Engineering Mechanics - Statics Chapter 6 F 3 5000 lb= a 9ft= b 12 ft= Solution F DJ 1 kip= Initial Guesses: A y 1 kip= F CD 1 kip= F CJ 1 kip= F KJ 1 kip= Given F 3 aF 2 4a()+ F 1 5a()+ A y 6a()− 0= A y − 2a()F 1 a+ F KJ b+ 0= F CD F KJ + a a 2 b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F CJ + 0= A y F 1 − F 2 − b a 2 b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F CJ − 0= F DJ − 0= A y F KJ F CJ F DJ F CD ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A y F KJ , F CJ , F DJ , F CD , () = A y F KJ F CJ F DJ F CD ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 9.5 11.25 3.125− 0 9.375− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ kip= Positive (T) Negative (C) 491 © 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 6 Determine the force in members EI and JI of the truss which serves to support the deck of a bridge. State if these members are in tension or compression. Units Used: kip 10 3 lb= Given: F 1 4000 lb= F 2 8000 lb= F 3 5000 lb= a 9ft= b 12 ft= Solution: Initial Guesses: G y 1 kip= F EI 1 kip= F JI 1 kip= Given F 1 − aF 2 2a− F 3 5a− G y 6a+ 0= G y 2aF 3 a− F JI b− 0= F EI F 3 − G y + 0= G y F JI F EI ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find G y F JI , F EI , () = G y F JI F EI ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 7.5 7.5 2.5− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kip= Positive (T) Negative (C) Problem 6-36 Determine the force in members BE, EF, and CB, and state if the members are in tension or compression. Units Used: kN 10 3 N= 492 Problem 6-35 © 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 6 Given: F 1 5kN= F 4 10 kN= F 2 10 kN= a 4m= F 3 5kN= b 4m= Solution: θ atan a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Inital Guesses F CB 1kN= F BE 1kN= F EF 1kN= Given F 1 F 2 + F BE cos θ () − 0= F CB − F EF − F BE sin θ () − F 3 − 0= F 1 − aF CB b+ 0= F CB F BE F EF ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find F CB F BE , F EF , () = F CB F BE F EF ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 5 21.2 25− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= Positive (T) Negative (C) Problem 6-37 Determine the force in members BF, BG, and AB, and state if the members are in tension or compression. Units Used: kN 10 3 N= Given: F 1 5kN= F 4 10 kN= 493 © 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 6 F 2 10 kN= a 4m= F 3 5kN= b 4m= Solution: θ atan a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Inital Guesses F AB 1kN= F BG 1kN= F BF 1kN= Given F 1 F 2 + F 4 + F BG cos θ () + 0= F 1 − 3aF 2 2a− F 4 a− F AB b+ 0= F BF − 0= F AB F BG F BF ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find F AB F BG , F BF , () = F AB F BG F BF ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 45 35.4− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= Positive (T) Negative (C) Problem 6-38 Determine the force developed in members GB and GF of the bridge truss and state if these members are in tension or compression. Given: F 1 600 lb= 494 © 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 6 F 2 800 lb= a 10 ft= b 10 ft= c 4ft= Solution: Initial Guesses A x 1lb= A y 1lb= F GB 1lb= F GF 1lb= Given