© 2011 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. Geometry: From the geometry of the figure, u 5 sin21 a 0.75 b 5 30° 1.5 Equations of Equilibrium: Applying the equations of equilibrium to the free-body diagram in Fig. (a), 1cSFy 5 0; FBA sin 30° 2 200(9.81) 5 0 FBA 3924 N 5 3.92 kN Ans. 1 SFx 5 0; S 3924 cos 30° 2 FBC 5 0 FBC 5 3398.28 N 5 3.40 kN Ans. 182 SM_CH04.indd 182 4/8/11 11:48:14 AM © 2011 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. Equations of Equilibrium: Applying the equations of equilibrium along the x and y axes to the free-body diagram in Fig. (a), 1cSFy 5 0; 3500 sin u 2 200(9.81) 5 0 u 5 34.10° 1 SFx 5 0; S 3500 cos 34.10° 2 FBC 5 0 FBC 5 2898.37 N 5 2.90 kN Ans. y 5 1.5 sin 34.10° 5 0.841 m 5 841 mm Ans. 183 SM_CH04.indd 183 4/8/11 11:48:15 AM © 2011 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. 2 SFx 5 0; 2T cos 30° 1 8 1 5 sin 45° 5 0 S T 5 13.32 5 13.3 kN Ans. 1cSFy 5 0; F 2 13.32 sin 30° 2 5 cos 45° 5 0 F 5 10.2 kN Ans. 2 SFx 5 0; 8 2T cos u 1 5 sin 45° 5 0 S 12 2 T sin u 2 5 cos 45° 5 0 1cSFy 5 0; Solving, T 5 14.3 kN Ans. u 5 36.3° Ans. 184 SM_CH04.indd 184 4/8/11 11:48:15 AM ©©2011 This 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved. Thismaterial materialis isprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. The300-lb 1500-N electrical transformer with center of •5–25. electrical transformer with center of gravity •4–5. The gravity G is supported byata pin at Aa and a smooth at G is at supported by a pin A and smooth pad atpad B. at B. Determine the horizontal and vertical components of Determine the horizontal and vertical components of reaction reaction and the B on the at the pinat A the and pin the A reaction of reaction the pad Bofonthe thepad transformer. transformer. 0.45 1.5 ftm A 0.9 3 ftm G B Equations of Equilibrium: From the free-body diagram of the transformer, Fig. a, NB and Ay can be obtained by writing the moment equation of equilibrium about point A and the force equation of equilibrium along the y axis. + ΣMA = 0; NB (0.9) – 1500(0.45) = 0 NB = 750 N +↑ΣFy = 0; Ans Ay – 1500 = 0 Ay = 1500 N Ans Using the result NB = 750 N and writing the force equation of equilibrium along the x axis, + → ΣFx = 0; 750 – Ax = 0 Ax = 750 N Ans 0.45 m 1500 N 0.9 m 334 185 SM_CH04.indd 185 05a Ch05a 319-360.indd 334 4/8/11 11:48:15 AM AM 6/12/09 8:43:11 © 2011 Pearson Education, Inc., Upper Saddle River, rights reserved. This material is protected under copyright laws they currently © 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. AllAll rights reserved. This material is protected under allall copyright laws as as they currently exist. portion material may reproduced, form means, without permission writing from publisher. exist. NoNo portion of of thisthis material may bebe reproduced, in in anyany form or or byby anyany means, without permission in in writing from thethe publisher. 5–26. 4–6. A skeletal diagram of a hand holding a load is shown in the upper figure. If the load and the forearm have masses of 2 kg and 1.2 kg, respectively, and their centers of mass are located at G1 and G2, determine the force developed in the biceps CD and the horizontal and vertical components of reaction at the elbow joint B. The forearm supporting system can be modeled as the structural system shown in the lower figure. D G1 C B A G2 D Equations of Equilibrium: From the free-body diagram of the structural system, Fig. a, FCD can be obtained by writing the moment equation of equilibrium about point B. � + ΣMB = 0; G1 C A 100 mm G2 135 mm 75 B 65 mm 2(9.81)(100 + 135 + 65) + 1.2(9.81)(135 + 65) –FCD sin 75° (65) = 0 FCD = 131.25 N = 131 N Ans Using the above result and writing the force equations of equilibrium along the x and y axes, + S ΣFx = 0; 131.25 cos 75° – Bx = 0 + ↑ΣFy = 0; 131.25 sin 75° – 2(9.81) – 1.2(9.81) – By = 0 Bx = 33.97 N = 34.0 N By = 95.38 N = 95.4 N Ans Ans 335 186 SM_CH04.indd 186 335 05a Ch05a 319-360.indd 4/8/11 11:48:16 6/12/09 8:43:15 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 5–27. As an airplane’s brakes are applied, the nose wheel 4–7. exerts two forces on the end of the landing gear as shown. Determine the horizontal and vertical components of reaction at the pin C and the force in strut AB. C 30 20 B 400 mm A 600 mm 2 kN 6 kN Equations of Equilibrium : The force in strut AB can be obtained directly by summing moments about point C. � + ΣMC = 0; 2(1) – 6(1 tan 20°) + FAB sin 50° (0.4) – FAB cos 50° (0.4 tan 20°) = 0 FAB = 0.8637 kN = 0.864 kN Ans Using the result FAB = 0.8637 kN and sum forces along x and y axes, we have, + ↑ΣFy = 0; 6 + 0.8637 cos 50° – Cy = 0 S ΣFx = 0; + 0.8637 sin 50° + 2 – Cx = 0 Cx = 2.66 kN Cy = 6.56 kN Ans Ans 336 187 SM_CH04.indd 187 05a Ch05a 319-360.indd 336 4/8/11 11:48:16 AM AM 6/12/09 8:43:18 © 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently © 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form oror byby any means, without permission in in writing from thethe publisher. *5–28. The 1.4-Mg drainpipe is held in the tines of the fork *4–8. lift. Determine the normal forces at A and B as functions of the blade angle u and plot the results of force (vertical axis) versus u (horizontal axis) for 0 … u … 90°. 0.4 m A B u + Q©Fx = 0; NA – 1.4(10)3 (9.81) sin u = 0 NA = 13.7 sin u kN Ans + a©Fy = 0; NB – 1.4(10)3 (9.81) cos u = 0 NB = 13.7 cos u kN Ans 337 188 188 337 05a SM_CH04.indd Ch05a 319-360.indd 4/8/118:43:20 11:48:17 6/12/09 AM AM ©©2011 This 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved. Thismaterial materialis isprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. •5–29. •4–9. The mass of 700 kg is suspended from a trolley which moves along the crane rail from d = 1.7 m to d = 3.5 m. Determine the force along the pin-connected knee strut BC (short link) and the magnitude of force at pin A as a function of position d. Plot these results of FBC and FA (vertical axis) versus d (horizontal axis). d A C 2m B + ΣMA = 0; 4 FBC (1.5) – 700(9.81)(d) = 0 5 FBC = 5722.5d S ©Fz = 0; + 1.5 m Ans 3 –Az + (5722.5d) = 0 5 Az = 3433.5d + ↑©Fy = 0; 4 –Ay + (5722.5d) – 700 (9.81) = 0 5 Ay = 4578d – 6867 FA = (3433.5d)2 + (4578d – 6867)2 Ans 338 189 SM_CH04.indd 189 05a Ch05a 319-360.indd 338 4/8/11 11:48:17 AM AM 6/12/09 8:43:21 © 2011 Pearson Education, Inc., Upper Saddle River, rights reserved. This material is protected under copyright laws they currently © 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. AllAll rights reserved. This material is protected under allall copyright laws as as they currently exist. portion material may reproduced, form means, without permission writing from publisher. exist. NoNo portion of of thisthis material may bebe reproduced, in in anyany form or or byby anyany means, without permission in in writing from thethe publisher. 4–10. If 5–30. If the the force force of of FF == 100 500 lb N is applied to the handle of the bar bender, determine the horizontal and vertical components of reaction at pin A and the reaction of the roller B on the smooth bar. C 1000 mm 40 in. F 60 Equations of Equilibrium: From the free-body diagram of the handle of the bar bender, Fig. a, Ay and NB can be obtained by writing the force equation of equilibrium along the y axis and the moment equation of equilibrium about point A, respectively. +↑ΣFy = 0; A 125 mm 5 in. Ay – 500 sin 30° = 0 Ans Ay = 250 N + ΣMA = 0; B NB cos 60° (125) – 500(1000) = 0 NB = 8000 N = 8.0 kN Ans Using the result NB = 8000 N and writing the force equation of equilibrium along the x axis, + → ΣFx = 0; Ax – 8000 + 500 cos 30° = 0 Ax = 7567.0 N = 7.567 kN Ans 500 N 12 5m m 1000 mm 339 190 SM_CH04.indd 190 339 05a Ch05a 319-360.indd 4/8/11 11:48:17 6/12/09 8:43:24 AM AM ©©2011 This 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved. Thismaterial materialis isprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 5–31. 4–11. If the force of the smooth roller at B on the bar bender is required to be 1.5 7.5 kip, kN, determine the horizontal and vertical components of reaction at pin A and the required magnitude of force F applied to the handle. C 1000 40 mm in. F 60 Equations of Equilibrium: From the free-body diagram of the handle of the bar bender, Fig. a, force F can be obtained by writing the moment equation of equilibrium about point A. + ΣMA = 0; B 7.5 cos 60° (125) – F (1000) = 0 F = 0.46875 kN A 125 mm 5 in. Ans Using the above result and writing the force equation of equilibrium along the x and y axis, + → ΣFx = 0; Ax + 0.46875 cos 30° – 7.5 = 0 Ans Ax = 7.094 kN +↑ΣFy = 0; Ay – 0.46875 sin 30° = 0 Ans Ay = 0.234 kN 12 5m m 1000 mm 340 191 SM_CH04.indd 191 05a Ch05a 319-360.indd 340 4/8/11 11:48:18 AM AM 6/12/09 8:43:26 © 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently © 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form oror byby any means, without permission in in writing from thethe publisher. *4–12. *5–32. The jib crane is supported by a pin at C and rod AB. If the load has a mass of 2 Mg with its center of mass located at G, determine the horizontal and vertical components of reaction at the pin C and the force developed in rod AB on the crane when x = 5 m. A 3.2 m C Equation of Equilibrium: Realizing that rod AB is a two-force member, it will exert a force FAB directed along its axis on the beam, as shown on the free-body diagram in Fig. a. From the free-body diagram, FAB can be obtained by writing the moment equation of equilibrium about point C. + ΣMC = 0; 4m 0.2 m B D G x 3 4 FAB (4) + FAB (0.2) – 2000(9.81)(5) = 0 5 5 FAB = 38 320.31 N = 38.3 kN Ans Using the above result and writing the force equations of equilibrium along the x and y axes. + S ©Fx = 0; + ↑©Fy = 0; 4 Cx – 38 320.31 = 0 5 Cx = 30 656.25 N = 30.7 kN Ans 3 38 320.31 – 2000(9.81) – Cy = 0 5 Cy = 3372.19 N = 3.37 kN Ans 341 192 192 341 05a SM_CH04.indd Ch05a 319-360.indd 4/8/118:43:29 11:48:18 6/12/09 AM AM ©©2011 This 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved. Thismaterial materialis isprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. •5–33. •4–13. The jib crane is supported by a pin at C and rod AB. The rod can withstand a maximum tension of 40 kN. If the load has a mass of 2 Mg, with its center of mass located at G, determine its maximum allowable distance x and the corresponding horizontal and vertical components of reaction at C. A 3.2 m C Equation of Equilibrium: Realizing that rod AB is a two-force member, it will exert a force FAB directed along its axis on the beam, as shown on the free-body diagram in Fig. a. From the free-body diagram, the distance x can be obtained by writing the moment equation of equilibrium about point C. + ΣMC = 0; 4m 0.2 m B D G x 3 4 40 000 (4) + 40 000 (0.2) – 2000(9.81)(x) = 0 5 5 x = 5.22 m Ans Writing the force equations of equilibrium along the x and y axes, + S ©Fx = 0; + ↑©Fy = 0; 4 Cx – 40 000 = 0 5 Cx = 32 000 N = 32 kN Ans 3 40 000 – 2000 (9.81) – Cy = 0 5 Cy = 4380 N = 4.38 kN Ans 342 193 SM_CH04.indd 193 05a Ch05a 319-360.indd 342 4/8/11 11:48:19 AM AM 6/12/09 8:43:32 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underall allcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. exist. 5–54. 4–14. The The uniform uniform rod rod AB AB has has aa weight weight of of 15 75 lb N and the spring is unstretched when u = 0°. If u = 30°, determine the stiffness k of the spring. 1.86m ft A u 30.9 ft m k Geometry : From triangle CDB, the cosine law gives B l= 0.7608 2 + 0.5196 2 – 2(0.7608) (0.5196) cos 120° = 1.1154 m Using the sine law, 0.45 m sin α sin 120° = 0.7608 1.1154 = 36.21° 75 N 0.45 m Equations of Equilibrium : The force in the spring can be obtained directly by summing moments about point A. + ΣMA = 0; 75 cos 30° (0.45) – Fsp cos 36.21° (0.9) = 0 0.9 cos 30° = 0.7608 m Fsp = 40.25 N 1.8 – Spring Force Formula : The spring stretches x = 1.1154 – 0.9 = 0.2154 m Fsp 40.25 k= = = 186.9 N/m 0.2154 x 0.9 m cos 30° 0.9 m Ans 0.9 tan 30° = 0.5196 m 5–55. The horizontal beam is supported by springs at its ends. Each spring has a stiffness of k = 5 kN>m and is originally unstretched so that the beam is in the horizontal position. Determine the angle of tilt of the beam if a load of 800 N is applied at point C as shown. 