Engineering Mechanics - Statics Chapter 5 Given: a 0.5 ft= b 3ft= c 4ft= d 4ft= Solution: ΣM B = 0; N A − ab+()Wb ccos θ()−()+ 0= As θ becomes smaller, N A goes to 0 so that, cos θ() b c = θ acos b c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 41.4 deg= Problem 5-47 The motor has a weight W. 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. 379 © 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: W 850 lb= a 0.5 ft= b 1ft= c 1.5 ft= θ 1 10 deg= θ 2 30 deg= θ 3 10 deg= Solution: Guesses F A 1lb= F B 1lb= F C 1lb= Given ΣM B = 0; F A cos θ 3 ()bWa+ F C cos θ 1 ()ac+()− 0= ΣF x = 0; F C sin θ 1 ()F B sin θ 2 ()− F A sin θ 3 ()− 0= ΣF y = 0; WF A cos θ 3 ()− F B cos θ 2 ()− F C cos θ 1 ()− 0= F A F B F C ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find F A F B , F C ,()= F A F B F C ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 432 0− 432 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= 380 © 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 boom supports the two vertical loads. Neglect the size of the collars at D and B and the thickness of the boom, and compute the horizontal and vertical components of force at the pin A and the force in cable CB. Given: F 1 800 N= F 2 350 N= a 1.5 m= b 1m= c 3= d 4= θ 30 deg= Solution: ΣM A = 0; F 1 − a cos θ() F 2 ab+( ) cos θ()− d c 2 d 2 + F CB ab+( ) sin θ()+ c c 2 d 2 + F CB ab+( ) cos θ()+ 0= F CB F 1 aF 2 ab+()+ ⎡ ⎣ ⎤ ⎦ cos θ() c 2 d 2 + d sin θ()ab+()c cos θ()ab+()+ = F CB 782 N= + → Σ F x = 0; A x d c 2 d 2 + F CB − 0= A x d c 2 d 2 + F CB = A x 625 N= + ↑ Σ F y = 0; A y F 1 − F 2 − c c 2 d 2 + F CB + 0= A y F 1 F 2 + c c 2 d 2 + F CB −= A y 681 N= Problem 5-49 The boom is intended to support two vertical loads F 1 and F 2 . If the cable CB can sustain a 381 Problem 5-48 © 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 maximum load T max before it fails, determine the critical loads if F 1 = 2F 2 . Also, what is the magnitude of the maximum reaction at pin A? Units Used: kN 10 3 N= Given: T max 1500 N= a 1.5 m= b 1m= c 3= d 4= θ 30deg= Solution: ΣM A = 0; F 1 2 F 2 = 2− F 2 a cos θ() F 2 ab+( ) cos θ()− d c 2 d 2 + T max ab+( ) sin θ()+ c c 2 d 2 + T max ab+( ) cos θ()+ 0= F 2 ab+()T max d sin θ() c cos θ()+ c 2 d 2 + cos θ()3 ab+() = F 2 724 N= F 1 2 F 2 = F 1 1.448 kN= + → ΣF x = 0; A x d c 2 d 2 + T max − 0= A x d c 2 d 2 + T max = A x 1.20 kN= + ↑ Σ F y = 0; A y F 2 − F 1 − c c 2 d 2 + T max + 0= 382 © 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 y F 2 F 1 + c c 2 d 2 + T max −= A y 1.27 kN= F A A x 2 A y 2 += F A 1.749 kN= Problem 5-50 The uniform rod of length L and weight W is supported on the smooth planes. Determine its position θ for equilibrium. Neglect the thickness of the rod. Solution: ΣM B = 0; W− L 2 cos θ() N A cos φθ−()L+ 0= N A W cos θ() 2 cos φθ−() = ΣM A = 0; W L 2 cos θ() N B cos ψθ+()L− 0= N B W cos θ() 2 cos ψθ+() = ΣF x = 0; N B sin ψ() N A sin φ()− 0= W cos θ() 2 cos ψθ+() sin ψ() W cos θ() 2 cos φθ−() sin φ()− 0= sin ψ()cos φθ−()sin φ()cos ψθ+()− 0= sin ψ()cos φ()cos θ() sin φ()sin θ()+ sin φ()cos ψ()cos θ() sin ψ()sin θ()−− 0= 2 sin ψ()sin φ()sin θ() sin φ()cos ψ() sin ψ()cos φ()− cos θ()= tan θ() sin φ()cos ψ() sin ψ()cos φ()− 2 sin ψ()sin φ() = cot ψ() cot φ()− 2 = θ atan cot ψ() cot φ()− 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = 383 © 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 Problem 5-51 The toggle switch consists of a cocking lever that is pinned to a fixed frame at A and held in place by the spring which has unstretched length δ. Determine the magnitude of the resultant force at A and the normal force on the peg at B when the lever is in the position shown. Given: δ 200 mm= k 5 N m = a 100 mm= b 300 mm= c 300 mm= θ 30 deg= Solution: Using the law of cosines and the