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Mechanics Engineering - 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., A ( B + D) = ( A B) + ( A D). Solution: Consider the three vectors; with A vertical. Note triangle obd is perpendicular to A. od = A ( B + D) = A ( B + D ) sin ( 3 ) ob = A B = A B sin ( 1 ) bd = A D = A B sin ( 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, A ( B + D) = A B + A D Note also, A = Axi + Ayj + A zk B = Bxi + B yj + B zK D = Dxi + Dyj + Dzk j k i Ay Az A ( B + D) = Ax B + D B + D B + D x y y z z x = (QED) Ay( Bz + Dz) - Az( By + Dy) i - Ax( Bz + Dz) - Az( Bx + Dx) j + Ax( By + Dy) - Ay( Bx - Dx) k ( Ay Bz - Az By) i - ( Ax Bz - Az Bx) j + ( Ax By - Ay Bx) k ... + ( Ay Dz - A z Dy) i - ( Ax Dz - A z Dx) j + ( A x Dy - A y Dx) 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. Engineering Mechanics - Statics Chapter 4 i j k i j k = Ax Ay Az + Ax Ay Az B B B D D D x y z x y z = ( A B) + ( A D) (QED) Problem 4-2 Prove the triple scalar product identity A ( B C) = ( A B) C. Solution: As shown in the figure Area = B ( C sin ( ) ) = B C Thus, Volume of parallelopiped is B C h But, h = A u Thus, Volume = A ( B C) Since A B C represents this same volume then A ( B C) = ( A B) C Also, LHS = A ( B C) (QED) B C = A B C B C i j k = ( A xi + A yj + Azk) Bx By Bz C C C x y z = Ax( By Cz - B z Cy) - Ay( Bx Cz - B z Cx) + Az( B x Cy - B y Cx) = Ax B y Cz - A x Bz Cy - Ay B x Cz + A y Bz Cx + Az Bx Cy - A z B y Cx RHS = ( A B) 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 j k A x Ay A z ( Cxi + Cyj + Czk) B B B x y z = Cx( Ay B z - Az By) - Cy( A x Bz - A z B x) + Cz( A x By - Ay B x) = Ax B y Cz - A x Bz Cy - Ay B x Cz + A y Bz Cx + Az Bx Cy - A z B y Cx LHS = RHS (QED) Thus, A B C = A B C Problem 4-3 Given the three nonzero vectors A, B, and C, show that if A ( B C) = 0, the three vectors must lie in the same plane. Solution: Consider, A ( B C) = A B C cos ( ) = = ( A cos ( ) ) B C h B C = BC h sin ( ) = volume of parallelepiped. If A ( B 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 1 = 30 deg 2 = 45 deg a = 5 in b = 13 in c = 3 in d = 6 in e = 3 in f = 6 in Solution: MRO =MO; MRO = F 1 cos ( 2 ) e - F1 sin ( 2 ) f - F2 cos ( 1 ) b - a - F2 sin ( 1 ) a 2 2 MRO = -858 lb in 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 = 1000 lb Given: F 1 = 40 lb F 2 = 60 lb b = 13 in c = 3 in d = 6 in e = 3 in f = 6 in 1 = 30 deg 2 = 45 deg a = 5 in Solution: MRP = MP; MRP = F1 cos ( 2 ) ( e + c) - F 1 sin ( 2 ) ( d + f) - F2 cos ( 1 ) + -F2 sin ( 1 ) ( a - c) 213 ( b - a + d ... 2 2 ) 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 MRP = -1165 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 = 60 deg = 30 deg a = 50 mm b = 300 mm Solution: M = F cos ( ) ( a + b sin ( ) ) - F sin ( ) ( b cos ( ) ) M F = cos ( ) ( a + b sin ( ) ) - sin ( ) ( b cos ( ) ) F = 77.6 N Problem 4-7 Determine the angle (0 <= <= 90 deg) so that the force F develops a clockwise moment M about point O. Given: F = 100 N = 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 = 30 deg Solution: Initial Guess b = 300 mm = 30 deg Given M = F cos ( ) ( a + b sin ( ) ) - F sin ( ) ( b cos ( ) ) = Find ( ) = 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 N Given: F B = 260 N a = 4m b = 3m c = 5m d = 2m Solution: Mo = F A sin ( ) d + FA cos ( ) c + F B f f +g Mo = 3.57 kN m 2 2 3 e = 2m f = 12 g = 5 = 30 deg F A = 400 N ( a + e) (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 N Given: F B = 260 N a = 4m b = 3m c = 5m d = 2m F A = 400 N Solution: Mp = F B g f +g 2 2 3 e = 2m f = 12 g = 5 = 30 deg b + FB f f +g 2 2 e - FA sin ( ) ( a - d) + FA cos ( ) ( b + c) Mp = 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 = 20 deg a = 30 mm b = 200 mm Scalar Solution MO = -F cos ( ) b + F sin ( ) a MO = -7.11 N m Vector Solution MO = 7.11 N m b -F sin ( ) MO = a -F cos ( ) 0 0 Problem 4-11 0 0 N m MO = -7.11 MO = 7.107 N m Determine the magnitude and directional sense of the resultant moment of the forces about point O. Units Used: kip = 10 lb Given: F 1 = 300 lb F 2 = 250 lb a = 6 ft b = 3 ft c = 4 ft d = 4 ft e = 10 ft f = 4 g = 3 3 = 30 deg = 30 deg 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: Mo = F 2 f f +g Mo = 2.42 kip ft 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: F1 = 4 N F2 = 8 N F3 = 6 N a = 0.2 m b = 0.35 m c = 0.25 m Solution: d = 0.15 m e = 20 mm f = 35 mm g = 15 mm 2 2 e sin ( ) + F2 g f +g 2 2 e cos ( ) + F1 sin ( ) a - F1 cos ( ) b positive means clockwise 1 = 30 deg 2 = 15 deg Positive means counterclockwise MA1 = F 1 cos ( 2 ) d + F1 sin ( 2 ) e MA2 = F 2 ( c + d) MA1 = 0.6 N m MA2 = 3.2 N m MA3 = F 3 cos ( 1 ) ( b + c + d) - F 3 sin ( 1 ) g MA3 = 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: F1 = 4 N 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 F2 = 8 N F3 = 6 N a = 0.2 m b = 0.35 m c = 0.25 m Solution: e = 20 mm f = 35 mm g = 15 mm 1 = 30 deg 2 = 15 deg Positive means clockwise MB1 = F 1 cos ( 2 ) ( a + b + c) - F 1 sin ( 2 ) e MB1 = 3.07 N m MB2 = F 2 ( a + b) MB3 = F 3 cos ( 1 ) a + F 3 sin ( 1 ) g MB2 = 4.4 N m MB3 = 1.084 N m Problem 4-14 Determine the moment of each force about the bolt located at A. Given: F B = 40 lb F C = 50 lb a = 2.5 ft b = 0.75 ft = 20 deg = 25 deg = 30 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: MB = FB cos ( ) a MB = 90.6 lb ft MC = F C cos ( ) ( a + b) MC = 141 lb 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 = 20 deg = 25 deg = 30 deg Solution: MA = FB cos ( ) a + FC cos ( ) ( a + b) MA = 195 lb 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. for -60 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. Engineering Mechanics - Statics Chapter 4 Units Used: kN = 10 N 3 Given: a = 16 mm F = 2.30 kN = ( -60 .. 80) Solution: MA ( ) = F ( a) cos ( deg) 50 N.m MA( ) 0 50 0 50 100 Problem 4-17 The Snorkel Co.produces the articulating boom platform that can support weight W. If the boom is in the position shown, determine the moment of this force about points A, B, and C. Units Used: kip = 10 lb 3 221 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: a = 3 ft b = 16 ft c = 15 ft 1 = 30 deg 2 = 70 deg W = 550 lb Solution: MA = W a MB = W( a + b cos ( 1 ) ) MC = W( a + b cos ( 1 ) - c cos ( 2 ) ) MA = 1.65 kip ft MB = 9.27 kip ft MC = 6.45 kip ft Problem 4-18 Determine the direction ( 0 180 ) of the force F so that it produces (a) the maximum moment about point A and (b) the minimum moment about point A. Compute the moment in each case. Given: F = 40 lb a = 8 ft b = 2 ft 222 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: The maximum occurs when the force is perpendicular to the line between A and the point of application of the force. The minimum occurs when the force is parallel to this line. ( a) MAmax = F a +b 2 2 MAmax = 329.848 lb ft b a = atan a a = 14.04 deg a = 76.0 deg MAmin = 0 lb ft a = 90 deg - a ( b) MAmin = 0 lb ft b b = atan a b = 14.04 deg b = 166 deg b = 180 deg - b Problem 4-19 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 1 = 20 deg a = 150 mm 2 = 60 deg Solution: MA = F cos ( 1 ) ( a) sin ( 2 ) - F sin ( 1 ) ( a) cos ( 2 ) MA = 7.71 N m Problem 4-20 The boom has length L, weight Wb, and mass center at G. If the maximum moment that can be developed by the motor at A is M, determine the maximum load W, having a mass center at G', that can be lifted. 223 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: L = 30 ft Wb = 800 lb a = 14 ft b = 2 ft = 30 deg M = 20 10 lb ft Solution: M = Wb ( L - a) cos ( ) + W ( L cos ( ) + b) W = M - Wb ( L - a) cos ( ) L cos ( ) + b W = 319 lb 3 Problem 4-21 The tool at A is used to hold a power lawnmower blade stationary while the nut is being loosened with the wrench. If a force P is applied to the wrench at B in the direction shown, determine the moment it creates about the nut at C. What is the magnitude of force F at A so that it creates the opposite moment about C ? Given: P = 50 N b = 300 mm c = 5 d = 12 = 60 deg a = 400 mm Solution: (a) MA = P sin ( ) b MA = 13.0 N m (b) MA - F d c +d 2 2 a=0 224 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 c2 + d2 F = MA da F = 35.2 N Problem 4-22 Determine the clockwise direction ( 0 deg 180 deg) of the force F so that it produces (a) the maximum moment about point A and (b) no moment about point A. Compute the moment in each case. Given: F = 80 lb a = 4 ft b = 1 ft Solution: (a) MAmax = F a + b b = atan 2 2 MAmax = 330 lb ft a = 14.0 deg a = 104 deg a = 90 deg + ( b) MAmin = 0 b b = atan a b = 14.04 deg Problem 4-23 The Y-type structure is used to support the high voltage transmission cables. If the supporting cables each exert a force F on the structure at B, determine the moment of each force about point A. Also, by the principle of transmissibility, locate the forces at points C and D and determine the moments. 