Engineering Mechanics - Statics Episode 3 Part 1 doc

Engineering Mechanics - Statics Chapter 8 μ pyg pp p can be held. Given: θ 20 deg= Solution: Paper: + ↑ Σ F y = 0; FF+ W− 0= F W 2 = F μ N= N W 2 μ = Cylinder: F' r W 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ r− 0= F' W 2 = N' W 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sin θ () − W 2 μ cos θ () − 0= N' W 2 sin θ () 1 μ cos θ () + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F' μ N'= W 2 μ W 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sin θ () 1 μ cos θ () + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = 1 μ sin θ () cos θ () += μ 1 cos θ () − sin θ () = μ 0.176= Problem 8-43 The crate has a weight W 1 and a center of gravity at G. If the coefficient of static friction between the crate and the floor is μ s , determine if the man of weight W 2 can push the crate to the left. The coefficient of static friction between his shoes and the floor is μ ' s . Assume the man exerts only a horizontal force on the crate. 801 © 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 8 W 1 300 lb= W 2 200 lb= μ s 0.2= μ ' s 0.35= a 4.5 ft= c 3ft= b 3.5 ft= d 4.5 ft= Solution: Σ F y = 0; N C W 1 − 0= N C W 1 = Σ F x = 0; μ s N C P− 0= P μ s N C = Σ Μ O = 0; W 1 − xPd+ 0= x Pd W 1 = Since x 0.90 ft= < a 4.50 ft= there will not be any tipping. Σ F y = 0; N m W 2 − 0= N m W 2 = N m 200.00 lb= Σ F x = 0; PF m − 0= F m P= F m 60.00 lb= F mmax μ ' s N m = F mmax 70.00 lb= Since F m 60.00 lb= < F mmax 70.00 lb= then the man can push the crate. Problem 8-44 The crate has a weight W 1 and a center of gravity at G. If the coefficient of static friction between the crate and the floor is μ s , determine the smallest weight of the man so that he can push the crate to the left. The coefficient of static friction between his shoes and the floor is μ ' s . Assume the man exerts only a horizontal force on the crate. 802 Given: © 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 8 W 1 300 lb= W 2 200 lb= μ s 0.2= μ ' s 0.35= a 4.5 ft= c 3ft= b 3.5 ft= d 4.5 ft= Solution: Σ F y = 0; N C W 1 − 0= N C W 1 = Σ F x = 0; μ s N C P− 0= P μ s N C = Σ Μ O = 0; W 1 − xPd+ 0= x Pd W 1 = Since x 0.90 ft= < a 4.50 ft= there will not be any tipping. Σ F x = 0; PF m − 0= F m P= F m 60.00 lb= F m μ ' s N m = N m F m μ ' s = N m 171.4 lb= Σ F y = 0; N m W 2 − 0= W 2 N m = W 2 171.4 lb= Problem 8-45 The wheel has weight W A and rests on a surface for which the coefficient of friction is μ B . A cord wrapped around the wheel is attached to the top of the homogeneous block of weight W C . 803 Given: © 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 8 If the coefficient of static friction at D is μ D determine the smallest vertical force that can be applied tangentially to the wheel which will cause motion to impend. Given: W A 20 lb= μ B 0.2= W C 30 lb= μ D 0.3= h 3ft= b 1.5 ft= Solution: Assume that slipping occurs at B, but that the block does not move. Guesses P 1lb= N B 1lb= F B 1lb= T 1lb= N D 1lb= F D 1lb= x 1ft= Given N B W A − P− 0= TF B − 0= PT− F B − () h 2 0= F B μ B N B = T− F D + 0= N D W C − 0= Th N D x− 0= P N B F B T N D F D x ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find PN B , F B , T, N D , F D , x, () = P N B F B T N D F D ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 13.33 33.33 6.67 6.67 30.00 6.67 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= x 0.67 ft= Now checke the assumptions F Dmax μ D N D = Since F D 6.67 lb= < F Dmax 