Engineering Mechanics - Statics Chapter 5 Given: a 3ft= b 2ft= c 1ft= d 3ft= e 9ft= T 300 lb= Solution: Initial guesses: F 10 lb= M Ax 10 lbft= M Az 10 lbft= A x 10 lb= A y 10 lb= A z 10 lb= Given A x A y A z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 0 TF− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ + 0= M Ax 0 M Az ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ a c− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 0 F− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ×+ e b− c− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 0 T ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ×+ 0= F A x A y A z M Ax M Az ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find FA x , A y , A z , M Ax , M Az , () = M Ax M Az ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0 0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lbft= A x A y A z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 0 600 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= F 900lb= Problem 5-70 The boom AB is held in equilibrium by a ball-and-socket joint A and a pulley and cord system as shown. Determine the x, y, z components of reaction at A and the tension in cable DEC. 401 © 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 5 Given: F 0 0 1500− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= a 5ft= b 4ft= cb= d 5ft= e 5ft= f 2ft= Solution: α atan a de+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = La 2 de+() 2 += β atan b L fL de+ − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = Guesses T BE 1lb= T DEC 1lb= A x 1lb= A y 1lb= A z 1lb= Given 2 T DEC cos β () T BE = 0 d 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F× 0 de+ 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 T BE − cos α () T BE sin α () ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ 0= A x A y A z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ F+ T BE 0 cos α () − sin α () ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ + 0= 402 © 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 5 A x A y A z T BE T DEC ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A x A y , A z , T BE , T DEC , () = T DEC 919lb= A x A y A z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 1.5 10 3 × 750 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ lb= Problem 5-71 The cable CED can sustain a maximum tension T max before it fails. Determine the greatest vertical force F that can be applied to the boom. Also, what are the x, y, z components of reaction at the ball-and-socket joint A? Given: T max 800 lb= a 5ft= b 4ft= cb= d 5ft= e 5ft= f 2ft= Solution: α atan a de+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = La 2 de+() 2 += β atan b L fL de+ − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = T DEC T max = Guesses T BE 1lb= F 1lb= A x 1lb= A y 1lb= A z 1lb= Given 2 T DEC cos β () T BE = 403 © 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 5 0 d 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 0 F− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ × 0 de+ 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 T BE − cos α () T BE sin α () ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ 0= A x A y A z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 0 F− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ + T BE 0 cos α () − sin α () ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ + 0= A x A y A z T BE F ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A x A y , A z , T BE , F, () = F 1306lb= A x A y A z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 1.306 10 3 × 653.197 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ lb= Problem 5-72 The uniform table has a