Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 05

5–1 Draw the free-body diagram of the dumpster D of the truck, which has a weight of 5000 lb and a center of gravity at G It is supported by a pin at A and a pin-connected hydraulic cylinder BC (short link) Explain the significance of each force on the diagram (See Fig 5–7b.) 1.5 m G D 1m 3m A B 20 30 C SOLUTION The Significance of Each Force: W is the effect of gravity (weight) on the dumpster Ay and Ax are the pin A reactions on the dumpster T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) FBC is the hydraulic cylinder BC reaction on the dumpster © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 5–2 Draw the free-body diagram of member ABC which is supported by a smooth collar at A, rocker at B, and short link CD Explain the significance of each force acting on the diagram (See Fig 5–7b.) SOLUTION 3m 60 A kN m B 45 4m D C 2.5 kN 6m The Significance of Each Force: NA is the smooth collar reaction on member ABC NB is the rocker support B reaction on member ABC FCD is the short link reaction on member ABC 2.5 kN is the effect of external applied force on member ABC T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) kN # m is the effect of external applied couple moment on member ABC © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 5–3 Draw the free-body diagram of the beam which supports the 80-kg load and is supported by the pin at A and a cable which wraps around the pulley at D Explain the significance of each force on the diagram (See Fig 5–7b.) D A B E SOLUTION C T force of cable on beam 2m 2m 1.5 m Ax, Ay force of pin on beam T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) 80(9.81)N force of load on beam © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 *5–4 Draw the free-body diagram of the hand punch, which is pinned at A and bears down on the smooth surface at B F ϭ lb 1.5 ft SOLUTION A B ft T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) 0.2 ft © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 5–5 Draw the free-body diagram of the uniform bar, which has a mass of 100 kg and a center of mass at G The supports A, B, and C are smooth 0.5 m 0.2 m 1.25 m C G 1.75 m SOLUTION A 0.1 m 30Њ T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) B © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 5–6 Draw the free-body diagram of the beam,which is pin supported at A and rests on the smooth incline at B 800 lb 800 lb 600 lb 600 lb 400 lb ft ft ft ft 0.6 ft 1.2 ft A 0.6 ft SOLUTION B T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 5–7 Draw the free-body diagram of the beam, which is pin connected at A and rocker-supported at B 500 N SOLUTION 800 Nиm B 5m A 4m T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) 8m © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 *5–8 Draw the free-body diagram of the bar, which has a negligible thickness and smooth points of contact at A, B, and C Explain the significance of each force on the diagram (See Fig 5–7b.) in 30 C in B A in SOLUTION 10 lb NA, NB, NC force of wood on bar 30 T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) 10 lb force of hand on bar © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 5–9 Draw the free-body diagram of the jib crane AB, which is pin connected at A and supported by member (link) BC C SOLUTION B 0.4 m A 4m 3m T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) kN © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 5–10 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 6m 2m 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; NB cos 30°(8) - 4(6) = NB = 3.464 kN = 3.46 kN Ans 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° = A x = 1.73 kN A y + 3.464 cos 30° - = T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) + c ©Fy = 0; Ans A y = 1.00 kN Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 5–79 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 45Њ ft A C ft SOLUTION ft B F1 = ( - 300 cos 45°j - 300 sin 45°k) ft 30Њ ft = { - 212.1j - 212.1k} lb 45Њ 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 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) = T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) ©Fx = 0; Ax = 633 lb Ans Ay = - 141 lb Ans Bx = - 721 lb Ans Bz = 895 lb Ans Cy = 200 lb Ans Cz = - 