Engineering Mechanics - Statics Chapter 3 b 2ft= Guesses δ 1ft= φ 10 deg= T 1lb= F 1lb= x 1ft= Given T− cos θ () F cos φ () + 0= Tsin θ () F sin φ () + P− 0= Fkx δ − () = asin θ () xsin φ () = acos θ () xcos φ () + ab+= δ φ T F x ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find δφ , T, F, x, () = T F ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 69.28 40.00 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= δ 2.66 ft= Problem 3-33 The flowerpot of mass M is suspended from three wires and supported by the hooks at B and C. Determine the tension in AB and AC for equilibrium. Given: M 20 kg= l 1 3.5 m= l 2 2m= l 3 4m= 161 © 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 3 l 4 0.5 m= g 9.81 m s 2 = Solution: Initial guesses: T AB 1N= T AC 1N= θ 10 deg= φ 10 deg= Given T AC − cos φ () T AB cos θ () + 0= T AC sin φ () T AB sin θ () + Mg− 0= l 1 cos φ () l 2 cos θ () + l 3 = l 1 sin φ () l 2 sin θ () l 4 += T AB T AC θ φ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find T AB T AC , θ , φ , () = θ φ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 53.13 36.87 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ deg= T AB T AC ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 156.96 117.72 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= Problem 3-3 4 A car is to be towed using the rope arrangement shown. The towing force required is P. Determine the minimum length l of rope AB so that the tension in either rope AB or AC does not exceed T. Hint: Use the equilibrium condition at point A to determine the required angle θ for attachment, then determine l using trigonometry applied to triangle ABC. Given: P 600 lb= T 750 lb= φ 30 deg= 162 © 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 3 d 4ft= Solution: The initial guesses T AB T= T AC T= θ 30 deg= l 2ft= Case 1: Assume T AC T= Given + → Σ F x = 0; T AC cos φ () T AB cos θ () − 0= + ↑ PT AC sin φ () − T AB sin θ () − 0= Σ F y = 0; l sin φ () d sin 180deg θ − φ − () = T AB θ l 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find T AB θ , l, () = T AB 687.39 lb= θ 19.11 deg= l 1 2.65 ft= Case 2: Assume T AB T= Given + → Σ F x = 0; T AC cos φ () T AB cos θ () − 0= + ↑ PT AC sin φ () − T AB sin θ () − 0= Σ F y = 0; l sin φ () d sin 180deg θ − φ − () = T AC θ l 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find T AC θ , l, () = T AC 840.83 lb= θ 13.85 deg= l 2 2.89 ft= l min l 1 l 2 , () = l 2.65ft= 163 © 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 3 Problem 3-35 Determine the mass of each of the two cylinders if they cause a sag of distance d when suspended from the rings at A and B. Note that s = 0 when the cylinders are removed. Given: d 0.5 m= l 1 1.5 m= l 2 2m= l 3 1m= k 100 N m = g 9.81 m s 2 = Solution: T AC kl 1 d+ () 2 l 2 2 + l 1 2 l 2 2 +− ⎡ ⎣ ⎤ ⎦ = T AC 32.84 N= θ atan l 1 d+ l 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 45deg= M T AC sin θ () g = M 2.37 kg= 164 © 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 3 The sling BAC is used to lift the load W with constant velocity. Determine the force in the sling and plot its value T (ordinate) as a function of its orientation θ , where 0 θ≤ 90°≤ . Solution: W 2Tcos θ () − 0= T 1 2 W cos θ () ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Problem 3-37 The lamp fixture has weight W and is suspended from two springs, each having unstretched length L and stiffness k. Determine the angle θ for equilibrium. Units Used: kN 10 3 N= Given: W 10 lb= L 4ft= k 5 lb ft = a 4ft= 165 Problem 3-36 © 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 3 The initial guesses: T 200 lb= θ 10 deg= Given Spring Tk a cos θ () L− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = + ↑ Σ F y = 0; 2Tsin θ () W− 0= T θ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find T θ , () = T 7.34lb= θ 42.97 deg= Problem 3-38 The uniform tank of weight W is suspended by means of a cable, of length l, which is attached to the sides of the tank and passes over the small pulley located at O. If the cable can be attached at either points A and B, or C and D, determine which attachment produces the least amount of tension in the cable.What is this tension? Given: W 200 lb= l 6ft= a 1ft= b 2ft= cb= d 2a= Solution: Free Body Diagram: By observation, the force F has to support the entire weight of the tank. Thus, F = W. The tension in 166 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 3 cable is the same throughout the cable. Equations of Equilibrium Σ F y = 0; W 2Tsin θ () − 0= Attached to CD θ 1 acos 2 a l ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 1 70.53 deg= Attached to AB θ 2 acos 2 b l ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 2 48.19 deg= We choose the largest angle (which will produce the smallest force) θ max θ 1 θ 2 , () = θ 70.53 deg= T 1 2 W sin θ () ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = T 106lb= Problem 3-39 A sphere of mass m s rests on the smooth parabolic surface. Determine the normal force it exerts on the surface and the mass m B of block B needed to hold it in the equilibrium position shown. Given: m s 4kg= a 0.4 m= b 0.4 m= θ 60 deg= g 9.81 m s 2 = Solution: k a b 2 = Geometry: The angle θ 1 which the surface make with the horizontal is to be determined first. 167 © 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 3 tan θ 1 () dy dx = 2 kx= evaluated at xa= θ 1 atan 2 ka()= θ 1 63.43 deg= Free Body Diagram : The tension in the cord is the same throughout the cord and is equal to the weight of block B, m B g. The initial guesses: m B 200 kg= F N 200 N= Given + → Σ F x = 0; m B gcos θ () F N sin θ 1 () − 0= + ↑ Σ F y = 0; m B gsin θ () F N cos θ 1 () + m s g− 0= m B F N ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find m B F N , () = F N 19.66 N= m B 3.58 kg= Problem 3-40 The pipe of mass M is supported at A by a system of five cords. Determine the force in each cord for equilibrium. Given: M 30 kg= c 3= d 4= g 9.81 m s 2 = θ 60 deg= Solution: Initial guesses: T AB 1N= T AE 1N= T BC 1N= T BD 1N= Given T AB sin θ () Mg− 0= T AE T AB cos θ () − 0= 168 © 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 3 T BD c c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ T AB sin θ () − 0= T BD d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ T AB cos θ () + T BC − 0= T AB T AE T BC T BD ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find T AB T AE , T BC , T BD , () = T AB T AE T BC T BD ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ 339.8 169.9 562.3 490.