Engineering Mechanics - Statics Chapter 5 Σ F z = 0; F− A z + 0= Σ M Ax = 0; Fc() B y b()− 0= Σ M Ay = 0; Fa() T B b()− 0= Σ M Az = 0; B y a() T B c()− 0= Solving, T B A x A y A z B y ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find T B A x , A y , A z , B y , () = T B A x A y A z B y ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 25 25 25− 50 25 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Problem 5-86 The member is supported by a square rod which fits loosely through a smooth square hole of the attached collar at A and by a roller at B. Determine the x, y, z components of reaction at these supports when the member is subjected to the loading shown. Given: M 50 lb ft⋅= F 20 40− 30− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= a 2ft= b 1ft= c 2ft= Solution: Initial Guesses A x 1lb= A y 1lb= 421 © 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 M Ax 1lbft= M Ay 1lbft= M Az 1lbft= B z 1lb= Given A x A y 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 0 B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + F+ 0= M Ax M Ay M Az ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 a 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 0 B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ 0 ab+ c− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F× 0 0 M− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ + ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ + 0= A x A y M Ax M Ay M Az B z ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A x A y , M Ax , M Ay , M Az , B z , () = A x A y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 20− 40 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= B z 30lb= M Ax M Ay M Az ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 110 40 110 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb ft⋅= Problem 5-87 The platform has mass M and center of mass located at G. If it is lifted using the three cables, determine the force in each of these cables. Units Used: Mg 10 3 kg= kN 10 3 N= g 9.81 m s 2 = 422 © 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: M 3Mg= 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 bF AC () a 2 b 2 + cF BC () a 2 c 2 + − 0= Mge F AC () a de+ a 2 b 2 + − F BC ad e+() a 2 c 2 + − 0= a a 2 c 2 + F BC bc+()Mgb− F DE b+ 0= F AC F BC F DE ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find F AC F BC , F DE , () = F AC F BC F DE ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find F AC F BC , F DE , () = F AC F BC F DE ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ kN= F BC 423 © 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 The platform has a mass of M and center of mass located at G. If it is lifted using the three cables, determine the force in each of the cables. Solve for each force by using a single moment equation of equilibrium. Units Used: Mg 1000 kg= kN 10 3 N= g 9.81 m s 2 = Given: M 2Mg= c 3m= a 4m= d 4m= b 3m= e 2m= Solution: r BC 0 c− a ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r AC 0 b a ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r AD e− d− b 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r BD d− e− c− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = First find F DE . F DE de+()Mgd− 0= F DE Mgd de+ = F DE 4.1 s 2 kN= Σ M y ' = 0; Next find F BC . Guess F BC 1kN= Given e 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 0 M− g ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ × ed+ c 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F BC r BC r BC ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×+ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ r AD 0= F BC Find F BC () =F BC Find F BC () = F BC kN=F BC Now find F AC . Guess F AC 1kN= 424 Problem 5-88 © 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 e 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 0 M− g ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ × ed+ b− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F AC r AC r AC ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×+ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ r BD 0= F AC Find F AC () =F AC Find F AC () = F