51.1°  PROBLEM 2.74 Solve Problem 2.73, assuming that point A is located 15° north of west and that the barrel of the gun forms an angle of 25° with the horizontal. PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.) SOLUTION Recoil force F = 400 N ∴ (a) F H= (400 N)cos25 ° = 362.52 N Fx = + F H cos15 ° = +(362.52 N)cos15 ° = +350.17 N Fx = +350 N  F y = − F sin 25° = −(400 N)sin 25 ° = −169.047 N Fy = −169.0 N  Fz = + F H sin15° = +(362.52 N)sin15° = +93.827 N (b) cosθ x = cosθ y = cosθ z = Fx +350.17 N = F 400 N Fy F = −169.047 N 400 N Fz +93.827 N = F 400 N F z = + 93.8 N  θx = 28.9°  θ y = 115.0°  θ z = 76.4°  PROBLEM 2.75 The angle between spring AB and the post DA is 30°. Knowing that the tension in the spring is 50 N, determine (a) the x, y, and z components of the force exerted on the circular plate at B, (b) the angles θ x, θy, and θ z defining the direction of the force at B. SOLUTION Fh = F cos 60° = (50 N) cos 60 ° Fh = 25.0 N Fx = −Fh cos 35 ° D F y =F sin 60 D F z = −F h sin 35 D Fx = ( −25.0 N)cos35 D D Fy = (50.0 N)sin 60 F z = ( −25.0 N)sin 35 Fx = −20.479 N Fy = 43.301 N F z = −14.3394 N (a)  20.5 N Fx = −  43.3 N Fy =  Fz = −  14.33 N (b) cos θ x = cos θ y = cos θ z = Fx −20.479 N = F 50 N Fy F = 43.301 N 50 N Fz -14.3394 N = F 50 N θx = 114.2°  θ y = 30.0°  θz = 106.7°  PROBLEM 2.76 The angle between spring AC and the post DA is 30°. Knowing that the tension in the spring is 40 N, determine (a) the x, y, and z components of the force exerted on the circular plate at C, (b) the angles θx, θy, and θz defining the direction of the force at C. SOLUTION Fh = F cos 60 ° = (40 N) cos 60° Fh = 20.0 N (a) Fx = Fh cos 35° = (20.0 N) cos 35° Fx = 16.3830 N Fy = F sin 60° = (40 N)sin 60° Fy = 34.641 N Fz = −Fh sin 35 ° = −(20.0 N)sin 35° Fz = −11.4715 N  (b) cos θ x = cos θ y = cos θ z = Fx 16.3830 N = F 40 N Fy F Fz F = 34.641 N 40 N = -11.4715 N 40 N F16.38 = N x  Fy =N 34.6  Fz 11.47 =− N  θ x = 65.8°  θ y = 30.0°  θz = 106.7° 