9.00 lb= then the block does not slip 804 © 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 8 Since x 0.67 ft= < b 2 0.75 ft= then the block does not tip. So our original assumption is correct. P 13.33 lb= Problem 8-46 Determine the smallest couple moment which can be applied to the wheel of weight W 1 that will cause impending motion. The cord is attached to the block of weight W 2 , and the coefficients of static friction are μ B and μ D . Given: W 1 20 lb= a 1.5 ft= W 2 30 lb= b 3ft= μ B 0.2= c 1.5 ft= μ D 0.3= Solution: For the wheel : Assume slipping occurs, Σ F y = 0; N B W 1 − 0= N B W 1 = N B 20.00 lb= Σ F x = 0; T μ B N B − 0= T μ B N B = T 4.00 lb= Σ M B = 0; MT2a− 0= MT2a= M 12.00 lb ft⋅= For block Σ F y = 0; N D W 2 − 0= N D W 2 = N D 30.00 lb= Σ F x = 0; F D T− 0= F D T= F D 4.00 lb= Σ M O = 0; Tb N D x− 0= xT b N D = x 0.40 ft= F Dmax μ D N D = F Dmax 9.00 lb= Since F D 4.00 lb= < F Dmax 9.00 lb= then the block doesn't slip 805 © 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 8 Since x 0.40 ft= < c 2 0.75 ft= then the block doesn't tip. Thus neither slipping nor tipping occurs for the block, and our assumption and answer are correct. Problem 8-47 The beam AB has a negligible mass and thickness and is subjected to a triangular distributed loading. It is supported at one end by a pin and at the other end by a post having a mass m p and negligible thickness. Determine the minimum force P needed to move the post. The coefficients of static friction at B and C are μ B and μ C respectively. Given: m p 50 kg= a 2m= μ B 0.4= b 400 mm= μ C 0.2= c 300 mm= w 800 N m = d 3= g 9.81 m s 2 = e 4= Solution: Member AB: Σ M A = 0; 1 2 wa ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 2a 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N B a+ 0= N B 1 3 wa= N B 533.33 N= 806 © 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 8 Post: Assume slipping occurs at C: F C μ C N C = The initial guesses are P 1N= N C 1N= F B 1N= Given e− d 2 e 2 + Pc F B bc+()+ 0= e d 2 e 2 + PF B − μ C N C − 0= d e 2 d 2 + PN C + N B − m p g− 0= P N C F B ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find PN C , F B , () = P 354.79 N= Now check to see if the post slips at B. F Bmax μ B N B = Since F B 122 N= < F Bmax 213 N= then our assumptions are correct P 355 N= Problem 8-48 The beam AB has a negligible mass and thickness and is subjected to a triangular distributed loading. It is supported at one end by a pin and at the other end by a post having a mass m p and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to P the post slips at both B and C simultaneously. 807 © 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 8 Given: m p 50 kg= P 150 N= w 800 N m = a 2m= b 400 mm= c 300 mm= d 3= e 4= Solution: Member AB: 1 2 − wa 2a 3 N B a+ 0= N B 1 3 wa= N B 533.33 N= Post: Guesses N C 1N= μ B 0.2= μ C 0.2= Given N C N B − P d d 2 e 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + m p g− 0= e d 2 e 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ P μ C N C − μ B N B − 0= e− d 2 e 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Pc μ B N B bc+()+ 0= N C μ B μ C ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find N C μ B , μ C , () = μ B μ C ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0.0964 0.0734 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = 808 © 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 8 Problem 8-49 The block of weight W is being pulled up the inclined plane of slope