Engineering Mechanics - Statics Chapter 6 F CD M 2 ga b+() 2 sin θ () a = F CD 6.5 kN= Σ F x = 0; E x F CD cos θ ()() = E x 5607N= Σ F y = 0; E y M 2 g 2 − F CD sin θ () + 0= E y M 2 g 2 F CD sin θ () −= E y 1766− N= F R E x 2 E y 2 += F R 5.879 kN= Problem 6-89 Determine the horizontal and vertical components of force at each pin. The suspended cylinder has a weight W. Given: W 80 lb= d 6ft= a 3ft= e 2ft= b 4ft= r 1ft= c 4ft= Solution: Guesses A x 1lb= B x 1lb= B y 1lb= F CD 1lb= E x 1lb= E y 1lb= Given E x B x − W+ 0= E y − B y + 0= 561 © 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 W− rE y a+ E x d− 0= B y − c c 2 d 2 + F CD + W− 0= B y − a c c 2 d 2 + F CD d+ Wd e+ r−()− 0= A x − B x + d c 2 d 2 + F CD + W− 0= A x B x B y F CD E x E y ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A x B x , B y , F CD , E x , E y , () = C x F CD d c 2 d 2 + = C y F CD c c 2 d 2 + = D x C x −= D y C y −= A x B x B y C x C y D x D y E x E y ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 160 80 26.667 160 106.667 160− 106.667− 8.694− 10 13− × 26.667 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= 562 © 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 The two-member frame is pin connected at C, D, and E. The cable is attached to A, passes over the smooth peg at B, and is attached to a load W. Determine the horizontal and vertical reactions at each pin. Given: a 2ft= b 1ft= c 0.75 ft= W 100 lb= Solution: d c b a 2 b+()= Initial guesses: C x 1lb= C y 1lb= D x 1lb= D y 1lb= E x 1lb= E y 1lb= Given D x − C x + W− 0= D y − C y + W− 0= C x − cC y b+ Wd+ Wa 2 b+()− 0= WC x − E x + 0= E y C y − 0= W− dC x c+ C y b+ 0= 563 Problem 6-90 562 © 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 C x C y D x D y E x E y ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find C x C y , D x , D y , E x , E y , () = C x C y D x D y E x E y ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 133 200 33 100 33 200 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Problem 6-91 Determine the horizontal and vertical components of force which the pins at A, B, and C exert on member ABC of the frame. Given: F 1 400 N= F 2 300 N= F 3 300 N= a 1.5 m= b 2m= c 1.5 m= d 2.5 m= f 1.5 m= g 2m= eab+ c+ d−= Solution: Guesses A y 1N= C x 1N= C y 1N= F BD 1N= F BE 1N= 564 © 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 gF 2 ab+()+ F 3 a+ A y fg+()− 0= F 1 gC y fg+()− 0= C x e 0= C x − fg+ e 2 fg+() 2 + F BD − fg+ d 2 fg+() 2 + F BE + 0= A y C y − e e 2 fg+() 2 + F BD − d d 2 fg+() 2 + F BE − 0= A y C x C y F BD F BE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A y C x , C y , F BD , F BE , () = B x fg+ e 2 fg+() 2 + − F BD fg+ d 2 fg+() 2 + F BE += B y e e 2 fg+() 2 + F BD d d 2 fg+() 2 + F BE += A y 657 N= B x B y ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 429 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= C x C y ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 229 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= 565 © 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 The derrick is pin-connected to the pivot at A. Determine the largest mass that can be supported by the derrick if the maximum force that can be sustained by the pin at A is F max . Units Used: kN 10 3 N= g 9.81 m s 2 = Mg 10 3 kg= Given: F max 18 kN= L 5m= θ 60 deg= Solution: AB is a two-force member. Require F AB F max = + ↑ Σ F y = 0; F AB sin θ () Mg 2 sin θ () − W− 0= M 2 F AB g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sin θ () sin θ () 2+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = M 5.439 1 s 2 Mg= Problem 6-93 Determine the required mass of the suspended cylinder if the tension in the chain wrapped around the freely turning gear is T. Also, what is the magnitude of the resultant force on pin A? Units Used: kN 10 3 N= g 9.8 m s 2 = 566 Problem 6-92 © 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: T 2kN= L 2ft= θ 30 deg= φ 45 deg= Solution: Σ M A = 0; 2− TLcos θ () Mgcos φ () Lcos θ () + Mgsin φ () Lsin θ () + 0= M 2 Tcos θ () cos φ () cos θ () sin φ () sin θ () + () g = M 1793 1 s 2 kg= + → Σ F x = 0; 2 TMgcos φ () − A x − 0= A x 2 TMgcos φ () −= + ↑ Σ F y = 0; Mgsin φ () A y − 0= A y Mgsin φ () = F A A x 2 A y 2 += F A 2.928 kN= 567 © 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 The tongs consist of two jaws pinned to links at A, B, C, and D. Determine the horizontal and vertical components of force exerted on the stone of weight W at F and G in order to lift it. Given: a 1ft= b 2ft= c 1.5 ft= d 1ft= W 500 lb= Solution: Guesses F x 1lb= F y 1lb= F AD 1lb= F BE 1lb= Given 2 F y W− 0= F AD bF x bc+()− 0= F AD F x − a a 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BE − 0= F y − d a 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BE + 0= F x F y F AD F BE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F x F y , F AD , F BE , () = G x G y ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F x F y ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F x F y G x G y ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 333 250 333 250 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Problem 6-95 Determine the force P on the cable if the spring is compressed a distance δ when the mechanism is in the position shown. The spring has a stiffness k. 568 Problem 6-94 © 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: δ 0.5 in= c 6in= k 800 lb ft = d 6in= a 24 in= e 4in= b 6in= θ 30 deg= Solution: F E k δ = F E 33.333 lb= The initial guesses are P 20 lb= B x 11 lb= B y 34 lb= F CD 34 lb= Given Σ M A = 0; B x bB y c+ F E ab+()− 0= Σ M D = 0; B y dPe− 0= + → Σ F x = 0; B x − F CD cos θ () + 0= Σ M B = 0; F CD sin θ () dPde+()− 0= F CD B x B y P ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F CD B x , B y , P, () = B x 135.398 lb= B y 31.269 lb= F CD 156.344 lb= P 46.903 lb= 569 © 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 The scale consists of five pin-connected members. Determine the load W on the pan EG if a weight F is suspended from the hook at A. Given: F 3lb= b 3in= a 5in= c 4in= d 6in= f 2in= e 8in= Solution: Guesses T C 10 lb= T D 10 lb= T G 10 lb= W 10 lb= Given Member ABCD: Σ M B = 0; Fa T C b− T D be+ d−()− 0= Member EG: Σ M G = 0; T C − eWc+ 0= Σ F y = 0; T G W− T C + 0= Member FH: Σ M H = 0; T D − df+()T G f+ 0= T C T D T G W ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find T C T D , T G , W, () = W 7.06 lb= Problem 6-97 The machine shown is used for formin g metal p lates. It consists of two to gg les A BC and D E F , 570 Problem 6-96 © 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 6 Given: N w1 = 50 0 m N w2 = 400 m N w3 = 600 m a = 3m b = 3m Solution: Guesses Ax = 1 N Ay = 1 N Cx = 1 N Cy = 1 N Given 1 Ay + Cy − w1 a − w2 a = 0 2 Ax a − 1 − Ax + Cx + w3 b = 0 2 − Ay a + ⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ = Find ( Ax , Ay , Cx , Cy) ⎜ Cx ⎟ ⎜C ⎟ ⎝ y⎠ 1 2a 1 b a w1 a − w3 b − w2 a = 0 2 3 2 3 2 1 a w1 a = 0 2 3 ⎛ Ax ⎞ ⎜ ⎟ ⎛ 1400 ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ 25 0... in writing from the publisher Engineering Mechanics - Statics Chapter 6 Given −M ( ) ( ) ( ) ( ) g cos θ 1 − F AB cos θ 2 + FCB 2 −M g sin θ 1 − FAB sin θ 2 + FCB 2 ⎛ FAB ⎞ ⎜ ⎟ = Find F , F ( AB CB) ⎜ FCB ⎟ ⎝ ⎠ ⎛ Cx ⎟ ⎜ ⎞ ⎜ Cy ⎟ ⎝ ⎠ b =0 2 2 a +b a 2 − Mg = 0 2 a +b = ⎛b⎞ ⎜ ⎟ 2 2 a a +b ⎝ ⎠ FCB ⎛ FAB ⎞ ⎛ 9.7 ⎞ ⎜ ⎟ Cx ⎟ = ⎜ 11 .53 ⎟ kN ⎜ ⎜ ⎟ ⎜ C ⎟ ⎝ 8. 65 ⎠ ⎝ y ⎠ Problem 6-1 03 The tower truss has a weight... Given: M = 1 25 kg d = 1m a = 1m e = 1m b = 2m f = 2m c = 2m g = 9.81 m 2 s Solution: Member GFE: ΣME = 0; −F FB⎡ ⎢ ⎤ b + M g ( a + b) = 0 2 2⎥ ⎣ ( c + d) + ( b − e) ⎦ F FB = M g c+d ⎡ a + b ⎤ ( c + d) 2 + ( b − e) 2 ⎢ ⎥ ⎣ b( c + d) ⎦ F FB = 1.94 kN ΣF x = 0; ⎡ ⎤=0 ⎣ ( c + d) + ( b − e) ⎦ b−e E x − FFB⎢ 2 2 ⎡ ⎤ b−e E x = F FB⎢ 2 2 ⎣ ( c + d) + ( b − e) ⎦ Member EDC: ΣΜc = 0; ⎛ ⎞ ⎟d = 0 2 2 ⎝ e +d ⎠... Statics Chapter 6 Solution: (a) Bar: + ↑Σ Fy = 0; 2 ⎛ F⎞ − 2 ⎛ W⎞ = 0 ⎜ ⎟ ⎜ ⎟ ⎝ 2 2 F = W F = 1 75 lb Man: + ↑Σ Fy = 0; NC − W − 2 ⎛ F⎞ = 0 ⎜ ⎟ ⎝ 2 NC = W + F NC = 350 lb ( b) Bar: + ↑ Σ F y = 0; 2 ⎛ W⎞ − 2 F = 0 ⎜ ⎟ 2 ⎝4⎠ F = W 2 Man: + ↑Σ Fy = 0; NC − W + 2 F = 87 .5 lb ⎛ F⎞ = 0 ⎜ ⎟ ⎝ 2 NC = W − F NC = 87 .5 lb 58 9 © 20 07 R C Hibbeler Published by Pearson Education, Inc., Upper Saddle River, NJ... without permission in writing from the publisher Engineering Mechanics - Statics F CG = F Chapter 6 a F CG = 100 N b Sector gear : ⎞d = 0 2 2⎟ ⎝ c +d ⎠ ΣMH = 0; F CG( d + e) − FAB⎛ ⎜ c ⎛ c2 + d2 ⎞ ⎟ F = 29 7. 62 N F AB = FCG ( d + e) ⎜ ⎝ c d ⎠ AB Table: ΣF y = 0; c ⎞ ⎜ 2 2 ⎟ − Fs = 0 ⎝ c +d ⎠ F AB⎛ F s = FAB c ⎞ ⎛ ⎜ 2 2⎟ ⎝ c +d ⎠ F s = 28 6 N Problem 6-1 18 The mechanism is used to hide kitchen appliances... permission in writing from the publisher Engineering Mechanics - Statics Chapter 6 Given: F = 5 lb b = 1 in a = 1 .5 in c = 3 in d = 0. 