Engineering Mechanics - Statics Episode 1 Part 10 potx

Engineering Mechanics - Statics Chapter 5 Solution: When tipping occurs, R c = 0 Σ M D = 0; W 2 − fW 1 c+ W B bc+()+ 0= W B W 2 fW 1 c− bc+ = W B 78.6 lb= Problem 5-28 The articulated crane boom has a weight W and mass center at G. If it supports a load L, determine the force acting at the pin A and the compression in the hydraulic cylinder BC when the boom is in the position shown. Units Used: kip 10 3 lb= Given: W 125 lb= 361 © 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 L 600 lb= a 4ft= b 1ft= c 1ft= d 8ft= θ 40 deg= Solution: Guesses A x 1lb= A y 1lb= F B 1lb= Given A x − F B cos θ () + 0= A y − F B sin θ () + W− L− 0= F B cos θ () bF B sin θ () c+ Wa− Ld c+()− 0= A x A y F B ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find A x A y , F B , () = F B 4.19 kip= A x A y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3.208 1.967 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kip= Problem 5-29 The device is used to hold an elevator door open. If the spring has stiffness k and it is compressed a distnace δ , determine the horizontal and vertical components of reaction at the pin A and the resultant force at the wheel bearing B. Given: k 40 N m = b 125 mm= δ 0.2 m= c 100 mm= a 150 mm= θ 30 deg= 362 © 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: F s k δ = Σ M A = 0; F s − aF B cos θ () ab+()+ F B sin θ () c− 0= F B F s a cos θ () ab+( ) sin θ () c− = F B 6.378 N= + → Σ F x = 0; A x F B sin θ () − 0= A x F B sin θ () = A x 3.189 N= + ↑ Σ F y = 0; A y F s − F B cos θ () + 0= A y F s F B cos θ () −= A y 2.477 N= Problem 5-30 Determine the reactions on the bent rod which is supported by a smooth surface at B and by a collar at A, which is fixed to the rod and is free to slide over the fixed inclined rod. Given: F 100 lb= M 200 lb ft= a 3ft= b 3ft= c 2ft= 363 © 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 d 3= e 4= f 12= g 5= Solution: Initial Guesses: N A 20 lb= N B 10 lb= M A 30 lb ft= Given Σ M A = 0; M A Fa− M− N B f f 2 g 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ab+()+ N B g f 2 g 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ c− 0= Σ F x = 0; N A e e 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N B g f 2 g 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= Σ F y = 0; N A d e 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N B f f 2 g 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + F− 0= N A N B M A ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find N A N B , M A , () = N A N B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 39.7 82.5 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= M A 106lb ft⋅= Problem 5-31 The cantilevered jib crane is used to support the load F. If the trolley T can be placed anywhere in the range x 1 x≤ x 2 ≤ , determine the maximum magnitude of reaction at the supports A and B. Note that the supports are collars that allow the crane to rotate freely about the vertical axis. The collar at B supports a force in the vertical direction, whereas the one at A does not. Units Used: kip 1000 lb= Given: F 780 lb= 364 © 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 4ft= b 8ft= x 1 1.5 ft= x 2 7.5 ft= Solution: The maximum occurs when x = x 2 Σ M A = 0; F− x 2 B x a+ 0= B x F x 2 a = B x 1.462 10 3 × lb= + → Σ F x = 0; A x B x − 0= A x B x = A x 1.462 10 3 × lb= + ↑ Σ F y = 0; B y F− 0= B y F= B y 780lb= F B B x 2 B y 2 += F B 1.657 kip= Problem 5-32 The uniform rod AB has weight W. Determine the force in the cable when the rod is in the position shown. Given: W 15 lb= L 5ft= 365 © 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 1 30 deg= θ 2 10 deg= Solution: Σ M A = 0; N B Lsin θ 1 θ 