Engineering Mechanics - Statics Chapter 4 b 4m= c 1.5 m= d 10.5 m= Solution: r AB b a− cd− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r OA 0 0 d ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r OB b a− c ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F v F r AB r AB = M O1 r OA F v ×= M O1 61.2 81.6 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Nm⋅= M O2 r OB F v ×= M O2 61.2 81.6 0− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Nm⋅= Problem 4-43 Determine the smallest force F that must be applied along the rope in order to cause the curved rod, which has radius r, to fail at the support C. This requires a moment to be developed at C of magnitude M. 241 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 4 Given: r 5ft= M 80 lb ft⋅= θ 60 deg= a 7ft= b 6ft= Solution: r AB b arsin θ () − r− cos θ () ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = u AB r AB r AB = r CB b a r− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = Guess F 1lb= Given r CB F u AB () × M= F Find F()= F 18.6 lb= Problem 4-44 The pipe assembly is subjected to the force F . Determine the moment of this force about point A. 242 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 4 Given: F 80 N= a 400 mm= b 300 mm= c 200 mm= d 250 mm= θ 40 deg= φ 30 deg= Solution: r AC bd+ a c− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F v F cos φ () sin θ () cos φ () cos θ () sin φ () − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = M A r AC F v ×= M A 5.385− 13.093 11.377 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Nm⋅= Problem 4-45 The pipe assembly is subjected to the force F . Determine the moment of this force about point B. 243 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 4 Given: F 80 N= a 400 mm= b 300 mm= c 200 mm= d 250 mm= θ 40 deg= φ 30 deg= Solution: r BC bd+ 0 c− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r BC 550 0 200− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ mm= F v F cos φ () sin θ () cos φ () cos θ () sin φ () − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = F v 44.534 53.073 40− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= M B r BC F v ×= M B 10.615 13.093 29.19 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Nm⋅= Problem 4-46 The x-ray machine is used for medical diagnosis. If the camera and housing at C have mass M and a mass center at G, determine the moment of its weight about point O when it is in the position shown. Units Used: kN 10 3 N= 244 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 4 Given: M 150 kg= a 1.2 m= b 1.5 m= θ 60 deg= g 9.81 m s 2 = Solution: M O b()− cos θ () a b( )sin θ () ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ 0 0 M− g ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ×= M O 1.77− 1.1− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN m⋅= Problem 4-47 Using Cartesian vector analysis, determine the resultant moment of the three forces about the base of the column at A. Units Used : kN 10 3 N= Given: F 1 400 300 120 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= F 2 100 100− 60− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= F 3 0 0 500− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= a 4m= b 8m= 245 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 4 c 1m= Solution: r AB 0 0 ab+ ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r A3 0 c− b ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = The individual moments M A1 r AB F 1 ×= M A2 r AB F 2 ×= M A3 r A3 F 3 ×= M A1 3.6− 4.8 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN m⋅= M A2 1.2 1.2 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN m⋅= M A3 0.5 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN m⋅= The total moment M A M A1 M A2 + M A3 += M A 1.9− 6 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN m⋅= Problem 4-48 A force F produces a moment M O about the origin of coordinates, point O. If the force acts at a point having the given x coordinate, determine the y and z coordinates. Units Used : kN 10 3 N= Given : F 6 2− 1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= M O 4 5 14− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN m⋅= x 1m= Solution: The initial guesses: y 1m= z 1m= 246 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 4 Given x y z ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F× M