Engineering Mechanics - Statics Chapter 7 Problem 7-19 Determine the normal force, shear force, and moment at a section passing through point C. Units Used: kN 10 3 N= Given: P 8 kN= c 0.75 m= a 0.75m= d 0.5 m= b 0.75m= r 0.1 m= Solution: Σ M A = 0; T− dr+()Pa b+ c+()+ 0= TP ab+ c+ dr+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = T 30kN= + → A x T= A x 30 kN= Σ F x = 0; + ↑ Σ F y = 0; A y P= A y 8 kN= + → Σ F x = 0; N C − T− 0= N C T−= N C 30 − kN= + ↑ Σ F y = 0; V C P+ 0= V C P−= V C 8 − kN= Σ M C = 0; M C − Pc+ 0= M C Pc= M C 6 kN m⋅= Problem 7-20 The cable will fail when subjected to a tension T max . Determine the largest vertical load P the frame will support and calculate the internal normal force, shear force, and moment at a section passing through point C for this loading. Units Used: kN 10 3 N= 641 © 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 7 Given: T max 2 kN= a 0.75 m= b 0.75 m= c 0.75 m= d 0.5 m= r 0.1 m= Solution: Σ M A = 0; T max − rd+()Pa b+ c+()+ 0= PT max dr+ ab+ c+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = P 0.533 kN= + → Σ F x = 0; T max A x − 0= A x T max = A x 2 kN= + ↑ A y P− 0= A y P= A y 0.533 kN= Σ F y = 0; + → N C A x −= N C 2 − kN= Σ F x = 0; N C − A x − 0= + ↑ Σ F y = 0; V C − A y + 0= V C A y = V C 0.533 kN= M C A y c= M C 0.400 kN m⋅= Σ M C = 0; M C − A y c+ 0= Problem 7-21 Determine the internal shear force and moment acting at point C of the beam. Units Used: kip 10 3 lb= Given: w 2 kip ft = a 9 ft= 642 © 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 7 Solution: Σ F x = 0; N C 0 = N C 0 = Σ F y = 0; wa 2 wa 2 − V C − 0= V C 0 = Σ M C = 0; M C wa 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a− wa 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 0= M C wa 2 3 = M C 54.00 kip ft⋅= Problem 7-22 Determine the internal shear force and moment acting at point D of the beam. Units Used: kip 10 3 lb= Given: w 2 kip ft = a 6 ft= b 9 ft= Solution: N D 0 = Σ F x = 0; Σ F y = 0; wb 2 w a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − V D − 0= V D wb 2 w a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= V D 5.00 kip= Σ M D = 0; M D wb 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a− wa b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 0= M D wb 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a wa 3 6b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= M D 46.00 kip ft⋅= 643 © 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 7 Problem 7-23 The shaft is supported by a journal bearing at A and a thrust bearing at B. Determine the internal normal force, shear force, and moment at (a) point C, which is just to the right of the bearing at A, and (b) point D, which is just to the left of the F 2 force. Units Used: kip 10 3 lb= Given: F 1 2500 lb= a 6 ft= F 2 3000 lb= b 12 ft= w 75 lb ft = c 2ft= Solution: Σ M B = 0; A y − bc+()F 1 ab+ c+()+ wb b 2 c+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + F 2 c+ 0= A y 1 2 2F 1 ab+ c+()wb 2 2cb+ () + 2F 2 c+ bc+ = A y 4514 lb= Σ F x = 0; B x 0 lb= Σ F y = 0; A y F 1 − wb− F 2 − B y + 0= B y A y − F 1 + wb+ F 2 += B y 1886 lb= Segment AC : F 1 aM C + 0= Σ M C = 0; M C F 1 − a= M C 15 − kip ft⋅= Σ F x = 0; N C 0 = N C 0 = F 1 − A y + V C − 0= Σ F y = 0; V C A y F 1 −= V C 2.01 kip= Segment BD: Σ M D = 0; M D − B y c+ 0= M D B y c= M D 3.77 kip ft⋅= Σ F x = 0; N D 0 = N D 0 = Σ F y = 0; V D F 2 − B y + 0= V D F 2 B y −= V D 1.11 kip= 644 © 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 7 Problem 7-24 The jack AB is used to straighten the bent beam DE using the arrangement shown. If the axial compressive force in the jack is P, determine the internal moment developed at point C of the top beam. Neglect the weight of the beams. Units Used: kip 10 3 lb= Given: P 5000 lb= a 2 ft= b 10 ft= Solution: Segment: Σ M C = 0; M C P 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b+ 0= M C P 2 − b= M C 25.00 − kip ft⋅= Problem 7-25 The jack AB is used to straighten the bent beam DE using the arrangement shown. If the axial compressive force in the jack is P, determine the internal moment developed at point C of the top beam. Assume that each beam has a uniform weight density γ . Units Used: kip 10 3 lb= Given: P 5000 lb= 645 © 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 7 γ 150 lb ft = a 2 ft= b 10 ft= Solution: Beam: + ↑ Σ F y = 0; P 2 γ ab+()− 2R− 0= R P 2 γ ab+()−= R 700 lb= Segment: Σ M C = 0; M C Rb+ γ ab+() ab+ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 0= M C R− b γ ab+() 2 2 −= M C 17.8 − kip ft⋅= Problem 7-26 Determine the normal force, shear force, and moment in the beam at sections passing through points D and E. Point E is just to the right of the F load. Units Used: kip 10 3 lb= Given: a 6 ft= w 1.5 kip ft = b 6 ft= c 4 ft= F 3 kip= d 4 ft= 646 © 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 7 Σ M B = 0; 1 2 wa b+() ab+ 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ A y ab+()− 0= A y 1 2 wa b+() ab+ 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ab+ = A y 3 kip= + → Σ F x = 0; B x 0 = + ↑ Σ F y = 0; B y A y + 1 2 wa b+()− 0= B y A y − 1 2 wa b+()+= B y 6 kip= + → Σ F x = 0; N D 0 = + ↑ A y 1 2 aw ab+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a− V D − 0= Σ F y = 0; V D A y 1 2 aw ab+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a−= V D 0.75 kip= Σ M D = 0; M D 1 2 aw ab+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a a 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + A y a− 0= M D 1 − 2 aw ab+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a a 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ A y a+= M D 13.5 kip ft⋅= + → N E 0 = Σ F x = 0; + ↑ Σ F y = 0; V E − F− B y − 0= V E F− B y −= V E 9 − kip= Σ M E = 0; M E B y c+ 0= M E B y − c= M E 24.0 − kip ft⋅= 647 Solution: © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 7 Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame. Units Used: kN 10 3 N= Given: w 1 200 N m = w 2 400 N m = a 2.5 m= b 3 m= c 6 m= Solution: Σ M A = 0; w 1 − c c 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 2 w 2 w 1 − () c 2c 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − a a 2 c 2 + F BC c () + 0= F BC w 1 c 2 2 w 2 w 1 − () c 2 3 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ a 2 c 2 + ac = F BC 2600 N= + → Σ F x = 0; A x c a 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BC = A x 2400 N= + ↑ Σ F y = 0; A y w 1 c− 1 2 w 2 w 1 − () c− a a 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BC + 0= A y w 1 c 1 2 w 2 w 1 − () c+ a a 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BC −= A y 800 N= + → Σ F x = 0; A x − N D + 0= N D A x = N D 2.40 kN= + ↑ Σ F y = 0; A y w 1 b− 1 2 w 2 w 1 − () b c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b− V D − 0= V D A y w 1 b− 1 2 w 2 w 1 − () b 2 c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= V D 50 N= Σ M D = 0; A y − bw 1 b b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 1 2 w 2 w 1 − () b c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b b 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + M D + 0= M D A y b() w 1 b 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 1 2 w 2 w 1 − () b 3 3c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= M D 1.35 kN m⋅= 648 Problem 7-27 © 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 7 Problem 7-28 Determine the normal force, shear force, and moment at sections passing through points E and F. Member BC is pinned at B and there is a smooth slot in it at C. The pin at C is fixed to member CD. Units Used: kip 10 3 lb= Given: M 350 lb ft⋅= w 80 lb ft = c 2 ft= F 500 lb= d 3 ft= θ 60 deg= e 2 ft= a 2 ft= f 4 ft= b 1 ft= g 2 ft= Solution: Σ M B = 0; 1− 2 wd 2d 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F sin θ () d− C y de+()+ 0= C y wd 2 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F sin θ () d+ de+ = C y 307.8 lb= + → Σ F x = 0; B x F cos θ () − 0= B x F cos θ () = B x 250 lb= + ↑ Σ F y = 0; B y 1 2 wd− F sin θ () − C y + 0= B y 1 2 wd Fsin θ () C y −+= B y 245.2 lb= + → Σ F x = 0; N E − B x − 0= N E B x −= N E 250 − lb= 649 © 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 7 + ↑ Σ F y = 0; V E B y − 0= V E B y = V E 245 lb= Σ M E = 0 M E − B y c− 0= M E B y − c= M E 490 − lb ft⋅= + → Σ F x = 0; N F 0 = N F 0 lb= N F 0.00 lb= + ↑ Σ F y = 0; C y − V F − 0= V F C y −= V F 308 − lb= Σ M F = 0; C y f() M F + 0= M F f− C y = M F 1.23 − kip ft⋅= Problem 7-29 The bolt shank is subjected to a tension F. Determine the internal normal force, shear force, and moment at point C. Given: F 80 lb= a 6 in= Solution: Σ F x = 0; N C F+ 0= N C F−= N C 80.00 − lb= Σ F y = 0; V C 0 = Σ M C = 0; M C Fa+ 0= M C F− a= M C 480.00 − lb in⋅= Problem 7-30 Determine the normal force, shear force, and moment acting at sections passing through points B and C on the curved rod. 650 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. [...]... publisher Engineering Mechanics - Statics Chapter 7 Units Used: kip = 103 lb Given: θ 2 = 30 deg F = 500 lb r = 2 ft a = 3 θ 1 = 45 deg b = 4 Solution: ΣF N = 0; ⎛ F b ⎞ sin ( θ ) − ⎛ F a ⎞ cos ( θ ) + N = 2 2 B ⎜ 2 2⎟ ⎜ 2 2⎟ ⎝ a +b ⎠ ⎝ a +b ⎠ ⎤ ⎡( a)cos ( θ 2 ) − b sin ( θ 2 )⎥ 2 2 ⎢ ⎥ a +b ⎣ ⎦ NB = F ⎢ ΣF V = 0; VB + ΣMB = 0; NB = 59.8 lb ⎛ F b ⎞ cos ( θ ) + ⎛ F a ⎞ sin ( θ ) = 2 2 ⎜ 2 2⎟ ⎜ 2 2⎟ a +b... Chapter 7 − Ay x + M1 ( x) = 0 M1 ( x) = A y x 1 kN⋅ m x2 = b − a , 1.01( b − a) b 1 Ay − w( x − a) − V 2 ( x) = 0 V 2 ( x) = ⎡A y − w( x − a)⎤ ⎣ ⎦ ⎛ x − a ⎞ + M ( x) = − Ay x + w( x − a) ⎜ ⎟ 2 ⎝ 2 ⎠ 2 ⎡ ⎢Ay x − w ( x − a) ⎥ 1 M2 ( x) = 2 ⎣ ⎦ kN⋅ m 0 kN Force (kN) 2 V1( x1) 0 V2( x2) 2 4 0 0.5 1 1.5 2 2.5 3 2. 5 3 3.5 4 x1 , x2 Distance (m) Moment (kN-m) 2 M1( x1) 1 M2( x2) 0 1 0 0.5 1 1.5 2 3.5 4 x1 , x2... −F MB + 0 ⎤ ⎡ b cos ( θ 2) + ( a)sin ( θ 2 )⎥ ⎢ 2 2 ⎢ ⎥ a +b ⎣ ⎦ 0 V B = −496 lb ⎛ F b ⎞ r sin ( θ ) + F⎛ a ⎞ ( r − r cos ( θ ) ) = 2 2 ⎜ 2 2⎟ ⎜ 2 2⎟ a +b ⎠ a +b ⎠ ⎝ ⎝ MB = F r ⎤ ⎡−b sin ( θ 2) − a + ( a)cos ( θ 2 )⎥ ⎢ 2 2 ⎢ ⎥ a +b ⎣ ⎦ 0 MB = −480 lb⋅ ft Also, ΣF x = 0; Fb − Ax + =0 2 a +b Ax = F ΣF y = 0; Ay − 2 ⎛ b ⎞ ⎜ 2 2⎟ ⎝ a +b ⎠ Fa 2 a +b Ax = 400.00 lb =0 2 655 © 20 07 R C Hibbeler Published... Engineering Mechanics - Statics x1 = 0 , 0.01a a Chapter 7 x2 = a , 1.01a L − a x3 = L − a , 1.01( L − a) L V 2 ( x) = 0 V 3 ( x) = P V 1 ( x) = lb Px M1 ( x) = M2 ( x) = lb⋅ ft Pa M3 ( x) = lb⋅ ft −P lb P( L − x) lb⋅ ft Force (lb) 1000 V1( x1) 500 V2( x2) 0 V3( x3) 500 1000 0 2 4 6 8 10 8 10 12 x1 x2 x3 , , ft ft ft Distance (ft) Moment (lb-ft) 4000 