Engineering Mechanics - Statics Chapter 3 pp p rings at C and D. Given: δ 1.5 ft= k 40 lb ft = d 2ft= F 60 lb= Solution: Initial guesses: F s 1lb= T 1lb= y 1ft= Given F y 2d T− 0= d 2 y 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 − d TF s − 0= F s kd d 2 y 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 −− δ + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = F s T y ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find F s T, y, () = F s T ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 76.92 97.55 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= y 2.46 ft= Problem 3-69 Cord AB of length a is attached to the end B of a spring having an unstretched length b. The other end of the spring is attached to a roller C so that the spring remains horizontal as it stretches. If a weight W is suspended from B, determine the angle θ of cord AB for equilibrium. 201 © 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 Given: a 5ft= b 5ft= k 10 lb ft = W 10 lb= Solution: Initial Guesses F BA 1lb= F sp 1lb= θ 30 deg= Given F sp F BA cos θ () − 0= F BA sin θ () W− 0= F sp ka acos θ () − () = F sp F BA θ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find F sp F BA , θ , () = F sp F BA ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 11.82 15.49 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= θ 40.22 deg= Problem 3-70 The uniform crate of mass M is suspended by using a cord of length l that is attached to the sides of the crate and passes over the small pulley at O. If the cord can be attached at either points A and B, or C and D, determine which attachment produces the least amount of tension in the cord and specify the cord tension in this case. 202 © 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 Given: M 50 kg= g 9.81 m s 2 = a 0.6 m= b 1.5 m= l 2m= c a 2 = d b 2 = Solution: Case 1 Attached at A and B Guess T 1N= Given Mg l 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 d 2 − l 2 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2T− 0= T 1 Find T()= T 1 370.78 N= Case 2 Attached at C and D Guess T 1N= Given Mg l 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 c 2 − l 2 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 2T− 0= T 2 Find T()= T 2 257.09 N= Choose the arrangement that gives the smallest tension. T min T 1 T 2 , () = T 257.09 N= Problem 3-71 The man attempts to pull the log at C by using the three ropes. Determines the direction θ in which he should p ull on his ro p e with a force P , so that he exerts a maximum 203 © 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 pp force on the log. What is the force on the log for this case ? Also, determine the direction in which he should pull in order to maximize the force in the rope attached to B.What is this maximum force? Given: P 80 lb= φ 150 deg= Solution: + → F AB P cos θ () + F AC sin φ 90 deg− () − 0= Σ F x = 0; + ↑ Σ F y = 0; P sin θ () F AC cos φ 90 deg− () − 0= F AC P sin θ () cos φ 90deg− () = In order to maximize FAC we choose sin θ () 1.