O = y z ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find yz,()= y z ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m= Problem 4-49 The force F creates a moment about point O of M O . If the force passes through a point having the given x coordinate, determine the y and z coordinates of the point. Also, realizing that M O = Fd, determine the perpendicular distance d from point O to the line of action of F . Given: F 6 8 10 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= M O 14− 8 2 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Nm⋅= x 1m= Solution: The initial guesses: y 1m= z 1m= Given x y z ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F× M O = y z ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find yz,()= y z ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m= d M O F = d 1.149 m= Problem 4-50 The force F produces a moment M O about the origin of coordinates, point O. If the force acts at a point having the given x-coordinate, determine the y and z coordinates. 247 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 4 Units Used: kN 10 3 N= Given: x 1m= F 6 2− 1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= M O 4 5 14− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN m⋅= Solution: Initial Guesses: y 1m= z 1m= Given M O x y z ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F×= y z ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find yz,()= y z ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m= Problem 4-51 Determine the moment of the force F about the Oa axis. Express the result as a Cartesian vector. Given: F 50 20− 20 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= a 6m= b 2m= c 1m= d 3m= e 4m= 248 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 4 r OF c b− a ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r Oa 0 e d ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = u Oa r Oa r Oa = M Oa r OF F× () u Oa ⋅ ⎡ ⎣ ⎤ ⎦ u Oa = M Oa 0 217.6 163.2 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Nm⋅= Problem 4-52 Determine the moment of the force F about the aa axis. Express the result as a Cartesian vector. Given: F 600 lb= a 6ft= b 3ft= c 2ft= d 4ft= e 4ft= f 2ft= Solution: F v F c 2 d 2 + e 2 + d− e c ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r d 0 c− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = u aa 1 a 2 b 2 + f 2 + b− f− a ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = M aa rF v × () u aa ⋅ ⎡ ⎣ ⎤ ⎦ u aa = M aa 441− 294− 882 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb ft⋅= 249 Solution: © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 4 Problem 4-53 Determine the resultant moment of the two forces about the Oa axis. Express the result as a Cartesian vector. Given: F 1 80 lb= F 2 50 lb= α 120 deg= β 60 deg= γ 45 deg= a 5ft= b 4ft= c 6ft= θ 30 deg= φ 30 deg= Solution: F 1v F 1 cos α () cos β () cos γ () ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = F 2v 0 0 F 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = r 1 b( )sin θ () b( )cos θ () c ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ = r 2 0 a()− sin φ () 0 ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ = u aa cos φ () sin φ () − 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = M aa r 1 F 1v × r 2 F 2v ×+ () u aa ⎡ ⎣ ⎤ ⎦ u aa = M aa 26.132 15.087− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb ft⋅= 250 © 2007 R. C. Hibbeler. Published by 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. [...]... deg F 2 = 8 kN θ 2 = 45 deg a = 0.3 m b = 1. 5 m c = 1. 8 m Solution: MR = ΣMO; ( a) MRa = ( F1 cos ( θ 1 ) + F 2 cos ( θ 2 ) ) c + ( F2 cos ( θ 2 ) − F 1 sin ( θ 1 ) ) a + −( F2 cos ( θ 2 ) + F 1 cos ( θ 1 ) ) ( b + c) MRa = −9.69 kN⋅ m MR = ΣMA; ( b) MRb = ( F2 sin ( θ 2 ) − F1 sin ( θ 1 ) ) a − ( F 2 cos ( θ 2 ) + F1 cos ( θ 1 ) ) b MRb = −9.69 kN⋅ m Problem 4-8 2 Two couples act on the beam as shown... publisher Engineering Mechanics - Statics Chapter 4 Given: F 1 = 15 0 N a = 300 mm b = 400 mm d = 600 mm Solution: Initial guesses: Given F = 1N P = 1N ⎞ ⎛ d ⎞ ⎛ 0 ⎞ ⎛ d ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛ −F 1 ⎟ ⎛ 0 ⎞ ⎛ F 1 ⎞ ⎜ 0 ⎟ × ⎜ 0 ⎟ + ⎜ b ⎟ × ⎜ −P ⎟ + ⎜ b ⎟ × ⎜ ⎜ ⎟ ⎜ ⎟ + 0 ×⎜ 0 ⎟ =0 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎝a⎠ ⎝F⎠ ⎝0⎠ ⎝ 0 ⎠ ⎝0⎠ ⎜ 0 ⎟ ⎝a⎠ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛F⎞ ⎜ ⎟ = Find ( F , P) ⎝P⎠ ⎛ F ⎞ ⎛ 75 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ P ⎠ ⎝ 10 0... Mechanics - Statics Chapter 4 Problem 4 -7 7 The ends of the triangular plate are subjected to three couples Determine the magnitude of the force F so that the resultant couple moment is M clockwise Given: F 1 = 600 N F 2 = 250 N a = 1m θ = 40 deg M = 400 N⋅ m Solution: Initial Guess Given F1 F = 1N ⎛ a ⎞ − F a − F ⎛ a ⎞ = −M ⎜ ⎟ 2 ⎜ ⎟ ⎝ 2 cos ( θ ) ⎠ ⎝ 2 cos ( θ ) ⎠ F = Find ( F) F = 830 N Problem 4 -7 8 Two... −cos ( θ ) ⎠ ⎛ cos ( θ ) ⎞ ⎛ sin ( θ ) ⎞ ⎜ 0 ⎟ + c⎜ 0 ⎟ r2 = b ⎜ ⎟ ⎜ ⎟ ⎝ sin ( θ ) ⎠ ⎝ −cos ( θ ) ⎠ M1z = ( r1 × F 1 ) ⋅ k M1z = 38.2 lb⋅ in M2z = ( r2 × F 2 ) ⋅ k ⎛0⎞ ⎜ ⎟ k = 0 ⎜ ⎟ 1 M2z = 14 .1 lb⋅ in Problem 4-6 9 Determine the magnitude and sense of the couple moment Units Used: 3 kN = 10 N 2 61 © 20 07 R C Hibbeler Published by Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This... writing from the publisher Engineering Mechanics - Statics Chapter 4 Given: ⎛8⎞ F = ⎜ −4 ⎟ N ⎜ ⎟ ⎝ 10 ⎠ a = 5m b = 3m c = 4m d = 2m e = 3m Solution: ⎛ −b − e ⎞ r = ⎜ c+d ⎟ ⎜ ⎟ ⎝ −a ⎠ M = r× F ⎛ 40 ⎞ M = ⎜ 20 ⎟ N⋅ m ⎜ ⎟ ⎝ −24 ⎠ Problem 4-8 7 Determine the couple moment Express the result as a Cartesian vector Given: F = 80 N a = 6m b = 10 m c = 10 m d = 5m e = 4m f = 4m 274 © 20 07 R C Hibbeler Published... Used: 3 kip = 10 lb Given: F = 15 0 lb a = 8 ft b = 6 ft c = 8 ft d = 6 ft e = 6 ft f = 8 ft Solution: MC = ΣMA; MC = F d 2 d + f MC = 312 0 lb⋅ ft f ( a + f) + F 2 2 d + f ( c + d) 2 MC = 3 .12 0 kip⋅ ft Problem 4 -7 2 If the couple moment has magnitude M, determine the magnitude F of the couple forces Given: M = 300 lb⋅ ft a = 6 ft b = 12 ft c = 1 ft d = 2 ft e = 12 ft f = 7 ft 263 © 20 07 R C Hibbeler... reproduced, in any form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 4 Solution: ⎛b⎞ ⎜ ⎟ rA = 0 ⎜ ⎟ ⎝0⎠ ⎛0⎞ ⎜ ⎟ j = 1 ⎜ ⎟ ⎝0⎠ rAB ⎛ −b + c ⎞ ⎜ a ⎟ = ⎜ ⎟ ⎝ d ⎠ Fv = F My = ( rA × F v ) ⋅ j rAB rAB ⎛ −9 .79 8 ⎞ ⎜ ⎟ F v = 9 .79 8 lb ⎜ ⎟ ⎝ 19 .596 ⎠ My = 78 .384 lb⋅ ft Problem 4-5 9 The lug nut on the wheel of the automobile is to be removed using the wrench... publisher Engineering Mechanics - Statics Chapter 4 Solution: u = ⎛ b ⎞ ⎜c + d⎟ ⎟ 2 2 2⎜ a + b + ( c + d) ⎝ − a ⎠ 1 F v = Fu ⎛−f − b⎞ r = ⎜ −d − c ⎟ ⎜ ⎟ ⎝ e+a ⎠ ⎛ −252.6 ⎞ M = ⎜ 67. 4 ⎟ N⋅ m ⎜ ⎟ ⎝ −252.6 ⎠ M = r × Fv Problem 4-8 8 If the resultant couple of the two couples acting on the fire hydrant is MR = { 15 i + 30j} N ⋅ m, determine the force magnitude P Given: a = 0.2 m b = 0 .15 0 m ⎛ 15 ⎞ M = ⎜... from the publisher Engineering Mechanics - Statics Chapter 4 Problem 4-6 5 If a torque or moment M is required to loosen the bolt at A, determine the force P that must be applied perpendicular to the handle of the flex-headed ratchet wrench Given: M = 80 lb⋅ in θ = 60 deg a = 10 in b = 0 .75 in Solution: M = P ⎡b + ( a)sin ( θ )⎤ ⎣ ⎦ P = M b + ( a)sin ( θ ) P = 8.50 lb Problem 4-6 6 The A-frame is being... Engineering Mechanics - Statics Chapter 4 Problem 4-8 0 If the couple moment acting on the pipe has magnitude M, determine the magnitude F of the vertical force applied to each wrench Given: M = 400 N⋅ m a = 300 mm b = 15 0 mm c = 400 mm d = 200 mm e = 200 mm Solution: ⎛0⎞ k = ⎜0⎟ ⎜ ⎟ 1 rAB ⎛ −e − b ⎞ = ⎜ −c + d ⎟ ⎜ ⎟ ⎝ 0 ⎠ Guesss F = 1N rAB × ( Fk) = M Given F = Find ( F) F = 992. 278 N Problem 4-8 1 Determine . publisher. Engineering Mechanics - Statics Chapter 4 Units Used: kN 10 3 N= Given: x 1m= F 6 2− 1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= M O 4 5 14 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN m⋅= Solution: Initial Guesses: y 1m= z 1m= Given. M O 1. 77 − 1. 1− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN m⋅= Problem 4-4 7 Using Cartesian vector analysis, determine the resultant moment of the three forces about the base of the column at A. Units Used : kN 10 3 N= Given: F 1 400 300 12 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= F 2 10 0 10 0− 60− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= F 3 0 0 500− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= a. without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 4 Given: M 15 0 kg= a 1. 2 m= b 1. 5 m= θ 60 deg= g 9. 81 m s 2 = Solution: M O b()− cos θ () a b( )sin θ () ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ 0 0 M−