225 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 Units Used: kip = 1000 lb Given: F = 275 lb a = 85 ft = 30 deg Solution: MA1 = F sin ( ) a MA2 = F sin ( ) a Also b = ( a)tan ( ) MA1 = 11.7 kip ft MA2 = 11.7 kip ft MA1 = 11.7 kip ft MA2 = 11.7 kip ft MA1 = F cos ( ) b MA2 = F cos ( ) b Problem 4-24 The force F acts on the end of the pipe at B. Determine (a) the moment of this force about point A, and (b) the magnitude and direction of a horizantal force, applied at C, which produces the same moment. Given: F = 70 N a = 0.9 m b = 0.3 m c = 0.7 m = 60 deg Solution: MA = F sin ( ) c + F cos ( ) a 226 (a) MA = 73.9 N m 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 (b) F C ( a) = MA FC = MA a F C = 82.2 N Problem 4-25 The force F acts on the end of the pipe at B. Determine the angles ( 0 180 ) of the force that will produce maximum and minimum moments about point A. What are the magnitudes of these moments? Given: F = 70 N a = 0.9 m b = 0.3 m c = 0.7 m Solution: MA = F sin ( ) c + F cos ( ) a d d MA = c F cos ( ) - a F sin ( ) = 0 For maximum moment c max = atan a max = 37.9 deg MAmax = 79.812 N m MAmax = F sin ( max) c + F cos ( max) a For minimum moment MA = F sin ( ) c + F cos ( ) a = 0 -a c min = 180 deg + atan min = 128 deg MAmin = 0 N m MAmin = F c sin ( min ) + F ( a) cos ( min ) Problem 4-26 The towline exerts force P at the end of the crane boom of length L. Determine the placement 227 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 g p x of the hook at A so that this force creates a maximum moment about point O. What is this moment? Unit Used: kN = 10 N Given: P = 4 kN L = 20 m 3 = 30 deg a = 1.5 m Solution: Maximum moment, OB BA Guesses Given x = 1m d = 1 m (Length of the cable from B to A) L cos ( ) + d sin ( ) = x a + L sin ( ) = d cos ( ) x = Find ( x , d) d Mmax = P L x = 23.96 m Mmax = 80 kN m Problem 4-27 The towline exerts force P at the end of the crane boom of length L. Determine the position of the boom so that this force creates a maximum moment about point O. What is this moment? Units Used: kN = 10 N 3 228 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: P = 4 kN x = 25 m L = 20 m a = 1.5 m Solution: Maximum moment, OB BA Guesses Given = 30 deg d = 1m (length of cable from B to A) L cos ( ) + d sin ( ) = x a + L sin ( ) = d cos ( ) = Find ( , d) d Mmax = P L = 33.573 deg Mmax = 80 kN m Problem 4-28 Determine the resultant moment of the forces about point A. Solve the problem first by considering each force as a whole, and then by using the principle of moments. 229 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 Units Used: kN = 10 N Given: F 1 = 250 N F 2 = 300 N F 3 = 500 N a = 2m b = 3m c = 4m d = 3 e = 4 3 1 = 60 deg 2 = 30 deg Solution Using Whole Forces: d = atan L = a + b - d e Geometry e c e e2 + d2 MA = -2.532 kN m MA = -F1 ( a)cos ( 2 ) - F2 ( a + b) sin ( 1 ) - F3 L Solution Using Principle of Moments: MA = -F1 cos ( 2 ) a - F2 sin ( 1 ) ( a + b) + F3 d d +e 2 2 c - F3 e d +e 2 2 ( a + b) MA = -2.532 10 N m 3 Problem 4-29 If the resultant moment about point A is M clockwise, determine the magnitude of F 3. 230 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 Units Used: kN = 10 N Given: M = 4.8 kN m a = 2 m F 1 = 300 N F 2 = 400 N b = 3m c = 4m d = 3 e = 4 3 1 = 60 deg 2 = 30 deg Solution: Initial Guess Given F3 = 1 N -M = -F 1 cos ( 2 ) a - F 2 sin ( 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 4-30 The flat-belt tensioner is manufactured by the Daton Co. and is used with V-belt drives on poultry and livestock fans. If the tension in the belt is F, when the pulley is not turning, determine the moment of each of these forces about the pin at A. 231 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 = 52 lb a = 8 in b = 5 in c = 6 in 1 = 30 deg 2 = 20 deg Solution: MA1 = F cos ( 1 ) ( a + c cos ( 1 ) ) - F sin ( 1 ) ( b - c sin ( 1 ) ) MA1 = 542 lb in MA2 = F cos ( 2 ) ( a - c cos ( 2 ) ) - F ( sin ( 2 ) ) ( b + c sin ( 2 ) ) MA2 = -10.01 lb in Problem 4-31 The worker is using the bar to pull two pipes together in order to complete the connection. If he applies a horizantal force F to the handle of the lever, determine the moment of this force about the end A. What would be the tension T in the cable needed to cause the opposite moment about point A. Given: F = 80 lb 1 = 40 deg 2 = 20 deg a = 0.5 ft b = 4.5 ft 232 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: MA = F( a + b) cos ( 1 ) MA = 306 lb ft Require MA = T cos ( 2 ) ( a) cos ( 1 ) + T sin ( 2 ) ( a) sin ( 1 ) T = ( a) ( cos ( 2 ) cos ( 1 ) + sin ( 2 ) sin ( 1 ) ) MA T = 652 lb Problem 4-32 If it takes a force F to pull the nail out, determine the smallest vertical force P that must be applied to the handle of the crowbar. Hint: This requires the moment of F about point A to be equal to the moment of P about A. Why? Given: F = 125 lb a = 14 in b = 3 in c = 1.5 in 1 = 20 deg 2 = 60 deg 233 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: MF = F sin ( 2 ) ( b) P ( a)cos ( 1 ) + ( c)sin ( 1 ) = MF MF = 325 lb in P = ( a) cos ( 1 ) + ( c) sin ( 1 ) MF P = 23.8 lb Problem 4-33 The pipe wrench is activated by pulling on the cable segment with a horizantal force F . Determine the moment MA produced by the wrench on the pipe at . Neglect the size of the pulley. Given: F = 500 N a = 0.2 m b = 0.5 m c = 0.4 m = 20 deg Solution: Initial Guesses = 20 deg MA = 1 N m Given c cos ( ) - a b - c sin ( ) = tan ( - ) MA = F c sin ( ) = Find ( , MA) MA = 84.161 deg MA = 199 N m 234 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 4-34 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 b = 7m c = 3m Solution: d = 4m e = 6m f = 2m rOA -c = -b a MO = rOA F 260 MO = 180 N m 510 Problem 4-35 Determine the moment of the force at A about point P. Express the result as a Cartesian vector. Given: a = 4m b = 7m c = 3m d = 4m e = 6m f = 2m 60 F = -30 N -20 Solution: rPA -c - d = -b - e a+ f MP = rPA F 440 MP = 220 N m 990 235 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 4-36 Determine the moment of the force F at A about point O. Express the result as a cartesian vector. Units Used: 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: 3 rAB b - g = c+d h - a rOA -b = -c a F1 = F rAB rAB MO = rOA F1 -84 MO = -8 kN m -39 Problem 4-37 Determine the moment of the force F at A about point P. Express the result as a Cartesian vector. Units Used: kN = 10 N 3 236 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 = 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 rPA -b - f = -c - e a F1 = F rAB rAB MO = rPA F 1 -116 16 kN m MO = -135 Problem 4-38 The curved rod lies in the x-y plane and has radius r. If a force F acts at its end as shown, determine the moment of this force about point O. Given: r = 3m a = 1m = 45 deg F = 80 N b = 2 m 237 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: rAC a = -r -b rAC 1 = -3 m -2 Fv = F rAC rAC 21.381 F v = -64.143 N -42.762 3 = 3 m 0 rOA r = r 0 rOA MO = rOA Fv -128.285 MO = 128.285 N m -256.571 Problem 4-39 The curved rod lies in the x-y plane and has a radius r. If a force F acts at its end as shown, determine the moment of this force about point B. Given: F = 80 N a = 1m c = 3m r = 3m 238 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 b = 2m = 45 deg Solution: rAC a = -c -b rAC rAC rBA Fv = F r cos ( ) = r - r sin ( ) 0 MB = rBA F v -37.6 90.7 N m MB = -154.9 Problem 4-40 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 Solution: rAB b = a 0 MA = rAB F -720 MA = 120 N m -660 239 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 4-41 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 = 30 deg Solution: ( a)sin ( ) r = ( a)cos ( ) 0 0 0 F = -W MA = r F -229 MA = 132 lb ft 0 Problem 4-42 The man pulls on the rope with a force F. Determine the moment that this force exerts about the base of the pole at O. Solve the problem two ways, i.e., by using a position vector from O to A, then O to B. Given: F = 20 N a = 3m 240 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 b = 4m c = 1.5 m d = 10.5 m Solution: rAB rOB b -a = c - d b = -a c rOA 0 = 0 d rAB rAB Fv = F MO1 = rOA F v MO1 61.2 = 81.6 N m 0 61.2 = 81.6 N m -0 MO2 = rOB Fv MO2 Problem 4-43 Determine the smallest force F that must be applied along the rope in order to cause the curved rod, which has radius r, to fail at the support C. This requires a moment to be developed at C of magnitude M. 241 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: r = 5 ft M = 80 lb ft = 60 deg a = 7 ft b = 6 ft Solution: b a - r sin ( ) = -r cos ( ) F = 1 lb rCB ( F uAB) = M F = Find ( F) F = 18.6 lb uAB = rAB rAB rAB rCB b = a -r Guess Given Problem 4-44 The pipe assembly is subjected to the force F . Determine the moment of this force about point A. 242 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 = 80 N a = 400 mm b = 300 mm c = 200 mm d = 250 mm = 40 deg = 30 deg Solution: rAC b + d a = -c cos ( ) sin ( ) F v = F cos ( ) cos ( ) -sin ( ) -5.385 MA = 13.093 N m 11.377 MA = rAC Fv Problem 4-45 The pipe assembly is subjected to the force F . Determine the moment of this force about point B. 243 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 = 80 N a = 400 mm b = 300 mm c = 200 mm d = 250 mm = 40 deg = 30 deg Solution: rBC b + d = 0 -c 550 rBC = 0 mm -200 44.534 F v = 53.073 N -40 10.615 MB = 13.093 N m 29.19 cos ( ) sin ( ) F v = F cos ( ) cos ( ) -sin ( ) MB = rBC F v Problem 4-46 The x-ray machine is used for medical diagnosis. If the camera and housing at C have mass M and a mass center at G, determine the moment of its weight about point O when it is in the position shown. Units Used: kN = 10 N 3 244 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: M = 150 kg a = 1.2 m b = 1.5 m = 60 deg g = 9.81 m s Solution: 2 -( b) cos ( ) 0 0 MO = a ( b)sin ( ) -M g -1.77 MO = -1.1 kN m 0 Problem 4-47 Using Cartesian vector analysis, determine the resultant moment of the three forces about the base of the column at A. Units Used : kN = 10 N Given: 3 400 F 1 = 300 N 120 100 F 2 = -100 N -60 0 0 N F3 = -500 a = 4m b = 8m 245 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 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 The total moment MA = MA1 + MA2 + MA3 1.2 MA2 = 1.2 kN m 0 -1.9 6 kN m MA = 0 0.5 MA3 = 0 kN m 0 Problem 4-48 A force F produces a moment MO about the origin of coordinates, point O. If the force acts at a point having the given x coordinate, determine the y and z coordinates. Units Used : Given: kN = 10 N 3 6 F = -2 kN 1 4 5 kN m MO = -14 x = 1m Solution: The initial guesses: y = 1m z = 1m 246 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 x y F = M O z Problem 4-49 y = Find ( y , z) z y 2 = m z 1 The force F creates a moment about point O of MO. If the force passes through a point having the given x coordinate, determine the y and z coordinates of the point. Also, realizing that MO = Fd, determine the perpendicular distance d from point O to the line of action of F . Given: 6 F = 8 N 10 -14 MO = 8 N m 2 x = 1m Solution: The initial guesses: Given y = 1m z = 1m x y F = M O z d = MO F y = Find ( y , z) z d = 1.149 m y 1 = m z 3 Problem 4-50 The force F produces a moment MO about the origin of coordinates, point O. If the force acts at a point having the given x-coordinate, determine the y and z coordinates. 