Engineering Mechanics - Statics Chapter 2 α R β R γ R ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos F R F R ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α R β R γ R ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 75.4 90 165.4 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= Problem 2-68 Determine the magnitude and coordinate direction angles of the resultant force. Given: F 1 350 N= α 60 deg= F 2 250N= β 60 deg= c 3= γ 45 deg= d 4= θ 30 deg= Solution: F 1v F 1 cos α () cos β () cos γ () − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = F 1v 175 175 247.5− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= F 2h F 2 d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F 2y F 2 c c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F 2v F 2h cos θ () F 2h − sin θ () F 2y ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = F 2v 173.2 100− 150 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= F R F 1v F 2v += F R 369.3 N= α R β R γ R ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos F R F R ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α R β R γ R ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 19.5 78.3 105.3 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= Problem 2-69 Determine the magnitude and coordinate direction angles of F 3 so that the resultant of the three 71 © 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 2 forces acts along the positive y axis and has magnitude F. Given: F 600 lb= F 1 180 lb= F 2 300 lb= α 1 30 deg= α 2 40 deg= Solution: Initial guesses: α 40 deg= γ 50 deg= β 50 deg= F 3 45 lb= Given F Rx = Σ F x ; 0 F 1 − F 2 cos α 1 () sin α 2 () + F 3 cos α () += F Ry = Σ F y ; FF 2 cos α 1 () cos α 2 () F 3 cos β () += F Rz = Σ F z ; 0 F 2 − sin α 1 () F 3 cos γ () += cos α () 2 cos β () 2 + cos γ () 2 + 1= F 3 α β γ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find F 3 α , β , γ , () = F 3 428lb= α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 88.3 20.6 69.5 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= Problem 2-70 Determine the magnitude and coordinate direction angles of F 3 so that the resultant of the three forces is zero. 72 © 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 2 Given: F 1 180 lb= α 1 30 deg= F 2 300 lb= α 2 40 deg= Solution: Initial guesses: α 40 deg= γ 50 deg= β 50 deg= F 3 45 lb= Given F Rx = Σ F x ; 0 F 1 − F 2 cos α 1 () sin α 2 () + F 3 cos α () += F Ry = Σ F y ; 0 F 2 cos α 1 () cos α 2 () F 3 cos β () += F Rz = Σ F z ; 0 F 2 − sin α 1 () F 3 cos γ () += cos α () 2 cos β () 2 + cos γ () 2 + 1= F 3 α β γ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find F 3 α , β , γ , () = F 3 250lb= α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 87.0 142.9 53.1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= Problem 2-71 Specify the magnitude F 3 and directions α 3 , β 3 , and γ 3 of F 3 so that the resultant force of the three forces is F R . Units Used: kN 10 3 N= Given: F 1 12 kN= c 5= F 2 10 kN= d 12= θ 30 deg= 73 © 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 2 F R 0 9 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= Solution: Initial Guesses: F 3x 1kN= F 3y 1kN= F 3z 1kN= Given F R F 3x F 3y F 3z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ F 1 0 cos θ () sin θ () − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ + F 2 c 2 d 2 + d− 0 c ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ += F 3x F 3y F 3z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find F 3x F 3y , F 3z , () = F 3 F 3x F 3y F 3z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = F 3 9.2 1.4− 2.2 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= F 3 9.6 kN= α 3 β 3 γ 3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos F 3 F 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α 3 β 3 γ 3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 