= Thus θ 90 deg= F AC P sin θ () cos φ 90deg− () = F AC 160.00 lb= Now let's find the force in the rope AB. F AB P− cos θ () F AC sin φ 90 deg− () += F AB P− cos θ () P sin θ () sin φ 90 deg− () cos φ 90 deg− () += F AB P sin θ () sin φ 90 deg− () cos θ () cos φ 90 deg− () − cos φ 90 deg− () = P− cos θφ + 90deg− () cos φ 90 deg− () = 204 © 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 In order to maximize the force we set cos θφ + 90 deg− ( ) 1−= θφ + 90 deg− 180 deg= θ 270 deg φ −= θ 120.00 deg= F AB P− cos θφ + 90 deg− () cos φ 90 deg− () = F AB 160.00 lb= Problem 3-72 The "scale" consists of a known weight W which is suspended at A from a cord of total length L. Determine the weight w at B if A is at a distance y for equilibrium. Neglect the sizes and weights of the pulleys. Solution: + ↑ Σ F y = 0; 2W sin θ () w− 0= Geometry h Ly− 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 −= 1 2 Ly−() 2 d 2 −= 205 © 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 w 2W 1 2 Ly−() 2 d 2 + Ly− 2 ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ = w 2W Ly− Ly−() 2 d 2 −= Problem 3-73 Determine the maximum weight W that can be supported in the position shown if each cable AC and AB can support a maximum tension of F before it fails. Given: θ 30 deg= F 600 lb= c 12= d 5= Solution: Initial Guesses F AB F= F AC F= WF= Case 1 Assume that cable AC reaches maximum tension Given F AC sin θ () d c 2 d 2 + F AB − 0= F AC cos θ () c c 2 d 2 + F AB + W− 0= W 1 F AB1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find WF AB , () = W 1 F AB1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1239.62 780.00 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= Case 2 Assume that cable AB reaches maximum tension 206 © 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 Given F AC sin θ () d c 2 d 2 + F AB − 0= F AC cos θ () c c 2 d 2 + F AB + W− 0= W 2 F AC2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find WF AC , () = W 2 F AC2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 953.55 461.54 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= W min W 1 W 2 , () = W 953.6 lb= Problem 3-74 If the spring on rope OB has been stretched a distance δ . and fixed in place as shown, determine the tension developed in each of the other three ropes in order to hold the weight W in equilibrium. Rope OD lies in the x-y plane. Given: a 2ft= b 4ft= c 3ft= d 4ft= e 4ft= f 4ft= x B 2− ft= y B 3− ft= z B 3ft= θ 30 deg= k 20 lb in = δ 2in= W 225 lb= 207 © 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 Solution: Initial Guesses F OA 10 lb= F OC 10 lb= F OD 10 lb= Given c a 2 b 2 + c 2 + F OA k δ x B x B 2 y B 2 + z B 2 + + d− d 2 e 2 + f 2 + F OC + F OD sin θ () + 0= b− a 2 b 2 + c 2 + F OA k δ y B x B 2 y B 2 + z B 2 + + f d 2 e 2 + f 2 + F OC + F OD cos θ () + 0= a a 2 b 2 + c 2 + F OA k δ z B x B 2 y B 2 + z B 2 + + e d 2 e 2 + f 2 + F OC + W− 0= F OA F OC F OD ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find F OA F OC , F OD , () = F OA F OC F OD ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 201.6 215.7 58.6 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= Problem 3-75 The joint of a space frame is subjected to four member forces. Member OA lies in the x - y plane and member OB lies in the y - z plane. Determine the forces acting in each of the members required for equilibrium of the joint. Given: F 4 200 lb= θ 40 