800 N C A B 1m 3m Equations of Equilibrium : The spring force at A and B can be obtained directly by summing moments about points B and A, respectively. + ΣMB = 0; 800(2) – FA (3) = 0 FA = 533.33 N + ΣMA = 0; FB (3) – 800(1) = 0 FB = 266.67 N Spring Formula : Applying ∆ = F , we have k 533.33 = 0.1067 m 5(103) 266.67 = 0.05333 m ∆B = 5(103) ∆A = Geometry : The angle of tilt a is 0.05333 a = tan–1 = 1.02° 3 Ans 360 194 SM_CH04.indd 194 05a Ch05a 319-360.indd 360 4/8/11 6/12/0911:48:19 8:44:21AM AM 4–15. The airstroke actuator at D is used to apply a force of 5–23. F = 200 N on the member at B. Determine the horizontal and vertical components of reaction at the pin A and the force of the smooth shaft at C on the member. C 15 600 mm B A 60 200 mm 600 mm D F Equations of Equilibrium: From the free-body diagram of member ABC, Fig. a, NC can be obtained by writing the moment equation of equilibrium about point A. � :21 AM ©©2011 This 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved. Thismaterial materialis isprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. + ΣMA = 0; 200 sin 60° (800) – NC (600 + 200 sin 15°) = 0 Ans NC = 212.60 N = 213 N Using this result and writing the force equations of equilibrium along the x and y axes, + S ΣFx = 0; + ↑ΣFy = 0; –Ax + 212.60 cos 15° – 200 cos 60° = 0 Ans Ax = 105 N –Ay – 212.60 sin 15° + 200 sin 60° = 0 Ans Ay = 118 N 332 195 SM_CH04.indd 195 05a Ch05a 319-360.indd 332 4/8/11 11:48:19 AM AM 6/12/09 8:43:05 © 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently © 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form oror byby any means, without permission in in writing from thethe publisher. *5–24. *4–16. The airstroke actuator at D is used to apply a force of F on the member at B. The normal reaction of the smooth shaft at C on the member is 300 N. Determine the magnitude of F and the horizontal and vertical components of reaction at pin A. C 15 600 mm B A 60 200 mm 600 mm D F Equations of Equilibrium: From the free-body diagram of member ABC, Fig. a, force F can be obtained by writing the moment equation of equilibrium about point A. � + ΣMA = 0; F sin 60° (800) – 300(600 + 200 sin 15°) = 0 Ans F = 282.22 N = 282 N Using this result and writing the force equations of equilibrium along the x and y axes, + S ΣFx = 0; + ↑ΣFy = 0; –Ax + 300 cos 15° – 282.22 cos 60° = 0 Ans Ax = 149 N –Ay – 300 sin 15° + 282.22 sin 60° = 0 Ans Ay = 167 N 333 196 196 333 05a SM_CH04.indd Ch05a 319-360.indd 4/8/118:43:08 11:48:20 6/12/09 AM AM ©©2011 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybe bereproduced, reproduced,ininany anyform formor orby byany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 4–17. Outriggers A and B are used to stabilize the crane *5–36. from overturning when lifting large loads. If the load to be lifted is 3 Mg, determine the maximum boom angle u so that the crane does not overturn. The crane has a mass of 5 Mg and center of mass at GC, whereas the boom has a mass of 0.6 Mg and center of mass at GB. 4.5 m GB 5m GC u 2.8 m 0.7 m + ©MA = 0; A B 2.3 m – 5(9.81)(2.3) + 3(9.81)(9.5 sin u – 0.7) + 0.6(9.81)(5 sin u – 0.7) = 0 u = 26.4° Ans 345 197 SM_CH04.indd 197 05a Ch05a 319-360.indd 345 4/8/11 6/12/09 11:48:20 8:43:39 AM AM 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. •5–61. •4–18. If spring BC is unstretched with u = 0° and the bell crank achieves its equilibrium position when u = 15°, determine the force F applied perpendicular to segment AD and the horizontal and vertical components of reaction at pin A. Spring BC remains in the horizontal postion at all times due to the roller at C. C k 2 kN/m B F 150 u 300 mm A D Spring Force Formula: From the geometry shown in Fig. a, the stretch of spring BC when the bell crank rotates u = 15° about point A is x = 0.3 cos 30° – 0.3 cos 45° = 0.04768 m. Thus, the force developed in spring BC is given by 400 mm Fsp = kx = 2000(0.04768) = 95.35 N Equations of Equilibrium: From the free-body diagram of the bell crank, Fig. b, F can be obtained by writing the moment equation of equilibrium about point A. + ΣMA = 0; 95.35 sin 45° (300) – F(400) = 0 Ans F = 50.57 N = 50.6 N Using the above result and writing the force equations of equilibrium along the x and y axes, + S ©Fx = 0; Ax – 50.57 sin 15° – 95.35 = 0 Ans Ax = 108.44 N = 108 N + ↑©Fy = 0; Ay – 50.57 cos 15° = 0 Ans Ay = 48.84 N = 48.8 N 366 198 SM_CH04.indd 198 05b Ch05b 361-400.indd 366 4/8/11 6/12/09 11:48:20 8:45:35 AM AM ©©2011 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybe bereproduced, reproduced,ininany anyform formororby byany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 5–38. 4–19. Spring CD remains in the horizontal position at all times due to the roller at D. If the spring is unstretched when u = 0° and the bracket achieves its equilibrium position when u = 30°, determine the stiffness k of the spring and the horizontal and vertical components of reaction at pin A. D k 0.6 m C 0.45 m B u F 300 N A Spring Force Formula: At the equilibrium position, the spring elongates x = 0.6 sin 30° m. Using the spring force formula, the force in spring CD is found to be Fsp = kx = 0.3k. Equations of Equilibrium: From the free-body diagram of the bracket, Fig. a, the stiffness k of spring CD and Ay can be obtained by writing the moment equation of equilibrium about point A and the force equation of equilibrium along the x axis, respectively. + ΣMA = 0; 0.3k cos 30° (0.6) – 300 cos 30° (0.45) – 300 sin 30° (0.6) = 0 Ans k = 1327.35 N/m = 1.33 kN/m + ↑ΣFy = 0; Ay – 300 = 0 Ans Ay = 300 N Using the result k = 1327.35 N / m and writing the force equation of equilibrium along the x axis, + S ©Fx = 0; Ax – 0.3(1327.35) = 0 Ans Ax = 398.21 N = 398 N 347 199 SM_CH04.indd 199 05a Ch05a 319-360.indd 347 4/8/11 6/12/09 11:48:21 8:43:43 AM AM 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. 5–39. Spring CD remains in the horizontal position at all *4–20. times due to the roller at D. If the spring is unstretched when u = 0° and the stiffness is k = 1.5 kN>m, determine the smallest angle u for equilibrium and the horizontal and vertical components of reaction at pin A. D k 0.6 m C 0.45 m B u Spring Force Formula: At the equilibrium position, the spring elongates x = 0.6 sin u. Using the spring force formula, the force in spring CD is found to be Fsp = kx = 1500 (0.6 sin u) = 900 sin u. F 300 N A Equations of Equilibrium: From the free-body diagram of the bracket, Fig. a, the equilibrium position u and Ay can be obtained by writing the moment equation of equilibrium about point A and the force equation of equilibrium along the y axis, respectively. + ΣMA = 0; 900 sin u cos u (0.60) – 300 sin u (0.6) – 300 cos u (0.45) = 0 540 sin u cos u – 180 sin u – 135 cos u = 0 Solving by trial and error yields Ans u = 23.083° = 23.1° + ↑ΣFy = 0; Ay – 300 = 0 Ans Ay = 300 N Using the result u = 23.083° and writing the force equation of equilibrium along the x axis, + S ©Fx = 0; Ax – 900 sin 23.083° = 0 Ans Ax = 352.86 N = 353 N 348 200 SM_CH04.indd 200 05a Ch05a 319-360.indd 348 4/8/11 6/12/09 11:48:21 8:43:45 AM AM ©©2011 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybe bereproduced, reproduced,ininany anyform formor orby byany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 4–21. The The platform platformassembly assemblyhas has a weight of 1.25 kN *5–40. a weight of 1.25 kN and and center of gravity intendedto to support support a center of gravity at at1.GIf it itis isintended 1. If placed at point point G22, determine the maximum load of 2 kN placed smallest counterweight W that should be placed at B in order to prevent the platform from tipping over. G2 0.6 m When tipping occurs, RC = 0 + ΣMD = 0; 1.8 m –2 (0.6) + 1.25 (0.3) + WB (2.1) = 0 Ans WB = 0.393 kN G1 2 kN 1.25 kN 2.4 m B 1.8 m 0.6 m C 2.4 m 0.3 m D 0.3 m 1.8 m 0.3 m 1.8 m 0.3 m 349 201 SM_CH04.indd 201 05a Ch05a 319-360.indd 349 4/8/11 6/12/09 11:48:22 8:43:48 AM AM 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. •4–22. Determine the horizontal and vertical components •5–41. of reaction at the pin A and the reaction of the smooth collar B on the rod. C 300 lb 1500 N 2250 N 450 lb B 30 A 1 ftm 0.3 0.6 2 ftm D 1.2 4 ftm 10.3 ft m Equations of Equilibrium: From the free-body diagram, Ay and NB can be obtained by writing the force equation of equilibrium along the y axis and the moment equation of equilibrium about point A. +↑ΣFy = 0; Ay – 1500 – 2250 = 0 Ans Ay = 3750 N + ΣMA = 0; NB (1.2 sin 30°) – 1500(0.3) – 2250(0.9) = 0 Ans NB = 4125 N Using the result NB = 4125 N and writing the force equation of equilibrium along the x axis, + → ΣFx = 0; Ax – 4125 = 0 Ans Ax = 4125 N 1500 N 2250 N 1.2 m 0.3 m 0.6 m 0.3 m 350 202 SM_CH04.indd 202 05a Ch05a 319-360.indd 350 4/8/11 6/12/09 11:48:22 8:43:50 AM AM © ©2011 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underall allcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybe bereproduced, reproduced,ininany anyform formor orby byany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 4–23. Determine the support reactions of roller A and the 5–42. smooth collar B on the rod. The collar is fixed to the rod AB, but is allowed to slide along rod CD. A 900 N 1m 1m C 45 2m B D 600 N m 45 Equations of Equilibrium: From the free-body diagram of the rod, Fig. a, NB can be obtained by writing the force equation of equilibrium along the y axis. + ↑ΣFy = 0; NB sin 45° – 900 = 0 NB = 1272.79 N = 1.27 kN Ans Using the above result and writing the force equation of equilibrium and the moment equation of equilibrium about point B, + S ©Fx = 0; 1272.79 cos 45° – Ax = 0 Ax = 900 N + ΣMB = 0; Ans –900(1) + 900(2) sin 45° – 600 +MB = 0 MB = 227 N · m Ans 351 203 SM_CH04.indd 203 05a Ch05a 319-360.indd 351 4/8/11 6/12/09 11:48:22 8:43:54 AM AM 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. 5–43. *4–24. The Theuniform uniformrod rodAB ABhas hasaaweight weightof of 15 75 lb. N. Determine the force in the cable when the rod is in the position shown. B 75 N 1.5 sin 40° 5 ft 30 C A 10 0.75 cos 40° + ΣMA = 0; + → ΣFx = 0; NB (1.5 sin 40°) – 75 (0.75 cos 40°) = 0 T cos 10° – 44.69 = 0 NB = 44.69 N T = 45.38 N T Ans · 4–25. Determine the horizontal and vertical components *5–44. of force at the pin A and the reaction at the rocker B of the curved beam. 500 N 200 N 10 15 2m A + ΣMA = 0; NB (4) – 200 cos 15° (2) – 500 cos 10° (2) = 0 NB = 342.79 = 343 N + ↑ΣFy = 0; Ans Ay – 500 cos 10° – 200 cos 15° + 342.79 = 0 Ay = 342.8 = 343 N + S ©Fx = 0; B Ans – Ax + 500 sin 10° – 200 sin 15° = 0 Ax = 35.1 N Ans 352 204 SM_CH04.indd 204 05a Ch05a 319-360.indd 352 4/8/11 6/12/09 11:48:23 8:43:57 AM AM ©©2011 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybe bereproduced, reproduced,ininany anyform formororby byany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. •5–45. •4–26. The floor crane and the driver have a total weight of 2500 lb with 12.5 kN with aa center center of of gravity gravity at at G. If the crane is required to lift the 2.5-kN 500-lb drum, drum, determine the normal reaction on both the wheels at A and both the wheels at B when the boom is in the position shown. F 3.6 12 m ft 0.93mft D C 30 1.8 m 6 ft G E A 2.2 ft 0.66 m1.4 0.42 ft m B 2.52 m 8.4 ft Equations of Equilibrium: From the free-body diagram of the floor crane, Fig. a, + ΣMB = 0; 12.5 (0.42 + 2.52) – 2.5 (4.5 cos 30° – 2.52) – NA (0.66 + 0.42 + 2.52) = 0 Ans NA = 9.252 kN +↑ΣFy = 0; 9.252 – 12.5 – 2.5 + NB = 0 Ans NB = 5.748 kN 4.5 m 12.5 kN 2.5 kN 2.52 m 0.42 m 0.66 m 353 205 SM_CH04.indd 205 05a Ch05a 319-360.indd 353 4/8/11 6/12/09 11:48:23 8:44:00 AM AM 2011 Pearson Education, Inc., Upper Saddle River, Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ.All exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. 5–46. 4–27. The floor crane and the driver have a total weight of 2500kN lb with a center of gravity at G. Determine the largest 12.5 weight of the drum that can be lifted without causing the crane to overturn when its boom is in the position shown. F 12 ft 3.6 m 0.93 m ft D C 30 6 ftm 1.8 G E A 2.2 ft 0.66 m1.40.42 ft m B 2.52 8.4 ftm Equations of Equilibrium: Since the floor crane tends to overturn about point B, the wheel at A will leave the ground and NA = 0. From the free-body diagram of the floor crane, Fig. a, W can be obtained by writing the moment equation of equilibrium about point B. + ΣMB = 0; 12.5 (0.42 + 2.52) – W(4.5 cos 30° – 2.52) = 0 Ans W = 26.69 kN 4.5 m 12.5 kN 2.52 m 0.42 m 0.66 m 354 206 SM_CH04.indd 206 05a Ch05a 319-360.indd 354 4/8/11 6/12/09 11:48:23 8:44:02 AM AM ©©2011 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybe bereproduced, reproduced,ininany anyform formor orby byany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. *4–28. The motor has 5–47. has aa weight weightofof4.25 850 kN. lb. Determine the force that each of the chains exerts on the supporting hooks at A, B, and C. Neglect the size of the hooks and the thickness of the beam. 0.3 m 4.25 kN 850 lb 0.15 0.5m ft 4.25 kN 0.15 m 0.45 m A C 10 + ΣMB = 0; + → ΣFx = 0; +↑ΣFy = 0; 0.45 1.5 ftm 0.3 1 ftm FA cos 10° (0.3) + 4.25 (0.15) – FC cos 10° (0.6) = 0 (1) FC sin 10° – FB sin 30° – FA sin 10° = 0 (2) 4.25 – FA cos 10° – FB cos 30° – FC cos 10° = 0 (3) B 10 30 Solving Eqs. (1), (2) and (3) yields: FA = 2.16 kN FB = 0 Ans FC = 2.16 kN 4–29. Determine the force P needed to pull the 50-kg roller *5–48. over the smooth step. Take u = 60°. P u 0.1 m 0.6 m B A 20 0.5 f = cos–1 = 33.56° 0.6 + ΣMB = 0; 50 (9.81) sin 20° (0.5)+ 50 (9.81) cos 20° (0.3317) – P cos 60° (0.5) – P sin 60° (0.3317) = 0 P = 441 N Ans 355 207 SM_CH04.indd 207 05a Ch05a 319-360.indd 355 4/8/11 6/12/09 11:48:24 8:44:06 AM AM 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. •5–49. Determine the magnitude and direction u of the •4–30. minimum force P needed to pull the 50-kg roller over the smooth step. P u 0.1 m 0.6 m B A 20 For Pmin, NA S 0, + ΣMB = 0; 0.5 f = cos–1 = 33.56° 0.6 50 (9.81) sin 20° (0.5) + 50 (9.81) cos 20° (0.3317) – P cos u (0.5) – P sin u (0.3317) = 0 236.75 – P cos u (0.5) – P sin u (0.3317) = 0 P= 236.75 (0.5 cos u + 0.3317 sin u) For Pmin; – 236.75 (–0.5 sin u + 0.3317 cos u) dP = =0 (0.5 cos u + 0.3317 sin u)2 du tan u = 0.3317 0.5 u = 33.6° Ans Pmin = 395 N Ans 356 208 SM_CH04.indd 208 05a Ch05a 319-360.indd 356 4/8/11 6/12/09 11:48:24 8:44:09 AM AM ©©2011 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybe bereproduced, reproduced,ininany anyform formororby byany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 4–31. The winch cable on a tow truck is subjected to a 5–50. force of T = 6 kN when the cable is directed at u = 60°. Determine the magnitudes of the total brake frictional force F for the rear set of wheels B and the total normal forces at both front wheels A and both rear wheels B for equilibrium. The truck has a total mass of 4 Mg and mass center at G. + S ©Fx = 0; A 2m F B 2.5 m T 1.5 m Ans – NA (4.5) + 4(103)(9.81) (2.5) – 6000 sin 60° (3) – 6000 cos 60° (1.5) = 0 NA = 17 336 N = 17.3 kN + ↑©Fy = 0; 1.25 m 3m u 6000 sin 60° – F = 0 F = 5196 N = 5.20 kN + ΣMB = 0; G Ans 17 336 – 4(103)(9.81) – 6000 cos 60° + NB = 0 NB = 24 904 N = 24.9 kN Ans 357 209 SM_CH04.indd 209 05a Ch05a 319-360.indd 357 4/8/11 6/12/09 11:48:25 8:44:12 AM AM 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced, anyform formororbybyany anymeans, means, withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. exist. No ininany without 5–51. Determine the minimum cable force T and critical *4–32. angle u which will cause the tow truck to start tipping, i.e., for the normal reaction at A to be zero. Assume that the truck is braked and will not slip at B. The truck has a total mass of 4 Mg and mass center at G.x G 1.25 m A 2m + ΣMB = 0; 3m u 2.5 m T F B 1.5 m 4(103)(9.81)(2.5) – T sin u (3) – T cos u (1.5) = 0 T= 65 400 (cos u + 2 sin u) dT –65 400 (sin u + 2 cos u) = =0 du (cos u + 2 sin u)2 – sin u + 2 cos u = 0 u = tan–1 2 = 63.43° = 63.4° T= Ans 65 400 = 29.2 kN (cos 63.43° + 2 sin 63.43°) Ans 4–33. *5–52. Three uniform books, each having a weight W and length a, are stacked as shown. Determine the maximum distance d that the top book can extend out from the bottom one so the stack does not topple over. a d Equilibrium : For top two books, the upper book will topple when the center of gravity of this book is to the right of point A. Therefore, the maximum distance from the right edge of this book to point A is a/2. Equation of Equilibrium : For top entire three books, the top two books will topple about point B. + ΣMB = 0; a W(a – d) – W d – = 0 2 3a d= 4 Ans 358 210 SM_CH04.indd 210 05a Ch05a 319-360.indd 358 4/8/11 6/12/09 11:48:25 8:44:15 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 4–34. 