© 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–1 The members of a truss are pin connected at joint O Determine the magnitudes of F1 and F2 for equilibrium Set u = 60° y kN F2 70Њ 30Њ x O SOLUTION + ©F = 0; : x 4 F2 sin 70° + F1 cos 60° - cos 30° - (7) = kN u F1 0.9397F2 + 0.5F1 = 9.930 + c ©Fy = 0; F2 cos 70° + sin 30° - F1 sin 60° - (7) = 0.3420F2 - 0.8660F1 = 1.7 Solving: F2 = 9.60 kN Ans F1 = 1.83 kN Ans Ans: F2 = 9.60 kN F1 = 1.83 kN 161 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–2 The members of a truss are pin connected at joint O Determine the magnitude of F1 and its angle u for equilibrium Set F2 = kN y kN F2 70Њ 30Њ x O SOLUTION + ©F = 0; : x 4 sin 70° + F1 cos u - cos 30° - (7) = kN u F1 F1 cos u = 4.2920 + c ©Fy = 0; cos 70° + sin 30° - F1 sin u - (7) = F1 sin u = 0.3521 Solving: u = 4.69° Ans F1 = 4.31 kN Ans Ans: u = 4.69° F1 = 4.31 kN 162 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–3 Determine the magnitude and direction u of F so that the particle is in equilibrium y kN 30Њ x kN 60Њ kN Solution u Equations of Equilibrium Referring to the FBD shown in Fig a, + ΣFx = 0; F sin u + - cos 60° - cos 30° = S (1) F sin u = 3.9282 F +cΣFy = 0; sin 30° - sin 60° - F cos u = (2) F cos u = 0.5359 Divide Eq (1) by (2), sin u = 7.3301 cos u sin u Realizing that tan u = , then cos u tan u = 7.3301 Ans u = 82.23° = 82.2° Substitute this result into Eq (1), F sin 82.23° = 3.9282 Ans F = 3.9646 kN = 3.96 kN Ans: u = 82.2° F = 3.96 kN 163 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *3–4 The bearing consists of rollers, symmetrically confined within the housing The bottom one is subjected to a 125-N force at its contact A due to the load on the shaft Determine the normal reactions NB and NC on the bearing at its contact points B and C for equilibrium 40° SOLUTION + c ©Fy = 0; NB 125 - NC cos 40° = Ans NC = 163.176 = 163 N + ©F = 0; : x NC C B A 125 N NB - 163.176 sin 40° = Ans NB = 105 N Ans: NC = 163 N NB = 105 N 164 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–5 The members of a truss are connected to the gusset plate If the forces are concurrent at point O, determine the magnitudes of F and T for equilibrium Take u = 90° y kN F A B SOLUTION f = 90° - tan - a b = 53.13° O + ©F = 0; : x T cos 53.13° - F a b = + c ©Fy = 0; - T sin 53.13° - Fa b = x u C T Solving, T = 7.20 kN Ans F = 5.40 kN Ans Ans: T = 7.20 kN F = 5.40 kN 165 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–6 The gusset plate is subjected to the forces of three members Determine the tension force in member C and its angle u for equilibrium The forces are concurrent at point O Take F = kN y kN F A B SOLUTION + ©F = 0; : x T cos f - 8a b = (1) + c ©Fy = 0; - a b - T sin f = (2) Rearrange then divide Eq (1) into Eq (2): O x u C T tan f = 0.656, f = 33.27° T = 7.66 kN Ans u = f + tan - a b = 70.1° Ans Ans: T = 7.66 kN u = 70.1° 166 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–7 