75 in e = 1 in θ = 20 deg Solution: From FBD (a) ΣME = 0; ⎡ ⎤b = 0 ⎣ c + ( d + e) ⎦ F ( b + c) − F CD⎢ d+e 2 2⎥ ⎡ c2 + ( d + e) 2 ⎥ ⎣ b( d + e) ⎦ F CD = F( b + c) ⎢ ΣF x = 0; ⎡ F CD = 39.693 lb ⎤ ⎥ 2 2 ⎣ c + ( d + e) ⎦ E x = F CD⎢ c E x = 34 .28 6 lb From FBD (b) ΣMB = 0; NA sin... publisher Engineering Mechanics - Statics Chapter 6 Problem 6-1 11 Determine the force created in the hydraulic cylinders EF and AD in order to hold the shovel in equilibrium The shovel load has a mass W and a center of gravity at G All joints are pin connected Units Used: 3 Mg = 10 kg 3 kN = 10 N Given: a = 0 . 25 m θ 1 = 30 deg b = 0 . 25 m θ 2 = 10 deg c = 1 .5 m θ 3 = 60 deg d = 2m W = 1 . 25 Mg e = 0 .5 m Solution:... publisher Engineering Mechanics - Statics Chapter 6 Problem 6-1 14 A man having weight W1 attempts to lift himself using one of the two methods shown Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C The platform has weight W2 Given: W1 = 1 75 lb W2 = 30 lb Solution: (a) Bar: + ↑Σ Fy = 0; 2 F 2 ( ) − W1 + W2 = 0 F = W1 + W2 F = 20 5 lb Man:... the publisher Engineering Mechanics - Statics F = Chapter 6 W1 + W2 2 F = 1 02 lb Man: + ↑ Σ F y = 0; NC − W1 + 2 ⎛F⎞ = 0 ⎜ ⎟ ⎝ 2 NC = W1 − F NC = 72. 5 lb Problem 6-1 15 The piston C moves vertically between the two smooth walls If the spring has stiffness k and is unstretched when θ = 0, determine the couple M that must be applied to AB to hold the mechanism in equilibrium Given: k = 15 lb in θ = 30... Mechanics - Statics Ax = 1 lb Ay = 1 lb Chapter 6 MA = 1 lb⋅ ft Given Bx + a ⎛ ⎞ ⎜ 2 2 ⎟ NC − F = 0 ⎝ a +b ⎠ By − b ⎛ ⎞ ⎜ 2 2 ⎟ NC − F = 0 ⎝ a +b ⎠ (F + Bx)a − (F + By)b = 0 F− a ⎛ ⎞ ⎜ 2 2 ⎟ NC − Ax = 0 ⎝ a +b ⎠ b ⎛ ⎞ ⎜ 2 2 ⎟ NC − Ay = 0 ⎝ a +b ⎠ −F 2 a + a ⎛ ⎞ ⎜ 2 2 ⎟ NC a + MA = 0 ⎝ a +b ⎠ ⎛ Bx ⎞ ⎜ ⎟ By ⎟ ⎜ ⎜N ⎟ ⎜ C ⎟ = Find B , B , N , A , A , M ( x y C x y A) ⎜ Ax ⎟ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎝ MA ⎠ NC = 20 lb . Engineering Mechanics - Statics Chapter 6 F CD M 2 ga b+() 2 sin θ () a = F CD 6 .5 kN= Σ F x = 0; E x F CD cos θ ()() = E x 56 07N= Σ F y = 0; E y M 2 g 2 − F CD sin θ () + 0= E y M 2 g 2 F CD sin θ () −=. from the publisher. Engineering Mechanics - Statics Chapter 6 Given M− 2 gcos θ 1 () F AB cos θ 2 () − F CB b a 2 b 2 + + 0= M− 2 gsin θ 1 () F AB sin θ 2 () − F CB a a 2 b 2 + + Mg− 0= F AB F CB ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Find. F BD fg+ d 2 fg+() 2 + F BE += B y e e 2 fg+() 2 + F BD d d 2 fg+() 2 + F BE += A y 657 N= B x B y ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 429 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= C x C y ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 22 9 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= 56 5 © 20 07 R. C. Hibbeler. Published