F 2 bF 1 b 2c+()+ A y 2 bc+()− 0= A x 0= A y F GB − 0= A y − bF GF a− 0= A x A y F GB F GF ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A x A y , F GB , F GF , () = A x A y F GB F GF ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 0 671.429 671.429 671.429− ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Positive (T) Negative (C) Problem 6-39 Determine the force members BC, FC, and FE, and state if the members are in tension or compression. Units Used: kN 10 3 N= 495 © 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 6 Given: F 1 6kN= F 2 6kN= a 3m= b 3m= Solution: θ atan a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Initial Guesses D y 1kN= F BC 1kN= F FC 1kN= F FE 1kN= Given F 1 − bF 2 2b()− D y 3b()+ 0= D y bF FE cos θ () a− 0= F FC − F BC F FE + () cos θ () − 0= F 2 − D y + F FE F BC + () sin θ () + 0= D y F BC F FC F FE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find D y F BC , F FC , F FE , () = D y F BC F FC F FE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 6 8.49− 0 8.49 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ kN= Positive (T) Negative (C) 496 © 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 6 Determine the force in members IC and CG of the truss and state if these members are in tension or compression. Also, indicate all zero-force members. Units Used: kN 10 3 N= Given: F 1 6kN= F 2 6kN= a 1.5 m= b 2m= Solution: By inspection of joints B, D, H and I. AB, BC, CD, DE, HI, and GI are all zero-force members. Guesses A y 1kN= F IC 1kN= F CG 1kN= F CJ 1kN= Given A y − 4a()F 1 2a()+ F 2 a+ 0= A y − 2a() b a 2 b 2 + F IC a− a a 2 b 2 + F IC b− 0= a− a 2 b 2 + F IC a a 2 b 2 + F CJ + 0= b− a 2 b 2 + F IC b a 2 b 2 + F CJ − F CG − 0= A y F IC F CG F CJ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A y F IC , F CG , F CJ , () = A y F IC F CG F CJ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 4.5 5.625− 9 5.625− ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ kN= Positive (T) Negative (C) Problem 6-41 Determine the force in members JE and GF of the truss and state if these members are in tension or compression. Also, indicate all zero-force members. 497 Problem 6-40 © 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 6 Units Used: kN 10 3 N= Given: F 1 6kN= F 2 6kN= a 1.5 m= b 2m= Solution: By inspection of joints B, D, H and I. AB, BC, CD, DE, HI, and GI are all zero-force members. Guesses E y 1kN= F JE 1kN= F GF 1kN= Given F 1 − 2a()F 2 3a()− E y 4a()+ 0= E y b a 2 b 2 + F JE + 0= a− a 2 b 2 + F JE F GF − 0= E y F JE F GF ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find E y F JE , F GF , () = E y F JE F GF ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 7.5 9.375− 5.625 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= Positive (T) Negative (C) Problem 6-42 Determine the force in members BC, HC, and HG. After the truss is sectioned use a single equation of equilibrium for the calculation of each force. State if these members are in tension or compression. Units Used: kN 10 3 N= 498 © 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 6 Given: F 1 2kN= F 4 5kN= a 5m= F 2 4kN= F 5 3kN= b 2m= F 3 4kN= c 3m= Solution: Guesses A