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© 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 5–10 Determine the components of the support reactions at the fixed support A on the cantilevered beam kN 30Њ SOLUTION 30Њ Equations of Equilibrium: From the free-body diagram of the cantilever beam, Fig a, Ax, Ay, and MA can be obtained by writing the moment equation of equilibrium about point A + ©F = 0; : x 1.5 m 1.5 m 1.5 m kN cos 30° - A x = Ans A x = 3.46 kN + c ©Fy = 0; A A y - - sin 30° = Ans A y = kN a+ ©MA = 0;MA - 6(1.5) - cos 30° (1.5 sin 30°) - sin 30°(3 + 1.5 cos 30°) = MA = 20.2 kN # m Ans Ans: Ax = 3.46 kN Ay = kN MA = 20.2 kN # m 397 © 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 5–11 Determine the reactions at the supports 400 N/m B A 3m 3m Solution Equations of Equilibrium NA and By can be determined directly by writing the moment equations of equilibrium about points B and A, respectively, by referring to the beam’s FBD shown in Fig a a+ ΣMB = 0; (400)(6)(3) - NA a b(6) = NA = 750 N a+ ΣMA = 0; By(6) - Ans (400)(6)(3) = Ans By = 600 N Using the result of NA to write the force equation of equilibrium along the x axis, + ΣFx = 0; 750 a b - Bx = S Ans Bx = 450 N Ans: NA = 750 N By = 600 N Bx = 450 N 398 © 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 *5–12 kN Determine the horizontal and vertical components of reaction at the pin A and the reaction of the rocker B on the beam B A 30Њ SOLUTION Equations of Equilibrium: From the free-body diagram of the beam, Fig a, NB can be obtained by writing the moment equation of equilibrium about point A a + ©MA = 0; 6m 2m NB cos 30°(8) - 4(6) = Ans NB = 3.464 kN = 3.46 kN Using this result and writing the force equations of equilibrium along the x and y axes, we have + ©F = 0; : x A x - 3.464 sin 30° = Ans A x = 1.73 kN + c ©Fy = 0; A y + 3.464 cos 30° - = Ans A y = 1.00 kN Ans: NB = 3.46 kN Ax = 1.73 kN Ay = 1.00 kN 399 © 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 5–13 Determine the reactions at the supports 900 N/m 600 N/m B A 3m 3m Solution Equations of Equilibrium NA and By can be determined directly by writing the moment equations of equilibrium about points B and A, respectively, by referring to the FBD of the beam shown in Fig a a+ ΣMB = 0; 600(6)(3) + (300)(3)(5) - NA(6) = Ans NA = 2175 N = 2.175 kN a+ ΣMA = 0; By(6) - (300)(3)(1) - 600(6)(3) = Ans By = 1875 N = 1.875 kN Also, Bx can be determined directly by writing the force equation of equilibrium along the x axis + ΣFx = 0; Bx = 0 S Ans Ans: NA = 2.175 kN By = 1.875 kN Bx = 400 © 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 5–14 Determine the reactions at the supports 800 N/m A 3m B 1m 3m Solution Equations of Equilibrium NA can be determined directly by writing the moment equation of equilibrium about point B by referring to the FBD of the beam shown in Fig a a+ ΣMB = 0; 800(5)(2.5) - NA(3) = Ans NA = 3333.33 N = 3.33 kN Using this result to write the force equations of equilibrium along the x and y axes, + ΣFx = 0; Bx - 800(5) a b = S Ans Bx = 2400 N = 2.40 kN + c ΣFy = 0; 3333.33 - 800 (5)a b - By = Ans By = 