Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 03

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 SOLUTION + ©F = 0; : x O F2 sin 70° + F1 cos 60° - cos 30° - (7) = 5 0.9397F2 + 0.5F1 = 9.930 kN + c ©Fy = 0; u F2 cos 70° + sin 30° - F1 sin 60° - (7) = F1 Solving: 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) 0.3420F2 - 0.8660F1 = 1.7 F2 = 9.60 kN Ans F1 = 1.83 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 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 SOLUTION + ©F = 0; : x O sin 70° + F1 cos u - cos 30° - (7) = 5 F1 cos u = 4.2920 + c ©Fy = 0; cos 70° + sin 30° - F1 sin u - u (7) = kN F1 F1 sin u = 0.3521 Solving: Ans F1 = 4.31 kN 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) u = 4.69° © 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 3–3 The lift sling is used to hoist a container having a mass of 500 kg Determine the force in each of the cables AB and AC as a function of u If the maximum tension allowed in each cable is kN, determine the shortest lengths of cables AB and AC that can be used for the lift The center of gravity of the container is located at G F A SOLUTION B Free-Body Diagram: By observation, the force F1 has to support the entire weight of the container Thus, F1 = 50019.812 = 4905 N θ θ 1.5 m 1.5 m Equations of Equilibrium: + ©F = 0; : x FAC cos u - FAB cos u = + c ©Fy = 0; 4905 - 2F sin u = FAC = FAB = F F = 52452.5 cos u6 N G 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) Thus, FAC = FAB = F = 52.45 cos u6 kN Ans If the maximum allowable tension in the cable is kN, then 2452.5 cos u = 5000 u = 29.37° From the geometry, l = 1.5 and u = 29.37° Therefore cos u l = 1.5 = 1.72 m cos 29.37° 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 C *3–4 Cords AB and AC can each sustain a maximum tension of 800 lb If the drum has a weight of 900 lb, determine the smallest angle u at which they can be attached to the drum B A u u C SOLUTION + c ©Fy = 0; 900 - 2(800) sin u = 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) u = 34.2° © 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 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 = 30° A O kN u 45 D B SOLUTION + ©F = 0; : x T T = 13.32 = 13.3 kN + c ©Fy = 0; C -T cos 30° + + sin 45° = Ans kN F F - 13.32 sin 30° - cos 45° = 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 = 10.2 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 3–6 The gusset plate is subjected to the forces of four members Determine the force in member B and its proper orientation u for equilibrium The forces are concurrent at point O Take F = 12 kN A O kN u 45 D SOLUTION B + ©F = 0; : x - T cos u + sin 45° = + c ©Fy = 0; 12 - T sin u - cos 45° = T C kN F Solving, Ans u = 36.3° 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) T = 14.3 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 3–7 The device shown is used to straighten the frames of wrecked autos Determine the tension of each segment of the chain, i.e., AB and BC, if the force which the hydraulic cylinder DB exerts on point B is 3.50 kN, as shown B A 3.50 kN D 450 mm C 400 mm 250 mm SOLUTION Equations of Equilibrium: A direct solution for FBC can be obtained by summing forces along the y axis + c ©Fy = 0; 3.5 sin 48.37° - FBC sin 60.95° = FBC = 2.993 kN = 2.99 kN Ans Using the result FBC = 2.993 kN and summing forces along x axis, we have 3.5 cos 48.37° + 2.993 cos 60.95° - FAB = 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 FAB = 3.78 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 *3–8 Two electrically charged pith balls, each having a mass of 0.2 g, are suspended from light threads of equal length Determine the resultant horizontal force of repulsion, F, acting on each ball if the measured distance between them is r = 200 mm ϩ ϩ ϩ A SOLUTION + ©F = 0; : x F - Ta + c ©Fy = 0; TB 50 mm 150 mm ϩ ϩ ϩ ؊F 150 mm F ϩ ϩ ϩ r ϭ 200 mm ϩ ϩ ϩ ϩ ϩ B 75 b = 150 21502 - 752 R - 0.2(9.81)(10 - 3) = 150 T = 