5–63. The cart supports the uniform crate having a mass of 85 kg. Determine the vertical reactions on the three casters at A, B, and C. The caster at B is not shown. Neglect the mass of the cart. B 0.1 m A 0.2 m 0.5 m 0.4 m 0.2 m 0.6 m C 0.35 m 0.35 m Equations of Equilibrium : The normal reaction NC can be obtained directly by summing moments about x axis. ©Mx = 0; ©My = 0; ©Mz = 0; NC (1.3) – 833.85 (0.45) = 0 NC = 288.64 N = 289 N Ans 833.85 (0.3) – 288.64 (0.35) – NA (0.7) = 0 NA = 213.04 N = 213 N Ans NA + 288.64 + 213.04 – 833.85 = 0 NB = 332 N Ans 368 211 SM_CH04.indd 211 05b Ch05b 361-400.indd 368 4/8/11 11:48:25 AM AM 6/12/09 8:45:39 © 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently © 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form or or byby any means, without permission in in writing from thethe publisher. 4–35. *5–64. The pole for a power line is subjected to the two cable forces of 60 lb, each force lying in a plane parallel to the x- y plane. If the tension in the guy wire AB is 80 lb, determine the x, y, z components of reaction at the fixed base of the pole, O. z 300 60 lbN 0.3 m A 45 1 ft 45 4 ftm 1.2 60 N lb 300 400 80 N lb 3 mft 10 B 3 ft 0.9 m O y x Equations of Equilibrium : ΣFx = 0; Ox + 300 sin 45° – 300 sin 45° = 0 Ox = 0 ΣFy = 0; Oy + 300 cos 45° + 300 cos 45° = 0 Oy = –424.3 N ΣFz = 0; Ans Oz – 400 = 0 Oz = 400 N ΣMx = 0; Ans Ans (MO)x + 400 (0.9) 0.3 m – 2[300 cos 45° (4.2)] = 0 (MO)x = 1422 N · m ΣMy = 0; 300 N Ans (MO)y + 300 sin 45° (4.2) 300 N 1.2 m – 300 sin 45° (4.2) = 0 (MO)y = 0 ΣMz = 0; Ans 400 N 0.9 m (MO)z + 300 sin 45° (0.3) 3m – 300 sin 45° (0.3) = 0 (MO)z = 0 Ans 369 212 212 369 05b SM_CH04.indd Ch05b 361-400.indd 4/8/118:45:41 11:48:26 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. •5–65. If P = 6 kN, x = 0.75 m and y = 1 m, determine *4–36. the tension developed in cables AB, CD, and EF. Neglect the weight of the plate. z B F P x D A y E x C 2m 2m y Equations of Equilibrium: From the free-body diagram, Fig. a, TCD and TEF can be obtained by writing the moment equation of equilibrium about the x and y axes, respectively. ΣMx = 0; TCD (2) – 6(1) = 0 Ans TCD = 3 kN ΣMy = 0; TEF (2) – 6(0.75) = 0 Ans TEF = 2.25 kN Using the above results and writing the force equation of equilibrium along the z axis, ΣMz = 0; TAB + 3 + 2.25 – 6 = 0 Ans TAB = 0.75 kN 370 213 SM_CH04.indd 213 05b Ch05b 361-400.indd 370 4/8/11 11:48:26 AM AM 6/12/09 8:45:43 ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently exist. NoNo portion of of this material may bebe reproduced, in in any form oror byby any means, without permission in in writing from thethe publisher. exist. portion this material may reproduced, any form any means, without permission writing from publisher. 5–66. 4–37. Determine the location x and y of the point of application of force P so that the tension developed in cables AB, CD, and EF is the same. Neglect the weight of the plate. z B F P x D A y E x C 2m 2m y Equations of Equilibrium: Equilibrium: From Fromthe thefree-body free-bodydiagram diagramofofthe the plate, Fig. plate, Fig. a, a, and writing the moment equations of equilibrium about the x9 y9 axes, axes, x¿ and y¿ ΣMxœ = 0; T(2 – y) – 2T(y) = 0 Ans y = 0.667 m ΣMyœ = 0; 2T(x) – T(2 – x) = 0 Ans x = 0.667 m 371 214 05b SM_CH04.indd Ch05b 361-400.indd 214 371 6/12/09 AM AM 4/8/118:45:45 11:48:26 ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. No ininany without exist. Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced, anyform formororbybyany anymeans, means, withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 5–67. 4–38. Due to an unequal distribution of fuel in the wing tanks, the centers of gravity for the airplane fuselage A and wings B and C are located as shown. If these kN, lb, WBW =B 40 components have have weights weights WW 45 000 = kN, 8000and lb, AA ==225 WC =W30 kN, determine the normal reactions of the wheels and the normal reactions of the 6000 lb, determine C = D, E, and theF ground. wheels D, F E,on and on the ground. z D B A C E 2.48mft ΣFx = 0; 40 (1.8) – RD (4.2) – 30 (2.4) + Rg (4.2) = 0 ΣFy = 0; 40 (1.2) + 225 (2.1) + 30 (1.2) – RF (8.1) = 0 ΣFz = 0; RD + Rg + RF – 400 – 30 – 225 = 0 F 1.86 m ft 8 ft 2.4 m x ft 1.86 m 4 ft m 1.2 6 mft 0.9 20 3 ftm y 40 kN Solving, RD = 113.15 kN Ans Rg = 113.15 kN Ans RF = 68.0 kN Ans 30 kN 225 kN 372 215 SM_CH04.indd 215 05b Ch05b 361-400.indd 372 4/8/11 11:48:27 AM AM 6/12/09 8:45:46 © 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently © 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form or or byby any means, without permission in in writing from thethe publisher. 4–39. *5–68. Determine the magnitude of force F that must be exerted on the handle at C to hold the 75-kg crate in the position shown. Also, determine the components of reaction at the thrust bearing A and smooth journal bearing B. z 0.1 m A B x 0.6 m 0.5 m y 0.2 m 0.1 m Equations of Equilibrium: From the free-body diagram, Fig. a, F, Bz, Az, and Ay can be obtained by writing the moment equation of equilibrium about the y, x, and x¿ axes and the force equation of equilibrium along the y axis. C F ©My = 0; –F(0.2) + 75(9.81)(0.1) = 0 Ans F = 367.875 N = 368 N ©Mx = 0; Bz (0.5 + 0.6) – 75(9.81)(0.6) = 0 ©Mx¿ = 0; ©Fy = 0; Ay = 0 Bz = 401.32 N = 401 N –Az(0.6 + 0.5) + 75(9.81)(0.5) = 0 Ans Az = 334.43 N = 334 N Ans Using the result F = 367.875 N and writing the moment equations of equilibrium about the z and z¿ axes, ©Mz = 0; –Bx(0.5 + 0.6) – 367.875(0.2 + 0.1 + 0.5 + 0.6) = 0 ©Mz¿ = 0; Bx = –468.20 N = –468 N Ax(0.6 + 0.5) – 367.875(0.2 + 0.1) = 0 Ans Ax = 100.33 N = 100 N Ans The negative signs indicate that Bx act in the opposite sense to that shown on the free-body diagram. 373 216 216 373 05b SM_CH04.indd Ch05b 361-400.indd 4/8/118:45:51 11:48:27 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. •5–69. The shaft is supported by three smooth journal *4–40. bearings at A, B, and C. Determine the components of reaction at these bearings. z 900 N 600 N 450 N 0.9 m 0.6 m x 0.9 m 500 N 0.9 m C 0.6 m A 0.9 m B 0.9 m y Equations of Equilibrium: From the free-body diagram, Fig. a, Cy and Cz can be obtained by writing the force equation of equilibrium along the y axis and the moment equation of equilibrium about the y axis. ΣFy = 0; Cy – 450 = 0 Ans Cy = 450 N ΣMy = 0; Cz(0.9 + 0.9) – 900(0.9) + 600(0.6) = 0 Ans Cz = 250 N Using the above results ΣMx = 0; Bz(0.9 + 0.9) + 250(0.9 + 0.9 + 0.9) + 450(0.6) – 900(0.9 + 0.9 + 0.9) – 600(0.9) = 0 Ans Bz = 1125 N = 1.125 kN ΣMx¿ = 0; 600(0.9) + 450(0.6) – 900(0.9) + 250(0.9) – Az (0.9 + 0.9) = 0 Ans Az = 125 N ΣMz = 0; –Bx(0.9 + 0.9) + 500(0.9) + 450(0.9) – 450(0.9 + 0.9) = 0 ΣFx = 0; Ax + 25 – 500 = 0 Ans Bx = 25 N Ans Ax = 475 N 374 217 SM_CH04.indd 217 05b Ch05b 361-400.indd 374 4/8/11 11:48:28 AM AM 6/12/09 8:45:53 © 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently © 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form or or byby any means, without permission in in writing from thethe publisher. 5–70. 4–41. Determine the tension in cables BD and CD and the x, y, z components of reaction at the ball-and-socket joint at A. z D 3m 300 N B x A 1.5 m 0.5 m C rBD = {– 1i + 1.5j + 3k) m; 1m y rBD = 3.50 m r TBD = TBD BD = – 0.2857 TBD i + 0.4286 TBD j + 0.8571 TBD k rBD In a similar manner, r TCD = TCD CD = – 0.2857 TCD i – 0.4286 TCD j + 0.8571 TCD k rCD Thus, using the components of TBD and TCD, the scalar equations of equilibrium become: ΣFx = 0; Ax – 0.2857 TBD – 0.2857 TCD = 0 ΣFy = 0; Ay + 0.4286 TBD – 0.4286 TCD = 0 ΣFz = 0; Az + 0.8571 TBD + 0.8571 TCD – 300 = 0 ΣMAx = 0; – (0.8571 TBD) (1.5) + (0.8571 TCD) (1.5) = 0 ΣMAy = 0; 300 (1) – (0.8571 TBD) (1.5) – (0.8571 TCD) (1.5) = 0 Solving TBD = TCD = 117 N Ans Ax = 66.7 N Ans Ay = 0 Ans Az = 100 N Ans 375 218 218 375 05b SM_CH04.indd Ch05b 361-400.indd 4/8/118:45:57 11:48:28 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 5–71. usedtotosupport support 1.25-kN 5–71.The Therod rodassembly assembly is used thethe 250-lb cylinder. 4–42. (= 125 kg) cylinder. Determine the Determine the components of components reaction at of thereaction ball-andat the ball-and-socket A, the smooth journal bearing socket joint A, the joint smooth journal bearing E, and the E, force anddeveloped the force developed along rodconnections CD. The connections along rod CD. The at C and Datare C and D are ball-and-socket joints. ball-and-socket joints. z D Equations of Equilibrium: Since rod CD is a two – force member, it exerts a force FDC directed along its axis as defined by uDC on rod BC, Fig. a. Expressing each of the forces indicated on the free – body diagram in Cartesian vector form, FA = Axi + Ay j + Azk x FE = Exi + Ezk C 1 ftm 0.3 A 0.3 1 ftm 0.3 1 ftm W = [–1.25k] kN F 0.31 m ft E 0.45 1.5 ftm y FDC = –FDCk Applying the force equation of equilibrium ΣF = 0; FA + FE + FDC + W = 0 (Axi + Ay j + Azk) + (Exi + Ezk) + (–FDC k) + (–1.25k) = 0 (Ax + Ex)i + (Ay)j + (Az + Ez – FDC)k = 0 Equating i, j, and k components, Ax + Ex = 0 (1) Ay = 0 (2) Az + Ez – FDC – 1.25 = 0 (3) In order to apply the moment equation of equilibrium about point A, the position vectors rAC, rAE, and rAF, Fig. a, must be determined first. rAC = [–0.3i + 0.3j] m rAE = [0.6j] m rAF = [0.45i + 0.9j] m Thus, 0.3 m 0.3 m E(0, 0.6, 0) m 0.3 m ΣMA = 0; (rAC × FDC) + (rAE × FE) + (rAF × W) = 0 (–0.3i + 0.3j) × (–FDCk) + (0.6j) × (Exi + Ezk) + (0.45i + 0.9j) × (–1.25k) = 0 0.45 m (–0.3 FDC + 0.6 Ez – 1.125)i + (0.5625 – 0.3 FDC)j + (–0.6Ex)k = 0 Equating i, j, and k components, –0.3 FDC + 0.6 Ez – 1.125 = 0 (4) 0.5625 – 0.3 FDC = 0 (5) –0.6 Ex = 0 (6) W = 1.25 kN Solving Eqs. (1) through (6), yields FDC = 1.875 kN Ans Ex = 0 Ans Ez = 2.8125 kN Ans Ax = 0 Ans Ay = 0 Ans Az = 0.3125 kN Ans 376 219 SM_CH04.indd 219 05b Ch05b 361-400.indd 376 4/8/11 11:48:29 AM AM 6/12/09 8:45:59 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form oror byby any means, without permission in in writing from thethe publisher. 4–43. *5–72. Determine the components of reaction acting at the smooth journal bearings A, B, and C. z 450 N C 0.6 m 300 N m 45 A 0.4 m x B 0.8 m 0.4 m y Equations of Equilibrium: From the free-body diagram of the shaft, Fig. a, Cy and Cz can be obtained by writing the force equation of equilibrium along the y axis and the moment equation of equilibrium about the y axis. ΣFy = 0; 450 cos 45° + Cy = 0 Cy = –318.20 N = –318 N ΣMy = 0; Ans Cz (0.6) – 300 = 0 Cz = 500 N Ans Using the above results and writing the moment equations of equilibrium about the x and z axes, ΣMx = 0; Bz (0.8) – 450 cos 45° (0.4) – 450 sin 45° (0.8 + 0.4) + (318.20)(0.4) + 500(0.8 + 0.4) = 0 Bz = –272.70 N = –273 N ΣMz = 0; Ans Bx (0.8) – (–318.20)(0.6) = 0 Bx = –238.65 N = 239 N Ans Finally, using the above results and writing the force equations of equilibrium along the x and z axes, ΣFx = 0; Ax + 238.5 = 0 Ax = –238.65 N = –239 N ΣFz = 0; Ans Az – (–272.70) + 500 – 450 sin 45° = 0 Az = 90.90 N = 90.9 N Ans The negative signs indicate that Cy, Bz and AX act in the opposite sense of that shown on the free-body diagram. 377 220 220 377 05b SM_CH04.indd Ch05b 361-400.indd 4/8/118:46:04 11:48:30 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. *4–44. Determine the force components acting on the ball•5–73. and-socket at A, the reaction at the roller B and the tension on the cord CD needed for equilibrium of the quarter circular plate. z D 350 N 200 N 2m 1m A C 60 200 N 3m B y x Equations of Equilibrium : The normal reaction NB and Az can be obtained directly by summing moments about the x and y axes respectively. ΣMx = 0; NB (3) – 200(3) – 200 (3 sin 60°) = 0 NB = –373.21 N = 373 N Ans ΣMy = 0; 350(2) + 200 (3 cos 60°) – Az (3) = 0 Az = 333.33 N = 333 N Ans ΣFx = 0; TCD + 373.21 + 333.33 – 350 – 200 – 200 = 0 TCD = 43.5 N Ans ΣFx = 0; Ax = 0 Ans ΣFy = 0; Ay = 0 Ans 378 221 SM_CH04.indd 221 05b Ch05b 361-400.indd 378 4/8/11 11:48:30 AM AM 6/12/09 8:46:07 © 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently © 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form or or byby any means, without permission in in writing from thethe publisher. 5–74. determine thethe x, y, theload loadhas hasaaweight weightofof200 200lb, kN, determine x, 4–45. IfIfthe zy,components of reaction at the ball-and-socket jointjoint A and z components of reaction at the ball-and-socket A the in each of the andtension the tension in each ofwires. the wires. z 2m ft 4m ft 2 ft m D F ft 33 m A B x G E 2m ft y ft 2m 22 m ft C Equations of Equilibrium : Expressing the forces indicated on the free-body diagram Fig. a, in Cartesian vector form. FA = Ax i + Ay j + Az k W = [–200k] kN kN FBD = FBDk FCD = FCD uCD = FCD � (4 – 4)i + (0 – 4)j + (3 – 0)k (4 – 4)2 + (0 – 4)2 + (3 – 0)2 � 3 4 = – FCD j + FCD k 5 5 FEF = FEF k Applying the force equation of equilibrium, ΣF = 0; FA + FBD + FCD + FEF + W = 0 4 3 (Axi + Ay j + Az k) + FBD k + – FCD j + FCD k + FEF k + (–200k) = 0 5 5 4 3 (Ax)i + Ay – FCD j + Az + FBD + FCD + FEF – 200 k = 0 5 5 Equating i, j, and k components, Ax = 0 4 Ay – FCD = 0 5 3 Az + FBD + FCD + FEF – 200 = 0 5 (1) (2) (3) In order to write the moment equation of equilibrium about point A, the position vectors rAB , rAG , rAC , and rAE must be determined first. rAB rAG rAC rAE = [4i] m = [4i + 2j]m m = [4i + 4j] m m m = [2i + 4j] m Thus, ΣMA = 0; (rAB * FBD) + (rAC * FCD) + (rAE * FEF) + (rAG * W) = 0 4 3 (4i) * (FBD k) + (4i + 4j) * – FCD j + FCD k + (2i + 4j) * FEP k) + (4i + 2j) * (–200k) 5 5 12 16 12 FCD – 2FEF + 800 j + – FCD k = 0 FCD + 4FEF – 400 i + –4FBD – 5 5 5 379 222 222 379 05b SM_CH04.indd Ch05b 361-400.indd 4/8/118:46:10 11:48:31 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. Equating i, j, and k components, 12 F + 4FEF – 400 = 0 5 CD 12 – 4FBD – F – 2FEF + 800 = 0 5 CD 16 – F =0 5 CD (4) (5) (6) Solving Eqs. (1) through (6), FCD = 0 Ans FEF = 100 kNkN Ans FBD = 150 kN kN Ans Ax = 0 Ans Ay = 0 Ans Az = 100 kN kN Ans The negative signs indicate that Az acts in the opposite sense to that on the free-body diagram. F(–2, 0, 2) m D(4, 0, 3) m B(4, 0, 0) m G(4, 2, 0) m W = 200 kN E(2, 4, 0) m C(4, 4, 0) m 380 223 SM_CH04.indd 223 05b Ch05b 361-400.indd 380 4/8/11 11:48:31 AM AM 6/12/09 8:46:13 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form oror byby any means, without permission in in writing from thethe publisher. 