133.33 N = 133 N Ans: NA = 3.33 kN Bx = 2.40 kN By = 133 N 401 © 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 5–15 Determine the reactions at the supports kN 2m B A kN 2m 2m kN 2m Solution Equations of Equilibrium Ay and NB can be determined by writing the moment equations of equilibrium about points B and A, respectively, by referring to the FBD of the truss shown in Fig a a+ ΣMB = 0; 8(2) + 6(4) - 5(2) - Ay(6) = Ans Ay = 5.00 kN a+ ΣMA = 0; NB(6) - 8(4) - 6(2) - 5(2) = Ans NB = 9.00 kN Also, Ax can be determined directly by writing the force equation of equilibrium along x axis + ΣFx = 0; 5 - Ax = 0 Ax = 5.00 kN S Ans Ans: Ay = 5.00 kN NB = 9.00 kN Ax = 5.00 kN 402 © 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 *5–16 Determine the tension in the cable and the horizontal and vertical components of reaction of the pin A The pulley at D is frictionless and the cylinder weighs 80 lb D A SOLUTION B ft C ft ft Equations of Equilibrium: The tension force developed in the cable is the same throughout the whole cable The force in the cable can be obtained directly by summing moments about point A a + ©MA = 0; + ©F = 0; : x T152 + T ¢ ≤ 1102 - 801132 = 25 T = 74.583 lb = 74.6 lb Ax - 74.583 ¢ 25 Ans ≤ = Ans Ax = 33.4 lb + c ©Fy = 0; 74.583 + 74.583 25 - 80 - By = Ans Ay = 61.3 lb Ans: T = 74.6 lb Ax = 33.4 lb Ay = 61.3 lb 403 © 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 5–17 The man attempts to support the load of boards having a weight W and a center of gravity at G If he is standing on a smooth floor, determine the smallest angle u at which he can hold them up in the position shown Neglect his weight ft u SOLUTION a + ©MB = 0; G ft - NA (3.5) + W(3 - cos u) = As u becomes smaller, NA : so that, A W(3 - cos u) = 0.5 ft B ft Ans u = 41.4° Ans: u = 41.4° 404 © 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 5–18 Determine the components of reaction at the supports A and B on the rod P L –– SOLUTION Equations of Equilibrium: Since the roller at A offers no resistance to vertical movement, the vertical component of reaction at support A is equal to zero From the free-body diagram, Ax, By, and MA can be obtained by writing the force equations of equilibrium along the x and y axes and the moment equation of equilibrium about point B, respectively + ©F = 0; : x Ax = + c ©Fy = 0; By - P = Pa B Ans Ans By = P a + ©MB = 0; A L –– L b - MA = MA = PL Ans Ans: Ax = By = P PL MA = 405 © 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 5–19 The man has a weight W and stands at the center of the plank If the planes at A and B are smooth, determine the tension in the cord in terms of W and u B SOLUTION a + ©MB = 0; L W a cos f b - NA(L cos f ) = + ©F = 0; : x T cos u -NB sin u = + c ©Fy = 0; T sin u +NB cos u + NA f W = L u A (1) W - W= (2) Solving Eqs (1) and (2) yields: T= NB = W sin u Ans W cos u Ans: W T= sin u 406 © 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 5–75 z Member AB is supported by a cable BC and at A by a square rod which fits loosely through the square hole in the collar fixed to the member as shown Determine the components of reaction