Engineering Mechanics - Statics Chapter 5 Given: M 5Mg= d 1.50 m= a 0.3 m= e 0.6 m= b 0.75 m= θ 1 20 deg= c 1m= θ 2 30 deg= g 9.81 m s 2 = Solution: Guesses F 1kN= N A 1kN= N B 1kN= Given N A N B + F sin θ 2 () + Mgcos θ 1 () − 0= F− cos θ 2 () Mgsin θ 1 () + 0= F cos θ 2 () aFsin θ 2 () b− Mgcos θ 1 () c− Mgsin θ 1 () e− N B cd+()+ 0= F N A N B ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find FN A , N B , () = F N A N B ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 19.37 13.05 23.36 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= Problem 5-17 The uniform bar has mass M and center of mass at G. The supports A, B, and C are smooth. Determine the reactions at the points of contact at A, B, and C. Given: M 100 kg= a 1.75 m= b 1.25 m= 351 © 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 c 0.5 m= d 0.2 m= θ 30 deg= g 9.81 m s 2 = Solution: The initial guesses: N A 20 N= N B 30 N= N C 40 N= Given Σ M A = 0; M− g cos θ () aMgsin θ () d 2 − N B sin θ () d+ N C ab+()+ 0= + ↑ Σ F y = 0; N B Mg− N C cos θ () + 0= + ↑ N A N C sin θ () − 0= Σ F y = 0; N C N B N A ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find N C N B , N A , () = N C N B N A ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 493 554 247 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= Problem 5-18 The beam is pin-connected at A and rocker-supported at B. Determine the reactions at the pin A and at the roller at B. Given: F 500 N= M 800 N m⋅= a 8m= 352 © 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 b 4m= c 5m= Solution: α atan c ab+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Σ M A = 0; F− a cos α () M− B y a+ 0= B y Fa Mcos α () + cos α () a = B y 642 N= + → Σ F x = 0; A x − F sin α () + 0= A x F sin α () = A x 192 N= + ↑ Σ F y = 0; A y − F cos α () − B y + 0= A y F− cos α () B y += A y 180 N= Problem 5-19 Determine the magnitude of the reactions on the beam at A and B. Neglect the thickness of the beam. Given: F 1 600 N= F 2 400 N= θ 15 deg= a 4m= b 8m= c 3= d 4= Solution: Σ M A = 0; B y ab+()F 2 cos θ () ab+()− F 1 a− 0= 353 © 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 B y F 2 cos θ ( ) ab+()F 1 a+ ab+ = B y 586 N= + → Σ F x = 0; A x F 2 sin θ () − 0= A x F 2 sin θ () = A x 104 N= + ↑ Σ F y = 0; A y F 2 cos θ () − B y + F 1 − 0= A y F 2 cos θ () B y − F 1 += A y 400 N= F A A x 2 A y 2 += F A 413 N= Problem 5-20 Determine the reactions at the supports. Given: w 250 lb ft = a 6ft= b 6ft= c 6ft= Solution: Guesses A x 1lb= A y 1lb= B y 1lb= Given A x 0= A y B y + wa b+()− 1 2 wc− 0= wa a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ wb b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 1 2 wc b c 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − B y b+ 0= 354 © 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 A y B y ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find A x A y , B y , () = A x A y B y ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 2750 1000 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= Problem 5-21 When holding the stone of weight W in equilibrium, the humerus H, assumed to be smooth, exerts normal forces F C and F A on the radius C and ulna A as shown. Determine these forces and the force F B that the biceps B exerts on the radius for equilibrium. The stone has a center of mass at G. Neglect the weight of the arm. Given: W 5lb= θ 75 deg= a 2in= b 0.8 in= c 14 in= Solution: Σ M B = 0; W− ca−()F A a+ 0= F A W ca− a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F A 30lb= + ↑ Σ F y = 0; F B sin θ () W− F A − 0= F B WF A + sin θ () = F B 36.2 lb= + → Σ F x = 0; F C F B cos θ () − 0= 355 © 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 C F B cos θ ( ) = F C 9.378 lb= Problem 5-22 The uniform door has a weight W and a center of gravity at G. Determine the reactions at the hinges if the hinge at A supports only a horizontal reaction on the door, whereas the hinge at B exerts both horizontal and vertical reactions. Given: W 100 lb= a 3ft= b 3ft= c 0.5 ft= d 2ft= Solution: Σ M B = 0; Wd A x ab+()− 0= A x W d ab+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = A x 33.3 lb= Σ F x = 0; B x A x = B x 33.3 lb= Σ F y = 0; B y W= B y 100lb= Problem 5-23 The ramp of a ship has weight W and center of gravity at G. Determine the cable force in CD needed to just start lifting the ramp, (i.e., so the reaction at B becomes zero). Also, determine the horizontal and vertical components of force at the hinge (pin) at A. 356 © 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= a 4ft= θ 30 deg= b 3ft= φ 20 deg= c 6ft= Solution: Σ M A = 0; F CD − cos θ () bc+( ) cos φ () F CD sin θ () bc+( ) sin φ () + Wccos φ () + 0= F CD Wccos φ () bc+( ) cos θ () cos φ () sin θ () sin φ () − () = F CD 195lb= + → Σ F x = 0; F CD sin θ () A x − 0= A x F CD sin θ () = A x 97.5 lb= + ↑ Σ F y = 0; A y W− F CD cos θ () + 0= A y WF CD cos θ () −= A y 31.2 lb= Problem 5-24 The drainpipe of mass M is held in the tines of the fork lift. Determine the normal forces at A and B as functions of the blade angle θ and plot the results of force (ordinate) versus θ (abscissa) for 0 θ ≤ 90 deg≤ . 