247 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 Units Used: kN = 10 N Given: x = 1m 3 6 F = -2 kN 1 4 5 kN m MO = -14 Solution: Initial Guesses: y = 1m z = 1m Given x MO = y F z y = Find ( y , z) z y 2 = m z 1 Problem 4-51 Determine the moment of the force F about the Oa axis. Express the result as a Cartesian vector. Given: 50 F = -20 N 20 a = 6m b = 2m c = 1m d = 3m e = 4m 248 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: rOF c = -b a rOa 0 = e d uOa = rOa rOa MOa = ( rOF F) uOa uOa MOa 0 = 217.6 N m 163.2 Problem 4-52 Determine the moment of the force F about the aa axis. Express the result as a Cartesian vector. Given: F = 600 lb a = 6 ft b = 3 ft c = 2 ft d = 4 ft e = 4 ft f = 2 ft Solution: Fv = -d e 2 2 2 c +d +e c F d r = 0 -c -441 Maa = -294 lb ft 882 uaa = -b -f 2 2 2 a +b + f a 1 Maa = ( r Fv) uaa uaa 249 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 4-53 Determine the resultant moment of the two forces about the Oa axis. Express the result as a Cartesian vector. Given: F 1 = 80 lb F 2 = 50 lb = 120 deg = 60 deg = 45 deg a = 5 ft b = 4 ft c = 6 ft = 30 deg = 30 deg Solution: F 1v cos ( ) = F1 cos ( ) cos ( ) F 2v 0 = 0 F2 ( b)sin ( ) r1 = ( b)cos ( ) c cos ( ) = -sin ( ) 0 0 -( a) sin ( ) r2 = 0 uaa Maa = ( r1 F1v + r2 F2v) uaa uaa 26.132 Maa = -15.087 lb ft 0 250 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 4-54 The force F is applied to the handle of the box wrench. Determine the component of the moment of this force about the z axis which is effective in loosening the bolt. Given: a = 3 in b = 8 in c = 2 in 8 F = -1 lb 1 Solution: 0 k = 0 1 c r = -b a Mz = ( r F ) k Mz = 62 lb in Problem 4-55 The force F acts on the gear in the direction shown. Determine the moment of this force about the y axis. Given: F = 50 lb a = 3 in 1 = 60 deg 2 = 45 deg 3 = 120 deg 251 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: 0 j = 1 0 0 r = 0 a -cos ( 3) F v = F -cos ( 2 ) My = ( r Fv ) j -cos ( 1) My = 75 lb in Problem 4-56 The RollerBall skate is an in-line tandem skate that uses two large spherical wheels on each skate, rather than traditional wafer-shape wheels. During skating the two forces acting on the wheel of one skate consist of a normal force F 2 and a friction force F1. Determine the moment of both of these forces about the axle AB of the wheel. Given: = 30 deg F 1 = 13 lb F 2 = 78 lb a = 1.25 in Solution: F1 0 F = F2 r = -a 0 0 cos ( ) ab = -sin ( ) 0 Problem 4-57 The cutting tool on the lathe exerts a force F on the shaft in the direction shown. Determine the moment of this force about the y axis of the shaft. Mab = ( r F ) ab Mab = 0 lb in 252 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 Units Used: kN = 10 N Given: 3 6 F = -4 kN -7 a = 30 mm = 40 deg Solution: cos ( ) 0 r = a sin ( ) 0 j = 1 0 My = ( r F) j My = 0.277 kN m Problem 4-58 The hood of the automobile is supported by the strut AB, which exerts a force F on the hood. Determine the moment of this force about the hinged axis y. Given: F = 24 lb a = 2 ft b = 4 ft c = 2 ft d = 4 ft 253 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: b rA = 0 0 0 j = 1 0 Problem 4-59 rAB -b + c a = d Fv = F rAB rAB -9.798 F v = 9.798 lb 19.596 My = ( rA F v ) j My = -78.384 lb ft The lug nut on the wheel of the automobile is to be removed using the wrench and applying the vertical force F at A. Determine if this force is adequate, provided a torque M about the x axis is initially required to turn the nut. If the force F can be applied at A in any other direction, will it be possible to turn the nut? 254 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 = 30 N M = 14 N m a = 0.25 m b = 0.3 m c = 0.5 m d = 0.1 m Solution: Mx = F c - b Mx = 12 N m Mx < M No 2 2 For Mxmax, apply force perpendicular to the handle and the x-axis. Mxmax = F c Mxmax = 15 N m Mxmax > M Yes Problem 4-60 The lug nut on the wheel of the automobile is to be removed using the wrench and applying the vertical force F . Assume that the cheater pipe AB is slipped over the handle of the wrench and the F force can be applied at any point and in any direction on the assembly. Determine if this force is adequate, provided a torque M about the x axis is initially required to turn the nut. Given: F 1 = 30 N M = 14 N m a = 0.25 m b = 0.3 m c = 0.5 m d = 0.1 m 255 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: Mx = F1 a+c c c -b 2 2 Mx = 18 N m Mx > M Yes Mxmax occurs when force is applied perpendicular to both the handle and the x-axis. Mxmax = F1 ( a + c) Mxmax = 22.5 N m Mxmax > M Yes Problem 4-61 The bevel gear is subjected to the force F which is caused from contact with another gear. Determine the moment of this force about the y axis of the gear shaft. Given: a = 30 mm b = 40 mm 20 8 N F = -15 256 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: -b r = 0 a Problem 4-62 0 j = 1 0 My = ( r F) j My = 0 N m The wooden shaft is held in a lathe. The cutting tool exerts force F on the shaft in the direction shown. Determine the moment of this force about the x axis of the shaft. Express the result as a Cartesian vector. The distance OA is a. Given: a = 25 mm = 30 deg -5 F = -3 N 8 Solution: 0 r = ( a)cos ( ) ( a)sin ( ) 1 i = 0 0 Mx = ( r F) i i 0.211 0 N m Mx = 0 Problem 4-63 Determine the magnitude of the moment of the force F about the base line CA of the tripod. Given: 50 F = -20 N -80 257 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 a = 4m b = 2.5 m c = 1m d = 0.5 m e = 2m f = 1.5 m g = 2m Solution: rCA -g = e 0 uCA = rCA rCA rCD b - g = e a MCA = ( rCD F) uCA MCA = 226 N m Problem 4-64 The flex-headed ratchet wrench is subjected to force P, applied perpendicular to the handle as shown. Determine the moment or torque this imparts along the vertical axis of the bolt at A. Given: P = 16 lb a = 10 in = 60 deg b = 0.75 in Solution: M = P b + ( a)sin ( ) M = 150.564 lb in 258 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 4-65 If a torque or moment M is required to loosen the bolt at A, determine the force P that must be applied perpendicular to the handle of the flex-headed ratchet wrench. Given: M = 80 lb in = 60 deg a = 10 in b = 0.75 in Solution: M = P b + ( a)sin ( ) P = b + ( a)sin ( ) M P = 8.50 lb Problem 4-66 The A-frame is being hoisted into an upright position by the vertical force F . Determine the moment of this force about the y axis when the frame is in the position shown. Given: F = 80 lb a = 6 ft b = 6 ft = 30 deg = 15 deg Solution: Using the primed coordinates we have 259 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 -sin ( ) j = cos ( ) 0 0 Fv = F 0 1 rAC -b cos ( ) a = 2 b sin ( ) My = ( rAC F v ) j My = 281.528 lb ft Problem 4-67 Determine the moment of each force acting on the handle of the wrench about the a axis. Given: -2 F 1 = 4 lb -8 b = 6 in c = 4 in d = 3.5 in 3 F 2 = 2 lb -6 = 45 deg Solution: cos ( ) 0 ua = sin ( ) cos ( ) sin ( ) 0 + ( c + d) 0 r1 = b sin ( ) -cos ( ) M1a = ( r1 F1 ) ua M1a = 30 lb in cos ( ) sin ( ) 0 + c 0 r2 = b sin ( ) -cos ( ) 260 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 M2a = ( r2 F2 ) ua M2a = 8 lb in Problem 4-68 Determine the moment of each force acting on the handle of the wrench about the z axis. Given: -2 F 1 = 4 lb -8 b = 6 in c = 4 in d = 3.5 in 3 F 2 = 2 lb -6 = 45 deg Solution: cos ( ) sin ( ) 0 + ( c + d) 0 r1 = b sin ( ) -cos ( ) M1z = ( r1 F 1 ) k M2z = ( r2 F 2 ) k cos ( ) sin ( ) 0 + c 0 r2 = b sin ( ) -cos ( ) 0 k = 0 1 M1z = 38.2 lb in M2z = 14.1 lb in Problem 4-69 Determine the magnitude and sense of the couple moment. Units Used: kN = 10 N 261 3 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 = 5 kN = 30 deg a = 0.5 m b = 4m c = 2m d = 1m Solution: MC = F cos ( ) ( a + c) + F sin ( ) ( b - d) MC = 18.325 kN m Problem 4-70 Determine the magnitude and sense of the couple moment. Each force has a magnitude F. Given: F = 65 lb a = 2 ft b = 1.5 ft c = 4 ft d = 6 ft e = 3 ft Solution: Mc = MB; MC = F ( d + a) + F e ( c + a) 2 2 2 2 c +e c +e c (Counterclockwise) MC = 650 lb ft Problem 4-71 Determine the magnitude and sense of the couple moment. 