4–46. If the cable can be subjected to a maximum tension 5–75. 300 kN, lb, determine the maximum force F which may be of 1.5 applied to the plate. Compute the x, y, z components of reaction at the hinge A for this loading. ΣMy = 0; 0.3 (F) – 1.5 (0.9) = 0 F = 4.5 kN 1.5 kN Ax = 0 Ans ΣFy = 0; Ay = 0 Ans ΣFz = 0; –4.5 + 1.5 + Az = 0; C x B MAx = 0 0.9 9 ftm Ans Ans *5–76. The by by a pin at Aat and cablea Themember memberis issupported supported a pin A aand BC. theIfload at DatisD300 lb, kN, determine thethe x, x, y, y, z cableIfBC. the load is 1.5 determine components z componentsofofreaction reactionatatthe thepin pinAA and and the the tension tension in cable B C. BC. z 0.3 m 1 ft C 3 6 2 TBC = TBC i – j + k 7 7 7 ΣFy = 0; 3 ftm A 0.3 Ans Az = 3 kN ΣMAz = 0; MAz = 0 ΣFx = 0; F 20.2 ft m 10.1 ft m y 0.3 m ΣMAx = 0; MAx + 4.5 (0.1) – 0.3 (1.5) = 0; 3 ftm 0.3 0.9 m Ans ΣFx = 0; z 0.6 2 ftm 3 Ax + TBC = 0 7 0.6 2 ftm 1.8 6 ftm 0.6 m B 0.6 m 2 ft x 1.8 m 6 Ay – TBC = 0 7 A 0.6 2 ftm 1.2 m ΣFz = 0; 2 Az – 1.5 + TBC = 0 7 ΣMx = 0; 2 –1.5 (1.8) + TBC (1.8) = 0 7 ΣMy = 0; 2 MAy – 1.5 (0.6) + TBC (1.2) = 0 7 ΣMz = 0; 3 6 MAz – TBC (1.8) + TBC (1.2) = 0 7 7 1.5 kN y D Solving, TBC = 5.25 kN Ans Ax = –2.25 kN Ans Ay = 4.5 kN Ans Az = 0 Ans MAy = –0.9 kN · m Ans MAz = –1.35 kN · m Ans 381 224 224 381 05b SM_CH04.indd Ch05b 361-400.indd 4/8/118:46:15 11:48:32 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. •8–1. Determine the minimum horizontal force P •4–47. required to hold the crate from sliding down the plane. The crate has a mass of 50 kg and the coefficient of static friction between the crate and the plane is ms = 0.25. P 30 Free – Body Diagram. When the crate is on the verge of sliding down the plane, the frictional force F will act up the plane as indicated on the free – body diagram of the crate shown in Fig. a. Equations of Equilibrium. ©Fy¿ = 0; N – P sin 30° – 50(9.81) cos 30° = 0 ©Fx¿ = 0; P cos 30° + 0.25 N – 50(9.81) sin 30° = 0 Solving P = 140 N N = 494.94 N Ans 674 225 SM_CH04.indd 225 08a Ch08a 674-717.indd 674 4/8/11 11:48:32 AM AM 6/12/09 9:04:54 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form oror byby any means, without permission in in writing from thethe publisher. *4–48. Determine theminimum minimumforce forcePPrequired required to to push 8–2. Determine the the crate up the plane. The crate has a mass of 50 kg and the coefficient of static friction between the crate and the plane is ms = 0.25. P 30 When the crate is on the verge of sliding down the plane, the frictional force F¿ will act down the plane as indicated on the free – body diagram of the crate shown in Fig. b. Thus, F = msN = 0.25 N and F¿ = msN ¿ = 0.25 N¿. By referring to Fig. b, ©Fy¿ = 0; N ¿ – P sin 30° – 50(9.81) cos 30° = 0 ©Fx¿ = 0; P cos 30° – 0.25 N ¿ – 50(9.81) sin 30° = 0 Solving, P = 474 N Ans N¿ = 661.92 N 675 226 226 675 08a SM_CH04.indd Ch08a 674-717.indd 4/8/119:04:57 11:48:33 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 4–49. A horizontal force of P = 100 N is just sufficient to 8–3. hold the crate from sliding down the plane, and a horizontal force of P = 350 N is required to just push the crate up the plane. Determine the coefficient of static friction between the plane and the crate, and find the mass of the crate. P 30 Free – Body Diagram. When the crate is subjected to a force of P = 100 N, it is on the verge of slipping down the plane. Thus, the frictional force F will act up the plane as indicated on the free – body diagram of the crate shown in Fig. a. When P = 350 N, it will cause the crate to be on the verge of slipping up the plane, and so the frictional force F acts down the plane as indicated on the free – body diagram of the crate shown in Fig. a. Thus, F = msN and F ¿ = msN¿. Equations of Equilibrium. +a©Fy ¿ = 0; N – 100 sin 30° – m(9.81) cos 30° = 0 +Q©Fx ¿ = 0; msN + 100 cos 30° – m(9.81) sin 30° = 0 Eliminating N, ms = 4.905 m – 86.603 8.496 m + 50 (1) Also, by referring to Fig. b, we can write +a©Fy ¿ = 0; N ¿ – m(9.81) cos 30° – 350 sin 30° = 0 +Q©Fx ¿ = 0; 350 cos 30° – m(9.81) sin 30° – msN¿ = 0 Eliminating N ¿, ms = 303.11 – 4.905 m 175 + 8.496 m (2) Solving Eqs. (1) and (2) yields m = 36.46 kg Ans ms = 0.256 676 227 SM_CH04.indd 227 08a Ch08a 674-717.indd 676 4/8/11 11:48:33 AM AM 6/12/09 9:04:59 2011 Pearson Education, Inc., Upper Saddle River, Allrights rightsreserved. reserved. Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All This exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. *8–4. 4–50. If the coefficient of static friction at A is ms = 0.4 and the collar at B is smooth so it only exerts a horizontal force on the pipe, determine the minimum distance x so that the bracket can support the cylinder of any mass without slipping. Neglect the mass of the bracket. 100 mm x B C 200 mm A Free – Body Diagram. The weight of cylinder tends to cause the bracket to slide downward. Thus, the frictional force FA must act upwards as indicated in the free – body diagram shown in Fig. a. Here the bracket is required to be on the verge of slipping so that FA = msNA = 0.4 NA. Equations of Equilibrium. + ↑©Fy = 0; +©MB = 0; 0.4NA – mg = 0 NA = 2.5 mg 2.5 mg (0.2) + 0.4 (2.5 mg)(0.1) – m(g)(x + 0.1) = 0 Ans x = 0.5 m Note: Since x is independent of the mass of the cylinder, the bracket will not slip regardless of the mass of the cylinder provided x > 0.5 m. 677 228 228 677 08aSM_CH04.indd Ch08a 674-717.indd 4/8/119:05:01 11:48:33 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. •4–51. The The 180-lb 90-kg man •8–5. man climbs climbs up the the ladder ladder and stops at the position shown after he senses that the ladder is on the verge of slipping. Determine the inclination u of the ladder if the coefficient of static friction between the friction pad A and the ground is ms = 0.4.Assume the wall at B is smooth.The center of gravity for the man is at G. Neglect the weight of the ladder. B G 10 3m ft u 0.9 3 ftm A Free – Body Diagram. Since the weight of the man tends to cause the friction pad A to slide to the right, the frictional force FA must act to the left as indicated on the free – body diagram of the ladder, Fig. a. Here, the ladder is on the verge of slipping. Thus, FA = s NA. Equations of Equilibrium. +↑ΣFy = 0; + ΣMB = 0; NA – 90 (9.81) = 0 90 (9.81) (3 cos ) – 3 cos NA = 882.9 N s (90) (9.81) (3 sin ) – 90 (9.81) (0.9) = 0 – 0.4 (3) sin = 0.9 Ans = 52° 90 (9.81) N 3m 0.9 m 678 229 SM_CH04.indd 229 08a Ch08a 674-717.indd 678 4/8/11 11:48:34 AM AM 6/12/09 9:05:03 © 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently © 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form oror byby any means, without permission in in writing from thethe publisher. The 90-kg man climbs 8–6. 180-lb man climbs the ladder *4–52.The The 90-kg man climbsup upthe theladder ladderand andstops stops at the position shown after he senses that the ladder is on the verge of slipping. Determine the coefficient of static friction between the friction pad at A and ground if the inclination of the ladder is u = 60° and the wall at B is smooth.The center of gravity for the man is at G. Neglect the weight of the ladder. B G 10 3m ft u 0.9 3 ftm A Free – Body Diagram. Since the weight of the man tends to cause the friction pad A to slide to the right, the frictional force FA must act to the left as indicated on the free – body diagram of the ladder, Fig. a. Here, the ladder is on the verge of slipping. Thus, FA = s NA. Equations of Equilibrium. +↑ΣFy = 0; + ΣMB = 0; NA – 90 (9.81) = 0 90 (9.81) (3 cos 60°) – 3 cos 60° – s NA = 882.9 N s (90) (9.81) (3 sin 60°) – 90 (9.81) (0.9) = 0 (3) sin 60° = 0.9 Ans s = 0.231 90 (9.81) 3m 0.9 m 679 230 230 679 08a SM_CH04.indd Ch08a 674-717.indd 4/8/119:05:06 11:48:34 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 4–53. The uniform thin N and a 8–7. thin pole pole has has aa weight weight of of 150 30 lb 7.8ft.m.IfIf placed against smooth length of 26 it it is is placed against thethe smooth wallwall andand on on the rough floor in the position m, will itit remain remain in the rough floor in the position , will d d= =103 ft this position when it is released? The coefficient of static µss ==0.3. friction is m 0.3. B 7.8ftm 26 A d + ΣMA = 0; 150 (1.5) – NB (7.2) = 0 150 N NB = 31.25 N + → ΣFx = 0; 31.25 – FA = 0 7.2 m FA = 31.25 N +↑ΣFy = 0; NA – 150 = 0 NA = 150 N (FA)max = 0.3 (150) = 45 N > 31.25 N Yes, the pole will remain stationary. 1.5 m 1.5 m Ans 680 231 SM_CH04.indd 231 08a Ch08a 674-717.indd 680 4/8/11 11:48:35 AM AM 6/12/09 9:05:09 © 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently © 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form oror byby any means, without permission in in writing from thethe publisher. 4–54. N and and a length *8–8. The uniform pole pole has has aa weight weight of of150 30 lb 7.8ft.m. Determine maximum distance d be it can be of 26 Determine the the maximum distance d it can placed placedthe from the smooth wall not slip. The coefficient of from smooth wall and notand slip. The coefficient of static static friction between the and floorthe and theispole µs = friction between the floor pole ms =is 0.3 . 0.3. B 26 7.8ftm A d 150 N 7.8 sin +↑ΣFy = 0; m NA – 150 = 0 3.9 cos m NA = 150 N FA = (FA)max = 0.3 (150) = 45 N + → ΣFx = 0; 150 N NB – 45 = 0 7.8 m NB = 45 N + ΣMA = 0; 150 (3.9 cos ) – 45 (7.8 sin ) = 0 = 59.04° d = 7.8 cos 59.04° = 4.013 m Ans 681 232 232 681 08a SM_CH04.indd Ch08a 674-717.indd 4/8/119:05:13 11:48:36 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. •4–55. If the coefficient of static friction at all contacting •8–9. surfaces is ms, determine the inclination u at which the identical blocks, each of weight W, begin to slide. A B u Free – Body Diagram. Here, we will assume that the impending motion of the upper block is down the plane while the impending motion of the lower block is up the plane. Thus, the frictional force F acting on the upper block acts up the plane while the friction forces F and F¿ acting on the lower block act down the plane as indicated on the free – body diagram of the upper and lower blocks shown in Figs. a and b, respectively. Since both block are required to be on the verge of slipping, then F = msN and F ¿ = msN¿. Equations of Equilibrium. Referring to Fig. a, +a©Fy ¿ = 0; N – W cos u = 0 N = W cos u +Q©Fx ¿ = 0; T + ms(W cos u) – W sin u = 0 T = W sin u – msW cos u Using these results and referring to Fig. b, +a©Fy ¿ = 0; N¿ – W cos u – W cos u = 0 N¿ = 2W cos u +Q©Fx ¿ = 0; 2(W sin u – msW cos u) – msW cos u – ms(2W cos u) – W sin u = 0 sin u – 5ms cos u = 0 Ans u = tan–1 5ms Since the analysis yields a positive u, the above assumption is correct. 682 233 SM_CH04.indd 233 08a Ch08a 674-717.indd 682 4/8/11 11:48:36 AM AM 6/12/09 9:05:16 © 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently © 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form or or byby any means, without permission in in writing from thethe publisher. 8–10. 20-lb ladder *4–56. The uniform 10-kg ladder rests rests on the rough floor for which the coefficient of static friction is ms = 0.8 and against the smooth wall at B. Determine the horizontal force P the man must exert on the ladder in order to cause it to move. B 1.5 5 ft m ft 2.48 m P 51.5 ft m A 1.8 6 ftm 10 (9.81) N Assume that the ladder tips about A : 10 (9.81) N NB = 0; + → ΣFx = 0; +↑ΣFy = 0; P – FA = 0 1.2 m + ΣMA = 0; 1.2 m –10 (9.81) + NA = 0 NA = 98.1 N 1.2 m 0.9 m 0.9 m 98.1 (0.9) – P (1.2) = 0 0.9 m P = 73.6 N Thus FA = 73.6 N (FA)max = 0.8 (10) (9.81) = 78.5 N > 73.6 N OK Ladder tips as assumed. P = 73.6 N Ans 683 234 234 683 08a SM_CH04.indd Ch08a 674-717.indd 4/8/119:05:18 11:48:36 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 8–11. The uniform 10-kg 20-lb ladder ladder rests rests on the rough floor 4–57. for which the coefficient of static friction is ms = 0.4 and against the smooth wall at B. Determine the horizontal force P the man must exert on the ladder in order to cause it to move. B 51.5 ft m ft 2.48 m P 51.5 ft m A 1.8 6 ftm Assume that the ladder slips at A : 10 (9.81) N FA = 0.4 NA +↑ΣFy = 0; 1.2 m NA – 10 (9.81) = 0 NA = 98.1 N 1.2 m FA = 0.4 (98.1) = 39.24 N + ΣMB = 0; P(1.2) – 98.1 (0.9) + 98.1 (1.8) – 39.24 (2.4) = 0 P = 4.905 N + → ΣFx = 0; 0.9 m Ans 0.9 m NB + 4.905 – 39.24 = 0 NB = 34.335 N > 0 OK The ladder will remain in contact with the wall. 684 235 SM_CH04.indd 235 08a Ch08a 674-717.indd 684 4/8/11 11:48:37 AM AM 6/12/09 9:05:20 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form oror byby any means, without permission in in writing from thethe publisher. 