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ N= Problem 3-41 The joint of a space frame is subjected to four forces. Strut OA lies in the x-y plane and strut OB lies in the y-z plane. Determine the forces acting in each of the three struts required for equilibrium. Units Used: kN 10 3 N= Given: F 2kN= θ 1 45 deg= θ 2 40 deg= Solution: Σ F x = 0; R− sin θ 1 () 0= R 0= 169 © 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 3 Σ F z = 0; P sin θ 2 () F− 0= P F sin θ 2 () = P 3.11 kN= Σ F y = 0; QPcos θ 2 () − 0= QPcos θ 2 () = Q 2.38kN= Problem 3-42 Determine the magnitudes of F 1 , F 2 , and F 3 for equilibrium of the particle. Units Used: kN 10 3 N= Given: F 4 800 N= α 60 deg= β 30 deg= γ 30 deg= c 3= d 4= Solution: The initial guesses: F 1 100 N= F 2 100 N= F 3 100 N= Given F 1 cos α () 0 sin α () ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 c 2 d 2 + c d− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ + F 3 cos γ () − sin γ () − 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ + F 4 0 sin β () cos β () − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ + 0= 170 © 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 ⎛ F1 ⎞ ⎜ ⎟ ⎜ F2 ⎟ = Find ( F1 , F2 , F3) ⎜F ⎟ ⎝ 3⎠ Chapter 3 ⎛ F1 ⎞ ⎛ 5. 60 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ F2 ⎟ = ⎜ 8 .55 ⎟ kN ⎜ F ⎟ ⎝ 9.44 ⎠ ⎝ 3⎠ Problem 3-4 4 Determine the magnitudes of F1, F 2 and F3 for equilibrium of the particle F = {- 9i - 8j - 5k} Units Used: 3 kN = 10 N Given: ⎛ −9 ⎞ F = ⎜ −8 ⎟ kN ⎜ ⎟ ⎝ 5 ⎠ a = 4m b = 2m c = 4m θ 1 = 30 deg θ 2 = 60 deg θ 3 = 13 5 deg θ.. .Engineering Mechanics - Statics ⎛ F1 ⎞ ⎜ ⎟ ⎜ F2 ⎟ = Find ( F1 , F2 , F3) ⎜F ⎟ ⎝ 3⎠ Chapter 3 ⎛ F1 ⎞ ⎛ 800 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ F2 ⎟ = ⎜ 14 7 ⎟ N ⎜ F ⎟ ⎝ 56 4 ⎠ ⎝ 3⎠ Problem 3-4 3 Determine the magnitudes of F1, F 2, and F3 for equilibrium of the particle Units Used: kN = 10 00 N Given: F 4 = 8 .5 kN F 5 = 2.8 kN α = 15 deg β = 30 deg c = 7 d = 24 Solution: Initial Guesses: F 1 = 1 kN F 2 = 1 kN F 3 = 1 kN Given... Engineering Mechanics - Statics Chapter 3 3 kN = 10 N Given: T = 15 kN e = 4m a = 2m f = 6m b = 10 m g = 6m c = 12 m h = 6m d = 2m i = 2m gravity = 9. 81 m 2 s Solution: The initial guesses: FB = T FC = T FD = T M = 1 kg Case 1: Assume that cable B reaches maximum tension Given Σ F x = 0; F B( f − i ) 2 2 + 2 2 ( f − i) + h + c Σ F y = 0; F B ( − h) 2 2 F B( −c) 2 2 ( f − i) + h + c ⎛ M1 ⎞ ⎜ ⎟ ⎜ FC1... ⎜ ⎝a⎠ ⎝ −a ⎠ ⎝ −a ⎠ ⎝ 1 ⎠ F AO ⎜ ⎛ W3 ⎞ ⎜ ⎟ ⎜ FAO3 ⎟ = Find ( W , FAO , FAB) ⎜F ⎟ ⎝ AB3 ⎠ Final Answer ⎛ W3 ⎞ ⎛ 857 .14 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAO3 ⎟ = ⎜ 18 57 .14 ⎟ N ⎜F ⎟ ⎝ AB3 ⎠ ⎝ 642.86 ⎠ W = min ( W1 , W2 , W3 ) W = 13 8.46 N Problem 3 -5 2 Determine the tension in cables AB, AC, and AD, required to hold the crate of weight W in equilibrium Given: W = 60 lb a = 6 ft b = 12 ft c = 8 ft 18 1 © 2007 R C Hibbeler Published... publisher Engineering Mechanics - Statics Chapter 3 Given ⎛ cos ( θ 2 ) cos ( θ 1) ⎞ ⎛ cos ( θ 3 ) ⎞ ⎜ ⎟ ⎜ ⎟ F 1 ⎜ −cos ( θ 2 ) sin ( θ 1 ) ⎟ + F 2 ⎜ cos ( θ 5 ) ⎟ + ⎜ ⎟ ⎜ cos θ ⎟ sin ( θ 2 ) ⎝ ⎠ ⎝ ( 4) ⎠ ⎛ F1 ⎞ ⎜ ⎟ ⎜ F2 ⎟ = Find ( F1 , F2 , F3) ⎜F ⎟ ⎝ 3⎠ ⎛a⎞ ⎜ c ⎟+F=0 ⎟ 2 2 2⎜ a + b + c ⎝ −b ⎠ F3 ⎛ F1 ⎞ ⎛ 8.26 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ F2 ⎟ = ⎜ 3.84 ⎟ kN ⎜ F ⎟ ⎝ 12 . 