262 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 Units Used: kip = 10 lb Given: F = 150 lb a = 8 ft b = 6 ft c = 8 ft d = 6 ft e = 6 ft f = 8 ft Solution: MC = MA; MC = F d d + f MC = 3120 lb ft 2 2 3 ( a + f) + F 2 f d + f 2 ( c + d) MC = 3.120 kip ft Problem 4-72 If the couple moment has magnitude M, determine the magnitude F of the couple forces. Given: M = 300 lb ft a = 6 ft b = 12 ft c = 1 ft d = 2 ft e = 12 ft f = 7 ft 263 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: ( f - d) ( b + e) e( f + a) - 2 2 2 2 ( f - d) + e ( f - d) + e M=F F = M e( f + a) ( f - d) + e 2 2 - ( f - d) ( b + e) ( f - d) + e 2 2 F = 108 lb Problem 4-73 A clockwise couple M is resisted by the shaft of the electric motor. Determine the magnitude of the reactive forces -R and R which act at supports A and B so that the resultant of the two couples is zero. Given: a = 150 mm = 60 deg M = 5 N m Solution: MC = -M + tan ( ) 2R a =0 R = M tan ( ) 2 a R = 28.9 N Problem 4-74 The resultant couple moment created by the two couples acting on the disk is MR. Determine the magnitude of force T. 264 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 Units Used: kip = 10 lb Given: 3 0 MR = 0 kip in 10 a = 4 in b = 2 in c = 3 in Solution: Initial Guess T = 1 kip Given a 0 -b 0 -b - c 0 0 T + 0 -T + 0 -T = M R 0 0 0 0 0 0 T = 0.909 kip T = Find ( T) Problem 4-75 Three couple moments act on the pipe assembly. Determine the magnitude of M3 and the bend angle so that the resultant couple moment is zero. Given: 1 = 45 deg M1 = 900 N m M2 = 500 N m 265 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: Initial guesses: Given + Mx = 0; + = 10 deg M3 = 10 N m M1 - M3 cos ( ) - M2 cos ( 1 ) = 0 M3 sin ( ) - M2 sin ( 1 ) = 0 My = 0; = Find ( , M3) M3 = 32.9 deg M3 = 651 N m Problem 4-76 The floor causes couple moments MA and MB on the brushes of the polishing machine. Determine the magnitude of the couple forces that must be developed by the operator on the handles so that the resultant couple moment on the polisher is zero. What is the magnitude of these forces if the brush at B suddenly stops so that MB = 0? Given: a = 0.3 m MA = 40 N m MB = 30 N m Solution: MA - MB a MA a MA - MB - F 1 a = 0 F1 = F 1 = 33.3 N MA - F 2 a = 0 F2 = F 2 = 133 N 266 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 4-77 The ends of the triangular plate are subjected to three couples. Determine the magnitude of the force F so that the resultant couple moment is M clockwise. Given: F 1 = 600 N F 2 = 250 N a = 1m = 40 deg M = 400 N m Solution: Initial Guess F = 1N F = Find ( F) F = 830 N Given F1 a - F a - F a = -M 2 2 cos ( ) 2 cos ( ) Problem 4-78 Two couples act on the beam. Determine the magnitude of F so that the resultant couple moment is M counterclockwise. Where on the beam does the resultant couple moment act? Given: M = 450 lb ft P = 200 lb a = 1.5 ft b = 1.25 ft c = 2 ft = 30 deg 267 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: MR = M M = F b cos ( ) + P a F = b cos ( ) M - Pa F = 139 lb The resultant couple moment is a free vector. It can act at any point on the beam. Problem 4-79 Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve the problem (a) using Eq. 4-13, and (b) summing the moment of each force about point O. Given: 0 F = 0 N 25 a = 300 mm b = 150 mm c = 400 mm d = 200 mm e = 200 mm Solution: ( a) rAB -e - b = -c + d 0 a = d 0 rOA M = rAB F -5 M = 8.75 N m 0 ( b) rOB a + b + e c = 0 -5 M = 8.75 N m 0 M = rOB F + rOA ( -F) 268 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 4-80 If the couple moment acting on the pipe has magnitude M, determine the magnitude F of the vertical force applied to each wrench. Given: M = 400 N m a = 300 mm b = 150 mm c = 400 mm d = 200 mm e = 200 mm Solution: 0 k = 0 1 -e - b = -c + d 0 F = 1N rAB ( Fk) = M F = Find ( F) F = 992.278 N rAB Guesss Given Problem 4-81 Determine the resultant couple moment acting on the beam. Solve the problem two ways: (a) sum moments about point O; and (b) sum moments about point A. Units Used: kN = 10 N 3 269 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 1 = 2 kN F 2 = 8 kN a = 0.3 m b = 1.5 m c = 1.8 m Solution: ( a) MR = MO; MRa = ( F1 cos ( 1 ) + F 2 cos ( 2 ) ) c + ( F2 cos ( 2 ) - F 1 sin ( 1 ) ) a ... + -( F2 cos ( 2 ) + F 1 cos ( 1 ) ) ( b + c) MRa = -9.69 kN m ( b) MR = MA; MRb = ( F2 sin ( 2 ) - F1 sin ( 1 ) ) a - ( F 2 cos ( 2 ) + F1 cos ( 1 ) ) b MRb = -9.69 kN m 1 = 30 deg 2 = 45 deg Problem 4-82 Two couples act on the beam as shown. Determine the magnitude of F so that the resultant couple moment is M counterclockwise. Where on the beam does the resultant couple act? Given: M = 300 lb ft a = 4 ft b = 1.5 ft P = 200 lb c = 3 d = 4 270 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= 2 c c +d 2 Fa + d c +d 2 2 Fb - Pb F = c +d 2 2 M + P b ca + d b F = 167 lb Resultant couple can act anywhere. Problem 4-83 Two couples act on the frame. If the resultant couple moment is to be zero, determine the distance d between the couple forces F 1. Given: F 1 = 80 lb F 2 = 50 lb a = 1 ft b = 3 ft c = 2 ft e = 3 ft f = 3 g = 4 = 30 deg Solution: g -F cos ( ) e + 2 2 2 F1 d = 0 g +f d = F2 F1 cos ( ) e g2 + f 2 g d = 2.03 ft 271 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 4-84 Two couples act on the frame. Determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 4-13) and (b) summing the moments of all the force components about point A. Given: F 1 = 80 lb F 2 = 50 lb a = 1 ft b = 3 ft Solution: ( a) M = ( r F) c = 2 ft d = 4 ft e = 3 ft f = 3 g = 4 = 30 deg e -sin ( ) 0 F -g 1 0 F -cos ( ) + d - f M = 2 2 2 0 0 0 f + g 0 ( b) M1 = 0 0 lb ft M= 126.096 Summing the moments of all force components aboout point A. g -g F b + 1 2 2 2 2 F 1 ( b + d) f +g f +g M2 = F 2 cos ( ) c - F 2 sin ( ) ( a + b + d) - F 2 cos ( ) ( c + e) + F 2 sin ( ) ( a + b + d) M = M1 + M2 M = 126.096 lb ft Problem 4-85 Two couples act on the frame. Determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 4 -13) and (b) summing the moments of all the force components about point B. 272 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 1 = 80 lb F 2 = 50 lb a = 1 ft b = 3 ft c = 2 ft Solution: ( a) M = ( r F) d = 4ft e = 3 ft f = 3 ft g = 4 ft = 30 deg e -sin ( ) 0 F -g 1 0 F -cos ( ) + d - f M = 2 2 2 0 0 0 f + g 0 ( b) Summing the moments of all force components about point B. M1 = g g 2 2 F1( a + d) - 2 2 F1 a f +g f +g 0 lb ft M= 0 126.096 M2 = F 2 cos ( ) c - F 2 cos ( ) ( c + e) M = M1 + M2 M = 126.096 lb ft Problem 4-86 Determine the couple moment. Express the result as a Cartesian vector. 273 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: 8 F = -4 N 10 a = 5m b = 3m c = 4m d = 2m e = 3m Solution: -b - e r = c+d -a M = r F 40 M = 20 N m -24 Problem 4-87 Determine the couple moment. Express the result as a Cartesian vector. Given: F = 80 N a = 6m b = 10 m c = 10 m d = 5m e = 4m f = 4m 274 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: u = b c + d 2 2 2 a + b + ( c + d) - a 1 F v = Fu -f - b r = -d - c e+a M = r Fv -252.6 M = 67.4 N m -252.6 Problem 4-88 If the resultant couple of the two couples acting on the fire hydrant is MR = { -15i + 30j} N m, determine the force magnitude P. Given: a = 0.2 m b = 0.150 m -15 M = 30 N m 0 F = 75 N Solution: Initial guess P = 1N Given -F a M = Pb P 0 = Find ( P) P = 200 N Problem 4-89 If the resultant couple of the three couples acting on the triangular block is to be zero, determine the magnitude of forces F and P . 275 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 1 = 150 N a = 300 mm b = 400 mm d = 600 mm Solution: F = 1N P = 1N Initial guesses: Given d 0 d 0 0 -F 1 0 F 1 0 0 + b -P + b + 0 0 =0 0 a F 0 0 0 0 a 0 F 75 = N P 100 F = Find ( F , P) P Problem 4-90 Determine the couple moment that acts on the assembly. Express the result as a Cartesian vector. Member BA lies in the x-y plane. 276 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: 0 0 N F = 100 a = 300 mm b = 150 mm c = 200 mm d = 200 mm = 60 deg Solution: -( c + d) sin ( ) - b cos ( ) r = -b sin ( ) + ( c + d)cos ( ) 0 M = r F 7.01 M = 42.14 N m 0.00 Problem 4-91 If the magnitude of the resultant couple moment is M, determine the magnitude F of the forces applied to the wrenches. Given: M = 15 N m a = 300 mm b = 150 mm c = 200 mm d = 200 mm = 60 deg 277 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: -( c + d) sin ( ) - b cos ( ) r = -b sin ( ) + ( c + d)cos ( ) 0 Guess Given F = 1N r ( Fk) = M 0 k = 0 1 F = Find ( F) F = 35.112 N Problem 4-92 The gears are subjected to the couple moments shown. Determine the magnitude and coordinate direction angles of the resultant couple moment. Given: M1 = 40 lb ft M2 = 30 lb ft 1 = 20 deg 2 = 15 deg 3 = 30 deg Solution: M1 cos ( 1) sin ( 2) M1 = M1 cos ( 1 ) cos ( 2 ) -M sin ( 1) 1 MR = M1 + M2 -M2 sin ( 3) M2 = M2 cos ( 3 ) 0 MR = 64.0 lb ft MR = acos MR 94.7 = 13.2 deg 102.3 278 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 4-93 Express the moment of the couple acting on the rod in Cartesian vector form. What is the magnitude of the couple moment? Given: 14 F = -8 N -6 a = 1.5 m b = 0.5 m c = 0.5 m d = 0.8 m Solution: d 0 M = a F + 0 ( -F) -c b -17 M = -9.2 N m -27.4 M = 33.532 N m Problem 4-94 Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve the problem (a) using Eq. 4-13, and (b) summing the moment of each force about point O. 279 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: a = 0.3 m b = 0.4 m c = 0.6 m -6 F = 2 N 3 Solution: (a) 0 M = c F -b 0 0 0 ( -F ) + c F M = -a -a - b 2.6 M = 2.4 N m 3.6 2.6 M = 2.4 N m 3.6 (b) Problem 4-95 A couple acts on each of the handles of the minidual valve. Determine the magnitude and coordinate direction angles of the resultant couple moment. Given: F 1 = 35 N = 60 deg F 2 = 25 N r1 = 175 mm r2 = 175 mm Solution: -F1 2 r1 -F2 2r2 cos ( ) M = 0 + -F2 2 r2 sin ( ) 0 0 -16.63 M = -7.58 N m 0 280 M = 18.3 N m 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 = acos M 155.496 = 114.504 deg 90 Problem 4-96 Express the moment of the couple acting on the pipe in Cartesian vector form. What is the magnitude of the couple moment? Given: F = 125 N a = 150 mm b = 150 mm c = 200 mm d = 600 mm Solution: c 0 M = a + b 0 0 F 37.5 M = -25 N m 0 M = 45.1 N m Problem 4-97 If the couple moment acting on the pipe has a magnitude M, determine the magnitude F of the forces applied to the wrenches. 