4–58. *8–12. The coefficients of static and kinetic friction between the drum and brake bar are ms = 0.4 and mk = 0.3, respectively. If M = 50 N # m and P = 85 N determine the horizontal and vertical components of reaction at the pin O. Neglect the weight and thickness of the brake. The drum has a mass of 25 kg. 300 mm 700 mm B O 125 mm 500 mm M P A Equations of Equilibrium: From FBD (b), +ΣMO = 0; 50 – FB (0.125) = 0 FB = 400 N From FBD (a), +ΣMA = 0; 85(1.00) + 400(0.5) – NB (0.7) = 0 NB = 407.14 N Friction: Since FB > (FB)max = msNB = 0.4(407.14) = 162.86 N, the drum slips at point B and rotates. Therefore, the coefficient of kinetic friction should be used. Thus, FB = mkNB = 0.3 NB. Equations of Equilibrium: From FBD (b), +ΣMA = 0; 85(1.00) + 0.3 NB (0.5) – NB (0.7) = 0 NB = 154.54 N From FBD (b), +↑ΣFy = 0; Oy – 245.25 – 154.54 = 0 Oy = 400 N Ans + 0.3(154.54) – Ox = 0 Ox = 46.4 N Ans S ΣFx = 0; 685 236 236 685 08a SM_CH04.indd Ch08a 674-717.indd 4/8/119:05:23 11:48:37 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 4–59. The coefficient of static friction between the drum •8–13. and brake bar is ms = 0.4. If the moment M = 35 N # m, determine the smallest force P that needs to be applied to the brake bar in order to prevent the drum from rotating. Also determine the corresponding horizontal and vertical components of reaction at pin O. Neglect the weight and thickness of the brake bar. The drum has a mass of 25 kg. 300 mm 700 mm B O 125 mm P 35 – FB (0.125) = 0 FB = 280 N +↑ΣFy = 0; Oy – 245.25 – 700 = 0 Oy = 945 N Ans + 280 – Ox = 0 Ox = 280 N Ans S ΣFx = 0; From FBD (a), +ΣMA = 0; A Equations of Equilibrium: From FBD (b), Equations of Equilibrium: From FBD (b), +ΣMO = 0; 500 mm M P(1.00) + 280(0.5) – NB (0.7) = 0 Friction: When the drum is on the verge of rotating, FB = msNB 280 = 0.4NB NB = 700 N Substituting NB = 700 N into Eq. (1) yields, P = 350 N Ans 686 237 SM_CH04.indd 237 08a Ch08a 674-717.indd 686 4/8/11 11:48:37 AM AM 6/12/09 9:05:25 2011 Pearson Education, Inc., Upper Saddle River, Allrights rightsreserved. reserved. Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All This exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. *4–60. 8–14. Determine the minimum coefficient of static friction between the uniform 50-kg spool and the wall so that the spool does not slip. 60 B A 0.6 m 0.3 m Free – Body Diagram. Here, the frictional force FA must act upwards to produce the counterclockwise moment about the center of mass of the spool, opposing the impending clockwise rotational motion caused by force T as indicated on the free – body diagram of the spool, Fig. a. Since the spool is required to be on the verge of slipping, then FA = msNA. Equations of Equilibrium. Referring to Fig. a, +©MA = 0; mg (0.6) – T cos 60° (0.3 cos 60° + 0.6) – T sin 60° (0.3 sin 60°) = 0 T = mg + S ©Fx = 0; +↑©Fy = 0; mg sin 60° – NA = 0 NA = 0.8660 mg ms(0.8660 mg) + mg cos 60° – mg = 0 Ans ms = 0.577 Note: Since ms is independent of the mass of the spool, it will not slip regardless of its mass provided ms > 0.577. 687 238 238 687 08aSM_CH04.indd Ch08a 674-717.indd 4/8/119:05:27 11:48:38 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 8–15. 4–61. The spool has a mass of 200 kg and rests against the wall and on the floor. If the coefficient of static friction at B is (ms)B = 0.3, the coefficient of kinetic friction is (mk)B = 0.2, and the wall is smooth, determine the friction force developed at B when the vertical force applied to the cable is P = 800 N. P 0.4 m G A 0.1 m B S ©Fx = 0; + FB – NA = 0 +c©Fy = 0; 800 – 200(9.81) + NB = 0 +©MO = 0; –800(0.1) + FB (0.4) = 0 FB = 200 N NB = 1162 N (FB)max = 0.3(1162) = 348.6 N > 200 N Thus, FB = 200 N Ans 688 239 SM_CH04.indd 239 08a Ch08a 674-717.indd 688 4/8/11 11:48:38 AM AM 6/12/09 9:05:29 © 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently © 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form or or byby any means, without permission in in writing from thethe publisher. 4–62. The 40-kg boy stands stands on on the the beam and pulls on the *8–16. 80-lb boy cord with a force large enough to just cause him to slip. If the coefficient of static friction between his shoes and the beam is (ms)D = 0.4, determine the reactions at A and B. 500 lb. N. Neglect the The beam is uniform and has a weight of 100 size of the pulleys and the thickness of the beam. 13 5 12 D A B C 60 1.5 5 ftm 1 ftm 0.3 0.9 3 ftm Equations of Equilibrium and Friction : When the boy is on the verge to slipping, then FD = ( s)DND = 0.4ND. From FBD (a), +↑ΣFy = 0; 5 ND – T – 40 (9.81) = 0 13 [1] + → ΣFx 12 0.4ND – T = 0 13 [2] = 0; 1.2 m 4 ft 40 (9.81) N ND = 470.88 N 500 N 1.95 m FD = 188.352 N Solving Eqs. [1] and [2] yields T = 204.048 N T = 204.048 N ND = 470.88 N Hence, FD = 0.4 (470.88) = 188.352 N. From FBD (b), + ΣMB = 0; 5 500 (1.95) + 470.88 (2.4) – 204.048 (3.9) 13 1.5 m T = 204.048 N 0.9 m 0.3 m 1.2 m + 204.048 (3.9) + 204.048 sin 30° (2.1) – Ay (1.2) = 0 Ans Ay = 2340.9 N = 2.34 kN + → ΣFx = 0; 12 Bx + 204.048 – 188.352 – 204.048 cos 30° = 0 13 Bx = 176.7 N +↑ΣFy = 0; Ans 5 2340.9 + 204.048 – 204.048 13 – 204.048 sin 30° – 470.88 – 500 – By = 0 By = 1142.4 N = 1.14 kN Ans 689 240 240 689 08a SM_CH04.indd Ch08a 674-717.indd 4/8/119:05:31 11:48:38 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. No ininany without exist. Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced, anyform formororbybyany anymeans, means, withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. •4–63. The 40-kg boy stands stands on on the the beam and pulls with a •8–17. 80-lb boy 200lb. N.IfIf(m (µs)s)DD == 0.4 0.4,, determine the frictional force force of 40 between his shoes and the beam and the reactions at A and 500 lb. N. Neglect B. The beam is uniform and has a weight of 100 the size of the pulleys and the thickness of the beam. 13 5 12 D A B C 60 1.5 m 5 ft 1 ftm 0.3 0.9 3 ftm 1.2 m 4 ft Equations of Equilibrium and Friction : From FBD (a), 40 (9.81) N +↑ΣFy = 0; 5 ND – 200 – 40 (9.81) = 0 13 ND = 469.32 N + → ΣFx 12 FD – 200 = 0 13 FD = 184.62 N = 0; 200 N Since (FD)max = (µs) ND = 0.4 (469.32) = 187.73 N > FD, then the boy does not slip. Therefore, the friction force developed is Ans FD = 184.62 N = 184.6 N From FBD (b), + ΣMB = 0; ND = 469.31 N 500 N 1.95 m 200 N FD = 184.62 N 5 500 (1.95) + 409.32 (2.4) – 200 (3.9) 13 + 200 (3.9) + 200 sin 30° (2.1) – Ay (1.2) = 0 Ans Ay = 2326.14 N = 2.326 kN + → ΣFx = 0; 1.2 m 12 Bx + 200 – 184.62 – 200 cos 30° = 0 13 Ans Bx = 173.21 N +↑ΣFy = 0; 0.9 m 1.5 m 0.3 m 200 N 200 N 5 2326.14+ 200 – 200 13 – 200 sin 30° – 469.32 – 500 – By = 0 Ans By = 1133.74 N = 1.133 kN 690 241 SM_CH04.indd 241 08a Ch08a 674-717.indd 690 4/8/11 11:48:39 AM AM 6/12/09 9:05:34 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form oror byby any means, without permission in in writing from thethe publisher. *4–64. The tongs are used to lift the 150-kg crate, whose 8–18. center of mass is at G. Determine the least coefficient of static friction at the pivot blocks so that the crate can be lifted. P 275 mm E 500 mm C 30 F H D 500 mm A G B 300 mm Free – Body Diagram. Since the crate is suspended from the tongs, P must be equal to the weight of the crate; i.e., P = 150(9.81) N as indicated on the free – body diagram of joint H shown in Fig. a. Since the crate is required to be on the verge of slipping downward, FA and FB must act upward so that FA = msNA and FB = msNB as indicated on the free – body diagram of the crate shown in Fig. c. Equations of Equilibrium. Referring to Fig. a, + S ©Fx = 0; +↑©Fy = 0; FHE cos 30° – FHF cos 30° = 0 FHE = FHF = F 150(9.81) – 2F sin 30° = 0 F = 1471.5 N Referring to Fig. b, +©MC = 0; 1471.5 cos 30° (0.5) + 1471.5 sin 30° (0.275) – NA (0.5) – msNA (0.3) = 0 0.5 NA + 0.3msNA = 839.51 (1) Due to the symmetry of the system and loading, NB = NA. Referring to Fig. c, +↑©Fy = 0; 2msNA – 150(9.81) = 0 (2) Solving Eqs. (1) and (2), yields NA = 1237.57 N Ans ms = 0.595 691 242 242 691 08a SM_CH04.indd Ch08a 674-717.indd 4/8/119:05:38 11:48:39 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 8–19. B have have aa weight weight of of 50 10 N lband and30 6 lb, N, 4–65. Two blocks A and B respectively. They are resting on the incline for which the coefficients of static friction are mA = 0.15 and mB = 0.25. Determine the incline angle u for which both blocks begin to slide. Also find the required stretch or compression in the connecting spring for this to occur. The spring has a stiffness of k == 40 2 lb>ft N/m.. k k= 402N/m lb/ft A u 50 N Equations of Equilibrium : Using the spring force formula, Fsp = kx = 40x. From FBD (a), Fsp = 40x ΣFx′ = 0; 40x + FA – 50 sin + ΣFy′ = 0; NA – 50 cos = 0 [2] ↑ ↑ + =0 B [1] From FBD (b), ΣFx′ = 0; Fs – 40x – 30 sin = 0 [3] + ΣFy′ = 0; NB – 30 cos = 0 [4] ↑ ↑ + Friction : If block A and B are on the verge to move, slipping would have to occur at point A and B. Hence, FA = sA NA = 0.15NA and FB = s B NB = 0.25NB. Substituting these values into Eqs. [1], [2], [3] and [4] and solving, we have = 10.62° NA = 49.14 N x = 0.046 m 30 N Fsp = 40x Ans NB = 29.49 N 692 243 SM_CH04.indd 243 08a Ch08a 674-717.indd 692 4/8/11 11:48:40 AM AM 6/12/09 9:05:40 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form oror byby any means, without permission in in writing from thethe publisher. 4–66. Two blocks A and B *8–20. B have have aa weight weight of of50 10N lband and30 6 lb, N, respectively. They are resting on the incline for which the coefficients of static friction are mA = 0.15 and mB = 0.25. Determine the angle u which will cause motion of one of the blocks. What is the friction force under each of the blocks when this occurs? The spring has a stiffness of k == 40 2 lb>ft N/m and is originally unstretched. kk = 40 N/m 2 lb/ft A u Equations of Equilibrium : Since Block A and B is either not moving or on the verge of moving, the spring force Fsp = 0. From FBD (a), ΣFx′ = 0; FA – 50 sin = 0 [1] + ΣFy′ = 0; NA – 50 cos [2] ↑ ↑ + =0 B 50 N From FBD (b), ΣFx′ = 0; FB – 30 sin = 0 [3] + ΣFy′ = 0; NB – 30 cos = 0 [4] ↑ ↑ + 30 N Friction : Assuming block A is on the verge of slipping, then FA = sA NA = 0.15NA [5] Solving Eqs. [1], [2], [3], [4] and [5] yields = 8.531° FA = 7.42 N NA = 49.45 N FB = 4.45 N NB = 29.67 N Since (FB)max = µsB NB = 0.25 (29.67) = 7.42 N > FB, block B does not slip. Therefore, the above assumption is correct. Thus = 8.53° FB = 4.45 N FA = 7.42 N Ans 693 244 244 693 08a SM_CH04.indd Ch08a 674-717.indd 4/8/119:05:42 11:48:40 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. •8–21. weigh 1000 200 lb N and and 150 750 lb, N, •4–67. Crates A and B weigh respectively. They are connected together with a cable and placed on the inclined plane. If the angle u is gradually increased, determine u when the crates begin to slide. The coefficients of static friction between the crates and the plane are mA = 0.25 and mB = 0.35. B D A C u Free – Body Diagram. Since both crates are required to be on the verge of sliding down the plane, the frictional forces FA and FB must act up the plane so that FA = ANA = 0.25NA and FB = B NB = 0.35NB as indicated on the free – body diagram of the crates shown in Figs. a and b. + ΣFy′ = 0; NA – 1000 cos + ΣFx′ = 0; FCD + 0.25 (1000 cos ) – 1000 sin ↑ ↑ Equations of Equilibrium. Referring to Fig. a, =0 NA = 1000 cos =0 (1) + ΣFy′ = 0; NB – 750 cos + ΣFx′ = 0; 0.35 (750 cos ) – FCD – 750 sin = 0 ↑ ↑ Also, by referring to Fig. b, =0 NB = 750 cos (2) Solving Eqs. (1) and (2), yields Ans = 16.3° FCD = 41.13 N 750 N 1000 N 694 245 SM_CH04.indd 245 08a Ch08a 674-717.indd 694 4/8/11 11:48:41 AM AM 6/12/09 9:05:44 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. 8–30. lb with center of weight of of 8000 40 kN *4–68. The tractor has aa weight gravity at G. Determine if it can push push the the 250-kg 550-lb log up the incline. The coefficient of static friction between the log and the ground is ms = 0.5, and between the rear wheels of the tractor and the ground msœ = 0.8. The front wheels are free to roll. Assume the engine can develop enough torque to cause the rear wheels to slip. C G 1.25 ftm 0.375 B 10 A 2.5 ftm 0.75 2.1 7 ftm 30.9 ft m 40 kN Log : ΣFy = 0; ↑ + P = 1633.5 N NC – 250 (9.81) cos 10° = 0 0.75 m NC = 2415.2 N ΣFx = 0; ↑ + 0.9 m 0.375 m 2.1 m –0.5 (2415.2) – 250 (9.81) sin 10° + P = 0 P = 1633.5 N Tractor : + ΣMg = 0; 250 (9.81) N 3 1633.5 (0.375) + 40 (10 ) (cos 10°) (0.9) + 40 (103) (sin 10°) (0.75) – NA (3) = 0 NA = 13.758 kN ΣFx = 0; ↑ + FA – 40 (103) sin 10° – 1633.5 = 0 FA = 8.579 kN (FA)max = 0.8 (13.758) = 11.006 kN > 8.579 kN Tractor can move log. Ans 702 246 SM_CH04.indd 246 08a Ch08a 674-717.indd 702 4/8/11 6/12/09 11:48:41 9:06:07 AM AM ©©2011 This 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved. Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybe bereproduced, reproduced,ininany anyform formororby byany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. weight of of 8000 40 kN 8–31. The tractor has aa weight lb with with center of 4–69. gravity at G. Determine the greatest weight of the log that can be pushed up the incline. The coefficient of static friction between the log and the ground is ms = 0.5, and between the rear wheels of the tractor and the ground msœ = 0.7. The front wheels are free to roll. Assume the engine can develop enough torque to cause the rear wheels to slip. C G B 10 A 0.375 1.25 ftm 2.5 ftm 0.75 3 ft 72.1 ft m 0.9 m 40 kN Tractor : + ΣMg = 0; 0.75 m 40 (cos 10°) (0.9) + 400 (sin 10°) (0.75) + P (0.375) – NA (3) = 0 0.9 m 0.375 m 2.1 m NA – P (0.125) = 13.55417 ΣFx = 0; ↑ + 0.7 FA – 40 sin 10° – P = 0 0.7 FA – P = 6.94593 NA = 13.902 kN P = 2.786 kN P = 2.786 kN + ΣFy = 0; NC – W cos 10° = 0 + ΣFx = 0; 2.786 – W sin 10° – 0.5 NC = 0 ↑ ↑ Log : NC = 4.119 kN W = 4.183 kN Ans 703 247 SM_CH04.indd 247 08a Ch08a 674-717.indd 703 4/8/11 6/12/09 11:48:42 9:06:09 AM AM 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. 8–26. 180 N lb (and90 rests 4–70. The refrigerator has a weight of 900 kg)on anda tile forfloor which If the rests floor on a tile for which 0.25 If theman man pushes µs =. 