C The man attempts to pull down the tree using the cable and small pulley arrangement shown If the tension in AB is 60 lb, determine the tension in cable CAD and the angle u which the cable makes at the pulley D θ 20° A B 30° SOLUTION + R©Fx¿ = 0; 60 cos 10° - T - T cos u = +Q©Fy¿ = 0; T sin u - 60 sin 10° = Thus, T(1 + cos u) = 60 cos 10° u T(2cos2 ) = 60 cos 10° 2T sin (1) u u cos = 60 sin 10° 2 (2) Divide Eq.(2) by Eq.(1) tan u = tan 10° Ans u = 20° T Ans 30.5 lb Ans: u = 20° T = 30.5 lb 167 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *3–8 The cords ABC and BD can each support a maximum load of 100 lb Determine the maximum weight of the crate, and the angle u for equilibrium D u B A Equations of Equilibrium Assume that for equilibrium, the tension along the length of rope ABC is constant Assuming that the tension in cable BD reaches the limit first Then, TBD = 100 lb Referring to the FBD shown in Fig a, C + ΣFx = 0; W a b - 100 cos u = S 13 13 12 Solution 100 cos u = 5W 13 +cΣFy = 0; 100 sin u - W - W a 100 sin u = 25 W 13 (1) 12 b = 13 (2) Divide Eq (2) by (1), sin u = cos u Realizing that tan u = sin u , cos u tan u = Ans u = 78.69° = 78.7° Substitute this result into Eq (1), 100 cos 78.69° = W 13 W = 50.99 lb = 51.0 lb 100 lb (O.K) Ans Ans: u = 78.7° W = 51.0 lb 168 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–9 Determine the maximum force F that can be supported in the position shown if each chain can support a maximum tension of 600 lb before it fails B 30Њ A C F Solution Equations of Equilibrium Referring to the FBD shown in Fig a, +cΣFy = 0; TAB a b - F sin 30° = 0 TAB = 0.625 F + ΣFx = 0; TAC + 0.625 F a b - F cos 30° = 0 TAC = 0.4910 F S Since chain AB is subjected to a higher tension, its tension will reach the limit first Thus, TAB = 600; 0.625 F = 600 Ans F = 960 lb Ans: F = 960 lb 169 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–10 The block has a weight of 20 lb and is being hoisted at uniform velocity Determine the angle u for equilibrium and the force in cord AB B 20Њ A u C Solution D F Equations of Equilibrium Assume that for equilibrium, the tension along the length of cord CAD is constant Thus, F = 20 lb Referring to the FBD shown in Fig a, + ΣFx = 0; 20 sin u - TAB sin 20° = S TAB = 20 sin u sin 20° (1) +cΣFy = 0; TAB cos 20° - 20 cos u - 20 = (2) Substitute Eq (1) into (2), 20 sin u cos 20° - 20 cos u = 20 sin 20° sin u cos 20° - cos u sin 20° = sin 20° Realizing that sin (u - 20°) = sin u cos 20° - cos u sin 20°, then sin (u - 20°) = sin 20° u - 20° = 20° Ans u = 40° Substitute this result into Eq (1) TAB = 20 sin 40° = 37.59 lb = 37.6 lb sin 20° Ans Ans: u = 40° TAB = 37.6 lb 170 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–53 z If the tension developed in each of the cables cannot exceed 300 lb, determine the largest weight of the crate that can be supported Also, what is the force