law of sines lc 2 ab+() 2 + 2 ca b+( ) cos 180 deg θ−()−= sin φ() c sin 180 deg θ−() l = φ asin c sin 180 deg θ−() l ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = φ 12.808 deg= F s ks= kl δ−()= F s kl δ−()= F s 2.3832 N= ΣM A = 0; F s − sin φ()ab+()N B a+ 0= N B F s sin φ() ab+ a = N B 2.11 N= ΣF x = 0; A x F s cos φ()− 0= A x F s cos φ()= A x 2.3239 N= ΣF y = 0; A y N B + F s sin φ()− 0= A y F s sin φ() N B −= A y 1.5850− N= F A A x 2 A y 2 += F A 2.813 N= 384 © 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 Problem 5-52 The rigid beam of negligible weight is supported horizontally by two springs and a pin. If the springs are uncompressed when the load is removed, determine the force in each spring when the load P is applied. Also, compute the vertical deflection of end C. Assume the spring stiffness k is large enough so that only small deflections occur. Hint: The beam rotates about A so the deflections in the springs can be related. Solution: ΣM A = 0; F B LF C 2 L+ P 3 2 L− 0= F B 2 F C + 1.5 P= Δ C 2 Δ B = F C k 2 F B k = F C 2F B = 5 F B 1.5 P= F B 0.3 P= F c 0.6 P= Δ C 0.6 P k = Problem 5-53 The rod supports a weight W and is pinned at its end A. If it is also subjected to a couple moment of M, determine the angle θ for equilibrium.The spring has an unstretched length δ and a stiffness k. 385 © 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: W 200 lb= M 100 lb ft= δ 2ft= k 50 lb ft = a 3ft= b 3ft= c 2ft= Solution: Initial Guess: θ 10 deg= Given ka b+( )sin θ() c+ δ− ⎡⎣ ⎤⎦ ab+( ) cos θ() Wacos θ()− M− 0= θ Find θ()= θ 23.2 deg= Problem 5-54 The smooth pipe rests against the wall at the points of contact A, B, and C. Determine the reactions at these points needed to support the vertical force F. Neglect the pipe's thickness in the calculation. Given: F 45 lb= 386 © 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 θ 30 deg= a 16 in= b 20 in= c 8in= Solution: Initial Guesses: R A 1lb= R B 1lb= R C 1lb= Given ΣM A = 0; F cos θ()ab+()F sin θ()c− R C b− R B c tan θ()+ 0= + ↑ Σ F y = 0; R C cos θ() R B cos θ()− F− 0= + → Σ F x = 0; R A R B sin θ()+ R C sin θ()− 0= R A R B R C ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find R A R B , R C ,()= R A R B R C ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 25.981 11.945 63.907 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= Problem 5-55 The rigid metal strip of negligible weight is used as part of an electromagnetic switch. If the stiffness of the springs at A and B is k, and the strip is originally horizontal when the springs are unstretched, determine the smallest force needed to close the contact gap at C. Units Used: mN 10 3− N= 387 © 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: a 50 mm= b 50 mm= c 10 mm= k 5 N m = Solution: Initial Guesses: F 0.5 N= y A 1mm= y B 1mm= Given cy A − ab+ y B y A − a = ky A ky B + F− 0= ky B aFab+()− 0= y A y B F ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find y A y B , F,()= y A y B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2− 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ mm= F 10 mN= Problem 5-56 The rigid metal strip of negligible weight is used as part of an electromagnetic switch. Determine the maximum stiffness k of the springs at A and B so that the contact at C closes when the vertical force developed there is F. Originally the strip is horizontal as shown. Units Used: mN 10 3− N= Given: a 50 mm= b 50 mm= c 10 mm= F 0.5 N= 388 © 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 Initial Guesses: k 1 N m = y A 1mm= y B 1mm= Given cy A − ab+ y B y A − a = ky A ky B + F− 0= ky B aFab+()− 0= y A y B k ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find y A y B , k,()= y A y B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2− 