2 + () W L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ cos θ 1 θ 2 + () − 0= N B W cos θ 1 θ 2 + () 2 sin θ 1 θ 2 + () = N B 8.938 lb= Σ F x = 0; Tcos θ 2 () N B − T N B cos θ 2 () = T 9.08 lb= Problem 5-33 The power pole supports the three lines, each line exerting a vertical force on the pole due to its weight as shown. Determine the reactions at the fixed support D. If it is possible for wind or ice to snap the lines, determine which line(s) when removed create(s) a condition for the greatest moment reaction at D. Units Used: kip 10 3 lb= 366 © 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 1 800 lb= W 2 450 lb= W 3 400 lb= a 2ft= b 4ft= c 3ft= Solution: + → Σ F x = 0; D x 0= + ↑ Σ F y = 0; D y W 1 W 2 + W 3 + () − 0= D y W 1 W 2 + W 3 += D y 1.65 kip= Σ M D = 0; W 2 − bW 3 c− W 1 a+ M D + 0= M D W 2 bW 3 c+ W 1 a−= M D 1.4 kip ft⋅= Examine all cases. For these numbers we require line 1 to snap. M Dmax W 2 bW 3 c+= M Dmax 3kip ft⋅= Problem 5-34 The picnic table has a weight W T and a center of gravity at G T . If a man weighing W M has a center of gravity at G M and sits down in the centered position shown, determine the vertical reaction at each of the two legs at B.Neglect the thickness of the legs. What can you conclude from the results? Given: W T 50 lb= 367 © 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 M 225 lb= a 6in= b 20 in= c 20 in= Solution: Σ M A = 0; 2 N B bc+()W M a+ W T b− 0= N B W T bW M a− 2 bc+() = N B 4.37− lb= Since N B has a negative sign, the table will tip over. Problem 5-35 If the wheelbarrow and its contents have a mass of M and center of mass at G, determine the magnitude of the resultant force which the man must exert on each of the two handles in order to hold the wheelbarrow in equilibrium. Given: M 60 kg= a 0.6 m= b 0.5 m= c 0.9 m= d 0.5 m= g 9.81 m s 2 = 368 © 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: Σ M B = 0; A y − bc+()Mgc+ 0= A y Mgc bc+ = A y 378.386 N= + → Σ F x = 0; B x 0N= B x 0= + ↑ A y Mg− 2 B y + 0= Σ F y = 0; B y Mg A y − 2 = B y 105.107 N= Problem 5-36 The man has weight W and stands at the center of the plank. If the planes at A and B are smooth, determine the tension in the cord in terms of W and θ . Solution: Σ M B = 0; W L 2 cos φ () N A Lcos φ () − 0= N A W 2 = Σ F x = 0; Tcos θ () N B sin θ () − 0= (1) 369 © 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 Σ F y = 0; Tsin θ ( ) N B cos θ () + N A + W− 0= (2) Solving Eqs. (1) and (2) yields: T W 2 sin θ () = N B W 2 cos θ () = Problem 5-37 When no force is applied to the brake pedal of the lightweight truck, the retainer spring AB keeps the pedal in contact with the smooth brake light switch at C. If the force on the switch is F , determine the unstretched length of the spring if the stiffness of the spring is k. Given: F 3N= k 80 N m = a 100 mm= b 50 mm= c 40 mm= d 10 mm= θ 30 deg= Solution: Σ M D = 0; F s bFcos θ () c− F sin θ () d− 0= F s F cos θ () c sin θ () d+ b = F s 2.378 N= F s kx= x F s k = x 29.73mm= L 0 ax−= L 0 70.3 mm= 370 © 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 Chapter 5 Given: W = 850 lb a = 0.5 ft b = 1 ft c = 1. 5 ft θ 1 = 10 deg θ 2 = 30 deg θ 3 = 10 deg Solution: Guesses F A = 1 lb F B = 1 lb F C = 1 lb Given ΣMB = 0; F A cos ( θ 3 ) b + W a − FC cos ( θ 1 ) ( a + c) = 0 ΣF x = 0; F C sin ( θ 1 ) − F B sin ( θ 2 ) − FA sin ( θ 3 ) = 0 ΣF y = 0; W − F A cos ( θ 3 ) − F B cos ( θ 2 ) − F C cos ( θ 1 ) = 0 ⎛ FA... Statics Chapter 5 Units Used: 3 kN = 10 N Given: F = 500 N a = 0 .15 m L = 3m Solution: The initial guesses w1 = 1 kN m w2 = 1 kN m Given + ↑Σ Fy = 0; ΣMA = 0; ⎛ w1 ⎞ ⎜ ⎟ = Find ( w1 , w2) ⎝ w2 ⎠ 1 1 w1 a − w2 a − F = 0 2 2 −F L − 1 2 1 ⎛ 2 a⎞ ⎟ + w2 a⎜ ⎟ = 0 ⎝ 3⎠ 2 ⎝ 3 ⎠ w1 a⎛ ⎜ a⎞ ⎛ w1 ⎞ ⎛ 413 ⎞ kN ⎜ ⎟=⎜ ⎟ ⎝ w2 ⎠ ⎝ 407 ⎠ m Problem 5-4 3 The upper portion of the crane boom consists of the jib AB, which... publisher Engineering Mechanics - Statics Chapter 5 Problem 5-4 5 The mobile crane has weight W1 and center of gravity at G1; the boom has weight W2 and center of gravity at G2 If the suspended load has weight W determine the normal reactions at the tracks A and B For the calculation, neglect the thickness of the tracks Units Used: 3 kip = 10 lb Given: W1 = 12 0000 lb a = 4 ft W2 = 30000 lb b = 6 ft W = 16 000... 