0.25. horizontally on the refrigerator in the direction shown, determine the smallest magnitude of horizontal force needed to move it. Also, if if the the man man has has aa weight weight of of 150 750 lb, N determine smallest the coefficient friction between his ( 75 kg),the determine smallestofcoefficient of friction shoes andhis theshoes floorand so that he does not slip. between the floor so that he does not slip. 0.9 m G 0.45 m 1.2 m 0.9 m A Equations of Equilibrium : From FBD (a), +↑ΣFy = 0; N – 900 = 0 + → ΣFx P–F=0 [1] 900 (x) – P (1.2) = 0 [2] = 0; + ΣMA = 0; N = 900 N 0.45 m Friction : Assuming the refrigerator is on the verge of slipping, then F = N = 0.25 (900) = 225 N. Substituting this value into Eqs. [1], and [2] and solving yields P = 225 N 900 N x = 0.3 m 1.2 m Since x < 0.45 m, the refrigerator does not tip. Therefore, the above assumption is correct. Thus Ans P = 225 N From FBD (b), P = 225 N +↑ΣFy = 0; Nm – 750 = 0 Nm = 750 N + → ΣFx Fm – 225 = 0 Fm = 225 N = 0; 750 N When the man is on the verge of slipping, then Fm = 225 = ′Nm s ′ (750) s ′ = 0.300 Ans s 698 248 SM_CH04.indd 248 08a Ch08a 674-717.indd 698 4/8/11 6/12/09 11:48:42 9:05:57 AM AM ©©2011 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybe bereproduced, reproduced,ininany anyform formor orby byany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 900 lb N (and90 kg)on and 8–27. rests a 4–71. The refrigerator has a weight of 180 restsfloor on afor tile floor for which 0.25.man Also, man has µs = the tile which hasthe a weight of 0.25. Also, a weight 750 N ( 75ofkg) andfriction the coefficient of static 150 lb andofthe coefficient static between the floor friction between shoes is µs = 0.6.onIfthe he and his shoes is the floor he his pushes horizontally 0.6. Ifand pushes horizontally on the refrigerator, if he can refrigerator, determine if he can movedetermine it. If so, does the move it. If so, does the refrigerator slip or tip? refrigerator slip or tip? 0.9 m G 0.45 m 1.2 m 0.9 m A Equations of Equilibrium : From FBD (a), 0.45 m +↑ΣFy = 0; N – 900 = 0 + → ΣFx P–F=0 [1] 900 (x) – P (1.2) = 0 [2] = 0; + ΣMA = 0; N = 900 N 900 N Friction : Assuming the refrigerator is on the verge of slipping, then F = N = 0.25 (900) = 225 N. Substituting this value into Eqs. [1], and [2] and solving yields P = 225 N 1.2 m x = 0.3 m Since x < 0.45 m, the refrigerator does not tip. Therefore, the above assumption is correct. Thus, the refrigerator slips. Ans 750 N P = 225 N From FBD (b), +↑ΣFy = 0; Nm – 750 = 0 Nm = 750 N + → ΣFx Fm – 225 = 0 Fm = 225 N = 0; Since (Fm)max = s′Nm = 0.6 (750) = 450 N > Fm, then the man does not slip. Thus, the man is capable of moving the refrigerator. Ans 699 249 SM_CH04.indd 249 08a Ch08a 674-717.indd 699 4/8/11 6/12/09 11:48:42 9:06:01 AM AM 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form oror byby any means, without permission in in writing from thethe publisher. *8–24. The drum has a weight of 500 100 N lb and rests on the *4–72. The floor for which the coefficient coefficient of ofstatic staticfriction frictionisismµss = 0.6. 0.6. If a ==0.6 m and and b == 0.9 m,determine determinethe the smallest smallest magnitude of 2 ft 3 ft, the force P that will cause impending motion of the drum. P 5 3 a 4 Assume that the drum tips : b x = 0.3 m + ΣMO = 0; 3 4 500 (0.3) + P (0.6) – P (0.9) = 0 5 5 P = 416.67 N + → ΣFx = 0; 4 –F – 416.67 = 0 5 500 N F = 333.33 kN +↑ΣFy = 0; 3 N – 500 – 416.67 = 0 5 0.9 m N = 750 N 0.3 m Fmax = 0.6 (750) = 450 N > 333.33 N 0.3 m OK Drum tips as assumed, P = 416.67 N Ans The drum has a weight of 500 100 N lb and rests on the •8–25. •4–73. The 0.5. If floor for which the coefficient coefficient of ofstatic staticfriction frictionisismµss = 0.5. 3 ft 4 ft, a ==0.9 m and and b == 1.2 m,determine determinethe the smallest smallest magnitude of the force P that will cause impending motion of the drum. P 3 5 a 4 Assume that the drum slips : b F = 0.5 N + → ΣFx = 0; +↑ΣFy = 0; 4 –0.5 N + P = 0 5 3 –P – 500 + N = 0 5 500 N P = 500 N N = 800 N + ΣMO = 0; 1.2 m 3 4 800 (x) + 500 (0.45) – 500 (1.2) = 0 5 5 x = 0.431 m < 0.45 m 0.45 m 0.45 m OK Drum slips as assumed, P = 500 N Ans 697 250 250 697 08a SM_CH04.indd Ch08a 674-717.indd 4/8/119:05:52 11:48:43 6/12/09 AM AM 0.5 cos u – 1.06 sin u = 0 0.5 u = tan–1 Ans = 25.3° 1.06 © 2011 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. •4–74. •8–77. The square threaded screw of the clamp has a mean diameter of 14 mm and a lead of 6 mm. If ms = 0.2 for the threads, and the torque applied to the handle is 1.5 N # m, determine the compressive force F on the block. 1.5 N m 1 6 Frictional Forces on Screw: Here, u = tan–1 = tan–1 = 7.768°, 2pr 2p(7) W = F and f, = tan–1m, = tan–i (0.2) = 11.310°. Applying Eq. 8 – 3, we have M = Wr tan (u + f) F 1.5 = F (0.007) tan (7.768° + 11.310°) F = 620 N Ans Note: Since f, > u, the screw is self-locking. It will not unscrew even if the moment is removed. F 745 08b Ch08b 718-761.indd 745 6/12/09 10:34:16 AM 251 SM_CH04.indd 251 4/8/11 11:48:44 AM 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. 8–78. 4–75. The device is used to pull the battery cable terminal C from the post of a battery. If the required pulling force is 85 MM that must be applied to the 425lb,N,determine determinethe thetorque torque that must be applied to handle on on thethe screw to to tighten the handle screw tightenit.it.The Thescrew screw has has square threads, a mean in., aalead in., and the mean diameter diameterof of0.2 5 mm, leadofof0.08 2 mm, coefficient of static friction is m 0.5. µss ==0.5. M 1 2 Frictional Forces on Screw : Here, = tan–1 = tan–1 = 7.256°, 2π r 2π (2.5) W = 425 N and s = tan–1 s = tan–1 (0.5) = 26.565°. Applying Eq. 8–3, we have M = Wr tan ( + ) = 425 (2.5) tan (7.256° + 26.565°) = 711.85 N · m = 0.712 N · m Note: Since s A Ans C B > , the screw is self-locking. It will not unscrew even if the moment is removed. 8–79. The jacking mechanism mechanism consists consistsofofaalink linkthat thathas has 9 a *4–76. a square-threaded screw with a mean diameter ofin. 12and mm square-threaded screw with a mean diameter of 0.5 a and aoflead of in., 5 mm, coefficientofofstatic static friction friction is lead 0.20 andand thethe coefficient µss ==0.4. m Determinethe thetorque torqueM M that that should should be applied to 0.4.Determine to start start lifting lifting the the6000-lb 30-kN load the screw to load acting acting at at the end of member ABC. 30 kNlb 6000 C B M 187.5 mm 7.5 in. 250 10mm in. D A 20 imm n. 500 15 mm in. 375 10 in. 250 mm 250 = tan–1 = 21.80° 625 + ΣMA = 0; –30 (875) + FBD cos 21.80° (250) + FBD sin 21.80° (500) = 0 30 kN FBD = 62.828 kN z = tan–1 (0.4) = 21.80° 5 = tan–1 = 7.555° 2π (6) 250 mm M = Wr tan ( + ) M = 62.828 (6) tan (7.555° + 21.80°) 500 mm 375 mm M = 212.02 kN · mm = 212.0 N · m 746 252 SM_CH04.indd 252 08b Ch08b 718-761.indd 746 4/8/11 11:48:45 6/12/09 10:34:18 AM AM ©©2011 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybe bereproduced, reproduced,ininany anyform formor orby byany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 4–77. Determine the magnitude of the horizontal force P *8–80. that must be applied to the handle of the bench vise in order to produce a clamping force of 600 N on the block. The single square-threaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction is ms = 0.25. 100 mm Here, M = P (0.1) P 7.5 L = tan–1 u = tan = 5.455° 2pr 2p(12.5) –1 fs = tan–1ms = tan–1(0.25) = 14.036° W = 600 N Thus M = Wr tan (fs + u) P(0.1) = 600(0.0125) tan (14.036° + 5.455°) P = 26.5 N Ans Note: Since fs > u, the screw is self-locking. •4–78. Determine the clamping force exerted on the •8–81. block if a force of P = 30 N is applied to the lever of the bench vise. The single square-threaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction is ms = 0.25. 100 mm P Here, M = 30(0.1) = 3 N · m 7.5 L u = tan–1 = tan–1 = 5.455° 2pr 2p(12.5) fs = tan–1ms = tan–1(0.25) = 14.036° W=F Thus M = Wr tan (fs + u) 3 = F (0.0125) tan (14.036° + 5.455°) F = 678 N Ans Note: Since fs > u, the screw is self-locking. 747 253 SM_CH04.indd 253 08b Ch08b 718-761.indd 747 4/8/11 11:48:46 6/12/09 10:34:21 AM AM 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. 8–82. 4–79. Determine the required horizontal force that must be applied perpendicular to the handle in order to develop a 900-N clamping force on the pipe. The single squarethreaded screw has a mean diameter of 25 mm and a lead of 5 mm. The coefficient of static friction is ms = 0.4. Note: The screw is a two-force member since it is contained within pinned collars at A and B. E C 200 mm 150 mm A B 200 mm D Referring to the free-body diagram of member ED shown in Fig. a, +©MD = 0; FAB(0.2) – 900(0.4) = 0 FAB = 1800 N L 5 Here, u = tan–1 = tan–1 = 3.643° 2pr 2p(12.5) fs = tan–1ms = tan–1(0.4) = 21.801° M = F (0.15); and W = FAB = 1800 N M = Wr tan (fs + u) F(0.15) = 1800(0.0125) tan (21.801° + 3.643°) F = 71.4 N Ans Note: Since fs > u, the screw is self-locking. 748 254 SM_CH04.indd 254 08b Ch08b 718-761.indd 748 4/8/11 11:48:47 6/12/09 10:34:23 AM AM ©©2011 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybe bereproduced, reproduced,ininany anyform formor orby byany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 8–83. *4–80. If the clamping force on the pipe is 900 N, determine the horizontal force that must be applied perpendicular to the handle in order to loosen the screw. The single square-threaded screw has a mean diameter of 25 mm and a lead of 5 mm. The coefficient of static friction is ms = 0.4. Note: The screw is a two-force member since it is contained within pinned collars at A and B. E C 200 mm 150 mm A B 200 mm D Referring to the free-body diagram of member ED shown in Fig. a, +©MD = 0; FAB(0.2) – 900(0.4) = 0 FAB = 1800 N L 5 Here, u = tan–1 = tan–1 = 3.643° 2p(12.5) 2pr fs = tan–1ms = tan–1(0.4) = 21.801° M = F (0.15); and W = FAB = 1800 N M = Wr tan (fs – u) F(0.15) = 1800(0.0125) tan (21.801° + 3.643°) F = 49.2 N Ans 749 255 SM_CH04.indd 255 08b Ch08b 718-761.indd 749 4/8/11 11:48:48 6/12/09 10:34:25 AM AM 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. *8–84. 4–81. The clamp provides pressure from several directions on the edges of the board. If the square-threaded screw has a lead of 3 mm, mean radius of 10 mm, and the coefficient of static friction is ms = 0.4, determine the horizontal force developed on the board at A and the vertical forces developed at B and C if a torque of M = 1.5 N # m is applied to the handle to tighten it further. The blocks at B and C are pin connected to the board. B 45 D A M 45 C fz = tan–1(0.4) = 21.801° 3 up = tan–1 = 2.734° 2p(10) M = W(r) tan (fs + up) 1.5 = Az (0.01) tan (21.801° + 2.734°) Ans Az = 328.6 N + S ©Fx = 0; 328.6 – 2T cos 45° = 0 T = 232.36 N By = Cy = 232.36 sin 45° = 164 N Ans 750 256 SM_CH04.indd 256 08b Ch08b 718-761.indd 750 4/8/11 11:48:50 6/12/09 10:34:27 AM AM ©©2011 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybe bereproduced, reproduced,ininany anyform formor orby byany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. •4–82. If the jack supports the 200-kg crate, determine the •8–85. horizontal force that must be applied perpendicular to the handle at E to lower the crate. Each single square-threaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction is ms = 0.25. C A E 45 45 45 45 100 mm B D The force in rod AB can be obtained by first analyzing the equilibrium of joint C followed by joint B. Referring to the free-body diagram of joint C shown in Fig. a, S ©Fx = 0; + FCA sin 45° – FCB sin 45° = 0 FCA = FCB = F +c©Fy = 0; 2F cos 45° – 200(9.81) = 0 F = 1387.34 N Using the result of F and referring to the free-body diagram of joint B shown in Fig. b, +c©Fy = 0; FBD sin 45° – 1387.34 sin 45° = 0 FBD = 1387.34 N + 1387.34 cos 45° +1387.34 cos 45° – FAB = 0 FAB = 1962 N S ©Fx = 0; L 5 = tan–1 Here, u = tan–1 = 5.455° 2p(12.5) 2pr fs = tan–1ms = tan–1(0.25) = 14.036° M = F (0.1) and W = FAB = 1962 N Since M must overcome the friction of two screws, M = 2[Wr tan (fs – u)] F(0.1) = 2[1962(0.0125) tan (14.036° – 5.455°)] Ans F = 74.0 N Note: Since fs > u, the screw are self-locking. fz = tan–1(0.4) = 21.801° 3 up = tan–1 = 2.734° 2p(10) M = W(r) tan (fs + up) 1.5 = Az (0.01) tan (21.801° + 2.734°) Ans Az = 328.6 N + S ©Fx = 0; 328.6 – 2T cos 45° = 0 T = 232.36 N By = Cy = 232.36 sin 45° = 164 N Ans 751 257 SM_CH04.indd 257 08b Ch08b 718-761.indd 751 4/8/11 11:48:51 6/12/09 10:34:30 AM AM 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. 8–86. If the jack is required to lift the 200-kg crate, 4–83. determine the horizontal force that must be applied perpendicular to the handle at E. Each single squarethreaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction is ms = 0.25. C A E 45 45 45 45 100 mm B D The force in rod AB can be obtained by first analyzing the equilibrium of joint C followed by joint B. Referring to the free-body diagram of joint C shown in Fig. a, S ©Fx = 0; + FCA sin 45° – FCB sin 45° = 0 FCA = FCB = F +c©Fy = 0; 2F cos 45° – 200(9.81) = 0 F = 1387.34 N Using the result of F and referring to the free-body diagram of joint B shown in Fig. b, +c©Fy = 0; FBD sin 45° – 1387.34 sin 45° = 0 FBD = 1387.34 N + 1387.34 cos 45° +1387.34 cos 45° – FAB = 0 FAB = 1962 N S ©Fx = 0; L 5 = tan–1 Here, u = tan–1 = 5.455° 2pr 2p(12.5) fs = tan–1ms = tan–1(0.25) = 14.036° M = F (0.1) and W = FAB = 1962 N Since M must overcome the friction of two screws, M = 2[Wr tan (fs + u)] F(0.1) = 2[1962(0.0125) tan (14.036° + 5.455°)] F = 174 N Ans Note: Since fs > u, the screws are self-locking. 