α using a force P . If P acts at the angle φ as shown, show that for slipping to occur, P = W sin( α + θ )/ cos( φ − θ ) where θ is the angle of friction; θ = tan -1 μ Solution: Let μ tan θ () = Σ F x = 0; P cos φ () W sin α () − μ N− 0= Σ F y = 0; NWcos α () − P sin φ () + 0= P cos φ () W sin α () − μ W cos α () P sin φ () − 0= ( − PW sin α () μ cos α () + cos φ () μ sin φ () + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = W sin α () tan θ () cos α () + cos φ () tan θ () sin φ () + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = PW sin α () cos θ () sin θ () cos α () + cos φ () cos θ () sin θ () sin φ () + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = PW sin αθ + () cos φθ − () ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = QED() Problem 8-50 Determine the angle φ at which P should act on the block so that the magnitude of P is as small as possible to begin pulling the block up the incline. What is the corresponding value of P? The block has weight W and the slope α is known. 809 © 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 8 Solution: Let μ tan θ () = Σ F x = 0; P cos φ () W sin α () − μ N− 0= Σ F y = 0; NWcos α () − P sin φ () + 0= P cos φ () W sin α () − μ W cos α () P sin φ () − 0= ( − PW sin α () μ cos α () + cos φ () μ sin φ () + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = W sin α () tan θ () cos α () + cos φ () tan θ () sin φ () + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = PW sin α () cos θ () sin θ () cos α () + cos φ () cos θ () sin θ () sin φ () + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = W sin αθ + () cos φθ − () = dP d φ W sin αθ + () sin φθ − () cos 2 φθ − () ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = 0= sin αθ + () sin φθ − () 0= sin φθ − () 0= φθ = PWsin αφ + () = Problem 8-51 Two blocks A and B, each having a mass M, are connected by the linkage shown. If the coefficient of static friction at the contacting surfaces is μ s determine the largest vertical force P that may be applied to pin C of the linkage without causing the blocks to move. Neglect the weight of the links. 810 © 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 8 of the wedge and the thickness of the beam Units Used: 3 kN = 10 N Given: F 1 = 2 kN a = 3m F 2 = 4 kN b = 2m F 3 = 4 kN c = 3m F 4 = 2 kN θ = 15 deg μ s = 0.25 Solution: Guesses N1 = 1 kN N2 = 1 kN P = 1 kN Given (F1 − N1)( a + b + c) + F2( b + c) + F3 c = 0 N2 cos ( θ ) − μ s N2 sin ( θ ) − N1 = 0 μ s N1 + μ s N2 cos ( θ ) + N2 sin ( θ ) − P = 0 ⎛ N1 ⎞... publisher Engineering Mechanics - Statics Chapter 8 Given: mo = 45 kg μ A = 0.2 a = 30 0 mm b = 400 mm r = 12 5 mm Solution: Guesses M = 1Nm NA = 1 N Bx = 1 N By = 1 N Given M − Bx b − By a = 0 B x − μ A NA = 0 NA − mo g − By = 0 B y r − μ A NA r = 0 ⎛M⎞ ⎜N ⎟ ⎜ A ⎟ = Find ( M , N , B , B ) A x y ⎜ Bx ⎟ ⎜ ⎟ ⎝ By ⎠ ⎛ NA ⎞ ⎛ 5 51. 81 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Bx ⎟ = ⎜ 11 0 .36 ⎟ N ⎜ B ⎟ ⎝ 11 0 .36 ⎠ ⎝ y⎠ M = 77 .3 N⋅ m Problem 8-6 0.. .Engineering Mechanics - Statics Chapter 8 Given: M = 6 kg μ s = 0.5 θ 1 = 30 deg θ 2 = 30 deg Solution: Guesses P = 1N NA = 1 N FA = 1 N F AC = 1 N NB = 1 N FB = 1 N F BC = 1 N Assume that A slips first Given F AC cos ( θ 2 ) − FBC = 0 F AC sin ( θ 2 ) − P = 0 NA − M g − F AC sin ( θ 2 ) = 0 F A − FAC cos ( θ 2 ) = 0 F BC cos ( θ 1 ) − M g sin ( θ 1 ) − F B = 0 −F BC sin ( θ 1 ) − M g cos ( θ 1 )... is smooth The single square-threaded screw has a mean radius of r1 and a lead of r2, and the