AC kN=F AC Problem 5-89 The cables exert the forces shown on the pole. Assuming the pole is supported by a ball-and-socket joint at its base, determine the components of reaction at A. The forces F 1 and F 2 lie in a horizontal plane. Given: F 1 140 lb= F 2 75 lb= θ 30 deg= a 5ft= b 10 ft= c 15 ft= Solution: The initial guesses are T BC 100 lb= T BD 100 lb= A x 100 lb= A y 100 lb= A z 100 lb= Given F 1 cos θ () F 2 + () cT BC a c a 2 b 2 + c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − T BD a c a 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= F 1 sin θ () cbT BC c a 2 b 2 + c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= 425 © 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 F 1 sin θ () + b T BC a 2 b 2 + c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= A y F 1 cos θ () − F 2 − T BD a a 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + a T BC a 2 b 2 + c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 0= A z c T BD a 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − c T BC a 2 b 2 + c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= T BC T BD A x A y A z ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find T BC T BD , A x , A y , A z , () = T BC T BD ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 131.0 509.9 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= A x A y A z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0.0− 0.0 588.7 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= Problem 5-90 The silo has a weight W, a center of gravity at G and a radius r. Determine the vertical component of force that each of the three struts at A, B, and C exerts on the silo if it is subjected to a resultant wind loading of F which acts in the direction shown. Given: W 3500 lb= F 250 lb= θ 1 30 deg= θ 2 120 deg= θ 3 30 deg= r 5ft= b 12 ft= c 15 ft= 426 © 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 Solution: Initial Guesses: A z 1lb= B z 2lb= C z 31 lb= Given Σ M y = 0; B z rcos θ 1 () C z rcos θ 1 () − F sin θ 3 () c− 0= [1] Σ M x = 0; B z − rsin θ 1 () C z rsin θ 1 () − A z r+ F cos θ 3 () c− 0= [2] Σ F z = 0; A z B z + C z + W= [3] Solving Eqs.[1], [2] and [3] yields: A z B z C z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find A z B z , C z , () = A z B z C z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 1600 1167 734 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= Problem 5-91 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 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. Units Used: kN 10 3 N= Given: M 250 N m⋅= a 400 mm= b 300 mm= c 250 mm= d 120 mm= θ 30 deg= 427 © 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 φ 20 deg= Solution: Initial Guesses: A y 1kN= A z 1kN= B y 1kN= B z 1kN= F CD 1kN= Given 0 A y A z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 B y B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0 F CD − cos φ () F CD sin φ () ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0= da− csin θ () ccos θ () ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 F CD − cos φ () F CD sin φ () ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ × a− b− 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 B y B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ M− 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ + 0= A y A z B y B z F CD ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A y A z , B y , B z , F CD , () = A y A z ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 573 208− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= B y B z ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 382 139− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= F CD 1.015 kN= Problem 5-92 If neither the pin at A nor the roller at B can support a load no greater than F max , determine the maximum intensity of the distributed load w, so that failure of a support does not occur. 