at A and the tension in the cable needed to hold the rod in equilibrium 1.5 m 400 N 3m A y C 200 N Solution x Force And Position Vectors The coordinates of points B and C are B(3, 0, - 1) m C(0, 1.5, 0) m, respectively TBC = TBC a B (0 - 3)i + (1.5 - 0)j + [0 - ( - 1)]k rBC b = TBC ả rBC 2(0 - 3)2 + (1.5 - 0)2 + [0 - ( -1)]2 = - TBC i + TBC j + T k 7 BC F = {200j - 400k} N FA = Ax i + Ay j MA = (MA)x i + (MA)y j + (MA)z k r1{3 i} m r2 = {1.5 j} m Equations of Equilibrium Referring to the FBD of member AB shown in Fig a, the force equation of equilibrium gives ΣF = 0; TBC + F + FA = a - TBC + Ax bi + a TBC + 200 + Ayb j + a TBC - 400bk = 7 Equating i, j and k components - T + Ax = 0 BC (1) T + 200 + Ay = 0 BC (2) T - 400 = 0 BC (3) 466 1m © 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 5–75 Continued The moment equation of equilibrium gives ΣMA = O; MA + r1 * F + r2 * TBC = i ( MA ) x i + ( MA ) y j + ( MA ) zk + † j 200 k † + - 400 i - TBC j 1.5 TBC k TBC T d i + ( MA ) y + 1200 j + c ( MA ) z + TBC + 600 d k = BC Equating i, j, and k components, ( MA ) x + TBC = 0 = c ( MA ) x + ( MA ) y + 1200 = 0 (4) (5) TBC + 600 = 0 Solving Eqs (1) to (6), ( MA ) z + (6) Ans TBC = 1400 N = 1.40 kN Ay = 800 N Ans Ax = 1200 N = 1.20 kN Ans ( MA ) x = 600 N # m Ans ( MA ) y = - 1200 N # m = 1.20 kN # m Ans ( MA ) z = - 2400 N # m = 2.40 kN # m Ans The negative signs indicate that Ay, ( MA ) x, ( MA ) y and ( MA ) z are directed in sense opposite to those shown in FBD Ans: TBC = 1.40 kN Ay = 800 N Ax = 1.20 kN ( MA ) x = 600 N # m ( MA ) y = 1.20 kN # m ( MA ) z = 2.40 kN # m 467 © 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 *5–76 z The member is supported by a pin at A and cable BC Determine the components of reaction at these supports if the cylinder has a mass of 40 kg 0.5 m B 1m A D Solution y 1m Force And Position Vectors The coordinates of points B, C and D are B(0, -0.5, 1) m, x C(3, 1, 0) m and D(3, - 1, 0) m, respectively FCB = FCB a (0 - 3)i + ( -0.5 - 1)j + (1 - 0)k rCB b = FCB c d rCB 2(0 - 3)2 + ( - 0.5 - 1)2 + (1 - 0)2 = - FCBi - FCBj + FCBk 7 W = { - 40(9.81)k} N = { -392.4k} N FA = Ax i + Ay j + Az k MA = ( MA ) x i + ( MA ) z k rAC = {3i + j} m rAD = {3i - j} m Equations of Equilibrium Referring to the FBD of the assembly shown in Fig a the force equation of equilibrium gives ΣF = 0; FCB + W + FA = 0; a - FCB + Ax bi + a - FCB + Ay bj + a FCB + Az - 392.4bk = 7 Equating i, j and k components - FCB + Ax = 0 - FCB + Ay = 0 FCB + Az - 392.4 = 0 The moment equation of equilibrium gives (1) (2) (3) ΣMA = 0; rAC * FCB + rAD * W + MA = i - FCB 1m j - FCB k i + †3 FCB j -1 k † + ( MA ) x i + ( MA ) Z k = - 392.4 c FCB + 392.4 + ( MA ) x d i + a - FCB + 1177.2bj + c - FCB + FCB + ( MA ) z d k = 7 7 468 C 3m © 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 *5–76 Continued Equating i, j and k components, F + 392.4 + ( MA ) x = 0 CB - FCB + 1177.2 = 0 - FCB + FCB + ( MA ) z = 0 7 Solving Eqs (1) to (6), (4) (5) (6) Ans FCB = 1373.4 N = 1.37 kN ( MA ) x = - 784.8 N # m = 785 N # m Ans ( MA ) z = 588.6 N # m = 589 N # m Ans Ax = 1177.2 N = 1.18 kN Ans Ay = 588.6 