281 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: M = 300 N m a = 150 mm b = 150 mm c = 200 mm d = 600 mm Solution: Initial guess: F = 1N Given c 0 a + b 0 = M 0 F F = Find ( F) F = 832.1 N Problem 4-98 Replace the force at A by an equivalent force and couple moment at point O. Given: F = 375 N a = 2m b = 4m c = 2m d = 1m = 30 deg 282 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: sin ( ) F v = F -cos ( ) 0 -a MO = b Fv 0 187.5 F v = -324.76 N 0 0 0 N m MO = -100.481 Problem 4-99 Replace the force at A by an equivalent force and couple moment at point P . Given: F = 375 N a = 2m b = 4m c = 2m d = 1m = 30 deg Solution: sin ( ) F v = F -cos ( ) 0 -a - c MP = b - d Fv 0 Problem 4-100 187.5 F v = -324.76 N 0 0 0 N m MP = 736.538 Replace the force system by an equivalent resultant force and couple moment at point O. 283 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 1 = 60 lb F 2 = 85 lb F 3 = 25 lb a = 2 ft b = 3 ft c = 6 ft d = 4 ft = 45 deg e = 3 f = 4 Solution: 0 -F1 + F = 0 cos ( ) -e - f + F sin ( ) 3 2 2 e + f 0 0 F2 F = 115.245 lb -33.322 F = -110.322 lb 0 -c 0 0 F -e d cos ( ) 2 0 -F + a - f + b F sin ( ) MO = 1 3 0 0 e2 + f 2 0 0 0 0 0 MO = 0 lb ft 480 MO = 480 lb ft Problem 4-101 Replace the force system by an equivalent resultant force and couple moment at point P. 284 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 1 = 60 lb F 2 = 85 lb F 3 = 25 lb a = 2 ft b = 3 ft c = 6 ft d = 4 ft f = 4 = 45 deg e = 3 Solution: 0 F = -F1 + 0 cos ( ) -e - f + F sin ( ) 3 2 2 e + f 0 0 F2 -33.322 F = -110.322 lb 0 F = 115.245 lb -c - d 0 -d F -e 0 cos ( ) 2 0 -F + a - f + b F sin ( ) MP = 1 3 0 0 e2 + f 2 0 0 0 0 0 0 lb ft MP = 921 MP = 921 lb ft Problem 4-102 Replace the force system by an equivalent force and couple moment at point O. Units Used: kip = 10 lb Given: F 1 = 430 lb a = 2 ft F 2 = 260 lb e = 5 ft 285 3 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 b = 8 ft c = 3 ft d = a Solution: f = 12 g = 5 = 60 deg -sin ( ) F R = F1 -cos ( ) + 0 g f 2 2 g + f 0 F2 -272 25 lb FR = 0 F R = 274 lb -d -sin ( ) e F g 2 b F -cos ( ) + 0 f MO = 1 2 2 0 0 0 g + f 0 0 0 kip ft MO = 4.609 Problem 4-103 Replace the force system by an equivalent force and couple moment at point P. Units Used: kip = 10 lb Given: F 1 = 430 lb a = 2 ft b = 8 ft c = 3 ft d = a Solution: F 2 = 260 lb e = 5 ft f = 12 g = 5 3 = 60 deg -sin ( ) F R = F1 -cos ( ) + 0 g f 2 2 g + f 0 F2 -272 25 lb FR = 0 F R = 274 lb 286 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 0 -sin ( ) d + e F g 2 c f MP = b + c F 1 -cos ( ) + 2 2 0 0 0 g + f 0 0 0 kip ft MP = 5.476 Problem 4-104 Replace the loading system acting on the post by an equivalent resultant force and couple moment at point O. Given: F 1 = 30 lb F 2 = 40 lb F 3 = 60 lb Solution: a = 1 ft b = 3 ft c = 2 ft d = 3 e = 4 0 1 F R = F1 -1 + F2 0 + 0 0 4 F R = -78 lb 0 -d -e 2 2 d +e 0 F3 F R = 78.1 lb 0 0 0 1 0 F -d 3 a + b + c F -1 + c F 0 + b + c -e MO = 1 2 2 2 0 0 0 0 0 d + e 0 0 0 lb ft MO = 100 Problem 4-105 Replace the loading system acting on the post by an equivalent resultant force and couple moment at point P. 287 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 1 = 30 lb F 2 = 40 lb F 3 = 60 lb a = 1 ft b = 3 ft c = 2 ft d = 3 e = 4 Solution: 0 1 -1 + F 0 + FR = F1 2 0 0 4 F R = -78 lb 0 -d -e 2 2 d +e 0 F3 F R = 78.1 lb 0 0 0 1 0 F -d 3 0 ft F -1 + -a - b F 0 + -a -e MP = 1 2 2 2 0 0 0 0 0 d + e 0 0 0 lb ft MP = 124 Problem 4-106 Replace the force and couple system by an equivalent force and couple moment at point O. Units Used: kN = 10 N 288 3 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: M = 8 kN m a = 3m b = 3m c = 4m d = 4m e = 5m Solution: = 60 deg f = 12 g = 5 F 1 = 6 kN F 2 = 4 kN FR = -cos ( ) g f + F -sin ( ) 2 2 2 f + g 0 0 F1 F R = 2.097 kN 0.308 F R = 2.074 kN 0 0 -c F g 0 -cos ( ) 1 0 + -e f + -d F -sin ( ) MO = 2 2 2 M 0 f + g 0 0 0 0 0 kN m MO = -10.615 Problem 4-107 Replace the force and couple system by an equivalent force and couple moment at point P. Units Used: kN = 10 N 3 289 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: M = 8 kN m a = 3m b = 3m c = 4m d = 4m e = 5m Solution: = 60 deg f = 12 g = 5 F 1 = 6 kN F 2 = 4 kN FR = -cos ( ) g f + F -sin ( ) 2 2 2 f + g 0 0 F1 0.308 F R = 2.074 kN 0 F R = 2.097 kN 0 -c - b F g -b -cos ( ) 1 f + -d F -sin ( ) MP = 0 + -e 2 2 2 M 0 f + g 0 0 0 0 kN m 0 MP = -16.838 Problem 4-108 Replace the force system by a single force resultant and specify its point of application, measured along the x axis from point O. Given: F 1 = 125 lb F 2 = 350 lb F 3 = 850 lb 290 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 a = 2 ft b = 6 ft c = 3 ft d = 4 ft Solution: F Ry = F 3 - F 2 - F 1 F Ry x = F3 ( b + c) - F 2 ( b) + F1 ( a) x = F3 ( b + c) - F 2 ( b) + F1 a FRy x = 15.5 ft F Ry = 375 lb Problem 4-109 Replace the force system by a single force resultant and specify its point of application, measured along the x axis from point P. Given: F 1 = 125 lb F 2 = 350 lb F 3 = 850 lb a = 2 ft b = 6 ft c = 3 ft d = 4 ft Solution: F Ry = F 3 - F 2 - F 1 F Ry = 375 lb F Ry x = F2 ( d + c) - F 3 ( d) + F1 ( a + b + c + d) x = F2 d + F2 c - F3 d + F1 a + F1 b + F1 c + F1 d F Ry (to the right of P) x = 2.47 ft 291 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 4-110 The forces and couple moments which are exerted on the toe and heel plates of a snow ski are F t, Mt, and F h, Mh, respectively. Replace this system by an equivalent force and couple moment acting at point O. Express the results in Cartesian vector form. Given: a = 120 mm b = 800 mm Solution: -50 F t = 80 N -158 FR = Ft + Fh -20 F h = 60 N -250 -6 Mt = 4 N m 2 -20 Mh = 8 N m 3 -70 F R = 140 N -408 r0Ft a = 0 0 MRP = ( r0Ft F t) + Mt + Mh -26 MRP = 31 N m 14.6 Problem 4-111 The forces and couple moments which are exerted on the toe and heel plates of a snow ski are F t, Mt, and F h, Mh, respectively. Replace this system by an equivalent force and couple moment 292 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 acting at point P. Express the results in Cartesian vector form. Given: a = 120 mm b = 800 mm -50 80 N Ft = -158 -6 Mt = 4 N m 2 -20 60 N Fh = -250 -20 8 N m Mh = 3 Solution: FR = Ft + Fh -70 F R = 140 N -408 -26 MP = 357.4 N m 126.6 b a + b 0 F + 0 F MP = Mt + Mh + h t 0 0 Problem 4-112 Replace the three forces acting on the shaft by a single resultant force. Specify where the force acts, measured from end B. 293 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 1 = 500 lb F 2 = 200 lb F 3 = 260 lb a = 5 ft b = 3 ft c = 2 ft d = 4 ft Solution: FR = e = 3 f = 4 g = 12 h = 5 -f 0 -e + F -1 + 2 2 2 e + f 0 0 F1 x = 1 ft h -g 2 2 g +h 0 F3 -300 F R = -740 lb 0 F R = 798 lb Initial guess: Given a F - f a + b 0 a + b + c F h -x 1 3 0 -e + 0 F -1 + 0 -g = 0 F 2 2 2 2 2 R 0 e + f 0 0 0 0 g + h 0 0 x = Find ( x) x = -7.432 ft Problem 4-113 Replace the three forces acting on the shaft by a single resultant force. Specify where the force acts, measured from end B. Given: F 1 = 500 lb F 2 = 200 lb F 3 = 260 lb a = 5 ft e = 3 294 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 b = 3 ft c = 2 ft d = 4 ft Solution: FR = f = 4 g = 12 h = 5 -f 0 -e + F -1 + 2 2 2 e + f 0 0 F1 x = 1ft h -g 2 2 g +h 0 F3 -300 F R = -740 lb 0 F R = 798 lb Initial guess: Given -b - c - d F - f -c - d 0 -d F h -x 1 3 -e + 0 F -1 + 0 -g = 0 F 0 2 2 2 2 2 0 e + f 0 0 0 0 g + h 0 0 x = Find ( x) x = 6.568 ft measured to the left of B Problem 4-114 Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB, measured from A. Given: F 1 = 300 lb F 2 = 200 lb F 3 = 400 lb F 4 = 200 lb Solution: F Rx = -F 4 F Ry = -F 1 - F 2 - F 3 F Rx = -200 lb F Ry = -900 lb M = 600 lb ft a = 3 ft b = 4 ft c = 2 ft d = 7 ft 295 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 = FRx + FRy 2 2 F = 922 lb = atan FRy FRx = 77.5 deg F Ry x = -F2 a - F 3 ( a + b) - F 4 c + M x = - F2 ( a) + F 3 ( a + b) + F 4 c - M F Ry x = 3.556 ft Problem 4-115 Replace the loading on the frame by a single resultant force. Specify where the force acts , measured from end A. Given: F 1 = 450 N F 2 = 300 N F 3 = 700 N a = 2m b = 4m c = 3m M = 1500 N m = 60 deg = 30 deg Solution: F Rx = F 1 cos ( ) - F 3 sin ( ) F Ry = -F 1 sin ( ) - F3 cos ( ) - F2 F = FRx + FRy 2 2 F Rx = -125 N F Ry = -1.296 10 N F = 1.302 10 N 3 3 1 = atan FRy FRx 1 = 84.5 deg F Ry( x) = -F1 sin ( ) a - F2 ( a + b) - F3 cos ( ) ( a + b + c) - M x = -F1 sin ( ) a - F2 ( a + b) - F3 cos ( ) ( a + b + c) - M F Ry x = 7.36 m Problem 4-116 Replace the loading on the frame by a single resultant force. Specify where the force acts , measured from end B. 296 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 1 = 450 N F 2 = 300 N F 3 = 700 N a = 2m b = 4m c = 3m M = 1500 N m = 60 deg = 30deg Solution: F Rx = F 1 cos ( ) - F 3 sin ( ) F Ry = -F 1 sin ( ) - F3 cos ( ) - F2 F = FRx + FRy 2 2 F Rx = -125 N F Ry = -1.296 10 N F = 1.302 10 N 3 3 1 = atan FRy FRx 1 = 84.5 deg F Ry x = F1 sin ( ) b - F3 cos ( ) c - M x = F1 sin ( ) b - F3 cos ( ) c - M F Ry x = 1.36 m (to the right) Problem 4-117 Replace the loading system acting on the beam by an equivalent resultant force and couple moment at point O. Given: F 1 = 200 N F 2 = 450 N M = 200 N m a = 0.2 m b = 1.5 m c = 2m d = 1.5 m 297 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 = 30 deg Solution: -sin ( ) 0 1 + F -cos ( ) FR = F1 2 0 0 -225 F R = -190 N 0 F R = 294 N b + c 0 b -sin ( ) 0 a F 1 + a F -cos ( ) + M 0 MO = 1 2 0 0 0 0 -1 0 0 N m MO = -39.6 Problem 4-118 Determine the magnitude and direction of force F and its placement d on the beam so that the loading system is equivalent to a resultant force FR acting vertically downward at point A and a clockwise couple moment M. Units Used: kN = 10 N Given: F 1 = 5 kN F 2 = 3 kN F R = 12 kN M = 50 kN m Solution: Initial guesses: F = 1 kN a = 3m b = 4m c = 6m e = 7 f = 24 3 = 30 deg 298 d = 2m 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 -e F + F cos ( ) = 0 2 2 1 e +f - f F - F sin ( ) - F = -F 2 R 2 2 1 e +f f 2 2 F1 a + F sin ( ) ( a + b - d) + F2 ( a + b) = M e +f F = Find ( F , , d) d F = 4.427 kN = 71.565 deg d = 3.524 m Problem 4-119 Determine the magnitude and direction of force F and its placement d on the beam so that the loading system is equivalent to a resultant force F R acting vertically downward at point A and a clockwise couple moment M. Units Used: kN = 10 N Given: F 1 = 5 kN F 2 = 3 kN F R = 10 kN a = 3m b = 4m c = 6m 3 M = 45 kN m e = 7 f = 24 Solution: Initial guesses: F = 1 kN = 30 deg d = 1m Given -e F + F cos ( ) = 0 2 2 1 e +f - f F - F sin ( ) - F = -F 2 R 2 2 1 e +f 299 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 2 F1 a + F sin ( ) ( a + b - d) + F2 ( a + b) = M e +f F = Find ( F , , d) d Problem 4-120 F = 2.608 kN = 57.529 deg d = 2.636 m Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB, measured