752 258 SM_CH04.indd 258 08b Ch08b 718-761.indd 752 4/8/11 11:48:52 6/12/09 10:34:33 AM AM ©©2011 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybe bereproduced, reproduced,ininany anyform formororby byany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. *4–84. The machine part is held in place using the 8–87. double-end clamp. The bolt at B has square threads with a mean radius of 4 mm and a lead of 2 mm, and the coefficient of static friction with the nut is ms = 0.5. If a torque of M = 0.4 N # m is applied to the nut to tighten it, determine the normal force of the clamp at the smooth contacts A and C. 90 mm 260 mm B A C f = tan–1(0.5) = 26.565° 2 = 4.550° u = tan–1 2p(4) M = W(r) tan (u + f) 0.4 = W (0.004) tan (4.550° + 26.565°) W = 165.67 N +©MA = 0; NC (350) – 165.67 (260) = 0 NC = 123.1 = 123 N +c©Fy = 0; Ans NA – 165.67 + 123.1 = 0 NA = 42.6 N Ans 4–85. Blocks A and *8–88. andBBweigh weigh250 50 lb 30 lb, N and 150 N, respectively. Using the coefficients of static friction indicated, determine the greatest weight of block D without causing motion. m B 0.5 mBA 20 0.6 A C D mAC 0.4 For block A and B : Assuming block B does not slip +↑ΣFy = 0; NC – (250 + 150)= 0 NC = 400 N + → ΣFx 0.4 (400) – TB = 0 TB = 160 N = 0; (250 + 150) N For block B : +↑ΣFy = 0; NB cos 20° + FB sin 20° – 150 = 0 [1] + → ΣFx FB cos 20° – NB sin 20° – 160 = 0 [2] = 0; Solving Eqs. [1] and [2] yields : FB = 201.65 N 150 N NB = 86.23 N TB = 160 N Since FB = 201.65 N > NB = 0.6 (86.23) = 51.74 N, slipping does occur between A and B. Therefore, the assumption is no good. Since slipping occurs, FB = 0.6 NB. +↑ΣFy = 0; NB cos 20° + 0.6NB sin 20° – 150 = 0 NB = 131.0 N + → ΣFx 0.6 (131.0) cos 20° – 131.0 sin 20° – TB = 0 TB = 29.055 N = 0; T2 = T1e Where T2 = WD, T1 = TB = 5.812 N, 150 N = 0.5π rad WD = 29.055e0.5(0.5π) = 63.73 N Ans 259 753 SM_CH04.indd 259 08b Ch08b 718-761.indd 753 4/8/11 11:48:53 6/12/09 10:34:36 AM AM 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. •4–86. Blocks A and B •8–89. N each, each, and D weighs B weigh weigh 375 75 lb 150 N. Using Using the the coefficients coefficients of of static static friction indicated, 30 lb. determine the frictional force between blocks A and B and between block A and the floor C. For the rope, T3 = T1 e , where T2 = 150 N, T1 = TB, and m B 0.5 mBA 20 0.6 A = 0.5π rad. C D mAC 0.4 150 = TB e0.5(0.5π) TB = 68.391 N Ans FC = 68.4 N For block B : +↑ΣFy = 0; NB cos 20° + FB sin 20° – 3.75 = 0 [1] + → ΣFx FB cos 20° – NB sin 20° – 68.391 = 0 [2] = 0; Solving Eqs. [1] and [2] yields : NB = 329.0 N Ans FB = 192.5 N Since FB = 192.5 N > NB = 0.6 (329.0) = 197.4 N, slipping between A and B does not occur. 8–90. 4–87. A cylinder having a mass of 250 kg is to be supported by the cord which wraps over the pipe. Determine the smallest vertical force F needed to support the load if the cord passes (a) once over the pipe, b = 180°, and (b) two times over the pipe, b = 540°. Take ms = 0.2. F Frictional Force on Flat Belt: Here, T1 = F and T2 = 250(9.81) = 2452.5 N Applying Eq. 8 – 6, we have a) If b = 180° = p rad T2 = T1emb 2452.5 = Fe0.2p Ans F = 1308.38 N = 1.31 kN b) If b = 540° = 3p rad T2 = T1emb 2452.5 = Fe0.2(3p) Ans F = 372.38 N = 372 kN 260 754 SM_CH04.indd 260 08b Ch08b 718-761.indd 754 4/8/11 11:48:53 6/12/09 10:34:39 AM AM ©©2011 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybe bereproduced, reproduced,ininany anyform formor orby byany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. *4–88. A cylinder having a mass of 250 kg is to be 8–91. supported by the cord which wraps over the pipe. Determine the largest vertical force F that can be applied to the cord without moving the cylinder. The cord passes (a) once over the pipe, b = 180°, and (b) two times over the pipe, b = 540°. Take ms = 0.2. F Frictional Force on Flat Belt: Here, T1 = 250(9.81) = 2452.5 N and T2 = F. Applying Eq. 8 – 6, we have a) If b = 180° = p rad T2 = T1emb F = 2452.5e0.2p Ans F = 4597.10 N = 4.60 kN b) If b = 540° = 3p rad T2 = T1emb F = 2452.5e0.2(3p) Ans F = 152.32 N = 16.2 kN 755 261 SM_CH04.indd 261 08b Ch08b 718-761.indd 755 4/8/11 11:48:54 6/12/09 10:34:41 AM AM 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. 4–89. The boat *8–92. boat has hasa aweight weightof of 500N lb 2500 ( and 250 is kg)held and in is position off the side of aside shipofby the spars atspars A andatB. man held in position off the a ship by the AA and B. having a weight of 130 oflb650 getsN in boat, wraps a boat, rope A man having a weight ( the 65 kg) gets in the around overhead at C, and ties itattoC,the end wraps aan rope aroundboom an overhead boom and tiesofitthe to boat as shown. If theasboat is disconnected from the spars, the end of the boat shown. If the boat is disconnected determine the minimum number of half turns the rope from the spars, determine the minimum number of must half make around the boom so thatthe theboom boatsocan turns the rope must make around thatbe thesafely boat lowered into the water at constant velocity. Also, what is the can be safely lowered into the water at constant velocity. normal forceisbetween the boat the man? coefficient Also, what the normal forceand between theThe boat and the of kinetic friction between thefriction rope and the boom is man? The coefficient of kinetic between the rope m . Hint:isThe requires that the requires normal force 0.15 and boom µs =problem 0.15. Hint: The problem that s =the between theforce man’s feet andthe theman’s boat be small possible. the normal between feetasand theasboat be as small as possible. C A B Frictional Force on Flat Belt : If the normal force between the man and the boat is equal to zero, then, T1 = 650 N and T2 = 2500 N. Applying Eq. 8–6, we have T 2 = T1 e 2500 = 650e0.15 = 8.980 rad 2500 N The least number of half turns of the rope required is d = 2.86 turns. Thus Use Ans n = 3 half turns 650 N Equations of Equilibrium : From FBD (a), +↑ΣFy = 0; T2 – Nm – 2500 = 0 T2 = Nm + 2500 T1 + Nm – 650 = 0 T1 = 650 – Nm From FBD (b), + → ΣFx = 0; Frictional Force on Flat Belt : Hence, = 3π rad. Applying Eq. 8–6, we have T 2 = T1 e Nm + 2500 = (650 – Nm)e0.15 Nm = 33.71 N Ans 756 262 SM_CH04.indd 262 08b Ch08b 718-761.indd 756 4/8/11 11:48:54 6/12/09 10:34:42 AM AM ©©2011 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybe bereproduced, reproduced,ininany anyform formor orby byany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. •8–93. The 50-kg 100-lb boy boy at at A A is is suspended from the cable •4–90. that passes over the quarter circular cliff rock. Determine if it is possible for for the the 92.5-kg 185-lb woman woman to to hoist hoist him him up; and if this is possible, what smallest force must she exert on the horizontal cable? The coefficient of static friction between the cable and the rock is ms = 0.2, and between the shoes of the woman and the ground msœ = 0.8. = π 2 92.5 (9.81) N T2 = T1 e = 50 (9.81) e0.2 +↑ΣFy = 0; π 2 = 671.55 N T2 = 671.55 N A N = 92.5 (9.81) = 0 N = 907.425 N + → ΣFx = 0; 671.55 – F = 0 F = 671.55 N Fmax = 0.8 (907.425) = 725.9 N > 671.55 N Yes, just barely. Ans 8–94. The 50-kg 100-lb boy boy at at A A is is suspended from the cable 4–91. that passes over the quarter circular cliff rock. What horizontal force must the woman at A exert on the cable in order to let the boy descend at constant velocity? The coefficients of static and kinetic friction between the cable and the rock are ms = 0.4 and mk = 0.35, respectively. = π 2 T2 = T1 e ; 50 (9.81) = T1 e0.35 T1 = 283.1 N A π 2 Ans 757 263 SM_CH04.indd 263 08b Ch08b 718-761.indd 757 4/8/11 11:48:55 6/12/09 10:34:44 AM AM 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. 8–95. *4–92. A 10-kg cylinder D, which is attached to a small pulley B, is placed on the cord as shown. Determine the smallest angle u so that the cord does not slip over the peg at C. The cylinder at E has a mass of 10 kg, and the coefficient of static friction between the cord and the peg is ms = 0.1. A u u C B E D Since pulley B is smooth, the tension in the cord between pegs A and C remains constant. Referring to the free-body diagram of the joint B shown in Fig. a, we have +c©Fy = 0; 2T sin u – 10(9.81) = 0 T= 49.05 sin u In the case where cylinder E is on the verge of ascending, T2 = T = 49.05 p and T1 = 10(9.81) N. Here, + u, Fig. b. Thus, sin u 2 T2 = T1emsb p 49.05 = 10(9.81)e0.1 2 sin u In + u p 0.5 = 0.1 + u sin u 2 Solving by trial and error, yields Ans u = 0.4221 rad = 24.2° 4–93. *8–96. A 10-kg cylinder , which is attached to a small pulley B, is placed on the cord as shown. Determine the largest angle u so that the cord does not slip over the peg at C. The cylinder at E has a mass of 10 kg, and the coefficient of static friction between the cord and the peg is ms = 0.1. A u u C B E D In the case where cylinder E is on the verge of descending, T2 = 10(9.81) N and T1 = 49.05 p . Here, + u. Thus, sin u 2 T2 = T1emsb 10(9.81) = 49.05 0.1 p e 2 sin u + u p In (2 sin u) = 0.1 + u 2 Solving by trial and error, yields u = 0.6764 rad = 38.8° Ans 758 264 SM_CH04.indd 264 08b Ch08b 718-761.indd 758 4/8/11 11:48:56 6/12/09 10:34:47 AM AM ©©2011 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybe bereproduced, reproduced,ininany anyform formororby byany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. •4–94. Determine the smallest lever force P needed to •8–97. prevent the wheel from rotating if it is subjected to a torque of M = 250 N # m. The coefficient of static friction between the belt and the wheel is ms = 0.3. The wheel is pin connected at its center, B. 400 mm B M A 200 mm 750 mm P +©MA = 0; –F(200) + P(950) = 0 F = 4.75 P T2 = T1emb 3p 2 F¿ = 4.75 Pe0.3 +©MB = 0; = 19.53 P –19.53 P (0.4) + 250 + 4.75 P (0.4) = 0 P = 42.3 N Ans 759 265 SM_CH04.indd 265 08b Ch08b 718-761.indd 759 4/8/11 11:48:56 6/12/09 10:34:49 AM AM 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. 8–98. •4–95. If a force of P = 200 N is applied to the handle of the bell crank, determine the maximum torque M that can be resisted so that the flywheel is not on the verge of rotating clockwise. The coefficient of static friction between the brake band and the rim of the wheel is ms = 0.3. P 900 mm 400 mm C A B 100 mm O M 300 mm Referring to the free-body diagram of the bell crank shown in Fig. a and the flywheel shown in Fig. b, +©MB = 0; TA (0.3) + TC (0.1) – 200(1) = 0 (1) +©MO = 0; TA (0.4) – TC (0.4) – M = 0 (2) By considering the friction between the brake band and the rim of wheel where b = 270° p = 1.5p rad and 180° TA > TC, we can write TA = TCemsb TA = TCe0.3(1.5p) TA = 4.1112 TC (3) Solving Eqs. (1), (2), and (3) yields M = 187 N · m TA = 616.67 N Ans TC = 150.00 N 760 266 SM_CH04.indd 266 08b Ch08b 718-761.indd 760 4/8/11 11:48:57 6/12/09 10:34:51 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. *4–96. The uniform *8–104. beam is supported by the 250-N uniform50-lb 25-kg beam is supported byrope the which is attached to thetoend wraps overover the rope which is attached the of endthe of beam, the beam, wraps rough peg,peg, andand is then connected totothe 500-N the rough is then connected the100-lb 50-kg block. block. If the coefficient of static friction between the beam and the block, and between the rope and the peg, is ms = 0.4, determine the maximum distance that the block can be placed from A and still remain in equilibrium. Assume the block will not tip. d 1 ftm 0.3 A 3m 10 ft 50 (9.81) N Block : +↑ΣFy = 0; N – 50 (9.81) = 0 N = 490.5 N + → ΣFx = 0; T1 – 0.4 (490.5) = 0 50 (9.81) N T1 = 196.2 N T2 = T1 e ; T2 = 196.2 e 196.2 N 0.4( π ) 2 367.8 N = 367.8 N 1.5 m System : + ΣMA = 0; 0.3 m 3 m 25 (9.81) N –490.5 (d) – 196.2 (0.3) – 25 (9.81) (1.5) + 367.8 (3) = 0 d = 1.380 m Ans 766 267 SM_CH04.indd 267 08c Ch08c 762-805.indd 766 4/8/11 11:48:57 AM AM 6/15/09 10:48:45 © 2011 Pearson Education, Inc., Upper Saddle River, rights reserved. This material is protected under copyright laws they currently © 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. AllAll rights reserved. This material is protected under allall copyright laws as as they currently exist. portion material may reproduced, form means, without permission writing from publisher. exist. NoNo portion of of thisthis material may bebe reproduced, in in anyany form or or byby anyany means, without permission in in writing from thethe publisher. •8–105. •4–97. The 80-kg man tries to lower the 150-kg crate using a rope that passes over the rough peg. Determine the least number of full turns in addition to the basic wrap (165°) around the peg to do the job. The coefficients of static friction between the rope and the peg and between the man’s shoes and the ground are ms = 0.1 and msœ = 0.4, respectively. 15 If the man is on the verge of slipping, F = ms¿ N = 0.4 N. Referring to the free-body diagram of the man shown in Fig. a, + → ©Fx = 0; 0.4N – T sin 15° = 0 +↑©Fy = 0; N + T cos 15° – 80(9.81) = 0 Solving, T = 486.55 N N = 314.82 N Using the result for T and considering the friction between the rope and the peg, where T2 = 150(9.81) N, T1 = T = 486.55 N � � 90° + 75° and b1 = n(2p) + p = (2n + 0.9167)p rad, Fig. b, 180° T2 = T1emsb 150(9.81) = 486.55e0.1(2n + 0.9167)p In 3.024 = 0.1(2n + 0.9167)p n = 1.303 Thus, the required number of full turns is Ans n=2 767 268 SM_CH04.indd 268 767 08c Ch08c 762-805.indd 4/8/11 11:48:58 6/15/09 10:48:47 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 8–106. 4–98. If the rope wraps three full turns plus the basic wrap (165°) around the peg, determine if the 80-kg man can keep the 300-kg crate from moving. The coefficients of static friction between the rope and the peg and between the man’s shoes and the ground are ms = 0.1 and msœ = 0.4, respectively. 15 If the man is on the verge of slipping, F = ms¿ N = 0.4 N. Referring to the free-body diagram of the man shown in Fig. a, + → ©Fx = 0; 0.4N – T sin 15° = 0 +↑©Fy = 0; N + T cos 15° – 80(9.81) = 0 Solving, T = 486.55 N N = 314.82 N Using the result for T and considering the friction between the rope and the peg, where T2 = 300(9.81) N, T1 = T = 486.55 N � � 90° + 75° and b = n(2p) + p = (2n + 0.9167)p rad, Fig. b, 180° T2 = T1emsb 300(9.81) = 486.55e0.1(2n + 0.9167)p In 6.049 = 0.1(2n + 0.9167)p n = 2.406 Ans Since n > 3, the man can hold the crate in equilibrium. 768 269 SM_CH04.indd 269 08c Ch08c 762-805.indd 768 4/8/11 11:48:58 AM AM 6/15/09 10:48:48 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form oror byby any means, without permission in in writing from thethe publisher. 8–107. The drive pulley B in a video tape recorder is on 4–99. the verge of slipping when it is subjected to a torque of 0.005 N m. If the coefficient of static friction between the tape and the drive wheel and between the tape and the fixed shafts A and C is 0.1, determine the tensions 1 and 2 developed in the tape for equilibrium. 