2.266(10 - 3) 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) F = 1.13 mN © 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 3–9 Determine the maximum weight of the flowerpot that can be supported without exceeding a cable tension of 50 lb in either cable AB or AC C B 30° SOLUTION A Equations of Equilibrium: + ©F = 0; : x FAC sin 30° - FAB a b = FAC = 1.20FAB + c ©Fy = 0; (1) FAC cos 30° + FAB a b - 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) 0.8660FAC + 0.8FAB = W Since FAC FAB , failure will occur first at cable AC with FAC = 50 lb Then solving Eqs (1) and (2) yields FAB = 41.67 lb W = 76.6 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 3–10 Determine the tension developed in wires CA and CB required for equilibrium of the 10-kg cylinder Take u = 40° B A u 30° C SOLUTION Equations of Equilibrium: Applying the equations of equilibrium along the x and y axes to the free-body diagram shown in Fig a, + ©F = 0; : x FCB cos 40° - FCA cos 30° = (1) + c ©Fy = 0; FCB sin 40° + FCA sin 30° - 10(9.81) = (2) Solving Eqs (1) and (2), yields FCB = 90.4 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) FCA = 80.0 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 3–63 The thin ring can be adjusted vertically between three equally long cables from which the 100-kg chandelier is suspended If the ring remains in the horizontal plane and z = 600 mm, determine the tension in each cable z 0.5 m C 120 D 120 120 B x z SOLUTION A Geometry: Referring to the geometry of the free-body diagram shown in Fig a, the lengths of cables AB, AC, and AD are all l = 20.52 + 0.62 = 20.61 m Equations of Equilibrium: Equilibrium requires ©Fx = 0; FAD ¢ 20.61 0.5 20.61 ≤ - FAC ¢ ≤ - 2BF¢ 0.5 cos 30° 20.61 0.5 sin 30° 20.61 ≤ = FAD = FAC = F ≤R = FAB = F 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) âFy = 0; FAB Â 0.5 cos 30° Thus, cables AB, AC, and AD all develop the same tension âFz = 0; 3F Â 0.6 20.61 - 100(9.81) = FAB = FAC = FAD = 426 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 y *3–64 The thin ring can be adjusted vertically between three equally long cables from which the 100-kg chandelier is suspended If the ring remains in the horizontal plane and the tension in each cable is not allowed to exceed kN, determine the smallest allowable distance z required for equilibrium z 0.5 m C 120 D 120 120 B x z SOLUTION A Geometry: Referring to the geometry of the free-body diagram shown in Fig a, the lengths of cables AB, AC, and AD are all l = 20.52 + z2 Equations of Equilibrium: Equilibrium requires ©Fy = 0; FAB £ 0.5 cos 30° 20.5 + z 0.5 20.5 + z ≥ - FAC £ ≥ - 2C F£ 0.5 cos 30° 20.52 + z2 0.5 sin 30° ≥ = FAD = FAC = F ≥S = FAB = F 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; FAD £ 20.52 + z2 Thus, cables AB, AC, and AD all develop the same tension ©Fz = 0; 3F £ z 20.52 + z2 ≥ - 100(9.81) = Cables AB, AC, and AD will also achieve maximum tension simultaneously Substituting F = 1000 N, we obtain 3(1000) £ z 20.52 + z2 ≥ - 100(9.81) = z = 0.1730 m = 173 mm 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 3–65 z The 80-lb chandelier is supported by three wires as shown Determine the force in each wire for equilibrium D 135Њ ft B ft 90Њ ft 135Њ C SOLUTION 2.4 ft ©Fx = 0; 1 F F cos 45° = 2.6 AC 2.6 AB ©Fy = 0; - ©Fz = 0; 2.4 2.4 2.4 F + F + F - 80 = 2.6 AC 2.6 AD 2.6 AB 1 F + F sin 45° = 2.6 AD 2.6 AB A 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) Solving, FAB = 35.9 lb Ans FAC = FAD = 25.4 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 3–66 z If each wire can sustain a maximum tension of 120 lb before it fails, determine the greatest weight of the chandelier the wires will support in the position shown D 135Њ ft B ft 90Њ ft 135Њ C SOLUTION 2.4 ft ©Fx = 0; 1 F F cos 45° = 2.6 AC 2.6 AB ©Fy = 0; - ©Fz = 0; 2.4 2.4 2.4 F + F + F - W = 2.6 AC 2.6 AD 2.6 AB 1 + F F sin 45° = 2.6 AD 2.6 AB (1) (2) A (3) 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) Assume