x 1kN= A y 1kN= F BC 1kN= F HC 1kN= F HG 1kN= d 1m= Given c ad+ b a = A x − 0= F 1 A y − () 4a()F 2 3a()+ F 3 2a()+ F 4 a()+ 0= F 1 A y − () a() A x c()+ F BC c()− 0= F 1 A y − () 2a()F 2 a()+ a a 2 b 2 + F HG c()+ b a 2 b 2 + F HG a()+ 0= A y F 1 − () d() F 2 ad+()− c a 2 c 2 + F HC ad+()+ a a 2 c 2 + F HC c()+ 0= A y A x F BC F HC F HG d ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A y A x , F BC , F HC , F HG , d, () = A x A y ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 8.25 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kN= F BC F HC F HG ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 10.417− 2.235 9.155 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= d 2.5 m= Positive (T) Negative (C) Problem 6-43 Determine the force in members CD, CF, and CG and state if these members are in tension or 499 © 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 6 compression. Units Used: kN 10 3 N= Given: F 1 2kN= F 4 5kN= a 5m= F 2 4kN= F 5 3kN= b 2m= F 3 4kN= c 3m= Solution: Guesses E y 1kN= F CD 1kN= F CF 1kN= F CG 1kN= F FG 1kN= F GH 1kN= Given F 2 − a() F 3 2a()− F 4 3a()− E y F 5 − () 4a()+ 0= F CD c() E y F 5 − () a()+ 0= F 4 − a() F 5 E y − () 2a()− a a 2 b 2 + F FG bc+()− 0= a a 2 b 2 + F FG a a 2 b 2 + F GH − 0= b a 2 b 2 + F FG F GH + () F CG + 0= F 5 E y − () ac b−() b F 4 a ac b−() b + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ + c a 2 c 2 + F CF 2 a ac b−() b + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ − 0= 500 © 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. [...]... the publisher Engineering Mechanics - Statics ⎛ Ay ⎞ ⎜ ⎟ ⎛ −200 ⎞ ⎜ Az ⎟ ⎜ −667 ⎟ ⎟ ⎜B ⎟ ⎜ ⎜ 0 ⎟ ⎜ x⎟ = lb ⎜ Bz ⎟ ⎜ 667 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ −300 ⎟ ⎜ Cy ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ Cz ⎠ Chapter 6 ⎛ FBA ⎞ ⎜ ⎟ ⎛ 167 ⎞ ⎟ FBC ⎟ ⎜ ⎜ ⎜ 250 ⎟ ⎜F ⎟ ⎜ BD ⎟ = ⎜ −731 ⎟ lb ⎜ FAD ⎟ ⎜ 786 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ −391 ⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎝ ⎠ FCD ⎠ ⎝ Positive (T) Negative (C) Problem 6-5 8 The space truss is supported by a ball-and-socket joint... publisher Engineering Mechanics - Statics ⎛ FAB ⎞ ⎜ ⎟ ⎛ −300 ⎞ ⎟ FAC ⎟ ⎜ ⎜ ⎜ 583 .095 ⎟ ⎜F ⎟ ⎜ AD ⎟ = ⎜ 333.333 ⎟ lb ⎜ FAE ⎟ ⎜ −666.667 ⎟ ⎟ ⎜ ⎟ ⎜ 0 ⎜ ⎟ ⎜ FBC ⎟ ⎜ 500 ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ FBE ⎠ Chapter 6 ⎛ FBF ⎞ ⎜ ⎟ ⎛ 0 ⎞ ⎟ FCD ⎟ ⎜ ⎜ ⎜ −300 ⎟ ⎜F ⎟ Positive (T) ⎜ CF ⎟ = ⎜ −300 ⎟ lb Negative (C) ⎜ FDE ⎟ ⎜ 0 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ 424.264 ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ −300 ⎟ ⎝ ⎠ F EF ⎠ ⎝ Problem 6-5 9 The space truss is supported by a ball-and-socket... by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 6 ⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ FHG ⎟ = Find ( Ax , Ay , FHG , FBG , FBC ) ⎜F ⎟ ⎜ BG ⎟ ⎜ FBC ⎟ ⎝ ⎠ ⎛ Ax ⎞ ⎜ ⎟ ⎛ 0 ⎞ ⎜ Ay ⎟ ⎜ 9 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FHG ⎟ = ⎜ −10.062 ⎟ kN ⎜ F ⎟ ⎜ 1 .80 3 ⎟ ⎜ BG ⎟ ⎜ ⎟ ⎝ 8 ⎠ ⎜ FBC ⎟ ⎝ ⎠ Positive (T) Negative (C) Problem 6-4 9 The skewed truss carries the load shown Determine the force... permission in writing from the publisher Engineering Mechanics - Statics Chapter 6 ⎛ FAB ⎞ ⎜ ⎟ FAD ⎟ ⎜ ⎜F ⎟ ⎜ AC ⎟ = Find F , F , F , F , F , B ( AB AD AC BC BD y) ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎟ ⎝ By ⎠ ⎛ By ⎞ ⎛ 566 ⎞ ⎜ ⎟ FAD ⎟ = ⎜ 584 ⎟ N ⎜ ⎜ ⎟ ⎜F ⎟ ⎝ 0 ⎠ ⎝ BD ⎠ ⎛ FAB ⎞ ⎛ 584 ⎞ ⎜ ⎟ FAC ⎟ = ⎜ −1133 ⎟ N ⎜ ⎜ ⎟ ⎜ F ⎟ ⎝ −142 ⎠ ⎝ BC ⎠ Positive (T), Negative (C) Problem 6-5 6 Determine the force in each member of... publisher Engineering Mechanics - Statics Chapter 6 ⎛ 0 ⎞ ⎜ ⎟ F CD + F BC + FAC +⎜ 0 ⎟ =0 CD BC AC ⎜ −FEC ⎟ ⎝ ⎠ CD −BC −AC ⎛ FAC ⎞ ⎜ ⎟ FAD ⎟ ⎜ ⎜F ⎟ ⎜ BC ⎟ = Find F , F , F , F , F , F ( AC AD BC BD CD EC ) ⎜ FBD ⎟ ⎜ ⎟ FCD ⎟ ⎜ ⎜ ⎟ ⎝ FEC ⎠ ⎛ FAC ⎞ ⎜ ⎟ ⎛ 221 ⎞ ⎟ FAD ⎟ ⎜ ⎜ ⎜ 343 ⎟ ⎜F ⎟ ⎜ BC ⎟ = ⎜ 1 48 ⎟ N ⎜ FBD ⎟ ⎜ 186 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ −397 ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ −295 ⎟ ⎝ ⎠ FEC ⎠ ⎝ Positive (T) Negative (C) Problem 6-6 1.. .Engineering Mechanics - Statics Chapter 6 ⎛ Ey ⎞ ⎜ ⎟ F CD ⎟ ⎜ ⎜F ⎟ ⎜ CF ⎟ = Find E , F , F , F , F , F ( y CD CF CG FG GH ) ⎜ FCG ⎟ ⎜ ⎟ ⎜ FFG ⎟ ⎜ ⎟ ⎝ FGH ⎠ ⎛ Ey ⎞ ⎜ ⎟ ⎛ 9.75 ⎞ ⎟ F CD ⎟ ⎜ ⎜ ⎜ −11.25 ⎟ ⎜F ⎟ ⎜ CF ⎟ = ⎜ 3.207 ⎟ kN Positive (T) ⎜ FCG ⎟ ⎜ −6 .8 ⎟ Negative (C) ⎟ ⎜ ⎟ ⎜ ⎜ FFG ⎟ ⎜ 9.155 ⎟ ⎜ ⎟ ⎜ 9.155 ⎟ ⎝ ⎠ FGH ⎠ ⎝ Problem 6-4 4 Determine the force in members... F 2 = 2 kN 5 08 © 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 6 a = 1.5 m b = 1m c = 2m d = 0 .8 m Solution:... writing from the publisher Engineering Mechanics - Statics Chapter 6 ⎛ FAB ⎞ ⎛ FBF ⎞ −300 ⎞ 0 ⎜ ⎟ ⎛ ⎜ ⎟ ⎞ ⎟ FCD ⎟ ⎛ ⎜ ⎟ FAC ⎟ ⎜ ⎜ ⎜ 971 .82 5 −500 ⎟ ⎟ ⎜F ⎟ ⎜ ⎜ ⎟ ⎜ Positive (T) ⎜ 1.121 × 10− 11 ⎜ FCF ⎟ ⎜ −300 ⎟ ⎟ ⎜ AD ⎟ = lb =⎜ lb Negative (C) ⎟ ⎟ ⎜ FAE ⎟ ⎜ −366.667 ⎜ FDE ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎟ 0 ⎜ FBC ⎟ ⎜ ⎜ FDF ⎟ ⎜ 424.264 ⎟ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ 500 ⎠ F ⎟ ⎝ −300 ⎠ F BE ⎠ ⎝ ⎝ ⎝ EF ⎠ Problem 6-6 0 Determine the force... publisher Engineering Mechanics - Statics Chapter 6 Given ⎛ ⎝ F⎜a + b⎞ ⎟ − Ay( 2a + b) = 0 2⎠ ⎛ b⎞ F CD c − Ay⎜ a + ⎟ = 0 ⎝ 2⎠ F CD + F JN cos ( φ ) + F KJ cos ( θ ) = 0 Ay + F JN sin ( φ ) + FKJ sin ( θ ) = 0 ⎛ Ay ⎞ ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎟ = Find ( Ay , FCD , FJN , FKJ ) ⎜ FJN ⎟ ⎜F ⎟ ⎝ KJ ⎠ Ay = 1.5 kip ⎛ FCD ⎞ ⎛ 2.625 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FJN ⎟ = ⎜ 0 ⎟ kip ⎜ F ⎟ ⎝ −3.023 ⎠ ⎝ KJ ⎠ Positive (T), Negative (C) Problem 6-4 8 Determine... 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 6 p socket joint at C pp j Given: M = 50 kg g = 9 .81 m 2 s a = 0.25 m b = 0.5 m c = 2m Solution: h = 2 2 b −a ⎛ −a ⎞ AB = ⎜ −c ⎟ ⎜ ⎟ ⎝h⎠ ⎛a⎞ AD = ⎜ −c ⎟ ⎜ ⎟ ⎝h⎠ ⎛ 2a ⎞ BD = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠ ⎛a⎞ BC = ⎜ 0 ⎟ ⎜ ⎟ . form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 F 2 80 0 lb= a 10 ft= b 10 ft= c 4ft= Solution: Initial Guesses A x 1lb= A y 1lb= F GB 1lb=. publisher. Engineering Mechanics - Statics Chapter 6 Determine the force in members IC and CG of the truss and state if these members are in tension or compression. Also, indicate all zero-force. by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 a 1.5 m= b 1m= c 2m= d 0 .8 m= Solution: θ atan a c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = φ atan a cb− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Initial