N = 589 N Ans Az = 0 Ans Ans: FCB ( MA ) x ( MA ) z Ax Ay Az 469 = = = = = = 1.37 kN 785 N # m 589 N # m 1.18 kN 589 N © 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 5–77 z The member is supported by a square rod which fits loosely through the smooth square hole of the attached collar at A and by a roller at B Determine the components of reaction at these supports when the member is subjected to the loading shown A x B 1m 2m y 2m Solution C Force And Position Vectors The coordinates of points B and C are B(2,0,0) m and C(3,0,-2) m 300 N FA = -Ax i - Ay j F = {300i + 500j - 400k} N NB = NB k MA = - ( MA ) x i + ( MA ) y j - ( MA ) z k rAB = {2i} m rAC = {3i - 2k} m Equations of Equilibrium Referring to the FBD of the member shown in Fig a, the force equation of equilibrium gives ΣF = 0; FA + F + NB = ( 300 - Ax ) i + ( 500 - Ay ) j + ( NB - 400 ) k = 470 500 N 400 N © 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 5–77 Continued Equating i, j and k components, 300 - Ax = 0 Ax = 300 N Ans 500 - Ay = 0 Ay = 500 N Ans NB - 400 = 0 Ans NB = 400 N The moment equation of equilibrium gives ΣMA = 0; MA + rAB * NB + rAC * F = i - ( MA ) x i + ( MA ) y j - ( MA ) z k + † 1000 - ( MA ) x i + ( MA ) y Equating i, j and k components, j 0 - 200 j + k i † + † 400 300 1500 j 500 - ( MA ) z k = 1000 - ( MA ) x = 0 ( MA ) x = 1000 N # m = 1.00 kN # m ( MA ) y - 200 = 0 ( MA ) y = 200 N # m 1500 - ( MA ) z = 0 ( MA ) z = 1500 N # m = k -2 † = - 400 1.50 kN # m Ans Ans Ans Ans: Ax = 300 N Ay = 500 N NB = 400 N ( MA ) x = 1.00 kN # m ( MA ) y = 200 N # m ( MA ) z = 1.50 kN # m 471 © 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 5–78 The bent rod is supported at A, B, and C by smooth journal bearings Compute the x, y, z components of reaction at the bearings if the rod is subjected to forces F1 = 300 lb and F2 = 250 lb F1 lies in the y–z plane The bearings are in proper alignment and exert only force reactions on the rod F1 z 45 ft A C ft SOLUTION ft B F1 = (- 300 cos 45°j - 300 sin 45°k) ft 30 ft = {- 212.1j - 212.1k} lb y 45 F2 = (250 cos 45° sin 30°i + 250 cos 45° cos 30°j - 250 sin 45°k) x F2 = {88.39i + 153.1j - 176.8k} lb ©Fx = 0; Ax + Bx + 88.39 = ©Fy = 0; Ay + Cy - 212.1 + 153.1 = ©Fz = 0; Bz + Cz - 212.1 - 176.8 = ©Mx = 0; -Bz (3) - Ay (4) + 212.1(5) + 212.1(5) = ©My = 0; Cz (5) + Ax (4) = ©Mz = 0; Ax (5) + Bx (3) - Cy (5) = Ax = 633 lb Ans Ay = - 141 lb Ans Bx = - 721 lb Ans Bz = 895 lb Ans Cy = 200 lb Ans Cz = - 506 lb Ans Ans: Ax = Ay = Bx = Bz = Cy = Cz = 472 633 lb - 141 lb - 721 lb 895 lb 200 lb - 506 lb © 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 5–79 The bent rod is supported at A, B, and C by smooth journal bearings Determine the magnitude of F2 which will cause the reaction C y at the bearing C to be equal to zero The bearings are in proper alignment and exert only force reactions on the rod Set F1 = 300 lb F1 z 45 ft A C ft SOLUTION ft B F1 = ( - 300 cos 45°j - 300 sin 45°k) ft = { -212.1j - 212.1k} lb 45 F2 = (F2 cos 45° sin 30°i + F2 cos 45° cos 30°j - F2 sin 45°k) x = {0.3536F2 i + 0.6124F2 j - 0.7071F2 k} lb ©Fx = 0; Ax + Bx + 0.3536F2 = ©Fy = 0; Ay + 0.6124F2 - 212.1 = ©Fz = 0; Bz + Cz - 0.7071F2 - 212.1 = ©Mx = 0; - Bz (3) - Ay (4) + 212.1(5) + 212.1(5) = ©My = 0; Cz (5) + Ax (4) = ©Mz = 0; Ax (5) + Bx (3) = 30 ft F2 Ax = 357 lb Ay = -200 lb Bx = -595 lb Bz = 974 lb Cz = - 286 lb Ans F2 = 674 lb Ans: F2 = 674 lb 473 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 *5–80 The bar AB is supported by two smooth collars At A the connection is with a ball-and-socket joint and at B it is a rigid attachment If a 50-lb load is applied to the bar, determine the x, y, z components of reaction at A and B D B z C Solution ft ft ft Ax + Bx = 0 (1) By + 50 = E 50 lb ft x By = - 50 lb Ans Az + Bz = 0 (2) ft A F y MBz = 0 Ans MBx + 50(6) = MBx = - 300 lb # ft Ans BCD = -9i + 3j BCD = -0.94868i + 0.316228j Require FB # uCD = (Bxi - 50j + Bzk) # ( -0.94868i + 0.316228j) = -0.94868Bx - 50(0.316228) = Ans Bx = - 16.667 = - 16.7 lb From Eq (1); Ans Ax = 16.7 lb Require MB # uCD = ( -300i + MByj) # ( - 0.94868i + 0.316228j) = 300(0.94868) + MBy(0.316228) = MBy = - 900 lb # ft Ans Ans: By = -50 lb MBz = MBx = -300 lb # ft Bx = -16.7 lb Ax = 16.7 lb 474 © 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 5–81 The rod has a weight of lb>ft If it is supported by a balland-socket joint at C and a journal bearing at D, determine the x, y, z components of reaction at these supports and the moment M that must be applied along the axis of the rod to hold it in the position shown z D 60Њ A 0.5 ft 45Њ x Solution y ΣFx = 0; Cx + Dx - 15 sin 45° = (1) ΣFy = 0; Cy + Dy = (2) C M B ft ft ΣFz = 0; Cz - 15 cos 45° = Ans Cz = 10.6 lb ΣMx = 0; - cos 45°(0.25 sin 60°) - Dy(2) = Dy = - 0.230 lb Ans Cy = 0.230 lb Ans ΣMy = 0; - (12 sin 45°)(1) - (3 sin 45°)(1) + (3 cos 45°)(0.25 cos 60°) From Eq (2); + Dx(2) = Dx = 5.17 lb Ans Cx = 5.44 lb Ans ΣMz = 0; - M + (3 sin 45°)(0.25 sin 60°) = From Eq (1); M = 0.459 lb # ft Ans Ans: Cz = 10.6 lb Dy = - 0.230 lb Cy = 0.230 lb Dx = 5.17 lb Cx = 5.44 lb M = 0.459 lb # ft 475 © 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 5–82 The sign has a mass of 100 kg with center of mass at G Determine the x, y, z components of reaction at the ball-andsocket joint A and the tension in wires BC and BD z 1m D 2m C SOLUTION 1m 2m Equations of Equilibrium: Expressing the forces indicated on the free-body diagram, Fig a, in Cartesian vector form, we have A FA = A xi + A yj + A zk x B W = {- 100(9.81)k} N = {- 981k} N FBD = FBDuBD = FBD ≥ FBC = FBCuBC = FBC ≥ G (- - 0)i + (0 - 2)j + (1 - 0)k 2(- - 0)2 + (0 - 2)2 + (1 - 0)2 ¥ = a- 2 FBDi - FBDj + FBDkb 3 1m (1 - 0)i + (0 - 2)j + (2 - 0)k 2 ¥ = a FBCi - FBCj + FBCkb 3 2(1 - 0) + (0 - 2) + (2 - 0) 2 Applying the forces equation of equilibrium, we have ©F = 0; FA + FBD + FBC + W = 2 1 2 (A xi + A yj + A zk) + a - FBDi - FBDj + FBDk b + a FBCi - FBCj + FBCkb + ( - 981 k) = 3 3 3 a Ax - 2 2 FBD + FBC bi + a A y - FBD - FBC b j + aA z + FBD + FBC - 981bk = 3 3 3 Equating i, j, and k components, we have Ax - F + FBC = BD (1) Ay - 2 F - FBC = BD (2) Az + FBD + FBC - 981 = 3 (3) In order to write the moment equation of equilibrium about point A, the position vectors rAG