FAC = 120 lb From Eq (1) 1 (120) F cos 45° = 2.6 2.6 AB FAB = 169.71 120 lb (N G!) Assume FAB = 120 lb From Eqs (1) and (2) 1 F (120)(cos 45°) = 2.6 AC 2.6 FAC = 84.853 lb 120 lb (O K!) - 1 F + (120) sin 45° = 2.6 AD 2.6 FAD = 84.853 lb 120 lb (O K!) Thus, W = 2.4 (F + FAD + FAB) = 267.42 = 267 lb 2.6 AC 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 ■3–67 The 80-lb ball is suspended from the horizontal ring using three springs each having an unstretched length of 1.5 ft and stiffness of 50 lb/ft Determine the vertical distance h from the ring to point A for equilibrium 1.5 ft 1.5 ft 120° 120° 120° h SOLUTION Equation of Equilibrium: This problem can be easily solved if one realizes that due to symmetry all springs are subjected to a same tensile force of Fsp Summing forces along z axis yields ©Fz = 0; 3Fsp cos g - 80 = A (1) Spring Force: Applying Eq 3−2, we have 75 1.5 - 1.5b = - 75 sin g sin g (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) Fsp = ks = k1l - l02 = 50 a Substituting Eq (2) into (1) yields 3a 75 - 75b cos g - 80 = sin g tan g = 45 11 - sin g2 16 Solving by trial and error, we have g = 42.4425° Geometry: h = 1.5 1.5 = = 1.64 ft tan g tan 42.4425° 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 *3–68 The pipe is held in place by the vise If the bolt exerts a force of 50 lb on the pipe in the direction shown, determine the forces FA and FB that the smooth contacts at A and B exert on the pipe FA 30 A FB SOLUTION B C + ©F = 0; : x FB - FA cos 60° - 50 a b = + c ©Fy = 0; -FA sin 60° + 50 a b = Ans FB = 57.3 lb 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) FA = 34.6 lb 50 lb © 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 3–69 When y is zero, the springs sustain a force of 60 lb Determine the magnitude of the applied vertical forces F and - F required to pull point A away from point B a distance of y = ft The ends of cords CAD and CBD are attached to rings at C and D F A t 2f k ϭ 40 lb/ft C ft B 2f t t 2f y D k ϭ 40 lb/ft SOLUTION Initial spring stretch: –F s1 = + c ©Fy = 0; F - a T b = 0; - Fs + ¢ F = T 23 ≤F = 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 60 = 1.5 ft 40 Fs = 1.732F Final stretch is 1.5 + 0.268 = 1.768 ft 40(1.768) = 1.732F F = 40.8 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 3–70 When y is zero, the springs are each stretched 1.5 ft Determine the distance y if a force of F = 60 lb is applied to points A and B as shown The ends of cords CAD and CBD are attached to rings at C and D F A t 2f k 40 lb/ft C f t B 2f t t 2f y D k 40 lb/ft SOLUTION + c ©Fy = 0; T sin u = 60 –F T sin u = 30 + ©F = 0; : x 2T cos u = Fsp Fsp tan u = 60 (1) Fsp = kx Substitute F in Eq 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) Fsp = 40(1.5 + - cos u) 40(1.5 + - cos u) tan u = 60 (3.5 - cos u) tan u = 1.5 3.5 tan u - sin u = 1.5 1.75 tan u - sin u = 0.75 By trial and error: u = 37.96° y = sin 37.96° y = 2.46 ft 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 3–71 Romeo tries to reach Juliet by climbing with constant velocity up a rope which is knotted at point A Any of the three segments of the rope can sustain a maximum force of kN before it breaks Determine if Romeo, who has a mass of 65 kg, can climb the rope, and if so, can he along with Juliet, who has a mass of 60 kg, climb down with constant velocity? B 60° A C SOLUTION + c ©Fy = 0; TAB sin 60° - 6519.812 = TAB = 736.29 N 2000 N + ©F = 0; : x TAC - 736.29 cos 60° = TAC = 368.15 N 2000 N Yes, Romeo can climb up the rope TAB sin 60° - 12519.812 = 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 TAB = 1415.95 N 2000 N + ©F = 0; : x TAC - 1415.95 cos 60° = TAC = 708 N 2000 N Also, for the vertical segment, T = 125(9.81) = 1226 N 2000 N Yes, Romeo and Juliet can climb down 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 *■3–72 Determine the magnitudes of forces F1, F2, and F3 necessary to hold the force F = - 9i - 8j - 5k6 kN in