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ mm= k 250 N m = Problem 5-57 Determine the distance d for placement of the load P for equilibrium of the smooth bar in the position θ as shown. Neglect the weight of the bar. Solution: + ↑ Σ F y = 0; R cos θ() P− 0= Σ M A = 0; P− d cos θ() R a cos θ() + 0= Rdcos θ() 2 R a cos θ() = d a cos θ() 3 = Problem 5-58 The wheelbarrow and its contents have mass m and center of mass at G. Determine the greatest 389 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 5 g angle of tilt θ without causing the wheelbarrow to tip over. Solution: Require point G to be over the wheel axle for tipping. Thus b cos θ() a sin θ()= θ tan 1− b a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Problem 5-59 Determine the force P needed to pull the roller of mass M over the smooth step. Given: M 50 kg= a 0.6 m= b 0.1 m= θ 60 deg= θ 1 20 deg= g 9.81 m s 2 = 390 © 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: φ acos ab− a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = φ 33.56 deg= ΣM B = 0, Mgsin θ 1 ()ab−()Mgcos θ 1 ()a sin φ ()+ P cos θ()ab−()P sin θ()a sin φ()−+ ... 0= PMg sin θ 1 ()ab−( ) cos θ 1 ()a sin φ ()+ cos θ() ab−( ) sin θ()a sin φ()+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = P 441 N= Problem 5-60 Determine the magnitude and direction θ of the minimum force P needed to pull the roller of mass M over the smooth step. Given: a 0.6 m= b 0.1 m= θ 1 20 deg= M 50 kg= g 9.81 m s 2 = Solution: For P min , N A tends to 0 φ acos ab− a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = φ 33.56 deg= ΣM B = 0 Mgsin θ 1 ()ab−()Mgcos θ 1 ()a sin φ ()+ P− cos θ()ab−() ⎡⎣ ⎤⎦ P sin θ()a sin φ()−+ ... 0= 391 © 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 P Mgsin θ 1 ()ab−( ) cos θ 1 ( )a sin φ ( )+ ⎡ ⎣ ⎤ ⎦ cos θ()ab−()a sin φ()sin θ()+ = For P min : dP dθ Mgsin θ 1 ()ab−( ) cos θ 1 ()a sin φ ()+ ⎡ ⎣ ⎤ ⎦ cos θ()ab−()a sin φ()sin θ()+ ⎡⎣ ⎤⎦ 2 a sin φ()cos θ() ab−( )sin θ()− ⎡⎣ ⎤⎦ = 0= which gives, θ atan sin φ() a ab− ⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 33.6 deg= P Mgsin θ 1 ()ab−( ) cos θ 1 ()a sin φ ()+ ⎡ ⎣ ⎤ ⎦ cos θ()ab−()a sin φ()sin θ()+ = P 395 N= Problem 5-61 A uniform glass rod having a length L is placed in the smooth hemispherical bowl having a radius r. Determine the angle of inclination θ for equilibrium. Solution: By Observation φ = θ. Equilibirium : ΣM A = 0; N B 2 rcos θ() W L 2 cos θ()− 0= N B WL 4 r = ΣF x = 0; N A cos θ() W sin θ()− 0= N A W tan θ()= 392 © 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 ΣF y = 0; W tan θ()sin θ() WL 4 r + W cos θ()− 0= sin θ() 2 cos θ() 2 − 1 2 cos θ() 2 −= L− 4 r cos θ()= 2 cos θ() 2 L 4 r cos θ()− 1− 0= cos θ() LL 2 128 r 2 ++ 16 r = θ acos LL 2 128 r 2 ++ 16 r ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Problem 5-62 The disk has mass M and is supported on the smooth cylindrical surface by a spring having stiffness k and unstretched length l 0 . The spring remains in the horizontal position since its end A is attached to the small roller guide which has negligible weight. Determine the angle θ to the nearest degree for equilibrium of the roller. Given: M 20 kg= k 400 N m = l 0 1m= r 2m= g 9.81 m s 2 = a 0.2 m= Guesses F 10 N= R 10 N= θ 30 deg= Solution: Given + → Σ F y = 0; R sin θ() Mg− 0= + ↑ Σ F x = 0; R cos θ() F− 0= Spring Fkra+( )cos θ() l 0 − ⎡ ⎣ ⎤ ⎦ = 393 © 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 F R θ ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Find FR, θ,()= F R ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 163.633 255.481 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= θ 50.171 deg= There is also another answer that we can find by choosing different starting guesses. Guesses F 200 N= R 200 N= θ 20 deg= Solution: Given + → Σ F y = 0; R sin θ() Mg− 