from A. Given: F 1 = 500 N a = 3m b = 2m c = 1m d = 2m e = 3m f = 3 g = 4 F 2 = 300 N F 3 = 250 N M = 400 N m = 60 deg Solution: F Rx = -F 3 - F1( cos ( ) ) 2 2 g +f g F Rx = -450 N F Ry = -F 2 - F 3 - F1 sin ( ) 2 2 f +g f 2 F Ry = -883.0127 N FR = FRx + FRy 2 F R = 991.066 N 1 = atan FRy FRx 1 = 62.996 deg 300 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 Rx( y) = M + F 1 cos ( ) a + F3 g g + f 2 2 ( b + a) - F2 ( d) - F 3 f 2 g +f 2 ( d + e) M + F 1 cos ( ) a + F3 y = g g + f 2 2 ( b + a) - F2 ( d) - F 3 -FRx f 2 2 ( d + e) g +f y = 1.78 m Problem 4-121 Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member CD, measured from end C. Given: F 1 = 500 N F 2 = 300 N F 3 = 250 N M = 400 N m a = 3m b = 2m c = 1m d = 2m e = 3m f = 3 g = 4 = 60 deg Solution: F Rx = -F 3 - F ( cos ( ) ) 1 g +f g 2 2 F Rx = -450 N F Ry = -F 2 - F 3 - F sin ( ) 1 f +g f 2 2 2 F Ry = -883.0127 N FR = FRx + FRy 2 F R = 991.066 N 301 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 1 = atan FRy FRx f 2 2 1 = 62.996 deg F Ry( x) = M - F2 ( d + c) - F 3 M - F2 ( d + c) - F 3 x = ( c + d + e) - F ( b) cos ( ) - F c sin ( ) 1 1 g +f f 2 ( c + d + e) - F ( b) cos ( ) - F c sin ( ) 1 1 g +f 2 F Ry x = 2.64 m Problem 4-122 Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant's line of action intersects member AB, measured from point A. Given: F 1 = 35 lb a = 2 ft F 2 = 20 lb b = 4 ft F 3 = 25 lb c = 3 ft = 30 deg d = 2 ft Solution: F Rx = F 1 sin ( ) + F3 F Ry = -F 1 cos ( ) - F 2 FR = FRx + FRy 2 2 F Rx = 42.5 lb F Ry = -50.311 lb F R = 65.9 lb 1 = atan FRy FRx 1 = -49.8 deg F Ry x = -F1 cos ( ) a - F 2 ( a + b) + F 3 ( c) 302 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 x = -F1 cos ( ) a - F 2 ( a + b) + F 3 ( c) FRy x = 2.099 ft Problem 4-123 Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant's line of action intersects member BC, measured from point B. Given: F 1 = 35 lb F 2 = 20 lb F 3 = 25 lb = 30 deg a = 2 ft b = 4 ft c = 3 ft d = 2 ft Solution: F Rx = F 1 sin ( ) + F3 F Ry = -F 1 cos ( ) - F 2 2 2 F Rx = 42.5 lb F Ry = -50.311 lb FR = FRx + FRy F R = 65.9 lb 1 = atan FRy FRx 1 = -49.8 deg F Rx y = F 1 cos ( ) b + F3 ( c) 303 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 y = F 1 cos ( ) b + F3 ( c) FRx y = 4.617 ft (Below point B) Problem 4-124 Replace the force system acting on the frame by an equivalent resultant force and couple moment acting at point A. Given: F 1 = 35 lb F 2 = 20 lb F 3 = 25 lb a = 2 ft b = 4 ft c = 3 ft d = 2 ft = 30 deg Solution: F Rx = F 1 sin ( ) + F3 F Ry = F 1 cos ( ) + F 2 2 2 F Rx = 42.5 lb F Ry = 50.311 lb FR = FRx + FRy F R = 65.9 lb 1 = atan FRy FRx 1 = 49.8 deg MRA = -106 lb ft MRA = -F1 cos ( ) a - F 2 ( a + b) + F 3 ( c) Problem 4-125 Replace the force and couple-moment system by an equivalent resultant force and couple moment at point O. Express the results in Cartesian vector form. 304 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 Units Used: kN = 10 N Given: 3 8 F = 6 kN 8 -20 M = -70 kN m 20 Solution: a = 3m b = 3m c = 4m d = 6m e = 5m f = 6m g = 5m FR = F -f MR = M + e F g 8 F R = 6 kN 8 -10 MR = 18 kN m -56 Problem 4-126 Replace the force and couple-moment system by an equivalent resultant force and couple moment at point P. Express the results in Cartesian vector form. Units Used: kN = 10 N Given: 3 8 F = 6 kN 8 -20 M = -70 kN m 20 a = 3m b = 3m c = 4m e = 5m f = 6m 305 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 d = 6m Solution: FR = F g = 5m -f e F MR = M + d + g 8 F R = 6 kN 8 -46 MR = 66 kN m -56 Problem 4-127 Replace the force and couple-moment system by an equivalent resultant force and couple moment at point Q. Express the results in Cartesian vector form. Units Used: kN = 10 N Given: 3 8 F = 6 kN 8 -20 M = -70 kN m 20 a = 3m b = 3m c = 4m d = 6m e = 5m f = 6m g = 5m Solution: FR = F 0 MR = M + e F g 8 F R = 6 kN 8 -10 MR = -30 kN m -20 Problem 4-128 The belt passing over the pulley is subjected to forces F1 and F 2. F 1 acts in the -k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in 306 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 p y Cartesian vector form. Given: F 1 = 40 N F 2 = 40 N q p p p r = 80 mm a = 300 mm = 0 deg Solution: F 1v 0 = F1 0 -1 -a r1 = r 0 -a r2 = -r sin ( ) r cos ( ) F 2v 0 = F2 -cos ( ) -sin ( ) F R = F1v + F2v MA = r1 F1v + r2 F2v 0 F R = -40 N -40 Problem 4-129 0 MA = -12 N m 12 The belt passing over the pulley is subjected to forces F1 and F 2. F 1 acts in the -k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in Cartesian vector form. 307 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 1 = 40 N F 2 = 40 N = 0 deg r = 80 mm a = 300 mm = 45 deg Solution: F 1v 0 = F1 0 -1 F 2v 0 = F2 -cos ( ) -sin ( ) -a r1 = r 0 -a r2 = -r sin ( ) r cos ( ) F R = F1v + F2v MA = r1 F1v + r2 F2v 0 F R = -28.28 N -68.28 0 MA = -20.49 N m 8.49 Problem 4-130 Replace this system by an equivalent resultant force and couple moment acting at O. Express the results in Cartesian vector form. Given: F 1 = 50 N F 2 = 80 N 308 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 3 = 180 N a = 1.25 m b = 0.5 m c = 0.75 m Solution: 0 0 0 FR = 0 + 0 + 0 F 1 -F2 -F 3 0 0 N FR = -210 a + c 0 a 0 a 0 b 0 + b 0 + 0 0 MO = F 1 0 -F 2 0 -F 3 0 -15 MO = 225 N m 0 Problem 4-131 Handle forces F 1 and F 2 are applied to the electric drill. Replace this system by an equivalent resultant force and couple moment acting at point O. Express the results in Cartesian vector form. Given: a = 0.15 m b = 0.25 m c = 0.3 m 6 F 1 = -3 N -10 0 F2 = 2 N -4 309 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: FR = F1 + F2 6 F R = -1 N -14 1.3 3.3 N m MO = -0.45 a 0 0 F + -b F MO = 1 2 c c Problem 4-132 A biomechanical model of the lumbar region of the human trunk is shown. The forces acting in the four muscle groups consist of F R for the rectus, FO for the oblique, FL for the lumbar latissimus dorsi, and F E for the erector spinae. These loadings are symmetric with respect to the y - z plane. Replace this system of parallel forces by an equivalent force and couple moment acting at the spine, point O. Express the results in Cartesian vector form. Given: F R = 35 N F O = 45 N F L = 23 N F E = 32 N e = 40 mm Solution: F Res = Fi; MROx = MOx; F Res = 2( FR + F O + F L + F E) MRO = -2FR a + 2F E c + 2F L b F Res = 270 N MRO = -2.22 N m a = 75 mm b = 45 mm c = 15 mm d = 50 mm f = 30 mm Problem 4-133 The building slab is subjected to four parallel column loadings.Determine the equivalent resultant force and specify its location (x, y) on the slab. 310 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 Units Used: kN = 10 N Given: F 1 = 30 kN F 2 = 40 kN F 3 = 20 kN F 4 = 50 kN a = 3m b = 8m c = 2m d = 6m e = 4m Solution: + 3 FR = Fx; FR = F1 + F2 + F3 + F4 F R = 140 kN MRx = Mx; -F R( y) = -( F4 ) ( a) - ( F1 ) ( a + b) - ( F2 ) ( a + b + c) y = F4 a + F1 a + F1 b + F2 a + F2 b + F2 c FR y = 7.14 m MRy = My; (FR)x = (F4)( e) + (F3)( d + e) + (F2)( b + c) x = F4 e + F3 d + F3 e + F2 b + F2 c FR x = 5.71 m Problem 4-134 The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x, y) on the slab. 311 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 Units Used: kN = 10 N Given: F 1 = 20 kN F 2 = 50 kN F 3 = 20 kN F 4 = 50 kN a = 3m b = 8m c = 2m d = 6m e = 4m Solution: FR = F1 + F2 + F3 + F4 F R x = F2 e + F1 ( d + e) + F 2 ( d + e) x = 2 F2 e + F1 d + F1 e + F2 d FR x = 6.43 m F R = 140 kN 3 -F R y = -F 2 a - F3 ( a + b) - F2 ( a + b + c) 2 F2 a + F3 a + F3 b + F2 b + F2 c FR y = y = 7.29 m Problem 4-135 The pipe assembly is subjected to the action of a wrench at B and a couple at A. Determine the magnitude F of the couple forces so that the system can be simplified to a wrench acting at point C. Given: a = 0.6 m 312 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 b = 0.8 m c = 0.25 m d = 0.7 m e = 0.3 m f = 0.3 m g = 0.5 m h = 0.25 m P = 60 N Q = 40 N Solution: Initial Guess Given F = 1N MC = 1 N m - M C - P ( c + h) 0 + 0 0 = 0 0 0 -F ( e + F = Find ( F , MC) MC a -Q + b 0 f) 0 0 F = 53.3 N MC = 30 N m Problem 4-136 The three forces acting on the block each have a magnitude F 1 = F2 = F3. Replace this system by a wrench and specify the point where the wrench intersects the z axis, measured from point O. Given: F 1 = 10 lb F2 = F1 F3 = F1 Solution: The vectors a = 6 ft b = 6 ft c = 2 ft 313 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 1v = b -a 2 2 b +a 0 F1 F 2v 0 = -F 2 0 M = 1 lb ft F 3v = -b a 2 2 b +a 0 F3 R y = 1 lb R z = 1 lb Place the wrench in the x - z plane. Guesses x = 1ft z = 1ft R x = 1 lb Rx Given Ry = F1v + F2v + F3v R z Rx 0 x Rx 0 b M 0 R + Ry = a F2v + a F1v + 0 F3v y 2 2 2 Rx + Ry + Rz R c z Rz 0 c z x z Rx M M Mv = R = Find ( x , z , M , Rx , Ry , Rz) Ry 2 2 2 x Rx + Ry + Rz R z Ry Rz Rx 0 0 x 0 Mv = -14.142 lb ft = ft Ry = -10 lb z 0.586 R 0 0 z Problem 4-137 Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(x, y) where its line of action intersects the plate. Units Used: kN = 10 N 3 314 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 A = 500 N F B = 800 N F C = 300 N a = 4m b = 6m Solution: FA FR = FC F B Guesses x = 1m FR FR F R = 0.9899 kN y = 1m M = 100 N m Given M x + y FR = 0 b 0 0 0 a F + a 0 C 0 0 0 FB x 1.163 = m y 2.061 M x = Find ( M , x , y) y M = 3.07 kN m Problem 4-138 Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(y, z) where its line of action intersects the plate. Given: F A = 80 lb F B = 60 lb F C = 40 lb a = 12 ft b = 12 ft 315 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: -F C F R = -F B -F A Guesses F R = 108 lb y = 1 ft FR FR z = 1 ft M = 1 lb ft Given M 0 + y FR = z 0 -FC 0 0 a -FB + a 0 0 b 0 0 y 0.414 = ft z 8.69 M y = Find ( M , y , z) z Problem 4-139 M = -624 lb ft The loading on the bookshelf is distributed as shown. Determine the magnitude of the equivalent resultant location, measured from point O. Given: w1 = 2 lb ft lb ft