357 © 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: Mg 10 3 kg= Given: M 1.4 Mg= a 0.4 m= g 9.81 m s 2 = Solution: θ 090 = N A θ () Mgsin θ deg () 10 3 = N B θ () Mgcos θ deg () 10 3 = 0 20406080100 0 5 10 15 Angle in Degrees Force in kN N A θ () N B θ () θ Problem 5-25 While slowly walking, a man having a total mass M places all his weight on one foot. Assuming that the normal force N C of the ground acts on his foot at C, determine the resultant vertical compressive force F B which the tibia T exerts on the astragalus B, and the vertical tension F A in the achilles tendon A at the instant shown. 358 © 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: M 80 kg= a 15 mm= b 5mm= c 20 mm= d 100 mm= Solution: N C Mg= N C 785 N= Σ M A = 0; F B − cN C cd+()+ 0= F B N C cd+ c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F B 4.71 kN= Σ F y = 0; F A F B − N C + 0= F A F B N C −= F A 3.92 kN= Problem 5-26 Determine the reactions at the roller A and pin B. 359 © 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 800 lbft= c 3ft= F 390 lb= d 5= a 8ft= e 12= b 4ft= θ 30 deg= Solution: Guesses R A 1lb= B x 1lb= B y 1lb= Given R A sin θ () B x + d e 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F− 0= R A cos θ () B y + e e 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F− 0= MR A cos θ () ab+()− B x c+ 0= R A B x B y ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find R A B x , B y , () = R A 105.1 lb= B x B y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 97.4 269 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= Problem 5-27 The platform assembly has weight W 1 and center of gravity at G 1 . If it is intended to support a maximum load W 2 placed at point G 2, ,determine the smallest counterweight W that should be placed at B in order to prevent the platform from tipping over. Given: W 1 250 lb= a 1ft= c 1ft= e 6ft= W 2 400 lb= b 6ft= d 8ft= f 2ft= 360 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. [...]... reproduced, in any form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 5 Solution: ΣMB = 0; − Ay ( b + c) + M g c = 0 Ay = Mgc b+c Ay = 378.3 86 N + → Σ Fx = 0; B x = 0N + Ay − M g + 2 By = 0 ↑Σ Fy = 0; By = Bx = 0 M g − Ay 2 B y = 105.107 N Problem 5-3 6 The man has weight W and stands at the center of the plank If the planes at A and B are... 10 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 = 60 0 lb a = 4 ft b = 1 ft... permission in writing from the publisher Engineering Mechanics - Statics Solution: ΣMA = 0; Chapter 5 F s = kδ −F s a + F B cos ( θ ) ( a + b) − F B sin ( θ ) c = 0 FB = Fs a cos ( θ ) ( a + b) − sin ( θ ) c F B = 6. 378 N + → Σ Fx = 0; Ax − F B sin ( θ ) = 0 Ax = FB sin ( θ ) Ax = 3.189 N + ↑Σ Fy = 0; Ay − F s + F B cos ( θ ) = 0 Ay = Fs − FB cos ( θ ) Ay = 2.477 N Problem 5-3 0 Determine the reactions on the... reaction at D Units Used: 3 kip = 10 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: W1 = 800 lb... Given: WT = 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 WM = 225 lb a = 6 in b = 20 in c =.. .Engineering Mechanics - Statics Chapter 5 Solution: When tipping occurs, R c = 0 ΣMD = 0; −W2 f + W1 c + WB ( b + c) = 0 WB = W2 f − W1 c b+c WB = 78 .6 lb Problem 5-2 8 The articulated crane boom has a weight W and mass center at G If it supports a load L, determine the force acting... permission in writing from the publisher Engineering Mechanics - Statics Chapter 5 Units Used: 3 kN = 10 N Given: F = 5 kN r = 0.1 m a = r b = 1.5 m c = 5m Solution: From pulley, tension in the hoist line is ΣMB = 0; T( a) − F( r) = 0 T = F r T = 5 kN a From the jib, ΣMA = 0; b+a −F ( c) + TBC c=0 2 c + ( b + a) 2 c + ( b + a) TBC = F + ↑Σ Fy = 0; 2 2 TBC = 16. 4 06 kN b+a b+a ⎤−F=0 ⎢ 2 2⎥ ⎣ c + ( b + a)... ⎦ θ = acos ⎢ θ = 26. 4 deg 377 © 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 5-4 5 The mobile crane... 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 = 4 ft b = 8 ft... x = x2 ΣMA = 0; −F x2 + B x a = 0 Bx = F x2 a 3 B x = 1. 462 × 10 lb + → Σ Fx = 0; + ↑Σ Fy = 0; FB = Ax − Bx = 0 By − F = 0 2 Bx + By Ax = Bx By = F 2 3 Ax = 1. 462 × 10 lb B y = 780 lb F B = 1 .65 7 kip Problem 5-3 2 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 = 5 ft 365 © 2007 R C Hibbeler Published by Pearson Education, Inc., . the publisher. 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 5-2 8 The articulated. the publisher. Engineering Mechanics - Statics Chapter 5 b 4m= c 5m= Solution: α atan c ab+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Σ M A = 0; F− a cos α () M− B y a+ 0= B y Fa Mcos α () + cos α () a = B y 64 2 N= + → Σ F x. means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 5 B y F 2 cos θ ( ) ab+()F 1 a+ ab+ = B y 5 86 N= + → Σ F x = 0; A x F 2 sin θ () − 0= A x F 2 sin θ () =