0= + ↑ Σ F x = 0; R cos θ() F− 0= Spring Fkra+( )cos θ() l 0 − ⎡ ⎣ ⎤ ⎦ = F R θ ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Find FR, θ,()= F R ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 383.372 430.66 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= θ 27.102 deg= Problem 5-63 Determine the x, y, z components of reaction at the fixed wall A.The force F 2 is parallel to the z axis and the force F 1 is parallel to the y axis. Given: a 2m= d 2m= b 1m= F 1 200 N= c 2.5 m= F 2 150 N= 394 © 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: ΣF x = 0; A x 0= ΣF x = 0; A y F 1 −= A y 200− N= ΣF x = 0; A z F 2 = A z 150 N= ΣΜ x = 0; M Ax F 2 − aF 1 d+= M Ax 100 N m⋅= ΣΜ y = 0; M Ay 0= ΣM z = 0; M Az F 1 c= M Az 500 N m⋅= Problem 5-64 The wing of the jet aircraft is subjected to thrust T from its engine and the resultant lift force L. If the mass of the wing is M and the mass center is at G, determine the x, y, z components of reaction where the wing is fixed to the fuselage at A. Units Used: Mg 10 3 kg= kN 10 3 N= g 9.81 m s 2 = 395 © 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 T 8kN= L 45 kN= M 2.1 Mg= a 2.5 m= b 5m= c 3m= d 7m= Solution: ΣF x = 0; A x − T+ 0= A x T= A x 8kN= ΣF y = 0; A y 0= A y 0= ΣF z = 0; A z − Mg− L+ 0= A z LMg−= A z 24.4 kN= ΣM y = 0; M y Ta()− 0= M y Ta= M y 20.0 kN m⋅= ΣM x = 0; Lb c+ d+()Mgb− M x − 0= M x Lb c+ d+()Mgb−= M x 572 kN m⋅= ΣM z = 0; M z Tb c+()− 0= M z Tb c+()= M z 64.0 kN m⋅= Problem 5-65 The uniform concrete slab has weight W. Determine the tension in each of the three parallel supporting cables when the slab is held in the horizontal plane as shown. 396 Given: 395 © 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: kip 10 3 lb= Given: W 5500 lb= a 6ft= b 3ft= c 3ft= d 6ft= Solution: Equations of Equilibrium : The cable tension T B can be obtained directly by summing moments about the y axis. ΣM y = 0; W d 2 T B d− 0= T B W 2 = T B 2.75 kip= T C aT B ab+()+ W ab+ c+ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= ΣM x = 0; T C 1 a W ab+ c+ 2 T B ab+()− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = T C 1.375 kip= ΣF z = 0; T A T B + T C + W− 0= T A T B − T C − W+= T A 1.375 kip= 397 © 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 air-conditioning unit is hoisted to the roof of a building using the three cables. If the tensions in the cables are T A , T B and T C , determine the weight of the unit and the location (x, y) of its center of gravity G. Given: T A 250 lb= T B 300 lb= T C 200 lb= a 5ft= b 4ft= c 3ft= d 7ft= e 6ft= Solution: ΣF z = 0; T A T B + T C + W− 0= WT A T B + T C += W 750 lb= ΣM y = 0; W x T A cd+()− T C d− 0= x T A cd+()T C d+ W = x 5.2 ft= ΣM x = 0; T A aT B ab+ e−()+ T C ab+()+ Wy− 0= y T A aT B ab+ e−()+ T C ab+()+ W = y 5.267 ft= Problem 5-67 The platform truck supports the three loadings shown. Determine the normal reactions on each of its three wheels. 398 Problem 5-66 © 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 1 380 lb= b 12 in= c 10 in= F 2 500 lb= d 5in= F 3 800 lb= e 12 in= a 8in= f 12 in= Solution: The initail guesses are F A 1lb= F B 1lb= F C 1lb= Given ΣM x = 0; F 1 cd+()F 2 bc+ d+()+ F 3 d+ F A ab+ c+ d+()− 0= ΣM y = 0; F 1 eF B e− F 2 f− F C f+ 0= ΣF y = 0; F B F C + F 2 − F A + F 1 − F 3 − 0= F A F B F C ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find F A F B , F C ,()= F A F B F C ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 663 449 569 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= Problem 5-68 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 components have weights W A , W B and W C , determine the normal reactions of the wheels D, E, and F on the ground. 399 © 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. dhanesh_h Mathcad - CombinedMathcads