deg= φ 45 deg= Solution: The initial guesses : F 1 200 lb= F 2 200 lb= F 3 200 lb= 208 © 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 Given Σ F y = 0; F 3 F 1 cos φ () + F 2 cos θ () − 0= Σ F x = 0; F 1 − sin φ () 0= Σ F z = 0; F 2 sin θ () F 4 − 0= F 1 F 2 F 3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find F 1 F 2 , F 3 , () = F 1 F 2 F 3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 311.1 238.4 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= 209 © 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 4 Problem 4-1 If A , B , and D are given vectors, prove the distributive law for the vector cross product, i.e., ABD+ () × AB× () AD× () += . Solution: Consider the three vectors; with A vertical. Note triangle obd is perpendicular to A . od ABD+ () ×= A BD+ () sin θ 3 () = ob AB×= A B sin θ 1 () = bd AD×= A B sin θ 2 () = Also, these three cross products all lie in the plane obd since they are all perpendicular to A. As noted the magnitude of each cross product is proportional to the length of each side of the triangle. The three vector cross - products also form a closed triangle o'b'd' which is similar to triangle obd. Thus from the figure, ABD+ () × AB× AD×+= (QED) Note also , A A x i A y j+ A z k+= B B x i B y j+ B z K+= D D x i D y j+ D z k+= ABD+ () × i A x B x D x + j A y B y D y + k A z B z D z + ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = = A y B z D z + () A z B y D y + () − ⎡ ⎣ ⎤ ⎦ i A x B z D z + () A z B x D x + () − ⎡ ⎣ ⎤ ⎦ j− A x B y D y + () A y B x D x − () − ⎡ ⎣ ⎤ ⎦ k+ = A y B z A z B y − () i A x B z A z B x − () j− A x B y A y B x − () k+ ⎡ ⎣ ⎤ ⎦ A y D z A z D y − () i A x D z A z D x − () j− A x D y A y D x − () k+ ⎡ ⎣ ⎤ ⎦ + 210 © 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. [...]... without permission in writing from the publisher Engineering Mechanics - Statics Chapter 4 Units Used: kip = 10 00 lb Given: F = 275 lb a = 85 ft θ = 30 deg Solution: MA1 = F sin ( θ ) a MA1 = 11 .7 kip⋅ ft MA2 = F sin ( θ ) a MA2 = 11 .7 kip⋅ ft Also b = ( a)tan ( θ ) MA1 = F cos ( θ ) b MA1 = 11 .7 kip⋅ ft MA2 = F cos ( θ ) b MA2 = 11 .7 kip⋅ ft Problem 4-2 4 The force F acts on the end of the pipe at B... without permission in writing from the publisher Engineering Mechanics - Statics Chapter 4 Given: a = 3 ft b = 16 ft c = 15 ft θ 1 = 30 deg θ 2 = 70 deg W = 550 lb Solution: MA = W a MA = 1. 65 kip⋅ ft MB = W( a + b cos ( θ 1 ) ) MB = 9.27 kip⋅ ft MC = W( a + b cos ( θ 1 ) − c cos ( θ 2 ) ) MC = 6. 45 kip⋅ ft Problem 4 -1 8 Determine the direction θ ( 0° ≤ θ ≤ 18 0°) of the force F so that it produces (a) the... permission in writing from the publisher Engineering Mechanics - Statics Chapter 4 Given: F = 52 lb a = 8 in b = 5 in c = 6 in θ 1 = 30 deg θ 2 = 20 deg Solution: MA1 = F cos ( θ 1 ) ( a + c cos ( θ 1 ) ) − F sin ( θ 1 ) ( b − c sin ( θ 1 ) ) MA1 = 542 lb⋅ in MA2 = F cos ( θ 2 ) ( a − c cos ( θ 2 ) ) − F ( sin ( θ 2 ) ) ( b + c sin ( θ 2 ) ) MA2 = 10 . 01 lb⋅ in Problem 4-3 1 The worker is using the bar to pull... atan ⎛ ⎟ ⎜ φ b = 14 .04 deg θ b = 18 0 deg − φ b θ b = 16 6 deg ⎝ a⎠ ( b) ⎝ a⎠ Problem 4 -1 9 The rod on the power control mechanism for a business jet is subjected to force F Determine the moment of this force about the bearing at A Given: F = 80 N θ 1 = 20 deg a = 15 0 mm θ 2 = 60 deg Solution: MA = F cos ( θ 1 ) ( a) sin ( θ 2 ) − F sin ( θ 1 ) ( a) cos ( θ 2 ) MA = 7. 71 N⋅ m Problem 4-2 0 The boom has... + F sin ( θ ) a MO = −7 .11 N⋅ m MO = 7 .11 N⋅ m Vector Solution ⎛ b ⎞ ⎛ −F sin ( θ ) ⎞ ⎜ ⎟ ⎜ ⎟ MO = a × −F cos ( θ ) ⎜ ⎟ ⎜ ⎟ 0 ⎝0⎠ ⎝ ⎠ ⎛ 0 ⎞ ⎜ 0 ⎟ N⋅ m MO = ⎜ ⎟ ⎝ −7 .11 ⎠ MO = 7 .10 7 N⋅ m Problem 4 -1 1 Determine the magnitude and directional sense of the resultant moment of the forces about point O Units Used: 3 kip = 10 lb Given: F 1 = 300 lb e = 10 ft F 2 = 250 lb f = 4 a = 6 ft g = 3 b = 3 ft θ = 30... publisher Engineering Mechanics - Statics Chapter 4 Units Used: 3 kN = 10 N Given: M = 4.8 kN⋅ m a = 2 m F 1 = 300 N b = 3m F 2 = 400 N c = 4m θ 1 = 60 deg d = 3 θ 2 = 30 deg e = 4 Solution: Initial Guess F3 = 1 N Given −M = −F 1 cos ( θ 2 ) a − F 2 sin ( θ 1 ) ( a + b) + F 3 F 3 = Find ( F3 ) e ⎛ d ⎞c − F ⎛ ⎞ ( a + b) 3 ⎜ ⎜ 2 2⎟ 2 2⎟ ⎝ d +e ⎠ ⎝ d +e ⎠ F 3 = 1. 593 kN Problem 4-3 0 The flat-belt tensioner... 13 in c = 3 in d = 6 in e = 3 in f = 6 in Solution: MRO =ΣMO; MRO = F 1 cos ( θ 2 ) e − F1 sin ( θ 2 ) f − F2 cos ( θ 1 ) b − a − F2 sin ( θ 1 ) a 2 MRO = −858 lb⋅ in 2 MRO = 858 lb⋅ in Problem 4-5 Determine the magnitude and directional sense of the resultant moment of the forces at A and B about point P Units Used: kip = 10 00 lb Given: F 1 = 40 lb b = 13 in F 2 = 60 lb c = 3 in θ 1 = 30 deg d = 6. .. form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 4 MRP = 11 65 lb⋅ in MRP = 1. 17 kip⋅ in Problem 4 -6 Determine the magnitude of the force F that should be applied at the end of the lever such that this force creates a clockwise moment M about point O Given: M = 15 N m φ = 60 deg θ = 30 deg a = 50 mm b = 300 mm Solution: M = F cos ( θ ) ( a... Engineering Mechanics - Statics Chapter 4 Problem 4-3 6 Determine the moment of the force F at A about point O Express the result as a cartesian vector Units Used: 3 kN = 10 N Given: F = 13 kN a = 6m b = 2.5 m c = 3m d = 3m e = 8m f = 6m g = 4m h = 8m Solution: rAB ⎛b − g⎞ ⎜ ⎟ = c+d ⎜ ⎟ ⎝h − a⎠ MO = rOA × F1 rOA ⎛ −b ⎞ ⎜ ⎟ = −c ⎜ ⎟ ⎝a⎠ F1 = F rAB rAB ⎛ −84 ⎞ ⎜ ⎟ MO = −8 kN⋅ m ⎜ ⎟ ⎝ −39 ⎠ Problem 4-3 7... Mechanics - Statics F2 = 8 N e = 20 mm F3 = 6 N f = 35 mm a = 0.2 m g = 15 mm b = 0.35 m θ 1 = 30 deg c = 0.25 m Chapter 4 θ 2 = 15 deg Solution: Positive means clockwise MB1 = F 1 cos ( θ 2 ) ( a + b + c) − F 1 sin ( θ 2 ) e MB1 = 3.07 N⋅ m MB2 = F 2 ( a + b) MB2 = 4.4 N⋅ m MB3 = F 3 cos ( θ 1 ) a + F 3 sin ( θ 1 ) g MB3 = 1. 084 N⋅ m Problem 4 -1 4 Determine the moment of each force about the bolt located at . publisher. Engineering Mechanics - Statics Chapter 4 θ 1 30 deg= θ 2 45 deg= a 5in= b 13 in= c 3in= d 6in= e 3in= f 6in= Solution: M RO = Σ M O ; M RO F 1 cos θ 2 () eF 1 sin θ 2 () f− F 2 cos θ 1 () b 2 a 2 −−. without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 4 M RP 11 65 − lb in⋅= M RP 1. 17 kip in⋅= Problem 4 -6 Determine the magnitude of the force F that should. publisher. Engineering Mechanics - Statics Chapter 4 F 2 8N= e 20 mm= F 3 6N= f 35 mm= a 0.2 m= g 15 mm= b 0.35 m= θ 1 30 deg= c 0.25 m= θ 2 15 deg= Solution: Positive means clockwise M B1 F 1 cos θ 2 () ab+