15.5 98.4 77.0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= Problem 2-72 The pole is subjected to the force F , which has components acting along the x,y,z axes as shown. Given β and γ , determine the magnitude of the three components of F . Units Used: kN 1000 N= Given: F 3kN= β 30 deg= γ 75 deg= Solution: cos α () 2 cos β () 2 + cos γ () 2 + 1= ( ) 74 © 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 2 α acos cos β () 2 − cos γ () 2 − 1+ ( ) = α 64.67 deg= F x F cos α () = F y F cos β () = F z F cos γ () = F x 1.28 kN= F y 2.60 kN= F z 0.8 kN= Problem 2-73 The pole is subjected to the force F which has components F x and F z . Determine the magnitudes of F and F y . Units Used: kN 1000 N= Given: F x 1.5 kN= F z 1.25 kN= β 75 deg= Solution: cos α () 2 cos β () 2 + cos γ () 2 + 1= F x F ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 cos β () 2 + F z F ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + 1= F F x 2 F z 2 + 1 cos β () 2 − = F 2.02 kN= F y F cos β () = F y 0.5 kN= Problem 2-74 The eye bolt is subjected to the cable force F which has a component F x along the x axis, a component F z along the z axis, and a coordinate direction angle β. Determine the magnitude of F . 75 © 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 2 Given: F x 60 N= F z 80− N= β 80 deg= Solution: F y F cos β () = F y F x 2 F z 2 + F y 2 + cos β () = F y F x 2 F z 2 + 1 cos β () 2 − cos β () = F y 17.6 N= FF x 2 F y 2 + F z 2 += F 102 N= Problem 2-75 Three forces act on the hook. If the resultant force F R has a magnitude and direction as shown, determine the magnitude and the coordinate direction angles of force F 3 . Given: F R 120 N= F 1 80 N= F 2 110 N= c 3= d 4= θ 30 deg= φ 45 deg= 76 © 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 2 Solution: F 1v F 1 c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ d 0 c ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F 1v 64 0 48 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= F 2v F 2 0 0 1− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F 2v 0 0 110− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= F Rv F R cos φ () sin θ () cos φ () cos θ () sin φ () ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = F Rv 42.4 73.5 84.9 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= F 3v F Rv F 1v − F 2v −= F 3v 21.6− 73.5 146.9 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= F 3v 165.6 N= α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos F 3v F 3v ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 97.5 63.7 27.5 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= Problem 2-76 Determine the coordinate direction angles of F 1 and F R . Given: F R 120 N= F 1 80 N= F 2 110 N= c 3= d 4= θ 30 deg= φ 45 deg= 77 © 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 2 Solution: F 1v F 1 c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ d 0 c ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F 1v 64 0 48 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= α 1 β 1 γ 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos F 1v F 1v ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α 1 β 1 γ 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 36.9 90 53.1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= F Rv F R cos φ () sin θ () cos φ () cos θ () sin φ () ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = F Rv 42.4 73.5 84.9 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= α R β R γ R ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos F Rv F Rv ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α R β R γ R ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 69.3 52.2 45 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= Problem 2-77 The pole is subjected to the force F , which has components acting along the x, y, z axes as shown. Given the magnitude of F and the