M1( x1) M2( x2) 20 00 M3( x3) 0 0 2 4 6 12 x1 x2... writing from the publisher Engineering Mechanics - Statics Chapter 7 ( 2 a + b) w ⎡ ( b − a)⎤ − Ay b = 0 ⎢ ⎥ 1 1 ⎣3 2 Ay = ⎦ w ( 2 a + b) ( b − a) 6b This problem requires V C = 0 Summing forces vertically for the section, we have + ↑Σ Fy = 0; w 1 ( 2a + b) ( b − a) − 2 6b ⎛a + ⎜ ⎝ b⎞ w ⎟ 2 2 =0 w w ( 2a + b) ( b − a) = ( 2a + b) 8 6b 4 ( b − a) = 3 b b = 4a a 1 = 4 b Problem 7- 3 6 The semicircular arch... from the publisher Engineering Mechanics - Statics Chapter 7 Problem 7- 3 2 Determine the normal force, shear force, and moment at a section passing through point D of the two-member frame Units Used: 3 kN = 10 N Given: kN m w = 0 .75 F = 4 kN a = 1.5 m d = 1.5 m b = 1.5 m e = 3 c = 2. 5 m f = 4 Solution: Σ MC = 0; −B x( c + d) + f ⎛ ⎞ ⎜ 2 2 ⎟F d = ⎝ e +f ⎠ fdF Bx = 2 2 0 B x = 1 .2 kN e + f ( c + d) Σ... Engineering Mechanics - Statics Chapter 7 g shown; (b) set M0 and L as given ( ) p Given: M0 = 500 N⋅ m L = 8m Solution: For 0 ≤ x≤ L 3 + ↑Σ Fy = 0; V1 = 0 ΣMx = 0; For L 3 ≤x≤ M1 = 0 2L 3 + ↑Σ Fy = 0; ΣMx = 0; For 2L 3 M2 = M0 ≤x≤L + ↑Σ Fy = 0; ΣMx = 0; ( b) V2 = 0 V3 = 0 M3 = 0 x1 = 0 , 0.01L L 3 x2 = L L 2L , 1.01 3 3 3 x3 = 2L 2L 3 , 3 V 1 ( x1 ) = 0 V 2 ( x2 ) = 0 V 3 ( x3 ) = 0 M1 ( x1 ) = 0 M2 ( x2 )... Engineering Mechanics - Statics ⎡ ( L − x) ⎣ 2 M ( x) = ⎢−w 2 Chapter 7 ⎤ − F ( L − x)⎥ 1 ⎦ kN⋅ m Force (kN) 40 V( x) 20 0 0 1 2 3 4 5 6 x Distance (m) Moment (kN-m) 0 50 M( x) 100 150 0 1 2 3 4 5 6 x Distance (m) Problem 7- 5 0 Draw the shear and moment diagrams for the beam Units Used: 3 kN = 10 N Given: a = 2m b = 4m w = 1.5 kN m Solution: ⎛ b − a⎞ − A b = ⎟ y ⎝ 2 ⎠ w( b − a) ⎜ 0 Ay = w ( b − a) 2b... publisher Engineering Mechanics - Statics Ay = F ΣMA = 0; −MA + MA = Chapter 7 ⎛ a ⎞ ⎜ 2 2⎟ ⎝ a +b ⎠ ⎛ F a ⎞ 2r = ⎜ 2 2⎟ ⎝ a +b ⎠ 2F r a 2 2 Ay = 300.00 lb 0 MA = 120 0 lb⋅ ft a +b ΣF x = 0; NC + Ax sin ( θ 1 ) + Ay cos ( θ 1 ) = 0 NC = − A x sin ( θ 1 ) − A y cos ( θ 1 ) ΣF y = 0; NC = −495 lb V C − Ax cos ( θ 1 ) + A y sin ( θ 1 ) = 0 V C = A x cos ( θ 1 ) − Ay sin ( θ 1 ) ΣMC = 0; V C = 70 .7 lb −MC... V1( x1) V2( x2) 0 V3( x3) 0.5 1 0 1 2 3 4 5 6 7 8 9 x1 , x2 , x3 Distance in m 600 Moment in N - m 400 M1( x1) M2( x2) 20 0 M3( x3) 0 0 1 2 3 4 5 6 7 8 9 x1 , x2 , x3 Distance in m Problem 7- 4 5 The beam will fail when the maximum shear force is V max or the maximum bending moment is Mmax Determine the magnitude M0 of the largest couple moments it will support 670 © 20 07 R C Hibbeler Published by Pearson . the publisher. Engineering Mechanics - Statics Chapter 7 A y F a a 2 b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = A y 300.00 lb= Σ M A = 0; M A − Fa a 2 b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2r+ 0= M A 2Fra a 2 b 2 + = M A 120 0 lb ft⋅= Σ F x . 0= F BC w 1 c 2 2 w 2 w 1 − () c 2 3 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ a 2 c 2 + ac = F BC 26 00 N= + → Σ F x = 0; A x c a 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BC = A x 24 00 N= + ↑ Σ F y = 0; A y w 1 c− 1 2 w 2 w 1 − () c− a a 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BC + 0= A y w 1 c 1 2 w 2 w 1 − () c+ a a 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F BC −=. publisher. Engineering Mechanics - Statics Chapter 7 Units Used: kip 10 3 lb= Given: F 500 lb= θ 2 30 deg= r 2 ft= a 3= θ 1 45 deg= b 4= Solution: Fb a 2 b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sin θ 2 () Fa a 2 b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ cos θ 2 () −