506 lb Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 y *5–80 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 F2 = {0.3536F2 i + 0.6124F2 j - 0.7071F2 k} lb 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) = T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) ©Fx = 0; 30 ft Ax = 357 lb Ay = - 200 lb Bx = -595 lb Bz = 974 lb Cz = - 286 lb F2 = 674 lb Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 y 5–81 z 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 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 ≥ (- - 0)i + (0 - 2)j + (1 - 0)k 2( -2 - 0) + (0 - 2) + (1 - 0) 2 ¥ = a- 2 F i - FBDj + FBDkb BD 3 1m 1m 2 ¥ = a FBCi - FBCj + FBCkb 3 2(1 - 0) + (0 - 2) + (2 - 0) (1 - 0)i + (0 - 2)j + (2 - 0)k T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) FBC = FBCuBC = FBC ≥ G 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 F + FBC b i + a A y - FBD - FBC b j + aA z + FBD + FBC - 981 b k = BD 3 3 Equating i, j, and k components, we have Ax - F + FBC = BD Ay - 2 F - FBC = BD Az + F + FBC - 981 = BD (1) (2) (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 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 b i + a FBD - FBC b k = 3 3 Equating i, j, and k components we have F + FBC - 981 = BC (4) Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by © 2013 F - FBC = (5) BCCopyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 y 5–81 (continued) Soving Eqs (1) through (5), yields Ans FBC = 588.6 N = 589 N Ans Ax = Ans Ay = 588.6 N = 589 N Ans Az = 490.5 N Ans T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) FBD = 294.3 N = 294 N © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 5–82 z Determine the tensions in the cables and the components of reaction acting on the smooth collar at A necessary to hold the 50-lb sign in equilibrium The center of gravity for the sign is at G ft C ft ft ft E SOLUTION TDE B A ft 2 = TDE a i - j + k b 3 ft ft G D y x TBC = TBC a -1 2 i - j + kb 3 1 T - TBC + Ax = DE ©Fz = 0; 2 TDE + TBC - 50 = 3 ©Fy = 0; 2 - TDE - TBC + Ay = 3 ©Mx = 0; (MA)x + 2 T (2) + TBC(2) - 50(2) = DE ©My = 0; (MA)y - 2 T (3) + TBC(2) + 50(0.5) = DE ©Mz = 0; 2 - TDE(2) - TDE(3) + TBC(2) + TBC(2) = 3 3 Solving; TDE = 32.1429 = 32.1 lb TBC = 42.8571 = 42.9 lb Ax = 3.5714 = 3.57 lb Ay = 50 lb (MA)x = ft 2.5 ft T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) ©Fx = 0; 2.5 ft ft (MA)y = -17.8571 = - 17.9 lb # ft Ans Ans Ans Ans Ans Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 5–83 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 z 200 mm ©Fy = 0; Ans 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 = Dy = 79.2 N ©Fz = 0; 80 N 6510.082 - 8010.082 + T10.152 - 5010.152 = T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) ©Fx = 0; y 80 mm A 65 N Cz = 87.0 N ©Mz = 0; 150 mm B T T = 58.0 N ©My = 0; θ C Equations of Equilibrium: ©Mx = 0; D 300 mm x SOLUTION 50 N 250 mm Ans Dz + 87.0 - 80 - 65 = Dz = 58.0 N Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 *5–84 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 SOLUTION z 200 mm ©Fy = 0; y 80 mm A 65 N 80 N 6510.082 - 8010.082 + T10.152 - 5010.152 = Ans 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 = Dy = 68.5 N ©Fz = 0; B T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) ©Fx = 0; 150 mm C T Cz = 77.57 N = 77.6 N ©Mz = 0; θ x T = 58.0 N ©My = 0; D 300 mm Equations of Equilibrium: ©Mx = 0; 50 N 250 mm Ans Dz + 77.57 + 50 sin 45° - 80 - 65 = Dz = 32.1 N Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 5–85 If the roller at B can sustain a maximum load of kN, determine the largest magnitude of each of the three forces F that can be supported by the truss A 2m 45 2m 2m B 2m SOLUTION Equations of Equilibrium: The unknowns Ax and Ay can be eliminated by summing moments about point A a + ©MA = 0; F F F F(6) + F(4) + F(2) - cos 45°(2) = Ans T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) F = 0.3536 kN = 354 N © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 5–86 Determine the normal reaction at the roller A and horizontal and vertical components at pin B for equilibrium of the member 10 kN 0.6 m 0.6 m A SOLUTION Equations of Equilibrium: The normal reaction NA can be obtained directly by summing moments about point B a + ©MA = 0; kN 0.8 m 10(0.6 + 1.2 cos 60°) + (0.4) - NA (1.2 + 1.2 cos 