weight W and is supported by the framework shown. Determine the smallest vertical force P that can be applied to its surface that will cause it to tip over. Where should this force be applied? Given: W 20 lb= a 3.5 ft= b 2.5 ft= c 3ft= e 1.5 ft= f 1ft= Solution: θ atan f e ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 33.69 deg= desin θ () = d 0.832ft= φ atan a 2 e− b 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = φ 11.31 deg= 404 © 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 5 d' a 2 e− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 += d' 1.275 ft= Tipping will occur about the g - g axis. Require P to be applied at the corner of the table for P min . Wd Pd'sin 90 deg φ − θ + () = PW d d' sin 90 deg φ − θ + () = P 14.1 lb= Problem 5-73 The windlass is subjected to load W. Determine the horizontal force P needed to hold the handle in the position shown, and the components of reaction at the ball-and-socket joint A and the smooth journal bearing B. The bearing at B is in proper alignment and exerts only force reactions perpendicular to the shaft on the windlass. Given: W 150 lb= a 2ft= b 2ft= c 1ft= d 1ft= e 1ft= f 0.5 ft= Solution: Σ M y = 0; Wf Pd− 0= P Wf d = P 75lb= Σ F y = 0; A y 0lb= A y 0= Σ M x = 0; W− aB z ab+()+ 0= 405 © 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 5 B z Wa ab+ = B z 75lb= Σ F z = 0; A z B z + W− 0= A z WB z −= A z 75lb= Σ M z = 0; B x ab+()ab+ c+ e+()P− 0= B x Pa b+ c+ e+() ab+ = B x 112lb= Σ F x = 0; A x B x − P+ 0= A x B x P−= A x 37.5 lb= Problem 5-74 A ball of mass M rests between the grooves A and B of the incline and against a vertical wall at C. If all three surfaces of contact are smooth, determine the reactions of the surfaces on the ball. Hint: Use the x, y, z axes, with origin at the center of the ball, and the z axis inclined as shown. Given: M 2kg= θ 1 10 deg= θ 2 45 deg= Solution: Σ F x = 0; F c cos θ 1 () Mgsin θ 1 () − 0= F c Mgtan θ 1 () ⋅= F c 0.32 kg m⋅= 406 © 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 5 Σ F y = 0; N A cos θ 2 () N B cos θ 2 () − 0= N A N B = Σ F z = 0; 2 N A sin θ 2 () Mgcos θ 1 () − F c sin θ 1 () − 0= N A 1 2 Mgcos θ 1 () ⋅ F c sin θ 1 () ⋅+ sin θ 2 () ⋅= N A 1.3 kg m⋅= NN A = N B = Problem 5-75 Member AB is supported by cable BC and at A by a square rod which fits loosely through the square hole at the end joint of the member as shown. Determine the components of reaction at A and the tension in the cable needed to hold the cylinder of weight W in equilibrium. Units Used: kip 10 3 lb= Given: W 800 lb= a 2ft= b 6ft= c 3ft= Solution: Σ F x = 0 F BC c c 2 b 2 + a 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0= F BC 0lb= Σ F y = 0 A y 0= A y 0lb= A y 0lb= 407 © 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 5 Σ F z = 0 A z W− 0= A z W= A z 800lb= Σ M x = 0 M Ax Wb− 0= M Ax Wb= M Ax 4.80 kip ft⋅= Σ M y = 0 M Ay 0lbft= M Ay 0lb ft⋅= Σ M z = 0 M Az 0lbft= M Az 0lb ft⋅= Problem 5-76 The pipe assembly supports the vertical loads shown. Determine the components of reaction at the ball-and-socket joint A and the tension in the supporting cables BC and BD. Units Used: kN 10 3 N= Given: F 1 3kN= d 2m= F 2 4kN= e 1.5 m= a 1m= g 1m= b 1.5 m= h 3m= c 3m= i 2m= fce−= j 2m= Solution: The initial guesses are: T BD 1kN= T BC 1kN= A x 1kN= A y 1kN= A z 1kN= The vectors 408 © 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 5 r 1 0 ab+ f+ d ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r 2 0 ab+ c+ d ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r BC i a− hg− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r BD j− a− hg− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r AB 0 a g ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = u BC r BC r BC = u BD r BD r BD = i 1 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = j 0 1 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = k 0 0 1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = Given A x iA y j+ A z k+ F 1 k− F 2 k− T BD u BD + T BC u BC + 0= r AB T BD u BD T BC u BC + () × r 1 F 1 − k () ×+ r 2 F 2 − k () ×+ 0= T BD T BC A x A y A z ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find T BD T BC , A x , A y , A z , () = T BD T BC ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 17 17 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kN= A x A y A z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 11.333 15.667− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= Problem 5-77 The hatch door has a weight W and center of gravity at G. If the force F applied to the handle at C has coordinate direction angles of α , β and γ , determine the magnitude of F needed to hold the door slightly open as shown. The hinges are in proper alignment and exert only force reactions on the door. Determine the components of these reactions if A exerts only x and z components of force and B exerts x, y, z force components. Given: W 80 lb= α 60 deg= 409 © 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 5 β 45 deg= γ 60 deg= a 3ft= b 2ft= c 4ft= d 3ft= Solution: Initial Guesses: A x 1lb= A z 1lb= F 1lb= B x 1lb= B y 1lb= B z 1lb= Given A x 0 A z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ B x B y B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + F cos α () cos β () cos γ () ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0 0 W− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ + 0= ab+ 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F cos α () cos β () cos γ () ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ × a c 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 0 W− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ×+ 0 cd+ 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ B x B y B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ 0= A x A z B x B y B z F ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A x A z , B x , B y , B z , F, () = A x A z ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 96.5− 13.7− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= B x B y B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 48.5 67.9− 45.7 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= F 96lb= 410 © 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. [...]... publisher Engineering Mechanics - Statics Chapter 5 c = 2 ft i = 2 ft d = 2 ft j = 2 ft e = 2. 5 ft k = 3 ft Solution: The initial guesses are: TBC = 10 lb Ax = 10 lb MAx = 10 lb⋅ ft TDE = 10 lb Ay = 10 lb MAy = 10 lb⋅ ft Given TDE ( a − k) 2 2 ( a − k) + i + j −i TDE 2 ( a − k) + i + j j TBC 2 2 2 2 TBC 2 2 2 + Ax = 0 + Ay = 0 −W=0 2 2 ( d − b) + i + c i 2 2 ( a − k) + i + j MAy − TDE k 2 ( d − b)... writing from the publisher Engineering Mechanics - Statics ⎛ Chapter 5 ⎞ ⎟=0 2 2 2 a +b +c ⎠ ⎝ Ax + F 1 sin ( θ ) − b⎜ TBC TBC ⎞ ⎞ + a⎛ ⎜ 2 2 2 = 0 ⎟ 2 2 ⎝ a +c ⎠ ⎝ a +b +c ⎠ ⎛ Ay − F 1 cos ( θ ) − F 2 + TBD⎜ a TBC ⎛ TBD ⎞ ⎛ ⎞ − c⎜ ⎟ ⎟=0 2 2 2 2 2 ⎝ a +c ⎠ ⎝ a +b +c ⎠ Az − c⎜ ⎛ TBC ⎞ ⎜ ⎟ ⎜ TBD ⎟ ⎜ Ax ⎟ = Find ( T , T , A , A , A ) BC BD x y z ⎜ ⎟ ⎜ Ay ⎟ ⎜ A ⎟ ⎝ z ⎠ ⎛ TBC ⎞ ⎛ 13 1.0 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ TBD ⎠... b) 2 (w2 − w1)( a + b) − F − W = 0 a+b 2 + (w2 − w1)( a + b) 3 ( a + b) − W 2 1 ⎛ w1 ⎞ ⎜ ⎟ = Find ( w1 , w2) ⎝ w2 ⎠ 2 a+b 2 − Fa = 0 ⎛ w1 ⎞ ⎛ 62. 7 ⎞ lb ⎜ ⎟=⎜ ⎟ ⎝ w2 ⎠ ⎝ 91. 1 ⎠ ft Problem 5-9 7 The uniform ladder rests along the wall of a building at A and on the roof at B If the ladder has a weight W and the surfaces at A and B are assumed smooth, determine the angle θ for equilibrium Given: a = 18 ... publisher Engineering Mechanics - Statics Chapter 5 i −TDE a 2 2 ( a − k) + i + j 2 + TBC b i 2 =0 2 2 ( d − b) + i + c ⎛ MAx ⎞ ⎜ ⎟ ⎜ MAy ⎟ ⎜ ⎟ ⎜ TBC ⎟ = Find ( M , M , T , T , A , A ) Ax Ay BC DE x y ⎜ TDE ⎟ ⎜ ⎟ ⎜ Ax ⎟ ⎜ Ay ⎟ ⎝ ⎠ ⎛ TBC ⎞ ⎛ 42. 857 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ TDE ⎠ ⎝ 32. 14 3 ⎠ ⎛ Ax ⎞ ⎛ 3.5 71 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Ay ⎠ ⎝ 50 ⎠ ⎛ MAx ⎞ ⎛ 2. 698 × 10 − 13 ⎞ ⎜ ⎟=⎜ ⎟ lb⋅ ft ⎝ MAy ⎠ ⎝ 17 .857 ⎠ Problem 5-8 3 The member... or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 5 Given: M = 3 Mg a = 4m b = 3m c = 3m d = 4m e = 2m Solution: The initial guesses are: F AC = 10 N F BC = 10 N F DE = 10 N Given b ( F AC) 2 2 a +b − c ( F BC) 2 =0 2 a +c M g e − ( F AC) a d+e 2 2 − FBC a +b a 2 2 a( d + e) 2 =0 2 a +c F BC( b + c) − M g b + FDE b = 0 a +c ⎛ FAC ⎞ ⎜ ⎟ FBC ⎟ =... without permission in writing from the publisher Engineering Mechanics - Statics Chapter 5 ⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ Bx ⎟ = Find ( A , A , B , B , C , C ) x y x z y z ⎜ Bz ⎟ ⎜ ⎟ ⎜ Cy ⎟ ⎜ Cz ⎟ ⎝ ⎠ ⎛ Ax ⎞ ⎛ 6 32. 883 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Ay ⎠ ⎝ 14 1.0 81 ⎠ ⎛ Bx ⎞ ⎛ − 7 21 .27 1 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Bz ⎠ ⎝ 895 . 21 5 ⎠ ⎛ Cy ⎞ ⎛ 20 0. 12 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ Cz ⎠ ⎝ −506.306 ⎠ Problem 5-8 0 The bent rod is supported at A, B, and C by smooth... permission in writing from the publisher Engineering Mechanics - Statics Chapter 5 Given: F = 350 lb e = 12 ft a = 4 ft f = 4 ft b = 5 ft g = 10 ft c = 4 ft h = 4 ft d = 2 ft i = 10 ft Solution: Initial Guesses: F BC = 1 lb Ay = 1 lb MAy = 1 lb⋅ ft Ax = 1 lb Az = 1 lb MAz = 1 lb⋅ ft Given ⎛ 0 ⎞ ⎛d⎞ ⎡ ⎛ g − d ⎞⎤ ⎜ ⎟ ⎜ ⎟ ⎢ F ⎜ e − c ⎟⎥ = 0 ⎜ MAy ⎟ + ⎜ c ⎟ × ⎢ ⎟⎥ 2 2 2 ⎜ MAz ⎟ ⎝ 0 ⎠ ⎣ ( g − d) + ( e − c)... 0; Ax = 0 ΣMA = 0; −w a By = a 2 1 6 − 1 2 w b⎛ a + ⎜ ⎝ ⎟ + By( a + b) = 0 3⎠ 2 w 3a + 3 a b + b Ay + By − w a − ΣF y = 0; b⎞ 2 B y = 7 kN a+b 1 2 Ay = −By + w a + wb=0 1 2 Ay = 11 kN wb Problem 5-9 4 Determine the normal reaction at the roller A and horizontal and vertical components at pin B for equilibrium of the member Units Used: 3 kN = 10 N Given: F 1 = 10 kN F 2 = 6 kN a = 0.6 m b = 0.6 m c =... 0 −W=0 2 2 ( d − b) + i + c i 2 2 ( a − k) + i + j MAy − TDE k 2 ( d − b) + i + c +c ( a − k) + i + j MAx + TDE j 2 ( d − b) + i + c 2 TDE 2 2 −i 2 TBC + ( −b + d) 2 j 2 2 ( a − k) + i + j 2 i + c TBC 2 − Wi = 0 2 2 2 2 ( d − b) + i + c d + TBC c 2 + W ( k − f) = 0 ( d − b) + i + c 416 © 20 07 R C Hibbeler Published by Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material... publisher Engineering Mechanics - Statics Chapter 5 Solution: Initial Guesses: Az = 1 lb B z = 2 lb Cz = 31 lb Given ΣMy = 0; B z r cos ( θ 1 ) − Cz r cos ( θ 1 ) − F sin ( θ 3 ) c = 0 [1] ΣMx = 0; −B z r sin ( θ 1 ) − Cz r sin ( θ 1 ) + Az r − F cos ( θ 3 ) c = 0 [2] ΣF z = 0; Az + Bz + Cz = W [3] Solving Eqs. [1] , [2] and [3] yields: ⎛ Az ⎞ ⎜ ⎟ ⎜ Bz ⎟ = Find ( Az , Bz , Cz) ⎜C ⎟ ⎝ z⎠ ⎛ Az ⎞ ⎛ 16 00 ⎞ . ft⋅= Given ak−() T DE ak−() 2 i 2 + j 2 + b− d+() T BC db−() 2 i 2 + c 2 + + A x + 0= i− T DE ak−() 2 i 2 + j 2 + i T BC db−() 2 i 2 + c 2 + − A y + 0= j T DE ak−() 2 i 2 + j 2 + c T BC db−() 2 i 2 + c 2 + + W−. publisher. Engineering Mechanics - Statics Chapter 5 c 2ft= i 2ft= d 2ft= j 2ft= e 2. 5 ft= k 3ft= Solution: The initial guesses are: T BC 10 lb= A x 10 lb= M Ax 10 lb ft⋅= T DE 10 lb= A y 10 lb= M Ay 10 . Used: kN 10 3 N= Given: F 1 3kN= d 2m= F 2 4kN= e 1. 5 m= a 1m= g 1m= b 1. 5 m= h 3m= c 3m= i 2m= fce−= j 2m= Solution: The initial guesses are: T BD 1kN= T BC 1kN= A x 1kN= A y 1kN= A z 1kN= The