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 yA = 1 mm yB = 1 mm Given c − yA a+b = yB − yA a ⎛ yA ⎞ ⎜ ⎟ ⎜ yB ⎟ = Find ( yA , yB , F) ⎜F⎟ ⎝ ⎠ k yA + k yB − F = 0 ⎛ yA ⎞ ⎛ −2 ⎞ ⎜ ⎟ = ⎜ ⎟ mm ⎝ yB ⎠ ⎝ 4 ⎠ k yB a − F( a + b) = 0 F = 10 mN Problem 5-5 6 The rigid metal strip of negligible weight is used as part of an... any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 5 maximum load Tmax before it fails, determine the critical loads if F1 = 2F2 Also, what is the magnitude of the maximum reaction at pin A? Units Used: 3 kN = 10 N Given: Tmax = 15 00 N a = 1. 5 m b = 1m c = 3 d = 4 θ = 30deg Solution: ΣMA = 0; F1 = 2 F2 −2 F 2 a cos ( θ ) − F2 ( a + b) cos ( θ ) + d 2 2... means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 5 ΣF y = 0; W tan ( θ ) sin ( θ ) + WL 4r − W cos ( θ ) = 0 sin ( θ ) − cos ( θ ) = 1 − 2 cos ( θ ) = 2 2 2 cos ( θ ) − 2 cos ( θ ) = L 4r L+ 2 −L 4r cos ( θ ) cos ( θ ) − 1 = 0 2 ⎛ L + L2 + 12 8 r2 ⎞ ⎟ 16 r ⎝ ⎠ 2 L + 12 8 r θ = acos ⎜ 16 r Problem 5-6 2 The disk has mass M and is supported on the smooth cylindrical... reproduced, in any form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 5 Solution: ΣF x = 0; Ax = 0 ΣF x = 0; Ay = −F1 Ay = −200 N ΣF x = 0; Az = F2 Az = 15 0 N ΣΜx = 0; MAx = −F2 a + F 1 d MAx = 10 0 N⋅ m ΣΜy = 0; MAy = 0 ΣMz = 0; MAz = F 1 c MAz = 500 N⋅ m Problem 5-6 4 The wing of the jet aircraft is subjected to thrust T from its engine and the resultant... in writing from the publisher Engineering Mechanics - Statics Chapter 5 Solution: φ = acos ⎛ ⎜ a − b⎞ ⎟ ⎝ a ⎠ φ = 33.56 deg ΣMB = 0, M g sin ( θ 1 ) ( a − b) + M g cos ( θ 1 ) a sin ( φ ) = 0 + P cos ( θ ) ( a − b) − P sin ( θ ) a sin ( φ ) ⎡sin ( θ 1 ) ( a − b) + cos ( θ 1 ) a sin ( φ )⎤ ⎥ ⎣ cos ( θ ) ( a − b) + sin ( θ ) a sin ( φ ) ⎦ P = M g⎢ P = 4 41 N Problem 5-6 0 Determine the magnitude and... components of reaction of the bolt B on the bracket Given: m1 = 15 kg c = 50 mm m2 = 4 kg d = 200 mm a = 60 mm e = 15 0 mm b = 40 mm g = 9. 81 m 2 s Solution: ΣMA = 0; B x a − m2 g d − m1 g( d + e) = 0 Bx = g + → Σ Fx = 0; m2 d + m1 ( d + e) B x = 989 N a Ax − Bx = 0 Ax = Bx + ↑Σ Fy = 0; Ax = 989 N B y − m2 g − m1 g = 0 B y = m2 g + m1 g B y = 18 6 N Problem 5-4 2 A cantilever beam, having an extended length L,... Engineering Mechanics - Statics Chapter 5 Problem 5-3 9 The worker uses the hand truck to move material down the ramp If the truck and its contents are held in the position shown and have weight W with center of gravity at G, determine the resultant normal force of both wheels on the ground A and the magnitude of the force required at the grip B Given: W = 10 0 lb e = 1. 5 ft a = 1 ft f = 0.5 ft b = 1. 5 . 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 Σ M B . publisher. Engineering Mechanics - Statics Chapter 5 Units Used: kN 10 3 N= Given: F 500 N= a 0 .15 m= L 3m= Solution: The initial guesses w 1 1 kN m = w 2 1 kN m = Given + ↑ Σ F y = 0; 1 2 w 1 a 1 2 w 2 a−. publisher. Engineering Mechanics - Statics Chapter 5 θ 1 30 deg= θ 2 10 deg= Solution: Σ M A = 0; N B Lsin θ 1 θ 2 + () W L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ cos θ 1 θ 2 + () − 0= N B W cos θ 1 θ 2 + () 2 sin θ 1 θ 2 + () = N B 8.938