60°) = NA = 8.00 kN + ©F = 0; : x Bx - cos 30° = + c ©Fy = 0; By + 8.00 - sin 30° - 10 = Bx = 5.20 kN B Ans Ans Ans T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) By = 5.00 kN 60 0.4 m © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 5–87 The symmetrical shelf is subjected to a uniform load of kPa Support is provided by a bolt (or pin) located at each end A and A¿ and by the symmetrical brace arms, which bear against the smooth wall on both sides at B and B¿ Determine the force resisted by each bolt at the wall and the normal force at B for equilibrium A¿ kPa B¿ A 1.5 m 0.15 m SOLUTION B 0.2 m Equations of Equilibrium: Each shelf’s post at its end supports half of the applied load, ie, 4000 (0.2) (0.75) = 600 N The normal reaction NB can be obtained directly by summing moments about point A a + ©MA = 0; NB (0.15) - 600(0.1) = 400 - Ax = + c ©Fy = 0; Ay - 600 = Ans Ax = 400 N T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) + ©F = 0; : x NB = 400 N Ay = 600 N The force resisted by the bolt at A is FA = 2A2x + A2y = 24002 + 6002 = 721 N Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 *5–88 z Determine the x and z components of reaction at the journal bearing A and the tension in cords BC and BD necessary for equilibrium of the rod C 3m A 3m 2m D SOLUTION x F1 = 5- 800k6 N 6m F2 = 5350j6 N FBC = FBC F2 - 3j + 4k2 B = 5- 0.6FBCj + 0.8FBCk6 N F1 { 800 k} N 13j + 4k2 T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) FBD = FBD 4m {350 j} N = 50.6FBDj + 0.8FBDk6 N ©Fx = 0; Ax = ©Fy = 0; 350 - 0.6FBC + 0.6FBD = ©Fz = 0; Az - 800 + 0.8FBC + 0.8FBD = ©Mx = 0; ©My = 0; ©Mz = 0; Ans (MA)x + 0.8FBD162 + 0.8FBC162 - 800162 = 800122 - 0.8FBC122 - 0.8FBD122 = (MA)z - 0.6FBC122 + 0.6FBD122 = FBD = 208 N Ans FBC = 792 N Ans Az = Ans (MA)x = Ans (MA)z = 700 N # m Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 y 5–89 The uniform rod of length L and weight W is supported on the smooth planes Determine its position u for equilibrium Neglect the thickness of the rod L u c f SOLUTION a + ©MB = 0; - Wa L cos ub + NA cos f (L cos u) + NA sin f (L sin u) = NA = W cos u cos (f - u) NB sin c - NA sin f = + c ©Fy = 0; NB cos c + NA cos f - W = (2) T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) + ©F = 0; : x (1) NB = W - NA cos f cos c (3) Substituting Eqs (1) and (3) into Eq (2): aW - W cos u sin f W cos u cos f b tan c = cos (f - u) cos (f - u) cos (f - u) tan c - cos u tan c cos f - cos u sin f = sin u (2 sin f tan c) - cos u (sin f - cos f tan c) = tan u = sin f - cos f tan c sin f tan c u = tan - a 1 cot c - cot f b 2 Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 5–90 z Determine the x, y, z components of reaction at the ball supports B and C and the ball-and-socket A (not shown) for the uniformly loaded plate lb/ft2 y A ft B ft SOLUTION ft W = (4 ft)(2 ft)(2 lb/ft2) = 16 lb ft x ©Fx = 0; Ax = Ans ©Fy = 0; Ay = Ans ©Fz = 0; Az + Bz + Cz - 16 = ©Mx = 0; 2Bz - 16(1) + Cz (1) = (2) ©My = 0; - Bz (2) + 16(2) - Cz(4) = (3) (1) T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) Solving Eqs (1)—(3): C Az = Bz = Cz = 5.33 lb Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 5–91 z Determine the x, y, z components of reaction at the fixed wall A The 150-N force is parallel to the z axis and the 200-N force is parallel to the y axis 150 N A x 2m y 1m SOLUTION 2.5 m ©Fx = 0; Ax = ©Fy = 0; Ay + 200 = Ay = - 200 N ©Fz = 0; 2m Ans Az - 150 = Az = 150 N 200 N Ans - 150(2) + 200(2) - (MA)x = T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) ©Mx = 0; Ans (MA)x = 100 N # m Ans ©My = 0; (MA)y = Ans ©Mz = 0; 200(2.5) - (MA)z = (MA)z = 500 N # m Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 *5–92 Determine the reactions at the supports A and B for equilibrium of the beam 400 N/m 200 N/m A B 4m SOLUTION 3m Equations of Equilibrium: The normal reaction of NB can be obtained directly by summing moments about point A + ©MA = 0; NB(7) - 1400(3.5) - 300(6) = NB = 957.14 N = 957 N Ag - 1400 - 300 + 957 = Az = Ag = 743 N Ans T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) + : ©Fx = 0; Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 ... prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding... prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s),... prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s),