coefficient of static friction is μs Units Used: 3 kN = 10 N Given: F = 35 N a = 12 5 mm r1 = 6 mm r2 = 8 mm μ s = 0.27 Solution: φ = atan ( μ s) ⎛ r2 ⎞ ⎟ ⎝ 2 π r1 ⎠ θ = atan ⎜ φ = 15 .11 deg θ = 11 .98 deg F 2 a = P r1 tan ( θ + φ ) P = 2F a ⎞ ⎛ ⎜ r tan ( θ + φ ) ⎟ ⎝ 1 ⎠ P = 2.85 kN 836 © 2007 R C Hibbeler Published... without permission in writing from the publisher Engineering Mechanics - Statics Chapter 8 Units Used: 3 kN = 10 N Given: μ s = 0 .15 θ = 5 deg a = 600 mm b = 500 mm c = 250 mm d = 3 e = 4 F 1 = 8 kN F 2 = 15 kN Solution: Guesses P = 1N NA = 1 N NB = 1 N Given −P + μ s NB + μ s NA cos ( θ ) − NA sin ( θ ) = 0 NB − NA cos ( θ ) − μ s NA sin ( θ ) = 0 F1 b + e ⎛ ⎛ d ⎞ ⎞ ⎜ 2 2 ⎟ F2 a + ⎜ 2 2 ⎟ F2( b +... without permission in writing from the publisher Engineering Mechanics - Statics Chapter 8 Given: WA = 600 lb WB = 15 0 lb WC = 500 lb μ s = 0 .3 μ's = 0.5 θ = 45 deg Solution: Assume all blocks slip together N1 = WA + WB + WC P 1 = μ's N1 P 1 = 625.00 lb Assume that block A does not move and block B moves up Guesses NC = 1 lb N' = 1 lb N'' = 1 lb P = 1 lb Given N'' + μ's NC − P = 0 NC − WC − WB − μ... permission in writing from the publisher Engineering Mechanics - Statics Chapter 8 L = 18 ft W = 15 lb a = 3 ft b = 1 ft μ = 0.5 μ' = 0 .3 d = 10 ft Solution: Board: 1 2⎛ γ ⎞ ⎟ + N d = 0 N = L ⎜ ⎟ N = 48.60 lb 2 ⎝ d⎠ ⎝2⎠ −L γ ⎛ ⎜ L⎞ To cause slipping of the board on the saw horse: P xb = μ N P xb = 24 .30 lb To cause slipping at the ground: P xg = μ' ( N + W) P xg = 19 .08 lb To cause tipping ( N + W)b −... mm r = 12 5 mm Solution: Assume no motion Guesses B x = 1 N B y = 1 N NA = 1 N Given FA = 1 N M − By a − Bx b = 0 NA − By − mo g = 0 Bx − FA = 0 By r − FA r = 0 ⎛ Bx ⎞ ⎜ ⎟ ⎜ By ⎟ = Find ( B , B , N , F ) x y A A ⎜ NA ⎟ ⎜ ⎟ ⎝ FA ⎠ ⎛ NA ⎞ ⎛ 512 .88 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ FA ⎠ ⎝ 71. 43 ⎠ Check assumption: F Amax = μ A NA Since F A = 71. 4 N < F Amax = 10 2.6 N then our assumption is good F A = 71. 4 N Problem 8-6 1 A block... 0 .3 d = 14 ft Solution: Board: ⎛ L ⎞ + N d = 0 N = 1 L2 ⎛ γ ⎞ N = 34 . 71 lb ⎟ ⎜ ⎟ 2 ⎝2⎠ ⎝ d⎠ −L γ ⎜ To cause slipping of the board on the saw horse: P xb = μ N P xb = 17 .36 lb To cause slipping at the ground: P xg = μ' ( N + W) P xg = 14 . 91 lb To cause tipping ( N + W)b − P xt a = 0 P xt = ( N + W)b a P xt = 16 .57 lb Choose the critical case P x = min ( P xb , Pxg , P xt) P x = 14 . 91 lb Problem 8-5 9... ⎛ NA ⎞ ⎜ ⎟ ⎛ 526 ⎞ ⎜ ⎟ ⎜ NB ⎟ ⎜ 795 ⎟ ⎜ NC ⎟ ⎜ 692 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FA ⎟ = ⎜ 15 8 ⎟ N ⎜ F ⎟ ⎜ 15 8 ⎟ ⎜ B⎟ ⎜ ⎟ ⎜ FC ⎟ ⎜ 15 8 ⎟ ⎜ ⎟ ⎝ 11 74 ⎠ ⎝P ⎠ F Bmax = μ B NB ⎛ FBmax ⎞ ⎛ 19 9 ⎞ ⎜ ⎟=⎜ ⎟N FCmax ⎠ ⎝ 277 ⎠ ⎝ F Cmax = μ C NC Since F B = 15 8 N < FBmax = 19 9 N and FC = 15 8 N < F Cmax = 277 N then our assumption is correct P = 11 74 N Problem 8-5 5 The concrete pipe at A rests on top of B and C If the coefficient of . publisher. Engineering Mechanics - Statics Chapter 8 Given: m p 50 kg= P 15 0 N= w 800 N m = a 2m= b 400 mm= c 30 0 mm= d 3= e 4= Solution: Member AB: 1 2 − wa 2a 3 N B a+ 0= N B 1 3 wa= N B 533 .33 N= Post: Guesses. from the publisher. Engineering Mechanics - Statics Chapter 8 M 6kg= μ s 0.5= θ 1 30 deg= θ 2 30 deg= Solution: Guesses N A 1N= F A 1N= F AC 1N= P 1N= N B 1N= F B 1N= F BC 1N= Assume that A slips. publisher. Engineering Mechanics - Statics Chapter 8 Given: M 1 75 kg= M 2 10 0 kg= μ A 0 .3= μ B 0.25= μ C 0.4= θ 30 deg= r 15 0 mm= g 9. 81 m s 2 = Solution: Initial guesses: N A 10 0 N= F A 10 N= P 10 0