angles α and γ , determine the magnitudes of the components of F . Given: F 80 N= α 60 deg= γ 45 deg= Solution: β acos 1 cos α () 2 − cos γ () 2 −− () = β 120deg= F x F cos α () = F y F cos β () = F z F cos γ () = F x 40 N= F y 40 N= F z 56.6 N= 78 © 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 2 Problem 2-78 Two forces F 1 and F 2 act on the bolt. If the resultant force F R has magnitude F R and coordinate direction angles α and β, as shown, determine the magnitude of F 2 and its coordinate direction angles. Given: F 1 20 lb= F R 50 lb= α 110 deg= β 80 deg= Solution: cos α () 2 cos β () 2 + cos γ () 2 + 1= γ acos 1 cos α () 2 − cos β () 2 −− () = γ 157.44 deg= Initial Guesses F 2x 1lb= F 2y 1lb= F 2z 1lb= Given F R cos α () cos β () cos γ () ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ F 1 0 0 1− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2x F 2y F 2z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ += F 2x F 2y F 2z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find F 2x F 2y , F 2z , () = F 2 F 2x F 2y F 2z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = F 2 17.1− 8.7 26.2− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= F 2 32.4 lb= α 2 β 2 γ 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos F 2 F 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α 2 β 2 γ 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 121.8 74.5 143.8 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= Problem 2-79 Given r 1 , r 2 , and r 3 , determine the magnitude and direction of r2r 1 r 2 − 3r 3 += . 79 © 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 2 Given: r 1 3 4− 3 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m= r 2 4 0 5− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m= r 3 3 2− 5 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m= Solution: r 2r 1 r 2 − 3r 3 += r 11 14− 26 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m= r 31.5 m= α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos r r ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 69.6 116.4 34.4 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= Problem 2-80 Represent the position vector r acting from point A(a, b, c) to point B(d, e, f) in Cartesian vector form. Determine its coordinate direction angles and find the distance between points A and B. Given: a 3m= b 5m= c 6m= d 5m= e 2− m= f 1m= Solution: r da− eb− fc− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r 2 7− 5− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m= r 8.8 m= 80 © 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. [...]... writing from the publisher Engineering Mechanics - Statics Chapter 2 F 2 = 50 lb d = 2. 5 ft a = 6 ft e = 12 b = 2 ft f = 5 Solution: rAC ⎛ −d ⎞ ⎜ ⎟ −c ⎟ = ⎜ ⎜ d⎟ ⎜e f ⎟ ⎝ ⎠ rAB ⎛b⎞ = ⎜ −c ⎟ ⎜ ⎟ ⎝ −a ⎠ F R = F1v + F2v F 1v = F1 F 2v = F2 ⎛ 26 .2 ⎞ F 1v = ⎜ −41.9 ⎟ lb ⎜ ⎟ ⎝ 62. 9 ⎠ rAC rAC ⎛ 6.1 ⎞ F 2v = ⎜ − 12. 1 ⎟ kg ⎜ ⎟ ⎝ −18 .2 ⎠ rAB rAB ⎛ − 12. 8 ⎞ F R = ⎜ −68.7 ⎟ lb ⎜ ⎟ ⎝ 22 .8 ⎠ F R = 73.5 lb ⎛α ⎞ ⎜... by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 2 Solution: rAD ⎛ −c ⎞ ⎜ 2 ⎟ ⎜ ⎟ ⎜ d ⎟ = ⎜ 2 ⎟ ⎜e b⎟ ⎜ − ⎟ 2 2⎠ ⎛ ⎜ ⎜ = ⎜ ⎜ ⎜ ⎜a + ⎝ rCD rAD ⎛ −1 ⎞ ⎜ 1 ⎟m = ⎜ ⎟ ⎝ −0.5 ⎠ rAD = 1.5 m rBD = −rAD rBD = 1.5 m ⎞ ⎟ 2 ⎟ d ⎟ 2 ⎟ b e⎟ − ⎟ 2 2⎠ c rCD ⎛1⎞ ⎜ ⎟ = 1 m ⎜ ⎟ ⎝1⎠ rCD = 1.7 m Problem 2- 9 0 Express force F as a Cartesian vector; then determine its... the publisher Engineering Mechanics - Statics Chapter 2 Solution: Write the vectors and unit vectors r1v ⎛ sin ( ε ) cos ( φ ) ⎟ ⎞ ⎜ = r1 ⎜ −sin ( ε ) sin ( φ ) ⎟ ⎜ cos ( ε ) ⎟ ⎝ ⎠ ⎛ 5.01 ⎞ r1v = ⎜ 2. 89 ⎟ m ⎜ ⎟ ⎝ 6.89 ⎠ r2v ⎛ cos ( α ) ⎟ ⎞ ⎜ = r2 ⎜ cos ( β ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠ ⎛ 3 ⎞ r2v = ⎜ 4 .24 ⎟ m ⎜ ⎟ ⎝ −3 ⎠ u1 = r1v r1v u2 = r2v r2v ⎛ 0.557 ⎞ u1 = ⎜ −0. 321 ⎟ ⎜ ⎟ ⎝ 0.766 ⎠ ⎛ 0.5 ⎞ u2 = ⎜ 0.707 ⎟... the publisher Engineering Mechanics - Statics Chapter 2 Solution: Find the position vectors, then the force vectors rDC ⎛a⎞ = ⎜ −b ⎟ ⎜ ⎟ ⎝ −e ⎠ rDA ⎛x⎞ = ⎜ y ⎟ ⎜ ⎟ ⎝ −e ⎠ rDB ⎛ −c ⎞ = ⎜d ⎟ ⎜ ⎟ ⎝ −e ⎠ F 1v = F1 F 2v = F2 F 3v = F3 ⎛ 28 2.4 ⎞ F 1v = ⎜ −317.6 ⎟ N ⎜ ⎟ ⎝ − 423 .5 ⎠ rDC rDC ⎛ 23 0.8 ⎞ F 2v = ⎜ 173.1 ⎟ N ⎜ ⎟ ⎝ 27 7 ⎠ rDA rDA ⎛ −191.5 ⎞ F 3v = ⎜ 127 .7 ⎟ N ⎜ ⎟ ⎝ −766 .2 ⎠ rDB rDB Find the resultant,... 500 N F 2 = 400 N F R = 1000 N a = 1m b = 2m c = 2m d = 3m Solution: Initial Guesses F3 = 1 N x = 1m y = 1m Given ⎛0⎞ ⎛ −c ⎞ ⎛a⎞ ⎛x ⎞ F3 F2 ⎞⎜ ⎟ ⎛ ⎞⎜ ⎟ ⎜ 0 ⎟ = ⎛ F1 ⎞ ⎜ 0 ⎟ + ⎛ −b + y FR ⎜ ⎟ ⎜ 2 2 ⎟⎜ ⎟ ⎜ 2 2 2 ⎟⎜ ⎟ ⎜ 2 2 2 ⎟⎜ ⎟ c + d ⎠ ⎝ −d ⎠ ⎝ a + b + d ⎠ ⎝ −d ⎠ ⎝ x + y + d ⎠ ⎝ −d ⎠ ⎝ −1 ⎠ ⎝ ⎛ F3 ⎞ ⎜ ⎟ ⎜ x ⎟ = Find ( F3 , x , y) ⎜ y ⎟ ⎝ ⎠ ⎛ x ⎞ ⎛ 1.9 ⎞ ⎜ ⎟=⎜ ⎟m ⎝ y ⎠ ⎝ 2. 4 ⎠ F 3 = 380 N 96 © 20 07 R... publisher Engineering Mechanics - Statics Chapter 2 y = 1 ft Given Fx = ⎛ x − a ⎞F ⎜ ⎟ ⎝ L ⎠ 2 2 2 2 L = ( x − a) + y + z ⎛x⎞ ⎜ ⎟ = Find ( x , y) ⎝ y⎠ ⎛ x ⎞ ⎛ 8.67 ⎞ ⎜ ⎟=⎜ ⎟ ft ⎝ y ⎠ ⎝ 1.89 ⎠ Problem 2- 1 03 Each of the four forces acting at E has magnitude F Express each force as a Cartesian vector and determine the resultant force Units used: 3 kN = 10 N Given: F = 28 kN a = 4m b = 6m c = 12 m Solution:... without permission in writing from the publisher Engineering Mechanics - Statics Chapter 2 Given: a = 400 b = 125 θ = 25 deg Solution: rAB ⎡a + b sin ( θ ) ⎤ ⎢ ⎥ = −( b cos ( θ ) ) mm ⎢ ⎥ 0 ⎣ ⎦ rAB = 467 mm Problem 2- 8 9 Determine the length of wires AD, BD, and CD The ring at D is midway between A and B Given: a = 0.5 m b = 1.5 m c = 2m d = 2m e = 0.5 m 86 © 20 07 R C Hibbeler Published by Pearson Education,... Engineering Mechanics - Statics Chapter 2 = ( Ax Bx + Ay By + Az Bz) + ( Ax Dx + Ay Dy + Az Dz) = ( A⋅ B) + ( A⋅ D) (QED) Problem 2- 1 08 Cable BC exerts force F on the top of the flagpole Determine the projection of this force along the z axis of the pole Given: F = 28 N a = 12 m b = 6m c = 4m Solution: rBC ⎛b⎞ ⎜ ⎟ = −c ⎜ ⎟ ⎝ −a ⎠ Fv = F ⎛0⎞ ⎜ ⎟ k = 0 ⎜ ⎟ ⎝1⎠ rBC rBC F z = −Fv k F z = 24 N Problem 2- 1 09.. .Engineering Mechanics - Statics Chapter 2 α = acos ⎛ ⎜ d − a⎞ α = 76.9 deg β = acos ⎛ ⎜ e − b⎞ β = 1 42 deg γ = acos ⎛ ⎜ f − c⎞ γ = 124 deg ⎟ ⎝ r ⎠ ⎟ ⎝ r ⎠ ⎟ ⎝ r ⎠ Problem 2- 8 1 A position vector extends from the origin to point A(a, b, c) Determine the angles α, β, γ which the tail of the vector makes with the x, y, z axes, respectively Given: a = 2m b = 3m c = 6m Solution: ⎛a⎞ r = ⎜b⎟ ⎜ ⎟ ⎝c ⎠ 2 ... ) ) + ( c cos ( φ ) ) 2 2 2 r=2m 87 © 20 07 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 2 ⎡ −( c cos ( φ ) )⎤ . F R cos α () cos β () cos γ () ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ F 1 0 0 1− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2x F 2y F 2z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ += F 2x F 2y F 2z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find F 2x F 2y , F 2z , () = F 2 F 2x F 2y F 2z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = F 2 17.1− 8.7 26 .2 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= F 2 32. 4 lb= α 2 β 2 γ 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos F 2 F 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α 2 β 2 γ 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 121 .8 74.5 143.8 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= Problem. deg= Solution: cos α () 2 cos β () 2 + cos γ () 2 + 1= F x F ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 cos β () 2 + F z F ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + 1= F F x 2 F z 2 + 1 cos β () 2 − = F 2. 02 kN= F y F cos β () = F y 0.5 kN= Problem 2- 7 4 The eye. publisher. Engineering Mechanics - Statics Chapter 2 r AD c− 2 d 2 e 2 b 2 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = r AD 1− 1 0.5− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m= r AD 1.5 m= r BD r AD −= r BD 1.5 m= r CD c 2 d 2 a b 2 + e 2 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ =