Engineering Mechanics - Statics Chapter 4 = i A x B x j A y B y k A z B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ i A x D x j A y D y k A z D z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + = AB× () AD× () + (QED) Problem 4-2 Prove the triple scalar product identity AB C× () ⋅ AB× () C⋅= . Solution: As shown in the figure Area B Csin θ ()() = BC×= Thus, Volume of parallelopiped is BC× h But, h Au BC× ⋅= A BC× BC× ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= Thus, Volume AB C× () ⋅= Since AB× C⋅ represents this same volume then AB C× () ⋅ AB× () C⋅= (QED) Also, LHS AB C× () ⋅= = A x i A y j+ A z k+ () i B x C x j B y C y k B z C z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = A x B y C z B z C y − () A y B x C z B z C x − () − A z B x C y B y C x − () + = A x B y C z A x B z C y − A y B x C z − A y B z C x + A z B x C y + A z B y C x − RHS AB× () C⋅= 211 © 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 = i A x B x j A y B y k A z B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ C x i C y j+ C z k+ () = C x A y B z A z B y − () C y A x B z A z B x − () − C z A x B y A y B x − () + = A x B y C z A x B z C y − A y B x C z − A y B z C x + A z B x C y + A z B y C x − Thus, LHS RHS= AB⋅ C× AB× C⋅= (QED) Problem 4-3 Given the three nonzero vectors A , B , and C , show that if AB C× () ⋅ 0= , the three vectors must lie in the same plane. Solution: Consider, AB C× () ⋅ A BC× cos θ () = = A cos θ ()() BC× = h BC× = BC h sin φ () = volume of parallelepiped. If AB C× () ⋅ 0= , then the volume equals zero, so that A , B , and C are coplanar. Problem 4-4 Determine the magnitude and directional sense of the resultant moment of the forces at A and B about point O. Given: F 1 40 lb= F 2 60 lb= 212 © 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 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 −− F 2 sin θ 1 () a−= M RO 858− lb in⋅= M RO 858lb 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 1000 lb= Given: F 1 40 lb= b 13 in= F 2 60 lb= c 3in= θ 1 30 deg= d 6in= θ 2 45 deg= e 3in= a 5in= f 6in= Solution: M RP = Σ M P ; M RP F 1 cos θ 2 () ec+()F 1 sin θ 2 () df+()− F 2 cos θ 1 () b 2 a 2 − d+ () − F 2 − sin θ 1 () ac−()+ = 213 © 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 M RP 1165− lb in⋅= M RP 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 Nm= φ 60 deg= θ 30 deg= a 50 mm= b 300 mm= Solution: MFcos θ () absin φ () + () F sin θ () bcos φ ()() −= F M cos θ () absin φ () + () sin θ () bcos φ ()() − = F 77.6 N= Problem 4-7 Determine the angle θ (0 <= θ <= 90 deg) so that the force F develops a clockwise moment M about point O. Given: F 100 N= φ 60 deg= 214 © 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 M 20 N m⋅= a 50 mm= θ 30 deg= b 300 mm= Solution: I nitial Guess θ 30 deg= Given MFcos θ () absin φ () + () F sin θ () bcos φ ()() −= θ Find θ () = θ 28.6 deg= Problem 4-8 Determine the magnitude and directional sense of the moment of the forces about point O. Units Used: kN 10 3 N= Given: F B 260 N= e 2m= a 4m= f 12= b 3m= g 5= c 5m= θ 30 deg= d 2m= F A 400 N= Solution: M o F A sin θ () dF A cos θ () c+ F B f f 2 g 2 + ae+()+= M o 3.57 kN m⋅= (positive means counterclockwise) 215 © 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 4-9 Determine the magnitude and directional sense of the moment of the forces about point P. Units Used: kN 10 3 N= Given: F B 260 N= e 2m= a 4m= f 12= b 3m= g 5= c 5m= θ 30 deg= d 2m= F A 400 N= Solution: M p F B g f 2 g 2 + bF B f f 2 g 2 + e+ F A sin θ () ad−()− F A cos θ () bc+()+= M p 3.15 kN m⋅= (positive means counterclockwise) Problem 4-10 A force F is applied to the wrench. Determine the moment of this force about point O. Solve the problem using both a scalar analysis and a vector analysis. 