PROBLEM 4.93 A 4 × 8-ft sheet of plywood weighing 40 lb has been temporarily propped against column CD. It rests at A and B on small wooden blocks and against protruding nails. Neglecting friction at all surfaces of contact, determine the reactions at A, B, and C. SOLUTION Free-Body Diagram: We have five unknowns and six equations of equilibrium. Plywood sheet is free to move in x direction, but equilibrium is maintained (ΣFx = 0). ΣM A = 0: rB /A × ( B y j + Bz k ) + rC /A × C k + rG /A × ( −40 lb) j = 0 i j 5 0 0 By k i j k i j k 0 + 4 4sin 60° −4cos 60° + 2 2sin 60° −2 cos 60° = 0 Bz 0 C −40 0 0 0 (4C sin 60° − 80 cos 60°) i + (−5Bz − 4C ) j + (5B y − 80)k = 0 Equating the coefficients of the unit vectors to zero, i: 4C sin 60° − 80 cos 60° = 0 j: −5Bz − 4C = 0 k: 5B y − 80 = 0 ΣFy = 0: Ay + B y − 40 = 0 ΣFz = 0: Az + Bz + C = 0 C = 11.5470 lb Bz = 9.2376 lb B y = 16.0000 lb Ay = 40 − 16.0000 = 24.000 lb Az = 9.2376 − 11.5470 = −2.3094 lb A = (24.0 lb)j − (2.31 lb)k ; B = (16.00 lb) j − (9.24 lb)k ; C = (11.55 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 450 PROBLEM 4.99 The 45-lb square plate shown is supported by three vertical wires. Determine the tension in each wire. SOLUTION Free-Body Diagram: ΣM B = 0: rC /B × TC j + rA/B × TA j + rG /B × (−45 lb) j = 0 [−(20 in.)i + (15 in.)k ] × TC j + (20 in.)k × TA j + [−(10 in.)i + (10 in.)k ] × [−(45 lb)j] = 0 −20TC k − 15TC i − 20TAi + 450k + 450i = 0 Equating to zero the coefficients of the unit vectors, k: −20TC + 450 = 0 TC = 22.5 lb i: −15(22.5) − 20TA + 450 = 0 TA = 5.625 lb ΣFy = 0: TA + TB + TC − 45 lb = 0 5.625 lb + TB + 22.5 lb − 45 lb = 0 TB = 16.875 lb TA = 5.63 lb; TB = 16.88 lb; TC = 22.5 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 458 PROBLEM 4.105 A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C. SOLUTION Free-Body Diagram: (ΣMAC = 0). Five unknowns and six equations of equilibrium, but equilibrium is maintained rB = 1.2k rA = 2.4k AD = −0.8i + 0.6 j − 2.4k AD = 2.6 m AE = 0.8i + 1.2 j − 2.4k AE = 2.8 m AD TAD = (−0.8i + 0.6 j − 2.4k ) AD 2.6 AE TAE = = (0.8i + 1.2 j − 2.4k ) AE 2.8 TAD = TAE ΣM C = 0: rA × TAD + rA × TAE + rB × (−3 kN) j = 0 i j k i j k TAD T 0 0 2.4 0 2.4 AE + 1.2k × (−3.6 kN) j = 0 + 0 2.6 2.8 0.8 1.2 −2.4 −0.8 0.6 −2.4 Equate coefficients of unit vectors to zero: i : − 0.55385TAD − 1.02857TAE + 4.32 = 0 (1) j : − 0.73846TAD + 0.68671TAE = 0 TAD = 0.92857TAE From Eq. (1): (2) −0.55385(0.92857)TAE − 1.02857TAE + 4.32 = 0 1.54286TAE = 4.32 TAE = 2.800 kN TAE = 2.80 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 465 PROBLEM 4.105 (Continued) From Eq. (2): TAD = 0.92857(2.80) = 2.600 kN TAD = 2.60 kN 0.8 0.8 (2.6 kN) + (2.8 kN) = 0 2.6 2.8 0.6 1.2 (2.6 kN) + (2.8 kN) − (3.6 kN) = 0 ΣFy = 0: C y + 2.6 2.8 2.4 2.4 (2.6 kN) − (2.8 kN) = 0 