equilibrium z F1 F2 60° 135° 60° SOLUTION ©Fx = 0; F1 cos 60°cos 30° + F2 cos 135° + F - = F - = ©Fy = 0; -F1 cos 60°sin 30° + F2 cos 60° + ©Fz = 0; F1 sin 60° + F2 cos 60° - F3 - = 60° y 30° F3 x F (4 m, m, –2 m) 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) 0.433F1 - 0.707F2 + 0.667F3 = -0.250F1 + 0.500F2 + 0.667F3 = 0.866F1 + 0.500F2 - 0.333F3 = Solving, F1 = 8.26 kN Ans F2 = 3.84 kN Ans F3 = 12.2 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 3–73 The man attempts to pull the log at C by using the three ropes Determine the direction u in which he should pull on his rope with a force of 80 lb, so that he exerts a maximum force on the log What is the force on the log for this case? Also, determine the direction in which he should pull in order to maximize the force in the rope attached to B What is this maximum force? D A C u B 150 SOLUTION FAB + 80 cos u - FAC sin 60° = + c ©Fy = 0; 80 sin u - FAC cos 60° = (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) + ©F = 0; : x (2) FAC = 160 sin u dFAC = 160 cos u = du u = 90° Ans FAC = 160 lb Ans From Eq (1), FAC sin 60° = FAB + 80 cos u Substitute into Eq (2), 80 sin u sin 60° = (FAB + 80 cos u)cos 60° FAB = 138.6 sin u - 80 cos u dFAB = 138.6 cos u + 80 sin u = du u = tan - c 138.6 d = 120° - 80 FAB = 138.6 sin 120° - 80 cos 120° = 160 lb 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 ■3–74 The ring of negligible size is subjected to a vertical force of 200 lb Determine the longest length l of cord AC such that the tension acting in AC is 160 lb Also, what is the force acting in cord AB? Hint: Use the equilibrium condition to determine the required angle u for attachment, then determine l using trigonometry applied to ¢ABC C u 40 l A B ft 200 lb SOLUTION Equations of Equilibrium: + ©F = 0; : x FAB cos 40° - 160 cos u = (1) + c ©Fy = 0; FAB sin 40° + 160 sin u - 200 = (2) Solving Eqs (1) and (2) and choosing the smallest value of u, yields u = 33.25° 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) FAB = 175 lb Geometry: Applying law of sines, we have l = sin 40° sin 33.25° l = 2.34 ft 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 3–75 Determine the maximum weight of the engine that can be supported without exceeding a tension of 450 lb in chain AB and 480 lb in chain AC C B 30 A SOLUTION + ©F = 0; : x FAC cos 30° - FAB = (1) + c ©Fy = 0; FAC sin 30° - W = (2) Assuming cable AB reaches the maximum tension FAB = 450 lb From Eq (1) FAC cos 30° - 450 = FAC = 519.6 lb 480 lb (No Good) 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) Assuming cable AC reaches the maximum tension FAC = 480 lb From Eq (1) 480 cos 30° - FAB = FAB = 415.7 lb 450 lb From Eq (2) 480 sin 30° - W = W = 240 lb (OK) 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 *3–76 Determine the force in each cable needed to support the 500-lb load D z ft y ft SOLUTION B C At C: © Fx = 0; FCA ¢ 210 ≤ - FCB ¢ © Fy = 0; - FCA ¢ © Fz = 0; - 500 + FCD a b = 210 210 ≤ - FCB ¢ ft ≤ = 210 A x ≤ + FCD a 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) Solving: ft ft FCD = 625 lb Ans FCA = FCB = 198 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 3–77 The joint of a space frame is subjected to four member forces Member OA lies in the x–y plane and member OB lies in the y–z plane Determine the forces acting in each of the members required for equilibrium of the joint z F1 A F3 O 45 y B 40 SOLUTION F2 x Equation of Equilibrium: ©Fx = 0; -F1 sin 45° = ©Fz = 0; F2 sin 40° - 200 = F1 = F2 = 311.14 lb = 311 lb 200 lb Ans Ans Using the results F1 = and F2 = 311.14 lb and then summing forces along the y axis, we have F3 - 311.14 cos 40° = F3 = 238 lb 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) ©Fy = 0; © 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 ... 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... 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... 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