428 © 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 Units Used: kN 10 3 N= Given: F max 6kN= a 3m= b 3m= Solution: The greatest reaction is at A. Require Σ M B = 0; F max − ab+()wa a 2 b+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 1 2 wb 2 3 b+ 0= w F max ab+() a a 2 b+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b 2 3 + = w 2.18 kN m = Problem 5-93 If the maximum intensity of the distributed load acting on the beam is w , determine the reactions at the pin A and roller B. Units Used: kN 10 3 N= Given: F 6kN= a 3m= b 3m= 429 © 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 w 4 kN m = Solution: Σ F x = 0; A x 0= Σ M A = 0; w− a a 2 1 2 wb a b 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − B y ab+()+ 0= B y 1 6 w 3a 2 3 ab+ b 2 + ab+ = B y 7kN= Σ F y = 0; A y B y + wa− 1 2 wb− 0= A y B y − wa+ 1 2 wb+= A y 11kN= Problem 5-94 Determine the normal reaction at the roller A and horizontal and vertical components at pin B for equilibrium of the member. Units Used: kN 10 3 N= Given: F 1 10 kN= F 2 6kN= a 0.6 m= b 0.6 m= c 0.8 m= d 0.4 m= θ 60 deg= 430 © 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. [...]... form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 5 Given 1 w1 ( a + b) + w1 ( a + 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... FA kA kB = 2.5 kN m 4 37 © 20 07 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 Problem 6-1 Determine the force... FEF ⎠ 4 47 © 20 07 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 ⎛ FBA ⎞ ⎜ ⎟ ⎛ −8 ⎞ ⎜ FAF ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 4.1 67 ⎟ ⎜... 20 07 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 BC = 4 2P F CD = 17 P 1 FCD + 3 3 = 1.886 P (C) = 1. 374 ... +P=0 2 ( −F BD − FAB + FBC Joint D (FCD − FAD) 4 ) 1 =0 2 =0 17 ( ) F BD − P − F AD + F CD Joint C −F BC 1 2 − F CD 4 1 =0 17 =0 17 ⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FAD ⎟ = Find ( FAB , FBC , FAD , FCD , FBD) ⎜F ⎟ ⎜ CD ⎟ ⎜ FBD ⎟ ⎝ ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛ −0. 471 ⎞ ⎜ FBC ⎟ ⎜ −1.886 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FAD ⎟ = ⎜ 1. 374 ⎟ lb ⎜ F ⎟ ⎜ 1. 374 ⎟ ⎜ CD ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎝ 1.6 67 ⎠ ⎝ ⎠ Now find the critical load P1 = P Tmax ( max F AB , F... ⎟ ⎜ F ⎟ ⎜ 333 ⎟ ⎜ EG ⎟ ⎜ −6 67 ⎟ ⎜ FAG ⎟ ⎜ − 471 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FDC ⎟ = ⎜ 6 67 ⎟ lb ⎜ F ⎟ ⎜ 6 67 ⎟ ⎟ ⎜ EC ⎟ ⎜ ⎜ FBG ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ −943 ⎟ ⎜ FDE ⎟ ⎜ ⎟ ⎟ ⎝ 471 ⎠ ⎜ ⎝ FCG ⎠ Positive means tension, Negative means compression Problem 6-1 1 Determine the force in each member of the truss and state if the members are in tension or compression Given: P 1 = 500 lb P 2 = 1500 lb 452 © 20 07 R C Hibbeler Published by Pearson... ⎟ ⎜ EG ⎟ ⎜ −11 67 ⎟ ⎜ FAG ⎟ ⎜ −1 179 ⎟ ⎟ ⎜ ⎟ ⎜ FDC ⎟ = ⎜ 11 67 ⎟ lb ⎜ ⎜ F ⎟ ⎜ 11 67 ⎟ ⎟ ⎜ EC ⎟ ⎜ ⎜ FBG ⎟ ⎜ 500 ⎟ ⎜ ⎟ ⎜ −1650 ⎟ FDE ⎟ ⎜ ⎜ ⎟ ⎟ ⎝ 471 ⎠ ⎜ ⎝ FCG ⎠ Positive means tension, Negative means compression Problem 6-1 2 Determine the force in each member of the truss and state if the members are in tension or compression Units Used: 3 kN = 10 N Given: P 1 = 10 kN P 2 = 15 kN 454 © 20 07 R C Hibbeler... + f) = 0 435 © 20 07 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 B x = −P ΣF x = 0; ⎛ e + f ⎞ B = −35 .7 lb ⎜ ⎟ x ⎝ b +... a + b⎠ F B = 266.6 67 N ΣMB = 0; F b − FA( a + b) = 0 436 © 20 07 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... 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 Problem 6-3 The truss, used to support a balcony, is subjected to the loading shown Approximate each joint as a pin and determine the force in each member State whether the members are . writing from the publisher. Engineering Mechanics - Statics Chapter 5 B x P− ef+ bc+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = B x 35 .7 lb= Σ F x = 0; A x B x + P− 0= A x B x − P+= A x 135 .7 lb= B y 0= Σ F y = 0; Σ F z . publisher. Engineering Mechanics - Statics Chapter 5 F A F b ab+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F A 533.333 N= Spring force formula: x A x B = F A k A F B k B = k B F B F A k A = k B 2.5 kN m = 4 37 © 20 07 R. C publisher. Engineering Mechanics - Statics Chapter 5 w 4 kN m = Solution: Σ F x = 0; A x 0= Σ M A = 0; w− a a 2 1 2 wb a b 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − B y ab+()+ 0= B y 1 6 w 3a 2 3 ab+ b 2 + ab+ = B y 7kN= Σ F y