and rAB must be determined first rAG = {1j} m rAB = {2j} m 476 1m 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 5–82 Continued Thus, ©M A = 0; rAB * (FBC + FBD) + (rAG * W) = 2 2 1 (2j) * c a FBC - FBD b i - a FBC + FBD b j + a FBC + FBD bk d + (1j) * ( - 981k) = 3 3 3 4 a FBC + FBD - 981 bi + a FBD - FBC b k = 3 3 Equating i, j, and k components we have F + FBC - 981 = BC (4) F - FBC = BC (5) Ans: FBD = 294 N FBC = 589 N Ax = Ay = 589 N Az = 490.5 N 477 © 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 5–83 z Both pulleys are fixed to the shaft and as the shaft turns with constant angular velocity, the power of pulley A is transmitted to pulley B Determine the horizontal tension T in the belt on pulley B and the x, y, z components of reaction at the journal bearing C and thrust bearing D if u = 0° The bearings are in proper alignment and exert only force reactions on the shaft 200 mm 250 mm u 300 mm C SOLUTION 80 mm A x ©Fx = 0; ©Fy = 0; 165 + 80210.452 - Cz 10.752 = Ans 150 + 58.0210.22 - Cy 10.752 = Cy = 28.8 N Ans Dx = Ans Dy + 28.8 - 50 - 58.0 = Ans Dy = 79.2 N ©Fz = 0; 80 N Ans Cz = 87.0 N ©Mz = 0; y 65 N 6510.082 - 8010.082 + T10.152 - 5010.152 = T = 58.0 N ©My = 0; 150 mm B T Equations of Equilibrium: ©Mx = 0; 50 N D Dz + 87.0 - 80 - 65 = Ans Dz = 58.0 N Ans: T = 58.0 N Cz = 87.0 N Cy = 28.8 N Dx = Dy = 79.2 N Dz = 58.0 N 478 © 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 *5–84 z Both pulleys are fixed to the shaft and as the shaft turns with constant angular velocity, the power of pulley A is transmitted to pulley B Determine the horizontal tension T in the belt on pulley B and the x, y, z components of reaction at the journal bearing C and thrust bearing D if u = 45° The bearings are in proper alignment and exert only force reactions on the shaft 200 mm 250 mm u 300 mm C SOLUTION 80 mm A x ©Fx = 0; ©Fy = 0; 165 + 80210.452 - 50 sin 45°10.22 - Cz 10.752 = Ans 58.010.22 + 50 cos 45°10.22 - Cy 10.752 = Cy = 24.89 N = 24.9 N Ans Dx = Ans Dy + 24.89 - 50 cos 45° - 58.0 = Ans Dy = 68.5 N ©Fz = 0; 80 N Ans Cz = 77.57 N = 77.6 N ©Mz = 0; y 65 N 6510.082 - 8010.082 + T10.152 - 5010.152 = T = 58.0 N ©My = 0; 150 mm B T Equations of Equilibrium: ©Mx = 0; 50 N D Dz + 77.57 + 50 sin 45° - 80 - 65 = Ans Dz = 32.1 N Ans: T = Cz = Cy = Dy = Dz = 479 58.0 N 77.6 N 24.9 N 68.5 N 32.1 N © 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 5–85 Member AB is supported by a cable BC and at A by a square rod which fits loosely through the square hole at the end joint of the member as shown Determine the components of reaction at A and the tension in the cable needed to hold the 800-lb cylinder in equilibrium z C ft A x FBC ft = FBC a i - j + k b 7 ©Fx = 0; B ft SOLUTION y FBC a b = FBC = Ans ©Fy = 0; Ay = Ans ©Fz = 0; Az = 800 lb Ans ©Mx = 0; (MA)x - 800(6) = (MA)x = 4.80 kip # ft Ans ©My = 0; (MA)y = Ans ©Mz = 0; (MA)z = Ans Ans: FBC = Ay = Az = 800 lb (MA)x = 4.80 kip # ft (MA)y = (MA)z = 480 ... 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. .. 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,
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