© 2016 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 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 obd is perpendicular to A od = ƒ A * (B + D) ƒ = ƒ A ƒ ƒ B + D ƒ sin u3 ob = ƒ A * B ƒ = ƒ A ƒ ƒ B ƒ sin u1 bd = ƒ A * D ƒ = ƒ A ƒ ƒ D ƒ sin u2 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) (QED) Note also, A = Ax i + Ay j + Az k B = Bx i + By j + Bz k D = Dx i + Dy j + Dz k A * (B + D) = i Ax Bx + Dx j Ay By + Dy k Az Bz + Dz = [A y (Bz + Dz) - A z(By + Dy)]i - [A x(Bz + Dz) - A z(Bx + Dx)]j + [A x(By + Dy) - A y(Bx + Dx)]k = [(A y Bz - A zBy)i - (A x Bz - A z Bx)]j + (A x By - A y Bx)k + [(A y Dz - A z Dy)i - (A x Dz - A z Dx)j + (A x Dy - A y Dx)k i = Ax Bx j Ay By k i Az + Ax Bz Dx j Ay Dy k Az Dz = (A * B) + (A * D) (QED) 228 © 2016 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 4–2 Prove the triple scalar A # (B : C) = (A : B) # C product identity SOLUTION As shown in the figure Area = B(C sin u) = |B * C| Thus, Volume of parallelepiped is |B * C||h| But, |h| = |A # u(B * C)| = ` A # a B * C b` |B * C| Thus, Volume = |A # (B * C)| Since |(A * B) # C| represents this same volume then A # (B : C) = (A : B) # C (QED) Also, LHS = A # (B : C) i = (A x i + A y j + A z k) # Bx Cx j By Cy k Bz Cz = A x (ByCz - BzCy) - A y (BxCz - BzCx) + A z (BxCy - ByCx) = A xByCz - A xBzCy - A yBxCz + A yBzCx + A zBxCy - A zByCx RHS = (A : B) # C i = Ax Bx j Ay By k A z # (Cx i + Cy j + Cz k) Bz = Cx(A y Bz - A zBy) - Cy(A xBz - A zBx) + Cz(A xBy - A yBx) = A xByCz - A xBzCy - A yBxCz + A yBzCx + A zBxCy - A zByCx Thus, LHS = RHS A # (B : C) = (A : B) # C (QED) 229 © 2016 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 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 u = (|A| cos u)|B * C| = |h| |B * C| = BC |h| sin f = volume of parallelepiped If A # (B * C) = 0, then the volume equals zero, so that A, B, and C are coplanar 230 © 2016 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 *4–4 Determine the moment about point A of each of the three forces acting on the beam F2 = 500 lb F1 = 375 lb A ft ft SOLUTION B 0.5 ft ft 30˚ F3 = 160 lb a + 1MF12A = - 375182 = - 3000 lb # ft = 3.00 kip # ft (Clockwise) Ans a + 1MF22A = - 500 a b 1142 = -5600 lb # ft = 5.60 kip # ft (Clockwise) Ans a + 1MF32A = - 1601cos 30°21192 + 160 sin 30°10.52 = - 2593 lb # ft = 2.59 kip # ft (Clockwise) Ans Ans: ( MF1 ) A = 3.00 kip # ft (Clockwise) ( MF2 ) A = 5.60 kip # ft (Clockwise) ( MF3 ) A = 2.59 kip # ft (Clockwise) 231 © 2016 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 4–5 Determine the moment about point B of each of the three forces acting on the beam F2 = 500 lb F1 = 375 lb A ft SOLUTION ft B 0.5 ft ft 30˚ F3 = 160 lb a + 1MF12B = 3751112 = 4125 lb # ft = 4.125 kip # ft (Counterclockwise) Ans a + 1MF22B = 500 a b 152 = 2000 lb # ft = 2.00 kip # ft (Counterclockwise) Ans a + 1MF32B = 160 sin 30°10.52 - 160 cos 30°102 = 40.0 lb # ft (Counterclockwise) Ans Ans: ( MF1 ) B = 4.125 kip # ftd ( MF2 ) B = 2.00 kip # ftd ( MF3 ) B = 40.0 lb # ftd 232 © 2016 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 4–6 The crowbar is subjected to a vertical force of P = 25 lb at the grip, whereas it takes a force of F = 155 lb at the claw to pull the nail out Find the moment of each force about point A and determine if P is sufficient to pull out the nail The crowbar contacts the board at point A 60Њ F O 20Њ in P 14 in A 1.5 in Solution a + MP = 25 ( 14 cos 20° + 1.5 sin 20° ) = 341 in # lb (Counterclockwise) c + MF = 155 sin 60°(3) = 403 in # lb (Clockwise) Since MF MP, P = 25 lb is not sufficient to pull out the nail. Ans Ans: MP = 341 in # lbd MF = 403 in # lbb Not sufficient 233 © 2016 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 4–7 Determine the moment of each of the three forces about point A F1 F2 250 N 30 300 N 60 A 2m 3m 4m SOLUTION The moment arm measured perpendicular to each force from point A is d1 = sin 60° = 1.732 m B d2 = sin 60° = 4.330 m F3 500 N d3 = sin 53.13° = 1.60 m Using each force where MA = Fd, we have a + 1MF12A = - 25011.7322 = - 433 N # m = 433 N # m (Clockwise) Ans a + 1MF22A = - 30014.3302 = - 1299 N # m = 1.30 kN # m (Clockwise) Ans a + 1MF32A = - 50011.602 = - 800 N # m = 800 N # m (Clockwise) Ans Ans: ( MF1 ) A = 433 N # mb ( MF2 ) A = 1.30 kN # mb ( MF3 ) A = 800 N # mb 234 © 2016 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 *4–8 Determine the moment of each of the three forces about point B F1 F2 250 N 30 300 N 60 A 2m 3m SOLUTION 4m The forces are resolved into horizontal and vertical component as shown in Fig a For F1, a + MB = 250 cos 30°(3) - 250 sin 30°(4) = 149.51 N # m = 150 N # m d Ans B For F2, F3 500 N a + MB = 300 sin 60°(0) + 300 cos 60°(4) = 600 N # m d Ans Since the line of action of F3 passes through B, its moment arm about point B is zero Thus Ans MB = Ans: MB = 150 N # md MB = 600 N # md MB = 235 © 2016 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 4–9 Determine the moment of each force about the bolt located at A Take FB = 40 lb, FC = 50 lb 0.75 ft B 2.5 ft 30 FC 20 A C FB 25 SOLUTION a +MB = 40 cos 25°(2.5) = 90.6 lb # ft d Ans a +MC = 50 cos 30°(3.25) = 141 lb # ftd Ans Ans: MB = 90.6 lb # ftb MC = 141 lb # ftd 236 © 2016 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 4–10 If FB = 30 lb and FC = 45 lb, determine the resultant moment about the bolt located at A 0.75 ft B 2.5 ft A C 30 FC 20 FB 25 SOLUTION a + MA = 30 cos 25°(2.5) + 45 cos 30°(3.25) = 195 lb # ft d Ans: MA = 195 lb # ftd 237 ... 2016 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,... 2016 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,... 2016 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,