216 © 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 Given: F 40 N= θ 20 deg= a 30 mm= b 200 mm= Scalar Solution M O F− cos θ () bFsin θ () a+= M O 7.11− Nm⋅= M O 7.11 N m⋅= Vector Solution M O b a 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F− sin θ () F− cos θ () 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ×= M O 0 0 7.11− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Nm⋅= M O 7.107 N m⋅= Problem 4-11 Determine the magnitude and directional sense of the resultant moment of the forces about point O. Units Used: kip 10 3 lb= Given: F 1 300 lb= e 10 ft= F 2 250 lb= f 4= a 6ft= g 3= b 3ft= θ 30 deg= c 4ft= φ 30 deg= d 4ft= 217 © 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 Solution: M o F 2 f f 2 g 2 + esin φ () F 2 g f 2 g 2 + ecos φ () + F 1 sin θ () a+ F 1 cos θ () b−= M o 2.42 kip ft⋅= positive means clockwise Problem 4-12 To correct a birth defect, the tibia of the leg is straightened using three wires that are attached through holes made in the bone and then to an external brace that is worn by the patient. Determine the moment of each wire force about joint A. Given: F 1 4N= d 0.15 m= 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 counterclockwise M A1 F 1 cos θ 2 () dF 1 sin θ 2 () e+= M A1 0.6 N m⋅= M A2 F 2 cd+()= M A2 3.2 N m⋅= M A3 F 3 cos θ 1 () bc+ d+()F 3 sin θ 1 () g−= M A3 3.852 N m⋅= Problem 4-13 To correct a birth defect, the tibia of the leg is straightened using three wires that are attached through holes made in the bone and then to an external brace that is worn by the patient. Determine the moment of each wire force about joint B. Given: F 1 4N= d 0.15 m= 218 © 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 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+ c+()F 1 sin θ 2 () e−= M B1 3.07 N m⋅= M B2 F 2 ab+()= M B2 4.4 N m⋅= M B3 F 3 cos θ 1 () aF 3 sin θ 1 () g+= M B3 1.084 N m⋅= Problem 4-14 Determine the moment of each force about the bolt located at A. Given: F B 40 lb= a 2.5 ft= α 20 deg= γ 30 deg= F C 50 lb= b 0.75 ft= β 25 deg= 219 © 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 Solution: M B F B cos β () a= M B 90.6 lb ft⋅= M C F C cos γ () ab+()= M C 141lb ft⋅= Problem 4-15 Determine the resultant moment about the bolt located at A. Given: F B 30 lb= F C 45 lb= a 2.5 ft= b 0.75 ft= α 20 deg= β 25 deg= γ 30 deg= Solution: M A F B cos β () aF C cos γ () ab+()+= M A 195lb ft⋅= Problem 4-16 The elbow joint is flexed using the biceps brachii muscle, which remains essentially vertical as the arm moves in the vertical plane. If this muscle is located a distance a from the pivot point A on the humerus, determine the variation of the moment capacity about A if the constant force developed by the muscle is F. Plot these results of M vs. θ for 60− θ ≤ 80≤ . 220 © 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. [...]... permission in writing from the publisher Engineering Mechanics - Statics Chapter 4 Problem 4- 3 4 Determine the moment of the force at A about point O Express the result as a Cartesian vector Given: ⎛ 60 ⎞ ⎜ ⎟ F = −30 N ⎜ ⎟ ⎝ −20 ⎠ a = 4m d = 4m b = 7m e = 6m c = 3m f = 2m Solution: rOA ⎛ −c ⎞ ⎜ ⎟ = −b ⎜ ⎟ ⎝a⎠ MO = rOA × F ⎛ 260 ⎞ ⎜ ⎟ MO = 180 N⋅ m ⎜ ⎟ ⎝ 510 ⎠ Problem 4- 3 5 Determine the moment of the force... 