developed along strut AD? ft ft B C ft 5.5 ft A SOLUTION Force Vectors: We can express each of the forces on the free-body diagram shown in Fig a in Cartesian vector form as ( -2 - 0)i + (- - 0)j + (1.5 - 0)k FAB = FAB C 2 2( -2 - 0) + (-6 - 0) + (1.5 - 0) (2 - 0)i + ( -6 - 0)j + (3 - 0)k FAC = FAC C 2(2 - 0)2 + ( -6 - 0)2 + (3 - 0)2 S = - S = x 2.5 ft ft y 12 F i F j + F k 13 AB 13 AB 13 AB FAC i - FAC j + FAC k 7 (0 - 0)i + [0 - ( -6)]j + [0 - (- 2.5)]k FAD = FAD C D 2(0 - 0)2 + [0 - ( - 6)]2 + [0 - ( -2.5)]2 S = 12 FAD j + FAD k 13 13 W = -Wk Equations of Equilibrium: Equilibrium requires gF = 0; FAB + FAC + FAD + W = ¢- 12 12 FAB i FAB j + FAB k ≤ + ¢ FAC i - FAC j + FAC k ≤ + ¢ FAD j + FAD k ≤ + ( -Wk) = 13 13 13 7 13 13 ¢- 12 12 FAB + FAC ≤ i + ¢- FAB - FAC + FAD ≤ j + ¢ FAB + FAC + FAD - W ≤ k = 13 13 13 13 13 Equating the i, j, and k components yields - F + FAC = 13 AB (1) - 12 12 F - FAC + F = 13 AB 13 AD (2) 3 FAB + FAC + FAD - W = 13 13 (3) Let us assume that cable AC achieves maximum tension first Substituting FAC = 300 lb into Eqs (1) through (3) and solving, yields FAB = 278.57 lb FAD = 557 lb Ans W = 407 lb Since FAB = 278.57 lb 300 lb, our assumption is correct Ans: FAD = 557 lb W = 407 lb 213 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–54 z Determine the tension developed in each cable for equilibrium of the 300-lb crate ft ft B ft C ft D ft x ft A ft SOLUTION Force Vectors: We can express each of the forces shown in Fig a in Cartesian vector form as FAB = FAB C FAC = FAC C (-3 - 0)i + (-6 - 0)j + (2 - 0)k S = - FAB i - FAB j + FAB k 7 2(- - 0) + (-6 - 0) + (2 - 0) 2 (2 - 0)i + ( - - 0)j + (3 - 0)k 2 2(2 - 0) + ( -6 - 0) + (3 - 0) (0 - 0)i + (3 - 0)j + (4 - 0)k FAD = FAD C 2 2(0 - 0) + (3 - 0) + (4 - 0) S = S = F i - FAC j + FAC k AC 7 F j + FAD k AD W = {-300k} lb Equations of Equilibrium: Equilibrium requires g F = 0; FAB + FAC + FAD + W = ¢ - FAB i - 6 F j + FAB k ≤ + ¢ FAC i - FAC j + FAC k ≤ + ¢ FAD j + FAD k ≤ + ( -300k) = AB 7 7 5 Equating the i, j, and k components yields - FAB + FAC = 7 6 - FAB - FAC + FAD = 7 F + FAC + FAD - 300 = AB (1) (2) (3) Solving Eqs (1) through (3) yields FAB = 79.2 lb FAC = 119 lb Ans FAD = 283 lb Ans: FAB = 79.2 lb FAC = 119 lb FAD = 283 lb 214 y © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–55 z Determine the maximum weight of the crate that can be suspended from cables AB, AC, and AD so that the tension developed in any one of the cables does not exceed 250 lb ft ft B ft C ft D ft x ft A ft SOLUTION Force Vectors: We can express each of the forces shown in Fig a in Cartesian vector form as (- - 0)i + ( -6 - 0)j + (2 - 0)k FAB = FAB C S = - FAB i - FAB j + FAB k 7 2( -3 - 0) + (-6 - 0) + (2 - 0) 2 (2 - 0)i + ( -6 - 0)j + (3 - 0)k FAC = FAC C 2 2(2 - 0) + ( - - 0) + (3 - 0) FAD = FAD C (0 - 0)i + (3 - 0)j + (4 - 0)k 2 2(0 - 0) + (3 - 0) + (4 - 0) S = S = FAC i - FAC j + FAC k 7 FAD j + FAD k 5 W = -WC k