w2 = 3.5 a = 2.75 ft b = 4 ft c = 1.5 ft Solution: Guesses Given R = 1 lb d = 1ft w1 b + w2 c = R w1 b a - b c - w2 c + b - a = -d R 2 2 R = 13.25 lb d = 0.34 ft R = Find ( R , d) d 316 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 4-140 Replace the loading by an equivalent resultant force and couple moment acting at point A. Units Used: kN = 10 N Given: w1 = 600 w2 = 600 a = 2.5 m b = 2.5 m Solution: F R = w1 a - w2 b MRA = w1 a FR = 0 N MRA = 3.75 kN m N m N m 3 2 a + b Problem 4-141 Replace the loading by an equivalent force and couple moment acting at point O. Units Used: kN = 10 N Given: w = 6 kN m 3 F = 15 kN M = 500 kN m a = 7.5 m b = 4.5 m 317 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: FR = 1 2 w( a + b) + F F R = 51.0 kN b MR = -M - 1 w a 2 a - 1 w b a + 2 3 2 3 - F( a + b) MR = -914 kN m Problem 4-142 Replace the loading by a single resultant force, and specify the location of the force on the beam measured from point O. Units Used: kN = 10 N Given: w = 6 kN m 3 F = 15 kN M = 500 kN m a = 7.5 m b = 4.5 m Solution: Initial Guesses: Given FR = 1 2 w( a + b) + F F R = 1 kN d = 1m -F R d = -M - 1 w a 2 a - 1 w b a + b - F( a + b) 2 3 2 3 F R = 51 kN d = 17.922 m FR = Find ( FR , d) d Problem 4-143 The column is used to support the floor which exerts a force P on the top of the column. The effect of soil pressure along its side is distributed as shown. Replace this loading by an 318 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 p g p g y equivalent resultant force and specify where it acts along the column, measured from its base A. Units Used: Given: P = 3000 lb w1 = 80 lb ft lb ft kip = 10 lb 3 w2 = 200 h = 9 ft Solution: F Rx = w1 h + F Rx = 1260 lb FR = 1 (w2 - w1)h 2 F Ry = P 2 FRx + P P 2 F R = 3.25 kip = atan F Rx F Rx y = = 67.2 deg 1 h h (w2 - w1)h 3 + w1 h 2 2 y = 1 2 w2 + 2 w1 h FRx 6 y = 3.86 ft Problem 4-144 Replace the loading by an equivalent force and couple moment at point O. Units Used: kN = 10 N Given: w1 = 15 kN m kN m 319 3 w2 = 5 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 d = 9m Solution: FR = 1 (w1 + w2)d 2 d 1 d + ( w1 - w2 ) d 2 2 3 F R = 90 kN MRO = 338 kN m MRO = w2 d Problem 4-145 Replace the distributed loading by an equivalent resultant force, and specify its location on the beam, measured from the pin at C. Units Used: kip = 10 lb Given: w = 800 a = 15 ft b = 15 ft lb ft 3 = 30 deg Solution: FR = w a + FR x = w a a 2 wb 2 F R = 18 kip a wb b + a + 2 2 3 + wb FR b a + 2 3 wa x = x = 11.7 ft Problem 4-146 The beam supports the distributed load caused by the sandbags. Determine the resultant force on the beam and specify its location measured from point A. 320 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 Units Used: kN = 10 N 3 Given: w1 = 1.5 w2 = 1 kN m a = 3m b = 3m c = 1.5 m kN m kN m w3 = 2.5 Solution: F R = w1 a + w2 b + w3 c MA = w1 a a 2 + w2 b a + F R = 11.25 kN c + w3 c a + b + 2 2 d = MA FR d = 4.05 m b MA = 45.563 kN m Problem 4-147 Determine the length b of the triangular load and its position a on the beam such that the equivalent resultant force is zero and the resultant couple moment is M clockwise. Units Used: kN = 10 N Given: w1 = 4 kN m w2 = 2.5 c = 9m kN m 3 M = 8 kN m Solution: Initial Guesses: Given -1 2 a = 1m w1 b + 1 2 w2 c = 0 b = 1m 321 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 1 2 w1 b a + 2b - w2 c = -M 3 2 3 a 1.539 = m b 5.625 1 2c a = Find ( a , b) b Problem 4-148 Replace the distributed loading by an equivalent resultant force and specify its location, measured from point A. Units Used: kN = 10 N Given: w1 = 800 w2 = 200 a = 2m b = 3m Solution: F R = w2 b + w1 a + x F R = w1 a a 2 + 1 1 2 N m N m 3 (w1 - w2)b F R = 3.10 kN b b (w1 - w2)ba + 3 + w2 ba + 2 2 w1 a x = a 1 b b + ( w1 - w2 ) b a + + w2 b a + 2 2 3 2 FR x = 2.06 m Problem 4-149 The distribution of soil loading on the bottom of a building slab is shown. Replace this loading by an equivalent resultant force and specify its location, measured from point O. Units Used: 322 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 kip = 10 lb Given: w1 = 50 lb ft lb ft lb ft 3 w2 = 300 w3 = 100 a = 12 ft b = 9 ft Solution : F R = w1 a + F R d = w1 a (w2 - w1)a + 2 (w2 - w3)b + w3 b 2 1 1 a 2 + F R = 3.9 kip (w2 - w1)a 2 1 2 2a 3 2 + b b (w2 - w3)ba + 3 + w3 ba + 2 2 1 d = 11.3 ft d = 3 w3 b a + 2 w3 b + w1 a + 2 a w2 + 3 b w2 a + w2 b 6FR 2 2 Problem 4-150 The beam is subjected to the distributed loading. Determine the length b of the uniform load and its position a on the beam such that the resultant force and couple moment acting on the beam are zero. Given: w1 = 40 w2 = 60 Solution: Initial Guesses: a = 1 ft b = 1ft lb ft lb ft c = 10ft d = 6 ft 323 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 1 2 w2 d - w1 b = 0 1 2 w2 d c + - w1 b a + 3 d b 2 =0 a = Find ( a , b) b a 9.75 = ft b 4.5 Problem 4-151 Replace the loading by an equivalent resultant force and specify its location on the beam, measured from point B. Units Used: kip = 10 lb Given: w1 = 800 w2 = 500 a = 12 ft b = 9 ft Solution: FR = 1 2 a w1 + 1 2 lb ft lb ft 3 (w1 - w2)b + w2 b F R = 10.65 kip a 1 1 b b F R x = - a w1 + ( w1 - w2 ) b + w2 b 3 2 2 3 2 a 1 b b 1 - a w1 + ( w1 - w2 ) b + w2 b 3 2 3 2 2 FR x = x = 0.479 ft ( to the right of B ) Problem 4-152 Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member AB, measured from A. 324 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: w1 = 200 w2 = 100 w3 = 200 a = 5m b = 6m Solution: F Rx = -w3 a F Ry = -1 (w1 + w2)b 2 a b 1 b - w2 b - ( w1 - w2 ) b 2 2 2 3 b - F Rx = -1000 N F Ry = -900 N N m N m N m - y FRx = w3 a a 2 w3 a y = - w2 b 2 -FRx (w1 - w2)b 3 2 1 b y = 0.1 m Problem 4-153 Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member BC, measured from C. Units Used: kN = 10 N 3 325 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: w1 = 200 w2 = 100 w3 = 200 a = 5m b = 6m Solution : F Rx = -w3 a F Ry = -1 2 F Rx = -1000 N F Ry = -900 N b 2 + N m N m N m (w1 + w2)b a 2 + w2 b -x F Ry = -w3 a (w1 - w2)b 2 1 2b 3 -w3 a x = a b 1 2b + w2 b + ( w1 - w2 ) b 2 2 2 3 -FRy x = 0.556 m FRx = 1.345 kN FRy Problem 4-154 Replace the loading by an equivalent resultant force and couple moment acting at point O. Units Used: kN = 10 N Given: w1 = 7.5 kN m 3 326 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 w2 = 20 a = 3m b = 3m kN m c = 4.5 m Solution: FR = (w2 - w1)c + w1 c + w1 b + 2 1 1 2 w1 a F R = 95.6 kN MRo = - c c b (w2 - w1)c 3 - w1 c 2 - w1 bc + 2 - 2 1 1 2 w1 a b + c + a 3 MRo = -349 kN m Problem 4-155 Determine the equivalent resultant force and couple moment at point O. Units Used: kN = 10 N Given: a = 3m wO = 3 kN m 2 3 x w ( x) = wO a Solution: a F R = w ( x) dx 0 a MO = w ( x) ( a - x) dx 0 F R = 3 kN MO = 2.25 kN m 327 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 4-156 Wind has blown sand over a platform such that the intensity of the load can be approximated by the function w = w0 x . Simplify this distributed loading to an equivalent resultant force and d 3 specify the magnitude and location of the force, measured from A. Units Used: kN = 10 N Given: w0 = 500 d = 10 m w ( x) = w0 Solution: F R = w ( x) dx 0 d x w ( x) dx 0 FR d 3 N m x d 3 F R = 1.25 kN d = d=8m Problem 4-157 Determine the equivalent resultant force and its location, measured from point O. 328 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: 2w0 L x F R = w0 sin dx = L 0 L d= x w sin x dx 0 L 0 L FR = L 2 Problem 4-158 Determine the equivalent resultant force acting on the bottom of the wing due to air pressure and specify where it acts, measured from point A. Given: a = 3 ft k = 86 lb ft 3 2 w ( x) = k x Solution: F R = w ( x) dx 0 a x w ( x) dx 0 FR a F R = 774 lb x = x = 2.25 ft Problem 4-159 Currently eighty-five percent of all neck injuries are caused by rear-end car collisions. To 329 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 y g y p j y alleviate this problem, an automobile seat restraint has been developed that provides additional pressure contact with the cranium. During dynamic tests the distribution of load on the cranium has been plotted and shown to be parabolic. Determine the equivalent resultant force and its location, measured from point A. Given: a = 0.5 ft w0 = 12 lb ft k = 24 lb ft 3 w ( x) = w0 + kx 2 Solution: a F R = w ( x) dx 0 x w ( x) dx 0 FR a F R = 7 lb x = x = 0.268 ft Problem 4-160 Determine the equivalent resultant force of the distributed loading and its location, measured from point A. Evaluate the integrals using Simpson's rule. Units Used: kN = 10 N Given: c1 = 5 c2 = 16 a = 3 3 330 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 b = 1 Solution: FR = 0 0 a+ b c1 x + c2 + x dx 2 F R = 14.9 a+ b x c1 x + FR c2 + x dx d = 2.27 2 d = Problem 4-161 Determine the coordinate direction angles of F, which is applied to the end A of the pipe assembly, so that the moment of F about O is zero. Given: F = 20 lb a = 8 in b = 6 in c = 6 in d = 10 in Solution: Require Mo = 0. This happens when force F is directed either towards or away from point O. c r = a + b d u = r r 0.329 u = 0.768 0.549 If the force points away from O, then = acos ( u) 70.774 = 39.794 deg 56.714 If the force points towards O, then 331 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 = acos ( -u) 109.226 = 140.206 deg 123.286 Problem 4-162 Determine the moment of the force F about point O. The force has coordinate direction angles , , . Express the result as a Cartesian vector. Given: F = 20 lb a = 8 in b = 6 in c = 6 in d = 10 in = 60 deg = 120 deg = 45 deg Solution: c r = a + b d Problem 4-163 cos ( ) F v = F cos ( ) cos ( ) M = r Fv 297.99 M = 15.147 lb in -200 Replace the force at A by an equivalent resultant force and couple moment at point P. Express the results in Cartesian vector form. Units Used : 332 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 kN = 10 N Given: a = 4m b = 6m c = 8m d = 4m 3 -300 F = 200 N -500 Solution : FR = F -300 F R = 200 N -500 -3.8 MP = -7.2 kN m -0.6 -a - c b F MP = d Problem 4-164 Determine the moment of the force FC about the door hinge at A. Express the result as a Cartesian vector. 