21 ⎠ ⎝ 3⎠ Problem 3-4 5 The three cables are used to... publisher Engineering Mechanics - Statics Chapter 3 ⎛ W3 ⎞ ⎜ ⎟ ⎜ FAB3 ⎟ = Find ( W , FAB , FAC) ⎜F ⎟ ⎝ AC3 ⎠ W = min ( W1 , W2 , W3 ) ⎛ W3 ⎞ ⎛ 378.2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAB3 ⎟ = ⎜ 16 9.7 ⎟ lb ⎜F ⎟ ⎝ AC3 ⎠ ⎝ 12 0 ⎠ W = 267.42 lb Problem 3-6 0 Determine the force in each cable used to lift the surge arrester of mass M at constant velocity Units Used: 3 kN = 10 N 3 Mg = 10 kg Given: M = 9 .50 Mg a = 2m b = 0 .5 m θ = 45 deg... publisher Engineering Mechanics - Statics Chapter 3 ⎛ FB ⎞ ⎜ ⎟ ⎜ FC ⎟ = Find ( FB , FC , FD) ⎜F ⎟ ⎝ D⎠ ⎛ FB ⎞ ⎛ 28 .13 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FC ⎟ = ⎜ 39.78 ⎟ kN ⎜ F ⎟ ⎝ 28 .13 ⎠ ⎝ D⎠ Problem 3-6 1 The cylinder of weight W is supported by three chains as shown Determine the force in each chain for equilibrium Given: W = 800 lb r = 1 ft d = 1 ft Solution: The initial guesses: F AB = 1 lb F AC = 1 lb F AD = 1 lb Given... Engineering Mechanics - Statics Chapter 3 Problem 3 -5 0 The lamp has mass ml and is supported by pole AO and cables AB and AC If the force in the pole acts along its axis, determine the forces in AO, AB, and AC for equilibrium Given: ml = 15 kg d = 1. 5 m a = 6m e = 4m b = 1. 5 m f = 1. 5 m c = 2m g = 9. 81 m 2 s Solution: The initial guesses: F AO = 10 0 N F AB = 200 N F AC = 300 N Given Equilibrium equations:... Pmax F AB = Tmax F AC = Tmax W = 300N Case 1 Assume the pole reaches maximum compression Given ⎛ c ⎞ F ⎛ −c − e ⎞ F ⎛ −c ⎞ ⎛0⎞ ⎟ + AB ⎜ b + f ⎟ + AC ⎜ b + d ⎟ + W⎜ 0 ⎟ = 0 −b ⎟ AC ⎜ ⎟ ⎜ ⎟ AO ⎜ ⎟ AB ⎜ ⎝a⎠ ⎝ −a ⎠ ⎝ −a ⎠ ⎝ 1 ⎠ F AO ⎜ ⎛ W1 ⎞ ⎜ ⎟ ⎜ FAB1 ⎟ = Find ( W , FAB , FAC) ⎜F ⎟ ⎝ AC1 ⎠ ⎛ W1 ⎞ ⎛ 13 8.46 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FAB1 ⎟ = ⎜ 10 3. 85 ⎟ N ⎜F ⎟ ⎝ AC1 ⎠ ⎝ 80.77 ⎠ 18 0 © 2007 R C Hibbeler Published by Pearson... Statics Chapter 3 d = 7 ft e = 7 ft θ 1 = 20 deg θ 2 = 35 deg Solution: At A: Initial guesses: F AC = 57 0 lb F AB = 50 0 lb Given + → ΣFx = 0; F AB cos ( θ 1 ) − F AC cos ( θ 2 ) = 0 + F AB sin ( θ 1 ) + FAC sin ( θ 2 ) − W = 0 ↑ ΣFy = 0; ⎛ FAC ⎞ ⎜ ⎟ = Find ( FAC , FAB) ⎝ FAB ⎠ At C: ⎛ FAC ⎞ ⎛ 57 4 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ FAB ⎠ ⎝ 50 0 ⎠ Initial Guesses F CD = 1 lb F CE = 1 lb F CF = 1 lb Given ⎛ cos ( θ 2 ) ⎞ ⎜ ⎟ F . removed. Given: d 0 .5 m= l 1 1 .5 m= l 2 2m= l 3 1m= k 10 0 N m = g 9. 81 m s 2 = Solution: T AC kl 1 d+ () 2 l 2 2 + l 1 2 l 2 2 +− ⎡ ⎣ ⎤ ⎦ = T AC 32.84 N= θ atan l 1 d+ l 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 45deg= M T AC sin θ () g = M. F 1 , F 2 , and F 3 for equilibrium of the particle. Units Used: kN 10 00 N= Given: F 4 8 .5 kN= F 5 2.8 kN= α 15 deg= β 30 deg= c 7= d 24= Solution: Initial Guesses: F 1 1kN= F 2 1kN= F 3 1kN= Given F 1 cos β () − 0 sin β () ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 c 2 d 2 + c− d− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ +. publisher. Engineering Mechanics - Statics Chapter 3 F 1 F 2 F 3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find F 1 F 2 , F 3 , () = F 1 F 2 F 3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 5. 60 8 .55 9.44 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= Problem 3-4 4 Determine