ΣFz = 0: C z − 2.6 2.8 ΣFx = 0: C x − Cx = 0 C y = 1.800 kN C z = 4.80 kN C = (1.800 kN) j + (4.80 kN)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 466 PROBLEM 4.113 A 100-kg uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE that passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust. SOLUTION rB/A (960 − 180)i = 780i Dimensions in mm 960 450 − 90 i + k 2 2 = 390i + 225k = 600i + 450k rG/A = rC/A T = Tension in cable DCE CD = −690i + 675 j − 450k CD = 1065 mm CE = 270i + 675 j − 450k CE = 855 mm T (−690i + 675 j − 450k ) 1065 T TCE = (270i + 675 j − 450k ) 855 W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N) j TCD = ΣM A = 0: rC/A × TCD + rC/A × TCE + rG/A × (−Wj) + rB/A × B = 0 i j k i j k T T 600 0 450 450 + 600 0 1065 855 270 675 −450 −690 675 −450 i + 390 0 j 0 −981 k i 225 + 780 0 0 j 0 k 0 =0 By Bz PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 481 PROBLEM 4.113 (Continued) Coefficient of i: −(450)(675) T T − (450)(675) + 220.73 × 103 = 0 1065 855 T = 344.64 N Coefficient of j: (−690 × 450 + 600 × 450) T = 345 N 344.64 344.64 + (270 × 450 + 600 × 450) − 780 Bz = 0 1065 855 Bz = 185.516 N Coefficient of k: (600)(675) 344.64 344.64 + (600)(675) − 382.59 × 103 + 780 By = 0 By = 113.178 N 1065 855 B = (113.2 N) j + (185.5 N)k ΣF = 0: A + B + TCD + TCE + W = 0 690 270 (344.64) + (344.64) = 0 1065 855 Coefficient of i: Ax − Ax = 114.5 N Coefficient of j: Ay + 113.178 + 675 675 (344.64) + (344.64) − 981 = 0 1065 855 Ay = 377 N Coefficient of k: Az + 185.516 − 450 450 (344.64) − (344.64) = 0 1065 855 Az = 141.5 N A = (114.5 N)i + (377 N) j + (144.5 N)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 482 PROBLEM 4.138 The frame ACD is supported by ball-and-socket joints at A and D and by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the frame supports at Point C a load of magnitude P = 268 N, determine the tension in the cable. SOLUTION Free-Body Diagram: AD (1 m)i − (0.75 m)k = AD 1.25 m = 0.8i − 0.6k λ AD = λ AD TBG = TBG = TBG TBH = TBH = TBH BG BG −0.5i + 0.925 j − 0.4k 1.125 BH BH 0.375i + 0.75 j − 0.75k 1.125 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 522 PROBLEM 4.138 (Continued) rB /A = (0.5 m)i; rC/A = (1 m)i; P = −(268 N)j To eliminate the reactions at A and D, we shall write ΣM AD = 0: λ AD ⋅ (rB /A × TBG ) + λ AD ⋅ (rB /A × TBH ) + λ AD ⋅ (rC /A × P) = 0 (1) Substituting for terms in Eq. (1) and using determinants, −0.6 −0.6 −0.6 0.8 0 0.8 0 0.8 0 TBG TBH + 0.5 + 1 0.5 0 0 0 0 0 0 =0 1.125 1.125 −0.5 0.925 −0.4 0.375 0.75 −0.75 0 −268 0 Multiplying all terms by (–1.125), 0.27750TBG + 0.22500TBH = 180.900 For this problem, (2) TBG = TBH = T (0.27750 + 0.22500)T = 180.900 T = 360 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 523

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