333 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 = 250 N b = 1m c = 2.5 m d = 1.5 m e = 0.5 m = 30 deg Solution: rCB c-e = b + d cos ( ) -d sin ( ) rAB 0 = b 0 Fv = F rCB rCB MA = rAB F v -59.7 0.0 N m MA = -159.3 Problem 4-165 Determine the magnitude of the moment of the force FC about the hinged axis aa of the door. 334 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 = 250 N b = 1m c = 2.5 m d = 1.5 m e = 0.5 m = 30 deg Solution: rCB c-e = b + d cos ( ) -d sin ( ) rAB 0 = b 0 Fv = F rCB rCB 1 ua = 0 0 Maa = ( rAB Fv) ua Maa = -59.7 N m Problem 4-166 A force F1 acts vertically downward on the Z-bracket. Determine the moment of this force about the bolt axis (z axis), which is directed at angle from the vertical. 335 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 1 = 80 N a = 100 mm b = 300 mm c = 200 mm = 15 deg Solution: -b r = a+c 0 sin ( ) 0 F = F1 -cos ( ) 0 k = 0 1 Mz = ( r F ) k Mz = -6.212 N m Problem 4-167 Replace the force F having acting at point A by an equivalent force and couple moment at point C. Units Used: Given: F = 50 lb a = 10 ft b = 20 ft c = 15 ft kip = 10 lb 3 336 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 d = 10 ft e = 30 ft Solution : rAB d = c -e rAB rAB Fv = F rCA 0 = a + b e 14.286 F R = 21.429 lb -42.857 -1.929 MR = 0.429 kip ft -0.429 FR = Fv MR = rCA Fv Problem 4-168 The horizontal force F acts on the handle of the wrench. What is the magnitude of the moment of this force about the z axis? Given: F = 30 N a = 50 mm b = 200 mm c = 10 mm = 45 deg Solution: sin ( ) F v = F -cos ( ) 0 rOA -c = b a 0 k = 0 1 337 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 Mz = ( rOA F v ) k Mz = -4.03 N m Problem 4-169 The horizontal force F acts on the handle of the wrench. Determine the moment of this force about point O. Specify the coordinate direction angles , , of the moment axis. Given: F = 30 N b = 200 mm Solution: c = 10 mm a = 50 mm = 45 deg sin ( ) F v = F -cos ( ) 0 MO = rOA Fv rOA -c = b a 1.06 MO = 1.06 N m -4.03 75.7 = 75.7 deg 159.6 MO = acos MO Problem 4-170 If the resultant couple moment of the three couples acting on the triangular block is to be zero, determine the magnitudes of forces F and P. 338 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 1 = 10 lb a = 3 in b = 4 in c = 6 in d = 3 in = 30 deg Solution: Initial Guesses: Given F = 1 lb P = 1 lb 0 0 -F c + 0 + 0 -P c 0 a = 0 2 2 a + b b F1 d F = Find ( F , P) P F 3 = lb P 4 339 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.
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Exam No. 1 Review 2/19/2008 4:36:00 PM Difference between civil and criminal law Know the 4 sources of criminal law: common US Constitution, Legislated Law, Case law Understand what due process is. Know the Bill of Rights...
UT Dallas >> BA >> 4347 (Spring, 2008)
Chapter 22 HW 22-1. Sales of $10M; Inventory Turnover Ratio of 2 Increase Ratio to 5 while sales stay the same. How much cash must be freed up? Inventory = 10,000,000/ 2 = $5M Inventory= 10/5 = $2M 5-2=3M freed up due to the 3M reduction in inventory...
UT Dallas >> BA >> 4347 (Spring, 2008)
Chapter 4 HW a. Ratios happened to be standardized numbers that help in comparing a company\'s performance to the overall market, pinpointing the weaknesses and strength sectors; The Five Major Catergories are: 1. Liquidity 2. Asset Management 3. De...
UT Dallas >> BA >> 4347 (Spring, 2008)
Chapter 22 HW 22-1. Sales of $10M; Inventory Turnover Ratio of 2 Increase Ratio to 5 while sales stay the same. How much cash must be freed up? Inventory = 10,000,000/ 2 = $5M Inventory= 10/5 = $2M 5-2=3M freed up due to the 3M reduction in inventory...
UT Dallas >> BA >> 4347 (Spring, 2008)
Chapter 3 HW Sales has increased by $2,402,000 from 06 to 07, but do to the expansion the COGS increased substantially cutting into NI, causing it to be a shortfall for the \'07 year. The expansion did increase assets for the 07\' year mainly from inc...
UT Dallas >> BA >> 4347 (Spring, 2008)
Chapter 5 HW 2. N= 12 I= 10%*1,000 = PMT=100 PV= $850 FV=1,000 I/Y=? =12.48% 4. a. Risk-Free Rate + Inflation Premium (IP) = 4% +(2+4/2) =7% for 2-year Trasury securities b. 4%+(2+4+4)/3) =7.33% for 3-year securities 6. Quoted Rate = Rf +IP+MRP IP = ...
UC Davis >> PSC >> 100 (Fall, 2006)
PSC 100 MIDTERM #1 SG CHAPTER 1 1. Cognitive map a. b. Idea proposed by Tolman What rats learn in maze isn\'t series of S-R connections, but a picture of internal representation of the stimulus situation c. 2. Take physical space & make it mental spa...
UC Davis >> PSC >> 100 (Fall, 2006)
EXAM #3 SG CH. 9 KNOWLEDGE o PROCEDURAL o Know how DECLARATIVE Info about the world use a bath tub Declarative: we k...
UC Davis >> PSC >> 100 (Fall, 2006)
Ch. 4 Attribute Frequency Model o o o Midterm #2 SG Suggests that a prototype represents the mode or most frequently experienced combination of attributes Prototype is synonymous w/ the \"best example\" of a set of patterns Prototype is a pattern tha...
UC Davis >> PSC >> 100 (Fall, 2006)
PSC 100 FINAL STUDY GUIDE CH. 1 1. COGNITIVE MODEL a. Metaphor based on observations & inferences drawn from observations i. b. Describes detection, storage, and use of information 2 major kinds i. Sequential 1. 1. 1. ii. Cognitive processes are br...
UC Davis >> SOC >> 1 (Fall, 2006)
Soc 1: Intro to Sociology Mindy Romero Fall ,06 Gender and Family Week Four According to the U.S. Census Bureau. Family: A group of two or more people who reside together and are related by birth, marriage, or adoption. Household: Includes all the ...
UC Davis >> SOC >> 1 (Fall, 2006)
Soc 1: Intro to Sociology Mindy Romero Fall `06 Education Week Five Education The social institution responsible for the systematic transmission of knowledge, skills, and cultural values within a formally organized structure Functionalist Perspective...
UC Davis >> SOC >> 1 (Fall, 2006)
Soc 1: Intro to Sociology Mindy Romero Fall ,06 Race and Ethnicity: Social Definitions Week Six Race Versus Ethnicity What is the difference? Race Versus Ethnicity Race: A category of people identified on the basis of physical characteristics that d...
UC Davis >> SOC >> 1 (Fall, 2006)
Soc 1: Intro to Sociology Fall,06 Mindy Romero Race and Ethnicity: Structural Conditions Week Seven Theories of Prejudice: Sociological Perspective Disagrees with psychological perspective Instead, examines factors outside the individual - focuses o...
UC Davis >> SOC >> 1 (Fall, 2006)
Soc 1: Intro to Sociology Fall ,06 Mindy Romero Class Stratification Week Seven Continued Social Stratification is. A system in which groups of people are divided into layers based on their relative power and prestige, and property Social stratific...
UC Davis >> SOC >> 1 (Fall, 2006)
Soc 1: Intro to Sociology Fall `06 Mindy Romero Global Inequality Week Nine The World is Stratified. Global stratification: The unequal distribution of wealth, power, and prestige around the world. Global income inequality is greater now than it has ...
UC Davis >> SOC >> 1 (Fall, 2006)
Soc 1: Intro to Sociology Social Movements: Theory Week Ten - 12/5/06 Social Movement: large numbers of people that organize to bring about (proactive) or resist (reactive) social change Social movement organization: Organization established to promo...
UC Irvine >> E >> 54 (Winter, 2008)
9-12 9.10 It is possible to have a Cu-Ag alloy, which at equilibrium consists of an phase of composition 4 wt% Ag-96 wt% Cu and a phase of composition 95 wt% Ag-5 wt% Cu. From Figure 9.7 a horizontal tie can be constructed across the + region at...
UC Irvine >> E >> 54 (Winter, 2008)
9-4 Phase Equilibria 9.4 In order for a system to exist in a state of equilibrium the free energy must be a minimum for some specified combination of temperature, pressure, and composition. Excerpts from this work may be reproduced by instructors f...
UC Irvine >> E >> 54 (Winter, 2008)
9-12 9.10 It is possible to have a Cu-Ag alloy, which at equilibrium consists of an phase of composition 4 wt% Ag-96 wt% Cu and a phase of composition 95 wt% Ag-5 wt% Cu. From Figure 9.7 a horizontal tie can be constructed across the + region at...
UC Irvine >> E >> 54 (Winter, 2008)
9-14 9.12 Upon cooling a 50 wt% Ni-50 wt% Cu alloy from 1400C and utilizing Figure 9.3a: (a) The first solid phase forms at the temperature at which a vertical line at this composition intersects the L( + L) phase boundary-i.e., at about 1320C. (b) ...
Texas State >> ENG >> 2360 (Spring, 2008)
1-14-08 America in 1865 Political context End of the civil war Lincoln was assonated 13th amendment 14th amendment Womens suffrage Economic growth in the north, some recovery in the south Corruption in government 1873 first failure of New York Stock...
UC Irvine >> E >> 54 (Winter, 2008)
ENGR 54 Principles of Materials Science and Engineering Homework Set #6 Solutions 9-4 Phase Equilibria 9.4 In order for a system to exist in a state of equilibrium the free energy must be a minimum for some specified combination of temperature, pre...
UC Irvine >> E >> 54 (Winter, 2008)
ENGR 54 Principles of Materials Science and Engineering Homework Set #7 Solutions 10-2 10.2 (a) This problem first asks that we rewrite the expression for the total free energy change for nucleation (analogous to Equation 10.1) for the case of a cu...
UC Irvine >> E >> 54 (Winter, 2008)
9-39 9.32 (a) This portion of the problem asks that we determine the mass fractions of and phases for an 80 wt% Sn-20 wt% Pb alloy (at 180C). In order to do this it is necessary to employ the lever rule using a tie line that extends entirely acros...
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