Equations of Equilibrium: Equilibrium requires g F = 0; FAB + FAC + FAD + W = ¢ - FAB i - 6 FAB j + FAB k ≤ + ¢ FAC i - FAC j + FAC k ≤ + ¢ FAD j + FAD k ≤ + ( -WC k) = 7 7 5 ¢ - FAB + 6 3 FAC ≤ i + ¢ - FAB - FAC + FAD ≤ j + ¢ FAB + FAC + FAD - WC ≤ k = 7 7 Equating the i, j, and k components yields - FAB + FAC = 7 6 - FAB - FAC + FAD = 7 FAB + FAC + FAD - WC = 7 (1) (2) (3) Assuming that cable AD achieves maximum tension first, substituting FAD = 250 lb into Eqs (2) and (3), and solving Eqs (1) through (3) yields FAB = 70 lb WC = 265 lb FAC = 105 lb Ans Since FAB = 70 lb 250 lb and FAC = 105 lb, the above assumption is correct Ans: WC = 265 lb 215 y © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *3–56 The 25-kg flowerpot is supported at A by the three cords Determine the force acting in each cord for equilibrium z C D B 60Њ 30Њ 45Њ SOLUTION 30Њ FAD = FAD (sin 30°i - cos 30° sin 60°j + cos 30° cos 60°k) = 0.5FADi - 0.75FADj + 0.4330FAD k FAC = FAC (- sin 30°i - cos 30° sin 60°j + cos 30° cos 60°k) A x = -0.5FAC i - 0.75FACj + 0.4330FACk FAB = FAB(sin 45°j + cos 45°k) = 0.7071FAB j + 0.7071FAB k F = -25(9.81)k = {-245.25k} N ©F = ; FAD + FAB + FAC + F = (0.5FAD i - 0.75FAD j) + 0.4330FAD k + (0.7071FAB j + 0.7071FAB k) + ( -0.5FACi - 0.75FACj + 0.4330FACk) + ( -245.25k) = (0.5FAD - 0.5FAC)i + ( - 0.75FAD + 0.7071FAB - 0.75FAC) j + (0.4330FAD + 0.7071FAB + 0.4330FAC - 245.25) k = Thus, ©Fx = 0; 0.5FAD - 0.5FAC = [1] ©Fy = 0; -0.75FAD + 0.7071FAB - 0.75FAC = [2] ©Fz = 0; 0.4330FAD + 0.7071FAB + 0.4330FAC - 245.25 = [3] Solving Eqs [1], [2], and [3] yields: FAD = FAC = 104 N Ans FAB = 220 N Ans: FAD = FAC = 104 N FAB = 220 N 216 y © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–57 If each cord can sustain a maximum tension of 50 N before it fails, determine the greatest weight of the flowerpot the cords can support z C D B 60Њ 30Њ 45Њ SOLUTION 30Њ FAD = FAD (sin 30°i - cos 30° sin 60°j + cos 30° cos 60°k) A = 0.5FAD i - 0.75FAD j + 0.4330FAD k y x FAC = FAC (- sin 30°i - cos 30° sin 60° j + cos 30° cos 60° k) = -0.5FAC i - 0.75FAC j + 0.4330FAC k FAB = FAB (sin 45° j + cos 45° k) = 0.7071FAB j + 0.7071FAB k W = -Wk ©Fx = 0; 0.5FAD - 0.5FAC = (1) FAD = FAC ©Fy = 0; - 0.75FAD + 0.7071FAB - 0.75FAC = (2) 0.7071FAB = 1.5FAC ©Fz = 0; 0.4330FAD + 0.7071FAB + 0.4330FAC - W = 0.8660FAC + 1.5FAC - W = 2.366FAC = W Assume FAC = 50 N then FAB = 1.5(50) = 106.07 N 50 N (N G!) 0.7071 Assume FAB = 50 N Then FAC = 0.7071(50) = 23.57 N 50 N (O K!) 1.5 Thus, Ans W = 2.366(23.57) = 55.767 = 55.8 N Ans: W = 55.8 N 217 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–58 Determine the tension developed in the three cables required to support the traffic light, which has a mass of 15 kg Take h = m z C 6m D A h Solution 4m ΣFx = 0; ΣFy = 0; ΣFz = 0; F - FAC + AB F - FAC AB F - 15(9.81) AC 4m 3m = e - i - j + kf 7 uAD = e i - j f 5 B 4m uAB = e i + j f 5 uAC 3m x 4m 6m 3m y F = AD