10 mm T1 A M 0.005 N m 10 mm B 10 mm C Here T3 must overcome T4 and M, so T3 > T4. Also b = p rad. Thus, T2 T3 = T4 emsb T3 = T4 e0.1(p) T3 = 1.3691T4 (1) Referring to the free-body diagram of pulley B in Fig. a, +©M0 = 0; 0.005 + T4 (0.01) – T3 (0.01) = 0 (2) Solving Eqs. (1) and (2), yields T4 = 1.3546 N T3 = 1.8546 N Using the result of T4 and considering the friction on the fixed shaft A, where T1 > T4 and b = p rad, T1 = T4 e msb = 1.3546e0.1p Ans = 1.85N Using the result of T3 and considering the friction on the fixed shaft C, where T3 > T2 and b = P rad, 2 T3 = T2emsb 1.8546 = T2e0.1(p/2) Ans T2 = 1.59 N 769 270 270 769 08c SM_CH04.indd Ch08c 762-805.indd 4/8/11 11:48:59 6/15/09 10:48:50 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. *4–100. *8–108. Determine Determinethethe maximum number 25-kg *8–108. maximum number of 50-lbof packages packages that can be the belt without that can be placed on placed the belton without causing the causing belt to the belt to drive slip atwheel the drive wheel which with is rotating with slip at the A which is A rotating a constant a constant angular velocity. B is free to rotate. angular velocity. Wheel B isWheel free to rotate. Also, findAlso, the find the corresponding M that must be corresponding torsional torsional moment moment M that must be supplied supplied A. The conveyor belt is pre-tensioned to wheel to A. wheel The conveyor belt is pre-tensioned with the with the 1500-N horizontal The coefficient of friction kinetic 300-lb horizontal force. Theforce. coefficient of kinetic friction between theand beltplatform and platform between the belt P isP misk =k =0.2 , and 0.2, and the coefficient of static friction between the belt and the rim of each wheel is ms == 0.35. 0.35. 0.5 ftm 0.15 0.5m ft 0.15 B P A P = 1500 300 lb N M The maximum tension T2 of the conveyor belt can be obtained by considering the equilibrium of the free – body diagram of the top belt shown in Fig. a, +↑ΣFy = 0; n(25 (9.81)) – N = 0 N = 245.25 (1) + → ΣFx 750 + 0.2 (245.25n) – T2 = 0 T2 = 750 + 49.05n (2) = 0; By considering the case when the drive wheel A is on the verge of slipping, where and T1 = 750 N, = π rad, T2 = 750 + 49.05n T2 = T1 e 750 + 49.05n = 750 e0.35(π) n = 30.62 Thus, the maximum allowable number of boxes on the belt is Ans n = 30 Substituting n = 30 into Eq. (2) gives T2 = 2221.5 N. Referring to the free – body diagram of the wheel A shown in Fig. b, + ΣMO = 0; M + 750 (0.15) – 2221.5 (0.15) = 0 Ans M = 220.7 N · m n (245.25) N 0.15 m T2 = 2221.5 N 750 N T1 = 750 N 770 271 SM_CH04.indd 271 08c Ch08c 762-805.indd 770 4/8/11 11:48:59 AM AM 6/15/09 10:48:52 2011 Pearson Education, Inc., Upper Saddle River, NJ.All Allrights rightsreserved. reserved.This Thismaterial materialisisprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ. exist. No portion this material may reproduced, any form any means, without permission writing from the publisher. exist. No portion ofof this material may bebe reproduced, inin any form oror byby any means, without permission inin writing from the publisher. 8–151. A roofer, having a mass of 70 kg, walks slowly in an 4–101. upright position down along the surface of a dome that has a radius of curvature of r = 20 m. If the coefficient of static friction between his shoes and the dome is ms = 0.7, determine the angle u at which he first begins to slip. 20 m u 60 +QΣFy' = 0; N – 70(9.81) cos u = 0 (1) R+ΣFx' = 0; 70(9.81) sin u – 0.7 N = 0 (2) Solving Eqs. (1) and (2) yields: Ans u = 35.0° N = 562.6 N *8–152. Column D is subjected to a vertical load of 40 kN. 8000 lb.ItItisissupported supportedon ontwo two identical identical wedges wedges A A and B for which the coefficient of static friction at the contacting surfaces between A and B and between B and C is s = 0.4. Determine the force P needed to raise the column and the equilibrium force P¿ needed to hold wedge A stationary. The contacting surface between A and D is smooth. 40 kN 8000 lb Wedge A : +↑ΣFy = 0; D N cos 10° – 0.4N sin 10° – 40 = 0 P N = 43.70 kN + → ΣFx = 0; 10 10 A P¿ C 40 kN 0.4 (43.70) cos 10° + 43.70 sin 10° – Pʹ = 0 Pʹ = 24.80 kN B Ans Wedge B : +↑ΣFy = 0; NC + 0.4 (43.70) sin 10° – 43.70 cos 10° = 0 N = 43.70 kN 0.4 (43.70) kN NC = 40 kN + → ΣFx = 0; P – 0.4 (40) – 43.70 sin 10° – 0.4 (43.70) cos 10° = 0 P = 40.80 kN Ans 804 272 SM_CH04.indd 272 08c Ch08c 762-805.indd 804 4/8/11 11:49:00 6/15/09 10:49:51 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. •5–89. Determine the horizontal and vertical components •4–102. of reaction at the pin A and the reaction at the roller B required to support the truss. Set F = 600 N. A 2m 45 2m F 2m F B 2m F Equations of Equilibrium : The normal reaction NB can be obtained directly by summing moments about point A. + ΣMA = 0; 600(6) + 600(4) + 600(2) – NB cos 45° (2) = 0 Ans NB = 5091.17 N = 5.09 kN + S ΣFx = 0; Ax – 5091.17 cos 45° = 0 Ans Ax = 3600 N = 3.60 kN + ↑ΣFy = 0; 5091.17 sin 45° – 3(600) – Ay = 0 Ans Ay = 1800 N = 1.80 kN 5–90. If the roller at B can sustain a maximum load of 4–103. 3 kN, determine the largest magnitude of each of the three forces F that can be supported by the truss. A 2m 45 2m F 2m F B 2m F Equations of Equilibrium : The unknowns Az and Ay can be eliminated by summing moments about point A. + ΣMA = 0; F(6) + F(4) + F(2) – 3 cos 45° (2) = 0 F = 0.3536 kN = 354 N Ans 394 273 SM_CH04.indd 273 05b Ch05b 361-400.indd 394 4/8/11 11:49:01 AM AM 6/12/09 8:46:46 © 2011 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. 4–105. 274 SM_CH04.indd 274 4/8/11 11:49:01 AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. 4–106. *5–92. The shaft assembly is supported by two smooth journal bearings A and B and a short link DC. If a couple moment is applied to the shaft as shown, determine the components of force reaction at the journal bearings and the force in the link. The link lies in a plane parallel to the y–z plane and the bearings are properly aligned on the shaft. z D 30 20 120 mm C 250 mm A x ©Mx = 0; Ans Ans –Ay(0.7) – 1015.43 cos 20° (0.42) = 0 Ay = 572.51 = 573 N ©Fy = 0; Ans –208.38 + 1015.43 sin 20° + Bz = 0 Bc = –139 N ©(MB)z = 0; y –Az(0.7) – 1015.43 sin 20° (0.42) = 0 Az = –208.38 = –208 N ©Fz = 0; 300 mm – 250 + FCD cos 20° (0.25 cos 30°) + FCD sin 20° (0.25 sin 30°) = 0 FCD = 1015.43 N = 1.02 kN ©(MB)y = 0; 250 N m 400 mm B Ans 572.51 – 1015.43 cos 20° + By = 0 By = 382 N Ans 396 275 SM_CH04.indd 275 05b Ch05b 361-400.indd 396 4/8/11 11:49:01 AM AM 6/12/09 8:46:50 2011 Pearson Education, Inc., Upper Saddle River, Allrights rightsreserved. reserved. Thismaterial materialis isprotected protectedunder underallallcopyright copyrightlaws lawsasasthey theycurrently currently ©© 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All This exist. portion this material may reproduced, any form any means, without permission writing from the publisher. exist. NoNo portion ofof this material may bebe reproduced, in in any form oror byby any means, without permission in in writing from the publisher. •4–107. •5–93. Determine the reactions at the supports A and B of the frame. 50 10 kN kip 35 kN 7 kip 25 kN 5 kip 2.4 8 ftm 1.8 6 ftm 1.8 m 6 ft 10 kN 2 kip A 8 ft 2.4 m 0.5 2.5 kip kN 6 ft 1.8 m B 25 kN 35 kN 2.4 m 50 kN 1.8 m 10 kN 1.8 m 2.4 m 2.5 kN + ΣMB = 0; + → ΣFx = 0; +↑ΣFy = 0; 25 (4.2) + 35 (1.8) + 2.5 (1.8) – 10 (1.8) – Ay (4.2) = 0 Ay = 36.786 kN = 36.8 kN Ans Bx – 2.5 = 0 Ans Bx = 2.5 kN 1.8 m By + 36.786 – 25 – 35 – 50 – 10 = 0 By = 83.2 kN Ans 397 276 276 397 05b SM_CH04.indd Ch05b 361-400.indd 4/8/118:46:51 11:49:02 6/12/09 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. *4–108.A A skeletal diagramofofthe thelower lowerleg legisisshown shown in the 5–94. skeletal diagram lower figure. Here it can be noted that this portion of the leg is lifted by the quadriceps muscle attached to the hip at A and to the patella bone at B. This bone slides freely over cartilage at the knee joint. The quadriceps is further extended and attached to the tibia at C. Using the mechanical system shown in the upper figure to model the lower leg, determine the tension in the quadriceps at C and the magnitude of the resultant force at the femur (pin), D, in order to hold the lower leg in the position shown. The lower leg has a mass of 3.2 kg and a mass center at G1; the foot has a mass of 1.6 kg and a mass center at G2. 75 mm 25 mm A 350 mm B C 75 A 300 mm D B G1 G2 C D + ΣMD = 0; T sin 18.43° (75) – 3.2(9.81)(425 sin 75°) –1.6(9.81)(725 sin 75°) = 0 T = 1006.82 N = 1.01 kN + ↑ΣFy = 0; Ans Dy + 1006.82 sin 33.43° – 3.2(9.81) – 1.6(9.81) = 0 Dy = –507.66 N + S ΣFx = 0; Dx – 1006.82 cos 33.43° = 0 Dz = 840.20 N FD = D2x + D2y = (–507.66)2 + 840.202 = 982 N Ans 398 277 SM_CH04.indd 277 05b Ch05b 361-400.indd 398 4/8/11 11:49:02 AM AM 6/12/09 8:46:53 © 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently © 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form or or byby any means, without permission in in writing from thethe publisher. N acts acts on on the the crankshaft. 4–109. A vertical force 5–95. force of of 400 80 lb Determine the horizontal equilibrium force P that must be applied to the handle and the x, y, z components of force at the smooth journal bearing A and the thrust bearing B. The bearings are properly aligned and exert only force reactions on the shaft. z 80 lb 400 N B 250in.mm 10 y 350 14 mm in. A 350 14 mm in. 6150 in.mm x 8 in. 200 mm 4100 in. mm P 400 N ΣMy = 0; P(200) – 400 (250) = 0 P = 500 N Ans ΣMz = 0; Bz(700) – 400 (350) = 0 Bz = 200 N Ans ΣMz = 0; –Bx(700) – 500 (250) = 0 Bx = –178.6 N Ans ΣFz = 0; As + (–178.6) – 500 = 0 As = 678.6 N Ans ΣFy = 0; By = 0 ΣFx = 0; As + 200 – 400 = 0 250 350 mm 350 mm 150 mm Ans As = 200 N mm 200 mm 100 mm Ans Negative sign indicates that Bx acts in the opposite sense to that shown on the FBD. 399 278 278 399 05b SM_CH04.indd Ch05b 361-400.indd 4/8/11 11:49:03 6/12/09 8:46:57 AM AM ©©2011 rights reserved. This material is is protected under allall copyright laws asas they currently 2010Pearson PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All All rights reserved. This material protected under copyright laws they currently exist. exist.No Noportion portionofofthis thismaterial materialmay maybebereproduced, reproduced,ininany anyform formororbybyany anymeans, means,without withoutpermission permissionininwriting writingfrom fromthe thepublisher. publisher. •4–110. The truck has a mass of 1.25 Mg and a center of •8–145. mass at G. Determine the greatest load it can pull if (a) the truck has rear-wheel drive while the front wheels are free to roll, and (b) the truck has four-wheel drive. The coefficient of static friction friction between betweenthe thewheels wheelsand andthe theground groundisisms s== 0.5 0.5,, œ and between the the crate crate and andthe theground, ground itisis m ′s = 0.4. 0.4. s = 800 mm G 600 mm A B 1.5 m 1m a) The truck with rear wheel drive. Equations of Equilibrium and Friction: It is required that rear wheels of the truck slip. Hence FA = ms NA = 0.5 NA. From FBD (a) +©MB = 0; 1.25(103)(9.81)(1) + T (0.6) – NA (2.5) = 0 [1] + S ©Fx = 0; 0.5 NA – T = 0 [2] Solving Eqs. [1] and [2] yields NA = 5573.86 N T = 2786.93 N Since the crate moves, Fc = ms¿Nc = 0.4 Nc. From FBD (c), +c©Fy = 0; Nc – W = 0 + 2786.93 – 0.4 W = 0 S ©Fx = 0; Nc = W W = 6967.33 N = 6.97 kN Ans b) The truck with four wheel drive. Equations of Equilibrium and Friction: It is required that rear wheels and front wheels of the truck slip. Hence FA = msNA = 0.5 NA and FB = msNB = 0.5 NB. From FBD (b), +©MB = 0; 1.25(103)(9.81)(1) + T (0.6) – NA (2.5) = 0 [3] +©MA = 0; NB (2.5) + T (0.6) – 1.25(103)(9.81)(1.5)= 0 [4] + S ©Fx = 0; 0.5 NA + 0.5 NB – T = 0 [5] Solving Eqs. [3], [4] and [5] yields NA = 6376.5 N NB = 5886.0 N T = 6131.25 N Since the crate moves, Fc = ms¿Nc = 0.4 Nc. From FBD (c), +c©Fy = 0; Nc – W = 0 + 6131.25 – 0.4 W = 0 S ©Fx = 0; Nc = W W = 15328.125 N = 15.3 kN Ans 798 279 SM_CH04.indd 279 08c Ch08c 762-805.indd 798 4/8/11 11:49:04 AM AM 6/15/09 10:49:43 © 2011 Pearson Education, Inc., Upper Saddle River, All rights reserved. This material protected under copyright laws they currently © 2010 Pearson Education, Inc., Upper Saddle River, NJ.NJ. All rights reserved. This material is is protected under allall copyright laws asas they currently exist. portion this material may reproduced, any form any means, without permission writing from publisher. exist. NoNo portion of of this material may bebe reproduced, in in any form oror byby any means, without permission in in writing from thethe publisher. 4–111. Solve Prob. 8–145 if the truck and crate are 8–146. traveling up a 10° incline. 800 mm G 600 mm A B 1.5 m 1m a) The truck with rear wheel drive. Equations of Equilibrium and Friction: It is required that the rear wheel of the truck slip hence FA = msNA = 0.5 NA. From FBD (a). +©MB = 0; +Q©Fx¿ = 0; 1.25(103)(9.81) cos 10°(1) + 1.25(103)(9.81) sin 10°(0.8) + T (0.6) – NA (2.5) = 0 [1] 0.5 NA – 1.25(103)(9.81) sin 10° – T = 0 [2] Solving Eqs. [1] and [2] yields NA = 5682.76 N T = 712.02 N Since the crate moves, FC = ms¿NC = 0.4NC. From FBD (c), a+©Fy¿ = 0; NC – W cos 10° = 0 +Q©Fx = 0; 712.02 – W sin 10° – 0.4(0.9848 W) = 0 NC = 0.9848 W W = 1254.50 N = 1.25 kN Ans b) The truck with four wheel drive. Equations of Equilibrium and Friction: It is required that rear wheels of the truck slip hence FA = msNA = 0.5 NA. From FBD (b). +©MB = 0; +©MA = 0; +Q©Fx = 0; 1.25(103)(9.81) cos 10°(1) + 1.25(103)(9.81) sin 10°(0.8) + T (0.6) – NA (2.5) = 0 [3] –1.25(103)(9.81) cos 10°(1.5) + 1.25(103)(9.81) sin 10°(0.8) + T (0.6) + NB (2.5) = 0 [4] 0.5 NA + 0.5 NB – 1.25(103)(9.81) sin 10° – T = 0 [5] Solving Eqs. [3], [4] and [5] yields NA = 6449.98 N NB = 5626.23 N T = 3908.74 N Since the crate moves, FC = ms¿NC = 0.4 NC . From FBD (c), a+©Fy¿ = 0; NC – W cos 10° = 0 NC = 0.9848 W + 3908.74 – W sin 10° – 0.4(0.9848 W) = 0 S ©Fx = 0; Ans W = 6886.79 N = 6.89 kN 799 280 280 799 08c SM_CH04.indd Ch08c 762-805.indd 4/8/11 11:49:04 6/15/09 10:49:45 AM AM