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 = 40 0 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... 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... publisher Engineering Mechanics - Statics Chapter 4 θ = 45 deg b = 2m Solution: rAC ⎛a⎞ ⎜ ⎟ = −c ⎜ ⎟ ⎝ −b ⎠ Fv = F rAC rBA rAC MB = rBA × F v ⎛ r cos ( θ ) ⎞ ⎜ ⎟ = r − r sin ( θ ) ⎜ ⎟ 0 ⎠ ⎝ ⎛ −37.6 ⎞ ⎜ 90.7 ⎟ N⋅ m MB = ⎜ ⎟ ⎝ −1 54. 9 ⎠ Problem 4- 4 0 The force F acts at the end of the beam Determine the moment of the force about point A Given: ⎛ 600 ⎞ ⎜ ⎟ F = 300 N ⎜ ⎟ ⎝ −600 ⎠ a = 1.2 m b = 0.2 m c = 0 .4 m... Engineering Mechanics - Statics Chapter 4 Problem 4- 4 1 The pole supports a traffic light of weight W Using Cartesian vectors, determine the moment of the weight of the traffic light about the base of the pole at A Given: W = 22 lb a = 12 ft θ = 30 deg Solution: ⎡ ( a)sin ( θ ) ⎤ ⎢ ⎥ r = ( a)cos ( θ ) ⎢ ⎥ ⎣ 0 ⎦ ⎛ 0 ⎞ ⎜ 0 ⎟ F = ⎜ ⎟ ⎝ −W ⎠ MA = r × F ⎛ −229 ⎞ ⎜ ⎟ MA = 132 lb⋅ ft ⎜ ⎟ ⎝ 0 ⎠ Problem 4- 4 2... publisher Engineering Mechanics - Statics Chapter 4 Given: r = 5 ft M = 80 lb⋅ ft θ = 60 deg a = 7 ft b = 6 ft Solution: rAB b ⎞ ⎛ ⎜ a − r sin ( θ ) ⎟ = ⎜ ⎟ ⎝ −r cos ( θ ) ⎠ Guess Given uAB = rAB rCB rAB ⎛b⎞ = ⎜a⎟ ⎜ ⎟ ⎝ −r ⎠ F = 1 lb rCB × ( F uAB) = M F = Find ( F) F = 18.6 lb Problem 4- 4 4 The pipe assembly is subjected to the force F Determine the moment of this force about point A 242 © 2007 R... without permission in writing from the publisher Engineering Mechanics - Statics Chapter 4 Given: F = 80 N a = 40 0 mm b = 300 mm c = 200 mm d = 250 mm θ = 40 deg φ = 30 deg Solution: rAC ⎛b + d⎞ ⎜ a ⎟ = ⎜ ⎟ ⎝ −c ⎠ MA = rAC × Fv ⎛ cos ( φ ) sin ( θ ) ⎞ ⎜ ⎟ F v = F ⎜ cos ( φ ) cos ( θ ) ⎟ ⎜ −sin ( φ ) ⎟ ⎝ ⎠ ⎛ −5.385 ⎞ ⎜ ⎟ MA = 13.093 N⋅ m ⎜ ⎟ ⎝ 11.377 ⎠ Problem 4- 4 5 The pipe assembly is subjected to the force... without permission in writing from the publisher Engineering Mechanics - Statics Chapter 4 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 = rPA × F 1 rPA ⎛ −b − f ⎞ ⎜ ⎟ = −c − e ⎜ ⎟ ⎝ a ⎠ F1 = F rAB rAB ⎛ −116 ⎞ ⎜ 16 ⎟ kN⋅ m MO = ⎜ ⎟ ⎝ −135 ⎠ Problem 4- 3 8 The curved rod lies in the x-y plane and has radius r If a force F acts at... 80 N a = 40 0 mm b = 300 mm c = 200 mm d = 250 mm θ = 40 deg φ = 30 deg Solution: rBC ⎛b + d⎞ = ⎜ 0 ⎟ ⎜ ⎟ ⎝ −c ⎠ ⎛ 550 ⎞ rBC = ⎜ 0 ⎟ mm ⎜ ⎟ ⎝ −200 ⎠ ⎛ cos ( φ ) sin ( θ ) ⎞ ⎜ ⎟ F v = F ⎜ cos ( φ ) cos ( θ ) ⎟ ⎜ −sin ( φ ) ⎟ ⎝ ⎠ MB = rBC × F v ⎛ 44 .5 34 ⎞ F v = ⎜ 53.073 ⎟ N ⎜ ⎟ ⎝ 40 ⎠ ⎛ 10.615 ⎞ MB = ⎜ 13.093 ⎟ N⋅ m ⎜ ⎟ ⎝ 29.19 ⎠ Problem 4- 4 6 The x-ray machine is used for medical diagnosis If the camera... publisher Engineering Mechanics - Statics Chapter 4 c = 1m Solution: rAB ⎛ 0 ⎞ ⎜ 0 ⎟ = ⎜ ⎟ ⎝a + b⎠ rA3 ⎛0⎞ ⎜ ⎟ = −c ⎜ ⎟ ⎝b⎠ The individual moments MA1 = rAB × F1 MA2 = rAB × F2 MA3 = rA3 × F3 ⎛ −3.6 ⎞ ⎜ ⎟ MA1 = 4. 8 kN⋅ m ⎜ ⎟ ⎝ 0 ⎠ ⎛ 1.2 ⎞ ⎜ ⎟ MA2 = 1.2 kN⋅ m ⎜ ⎟ ⎝ 0 ⎠ ⎛ 0.5 ⎞ ⎜ ⎟ MA3 = 0 kN⋅ m ⎜ ⎟ ⎝ 0 ⎠ The total moment MA = MA1 + MA2 + MA3 ⎛ −1.9 ⎞ ⎜ 6 ⎟ kN⋅ m MA = ⎜ ⎟ ⎝ 0 ⎠ Problem 4- 4 8 A force . writing from the publisher. Engineering Mechanics - Statics Chapter 4 Solution: M B F B cos β () a= M B 90.6 lb ft⋅= M C F C cos γ () ab+()= M C 141 lb ft⋅= Problem 4- 1 5 Determine the resultant. means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 4 M RP 1165− lb in⋅= M RP 1.17 kip in⋅= Problem 4- 6 Determine the magnitude of the force F that should. form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 4 Problem 4- 9 Determine the magnitude and directional sense of the moment of the