F = AD = FAB = 441 N Ans FAC = 515 N Ans FAD = 221 N Ans Ans: FAB = 441 N FAC = 515 N FAD = 221 N 218 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–59 z Determine the tension developed in the three cables required to support the traffic light, which has a mass of 20 kg Take h = 3.5 m C 6m D A h 3m B 4m Solution uAB = uAC = uAD = 4m 3i + j + 0.5k 2 23 + + (0.5) 3i + j + 0.5k = 225.25 - 6i - j + 2.5k 2( - 6)2 + ( - 3)2 + 2.52 4i - j + 0.5k = 242 + ( - 3)2 + 0.52 ΣFx = 0; ΣFy = 0; ΣFz = 0; Solving, 225.25 225.25 0.5 225.25 FAB FAB FAB + = 4m 3m x - 6i - j + 2.5k 4m 6m 3m y 251.25 4i - j + 0.5k 225.25 251.25 251.25 2.5 251.25 FAC + FAC FAC + 225.25 225.25 0.5 225.25 FAD = FAD = FAD - 20(9.81) = FAB = 348 N Ans FAC = 413 N Ans FAD = 174 N Ans Ans: FAB = 348 N FAC = 413 N FAD = 174 N 219 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *3–60 The 800-lb cylinder is supported by three chains as shown Determine the force in each chain for equilibrium Take d = ft z D 135 ft B y 135 90 C SOLUTION FAD = FAD £ FAC = FAC £ FAB = FAB ¢ -1j + 1k 2( -1)2 + 12 1i + 1k 212 + 12 d ≥ = - 0.7071FADj + 0.7071FADk x A ≥ = 0.7071FACi + 0.7071FACk - 0.7071i + 0.7071j + 1k 2( -0.7071)2 + 0.70712 + 12 ≤ = -0.5FAB i + 0.5FAB j + 0.7071FAB k F = {-800k} lb ©F = 0; FAD + FAC + FAB + F = ( - 0.7071FADj + 0.7071FADk) + (0.7071FACi + 0.7071FACk) + ( -0.5FAB i + 0.5FAB j + 0.7071FAB k) + (- 800k) = (0.7071FAC - 0.5FAB) i + (- 7071FAD + 0.5FAB)j + (0.7071FAD + 0.7071FAC + 0.7071FAB - 800) k = ©Fx = 0; 0.7071FAC - 0.5FAB = (1) ©Fy = 0; -0.7071FAD + 0.5FAB = (2) ©Fz = 0; 0.7071FAD + 0.7071FAC + 0.7071FAB - 800 = (3) Solving Eqs (1), (2), and (3) yields: FAB = 469 lb Ans FAC = FAD = 331 lb Ans: FAB = 469 lb FAC = FAD = 331 lb 220 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–61 Determine the tension in each cable for equilibrium z 800 N A D 3m 5m C 4m 2m O 5m Solution 4m ΣFy = 0; FABa 4 b - FAC a b - FADa b = 157 138 166 B (1) b + FAC a b - FADa b = 157 138 166 ΣFz = 0; - FABa y x Equations of Equilibrium Referring to the FBD shown in Fig a, ΣFx = 0; FABa 4m (2) 5 b - FAC a b - FADa b + 800 = 157 138 166 (3) Solving Eqs (1), (2) and (3) FAC = 85.77 N = 85.8 N Ans FAB = 577.73 N = 578 N Ans FAD = 565.15 N = 565 N Ans Ans: FAC = 85.8 N FAB = 578 N FAD = 565 N 221 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–62 If the maximum force in each rod can not exceed 1500 N, determine the greatest mass of the crate that can be supported z C B 3m 2m 2m 1m A 2m 2m O 3m y 1m 3m Solution Equations of Equilibrium Referring to the FBD shown in Fig a, x b - FOC a b + FOB a b = 0 (1) ΣFx = 0; FOAa 114 122 ΣFy = 0; - FOAa ΣFz = 0; FOAa 2 b + FOC a b + FOB a b = 114 122 (2) b + FOC a b - FOB a b - m(9.81) = 114 122 (3) Solving Eqs (1), (2) and (3), FOC = 16.95m FOA = 15.46m FOB = 7.745m Since link OC subjected to the greatest force, it will reach the limiting force first, that is FOC = 1500 N Then 1500 = 16.95 m m = 88.48 kg = 88.5 kg Ans Ans: m = 88.5 kg 222 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–63 The crate has a mass of 130 kg Determine the tension developed in each cable for equilibrium z A x 1m 1m 4m B 3m C D 2m 1m y Solution Equations of Equilibrium Referring to the FBD shown in Fig a, ΣFx = 0; FADa 2 b - FBD a b - FCD a b = 16 16 ΣFy = 0; - FADa ΣFz = 0; FADa (1) 1 b - FBD a b + FCD a b = 16 16 (2) 1 b + FBD a b + FCD a b - 130(9.81) = 16 16 (3) Solving Eqs (1), (2) and (3) FAD = 1561.92 N = 1.56 kN Ans FBD = 520.64 N = 521 N Ans FCD = 1275.3 N = 1.28 kN Ans Ans: FAD = 1.56 kN FBD = 521 N FCD = 1.28 kN 223 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *3–64 If cable AD is tightened by a turnbuckle and develops a tension of 1300 lb, determine the tension developed in cables AB and AC and the force developed along the antenna tower AE at point A z A 30 ft C 10 ft B E 15 ft SOLUTION x 12.5 ft 10 ft D 15 ft y Force Vectors: We can express each of the forces on the free-body diagram shown in Fig a in Cartesian vector form as FAB = FAB C 2(10 - 0) + ( -15 - 0) + ( - 30 - 0) FAC = FAC C S = - FAC i - FAC j - FAC k 7 2( - 15 - 0) + ( -10 - 0) + ( -30 - 0) FAD = FAD C (10 - 0)i + ( - 15 - 0)j + ( -30 - 0)k 2 S = F i - FAB j - FAB k AB 7 (- 15 - 0)i + ( -10 - 0)j + ( -30 - 0)k 2 (0 - 0)i + (12.5 - 0)j + ( -30 - 0)k 2(0 - 0)2 + (12.5 - 0)2 + ( -30 - 0)2 S = {500j - 1200k} lb FAE = FAE k Equations of Equilibrium: Equilibrium requires g F = 0; ¢ FAB i ¢ FAB FAB + FAC + FAD + FAE = 6 FAB j - FAB k ≤ + ¢- FAC i - FAC j - FAC k ≤ + (500j - 1200k) + FAE k = 7 7 3 6 F ≤ i + ¢- FAB - FAC + 500 ≤ j + ¢ - FAB - FAC + FAE - 1200 ≤ k = AC 7 7 Equating the i, j, and k components yields FAB - FAC = 7 - FAB - FAC + 500 = 7 6 - FAB - FAC + FAE - 1200 = 7 (1) (2) (3) Solving Eqs (1) through (3) yields Ans Ans Ans FAB = 808 lb FAC = 538 lb FAE = 2354 lb = 2.35 kip Ans: FAB = 808 lb FAC = 538 lb FAE = 2.35 kip 224 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–65 If the tension developed in either cable AB or AC cannot exceed 1000 lb, determine the maximum tension that can be developed in cable AD when it is tightened by the turnbuckle Also, what is the force developed along the antenna tower at point A? z A 30 ft C 10 ft SOLUTION B Force Vectors: We can express each of the forces on the free-body diagram shown in Fig a in Cartesian vector form as x FAB = FAB C FAC = FAC C FAD = FC (10 - 0)i + (- 15 - 0)j + (- 30 - 0)k 2(10 - 0)2 + ( -15 - 0)2 + (- 30 - 0)2 S = ( - 15 - 0)i + ( -10 - 0)j + (-30 - 0)k 2(- 15 - 0)2 + (-10 - 0)2 + ( - 30 - 0)2 (0 - 0)i + (12.5 - 0)j + ( -30 - 0)k 2(0 - 0)2 + (12.5 - 0)2 + (-30 - 0)2 S = E 15 ft 12.5 ft 10 ft D 15 ft y F i - FAB j - FAB k AB 7 S = - FAC i - FAC j - FAC k 7 12 Fj Fk 13 13 FAE = FAE k Equations of Equilibrium: Equilibrium requires g F = 0; ¢ FAB i ¢ FAB FAB + FAC + FAD + FAE = 12 6 FAB j - FAB k ≤ + ¢ - FAC i - FAC j - FAC k ≤ + ¢ Fj F k ≤ + FAE k = 7 7 13 13 3 6 12 F ≤ i + ¢ - FAB - FAC + F ≤ j + ¢ - FAB - FAC F + FAE ≤ k = AC 7 13 7 13 Equating the i, j, and k components yields F - FAC = AB - FAB - FAC + F = 7 13 12 F + FAE = - FAB - FAC 7 13 (1) (2) (3) Let us assume that cable AB achieves maximum tension first Substituting FAB = 1000 lb into Eqs (1) through (3) and solving yields FAC = 666.67 lb FAE = 2914 lb = 2.91 kip Ans F = 1610 lb = 1.61 kip Since FAC = 666.67 lb 1000 lb, our assumption is correct Ans: FAE = 2.91 kip F = 1.61 kip 225 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–66 Determine the tension developed in cables AB, AC, and AD required for equilibrium of the 300-lb crate z B C ft ft Force Vectors: We can express each of the forces on the free-body diagram shown in Fig (a) in Cartesian vector form as FAB = FAB C FAC = FAC C A ft (-2 - 0)i + (1 - 0)j + (2 - 0)k 2(-2 - 0)2 + (1 - 0)2 + (2 - 0)2 S = - ft ft ft SOLUTION ft y D x 2 F i + FAB j + FAB k AB 3 (-2 - 0)i + (- - 0)j + (1 - 0)k 2 S = - FAC i - FAC j + FAC k 3 2(-2 - 0) + (- - 0) + (1 - 0) 2 FAD = FAD i W = [- 300k] lb Equations of Equilibrium: Equilibrium requires ©F = 0; FAB + FAC + FAD + W = 2 2 a - FAB i + FAB j + FAB k b + a - FAC i - FAC j + FACkb + FAD i + ( - 300k) = 3 3 3 2 2 a - FAB - FAC + FAD b i + a FAB - FAC b j + a FAB + FAC - 300b k = 3 3 3 Equating the i, j, and k components yields 2 - FAB - FAC + FAD = 3 (1) - FAC = F AB (2) F + FAC - 300 = AB (3) Solving Eqs (1) through (3) yields FAB = 360 lb Ans FAC = 180 lb Ans FAD = 360 lb Ans Ans: FAB = 360 lb FAC = 180 lb FAD = 360 lb 226 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3–67 Determine the maximum weight of the crate so that the tension developed in any cable does not exceed 450 lb z B C ft ft Force Vectors: We can express each of the forces on the free-body diagram shown in Fig (a) in Cartesian vector form as FAB = FAB C FAC = FAC C A ft ( - - 0)i + (1 - 0)j + (2 - 0)k 2( -2 - 0)2 + (1 - 0)2 + (2 - 0)2 S = - ft ft ft SOLUTION ft y D x 2 FAB i + FAB j + FAB k 3 (- - 0)i + (- - 0)j + (1 - 0)k 2 S = - FAC i - FAC j + FAC k 3 2(-2 - 0) + ( -2 - 0) + (1 - 0) 2 FAD = FAD i W = -Wk Equations of Equilibrium: Equilibrium requires ©F = 0; FAB + FAC + FAD + W = 2 2 a - FAB i + FAB j + FAB k b + a - FAC i - FAC j + FAC kb + FAD i + (- Wk) = 3 3 3 2 2 a - FAB - FAC + FAD bi + a FAB - FAC bj + a FAB + FAC - Wbk = 3 3 3 Equating the i, j, and k components yields 2 - FAB - FAC + FAD = 3 (1) F - FAC = AB (2) F + FAC - W = AB (3) Let us assume that cable AB achieves maximum tension first Substituting FAB = 450 lb into Eqs (1) through (3) and solving, yields FAC = 225 lb FAD = 450 lb Ans W = 375 lb Since FAC = 225 lb 450 lb, our assumption is correct Ans: W = 375 lb 227 ... Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by. .. 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced,... 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced,