Engineering Mechanics - Statics Episode 1 Part 2 pps

Engineering Mechanics - Statics Chapter 2 + ↑ F Ry = SF y ; F RY F 1 − c c 2 d 2 + F 2 cos φ () − F 3 cos θ () += F RX 162.8− N= F RY 520.9− N= F R F RX 2 F RY 2 += F R 546 N= α atan F RY F RX ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α 72.64 deg= βα 180 deg+= β 252.6 deg= Problem 2-35 Three forces act on the bracket. Determine the magnitude and direction θ of F 1 so that the resultant force is directed along the positive x' axis and has a magnitude of F R . Units Used: kN 10 3 N= Given: F R 1kN= F 2 450 N= F 3 200 N= α 45 deg= β 30 deg= Solution: + → F Rx = SF x ; F R cos β () F 3 F 2 cos α () + F 1 cos θβ + () += + ↑ F Ry = SF y ; F R − sin β () F 2 sin α () F 1 sin θβ + () −= 41 © 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 1 cos θβ + () F R cos β () F 3 − F 2 cos α () −= F 1 sin θβ + () F 2 sin α () F R sin β () += θ atan F 2 sin α () F R sin β () + F R cos β () F 3 − F 2 cos α () − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ β −= θ 37deg= F 1 F R cos β () F 3 − F 2 cos α () − () 2 F 2 sin α () F R sin β () + () 2 += F 1 889 N= Problem 2-36 Determine the magnitude and direction, measured counterclockwise from the x ' axis, of the resultant force of the three forces acting on the bracket. Given: F 1 300 N= F 2 450 N= F 3 200 N= α 45 deg= β 30 deg= θ 20 deg= Solution: F Rx F 1 cos θβ + () F 3 + F 2 cos α () += F Rx 711.03 N= F Ry F 1 − sin θβ + () F 2 sin α () += F Ry 88.38 N= 42 © 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 F Rx 2 F Ry 2 += F R 717 N= φ (angle from x axis) φ atan F Ry F Rx ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = φ 7.1 deg= φ ' (angle from x' axis) φ ' βφ += φ ' 37.1deg= Problem 2-37 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. Given: F 1 800 N= F 2 600 N= θ 40 deg= c 12= d 5= Solution: + → F Rx = Σ F x ; F Rx F 1 cos θ () F 2 c c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= F Rx 59 N= + ↑ F Ry = Σ F y ; F Ry F 1 sin θ () F 2 d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ += F Ry 745 N= F R F Rx 2 F Ry 2 += F R 747 N= 43 © 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 θ atan F Ry F Rx ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 85.5 deg= Problem 2-38 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. Units Used: kN 10 3 N= Given: F 1 30 kN= F 2 26 kN= θ 30 deg= c 5= d 12= Solution: + → F Rx = Σ F x ; F Rx F 1 − sin θ () c c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F 2 −= F Rx 25− kN= + ↑ F Ry = Σ F y ; F Ry F 1 − cos θ () d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F 2 += F Ry 2− kN= F R F Rx 2 F Ry 2 += F R 25.1 kN= φ atan F Ry F Rx ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = φ 4.5 deg= β 180 deg φ += β 184.5 deg= 44 © 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 2-39 Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis. Given: F 1 60 lb= F 2 70 lb= F 3 50 lb= θ 1 60 deg= θ 2 45 deg= c 1= d 1= Solution: θ 3 atan d c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F Rx F 1 − cos θ 3 () F 2 sin θ 1 () −= F Rx 103− lb= F Ry F 1 sin θ 3 () F 2 cos θ 1 () − F 3 −= F Ry 42.6− lb= F R F Rx 2 F Ry 2 += F R 111.5 lb= θ 180 deg atan F Ry F Rx ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ += θ 202deg= Problem 2-40 Determine the magnitude of the resultant force F R = F 1 + F 2 and its direction, measured counterclockwise from the positive x axis by summing the rectangular or x, y components of the forces to obtain the resultant force. 45 © 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 1 600 N= F 2 800 N= F 3 450 N= θ 1 60 deg= θ 2 45 deg= θ 3 75 deg= Solution: + → F Rx = Σ F x ; F Rx F 1 cos θ 2 () F 2 sin θ 1 () −= F Rx 268.556− N= + ↑ F Ry = Σ F y ; F Ry F 1 sin θ 2 () F 2 cos θ 1 () += F Ry 824.264 N= F R F Rx 2 F Ry 2 += F R 867 N= θ 180 deg atan F Ry F Rx ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= θ 108deg= Problem 2-41 Determine the magnitude and direction of the resultant F R = F 1 + F 2 + F 3 of the three forces by summing the rectangular or x, y components of the forces to obtain the resultant force. Given: F 1 30 N= F 2 20 N= 46 © 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 3 50 N= θ 20 deg= c 3= d 4= Solution: F Rx F 1 − d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F 2 sin θ ()() − F 3 += F Rx 19.2 N= F Ry F 1 c c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F 2 cos θ () −= F Ry 0.8− N= F R F Rx 2 F Ry 2 += F R 19.2 N= θ atan F Ry − F Rx ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 2.4 deg= Problem 2-42 Determine the magnitude and orientation, measured counterclockwise from the positive y axis, of the resultant force acting on the bracket. Given: F A 700 N= 47 © 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 B 600 N= θ 20 deg= φ 30 deg= Solution: Scalar Notation: Suming the force components algebraically, we have F Rx = Σ F x ; F Rx F A sin φ () F B cos θ () −= F Rx 213.8− N= F Ry = Σ F y ; F Ry F A cos φ () F B sin θ () += F Ry 811.4 N= The magnitude of the resultant force F R is F R F Rx 2 F Ry 2 += F R 839 N= The directional angle θ measured counterclockwise from the positive x axis is θ atan F Rx F Ry ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 14.8 deg= Problem 2-43 Determine the magnitude and direction, measured counterclockwise from the positive x' axis, of the resultant force of the three forces acting on the bracket. Given: F 1 300 N= 48 © 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 2 200 N= F 3 180 N= θ 10 deg= θ 1 60 deg= c 5= d 12= Solution: + → F Rx = Σ F x ; F Rx F 1 sin θ 1 θ + () d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F 3 −= F Rx 115.8 N= + ↑ F Ry = Σ F y ; F Ry F 1 cos θ 1 θ + () F 2 + c c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F 3 += F Ry 371.8 N= F R F Rx 2 F Ry 2 += F R 389 N= φ atan F Ry F Rx ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = φ 72.7 deg= ψφ 90 deg θ 1 − () −= ψ 42.7 deg= Problem 2-44 Determine the x and y components of F 1 and F 2 . 49 © 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 1 200 N= F 2 150 N= θ 45 deg= φ 30 deg= Solution: F 1x F 1 sin θ () = F 1x 141.4 N= F 1y F 1 cos θ () = F 1y 141.4 N= F 2x F 2 − cos φ () = F 2x 129.9− N= F 2y F 2 sin φ () = F 2y 75 N= Problem 2-45 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. Given: F 1 200 N= F 2 150 N= θ 45 deg= φ 30 deg= 50 Given: © 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. [...]... without permission in writing from the publisher Engineering Mechanics - Statics Chapter 2 F R = F 1 + F2 + F − 2FF1 sin ( θ ) − 2F2 Fcos( θ ) 2 2FR 2 dF R dF 2 = 2F − 2F1 sin ( θ ) − 2F2 cos ( θ ) If F is a minimum, then FR = 2 ⎞ ⎛ dFR = 0⎟ ⎜ ⎝ dF ⎠ F = F 1 sin ( θ ) + F2 cos ( θ ) (F1 − F sin(θ ) )2 + (F cos (θ ) − F2 )2 F = 5.96 kN F R = 2. 3 kN Problem 2- 5 2 Express each of the three forces acting on the... 2 ⎞ ⎜ ⎟ ⎛ F2 ⎞ ⎜ β 2 ⎟ = acos ⎜ ⎟ ⎝ F2 ⎠ ⎜γ ⎟ ⎝ 2 ⎛ α 2 ⎞ ⎛ 11 3.9 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β 2 ⎟ = ⎜ 14 9.5 ⎟ deg ⎜ γ ⎟ ⎝ 72. 3 ⎠ ⎝ 2 Problem 2- 5 8 Express each force in Cartesian vector form Units Used: 3 kN = 10 N Given: F 1 = 5 kN F 2 = 2 kN θ 1 = 60 deg θ 2 = 60 deg θ 3 = 45 deg Solution: F 1v ⎛ cos ( θ 2) ⎞ ⎜ ⎟ = F1 ⎜ cos ( θ 3 ) ⎟ ⎜ cos θ ⎟ ⎝ ( 1) ⎠ ⎛ 2. 5 ⎞ F 1v = ⎜ 3.5 ⎟ kN ⎜ ⎟ ⎝ 2. 5 ⎠ F 2v ⎛0⎞ = F2 ⎜ 1. .. shown Given: F 1 = 400 N F 2 = 300 N F 3 = 20 0 N α 2 = 60 deg γ 2 = 60 deg c = 3 d = 4 Solution: cos ( α 2 ) + cos ( β 2 ) + cos ( γ 2 ) = 1 2 2 2 Solving for the positive root, 2 2 β 2 = acos ⎛ 1 − cos ( α 2 ) − cos ( γ 2 ) ⎞ ⎝ ⎠ F 1v ⎛ F1 ⎞ ⎜ ⎟ = ⎜ 0 ⎟ ⎜0 ⎟ ⎝ ⎠ F 2v β 2 = 45 deg ⎛ cos ( α 2) ⎞ ⎜ ⎟ = F2 ⎜ cos ( β 2 ) ⎟ ⎜ cos γ ⎟ ⎝ ( 2) ⎠ F 3v = ⎛0⎞ ⎜ −d ⎟ ⎟ 2 2⎜ c +d ⎝ c ⎠ F3 69 © 20 07 R C Hibbeler... −cos ( γ ) ⎟ ⎝ ⎠ F 2h = F2 ⎛ d ⎞ ⎜ 2 2⎟ ⎝ c +d ⎠ F 2v ⎛ F2h cos ( θ ) ⎞ ⎜ ⎟ = ⎜ −F 2h sin ( θ ) ⎟ ⎜ F ⎟ 2y ⎝ ⎠ F R = F1v + F2v ⎛ αR ⎞ ⎜ ⎟ ⎛ FR ⎞ ⎜ β R ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ R⎠ ⎛ 17 5 ⎞ F 1v = ⎜ 17 5 ⎟ N ⎜ ⎟ ⎝ 24 7.5 ⎠ F 2y = F 2 ⎛ c ⎞ ⎜ 2 2⎟ ⎝ c +d ⎠ ⎛ 17 3 .2 ⎞ F 2v = ⎜ 10 0 ⎟ N ⎜ ⎟ ⎝ 15 0 ⎠ F R = 369.3 N ⎛ α R ⎞ ⎛ 19 .5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β R ⎟ = ⎜ 78.3 ⎟ deg ⎜ γ ⎟ ⎝ 10 5.3 ⎠ ⎝ R⎠ Problem 2- 6 9 Determine the... permission in writing from the publisher Engineering Mechanics - Statics Chapter 2 Units Used: 3 kN = 10 N Given: F = 25 00 N F 1 = 15 00 N F 2 = 600 N θ 1 = 60 deg θ 2 = 45 deg Solution: Initial Guesses: F Rx = 10 0 N F Ry = 10 0 N P = 10 0 N Given + → F Rx = ΣF x; F Rx = P + F 2 cos ( θ 2 ) + F1 cos ( θ 1 + θ 2 ) + F Ry = ΣF y; F Ry = F2 sin ( θ 2 ) + F1 sin ( θ 1 + θ 2 ) ↑ F= 2 F Rx + F Ry ⎛ FRx ⎞ ⎜ ⎟ ⎜ FRy ⎟... of the resultant force Given: F 1 = 15 0 lb θ = 60 deg F 2 = 27 5 lb c = 4 F 3 = 75 lb d = 3 Solution: Find the components of each force ⎛ ⎞ ⎟ 2 2 ⎝ c +d ⎠ F 1x = F 1 ⎜ F 1v = d ⎛ F1x ⎞ ⎜ ⎟ ⎝ F1y ⎠ F 2x = 0 lb F 2v = ⎛ F2x ⎞ ⎜ ⎟ ⎝ F2y ⎠ ⎛ ⎞ ⎟ 2 2 ⎝ c +d ⎠ F 1y = F 1 ⎜ F 1v = −c ⎛ 90 ⎞ ⎜ ⎟ lb ⎝ 12 0 ⎠ F 2y = −F 2 F 2v = ⎛ 0 ⎞ ⎜ ⎟ lb ⎝ 27 5 ⎠ 55 © 20 07 R C Hibbeler Published by Pearson Education, Inc.,... Engineering Mechanics - Statics Chapter 2 Given: F 1 = 400 lb F 2 = 600 lb θ 1 = 45 deg θ 2 = 60 deg θ 3 = 60 deg θ 4 = 45 deg θ 5 = 30 deg Solution: F 1v ⎛ −cos ( θ 2) ⎞ ⎜ ⎟ = F1 ⎜ cos ( θ 3 ) ⎟ ⎜ cos θ ⎟ ( 1) ⎠ ⎝ ⎛ 20 0 ⎞ F 1v = ⎜ 20 0 ⎟ lb ⎜ ⎟ ⎝ 28 2.8 ⎠ F 2v ⎛ cos ( θ 5) cos ( θ 4 ) ⎞ ⎜ ⎟ = F2 ⎜ cos ( θ 5 ) sin ( θ 4 ) ⎟ ⎜ −sin θ ( 5) ⎟ ⎝ ⎠ ⎛ 367.4 ⎞ F 2v = ⎜ 367.4 ⎟ lb ⎜ ⎟ ⎝ −300 ⎠ Problem 2- 6 1. .. Engineering Mechanics - Statics Chapter 2 Problem 2- 6 3 Specify the coordinate direction angles of F 1 and F 2 and express each force as a cartesian vector Given: F 1 = 80 lb F 2 = 13 0 lb φ = 30 deg θ = 40 deg Solution: F 1v ⎛ cos ( φ ) cos ( θ ) ⎟ ⎞ ⎜ = F1 ⎜ −cos ( φ ) sin ( θ ) ⎟ ⎜ ⎟ sin ( φ ) ⎝ ⎠ ⎛ 1 ⎞ ⎜ ⎟ ⎛ F1v ⎞ ⎜ β 1 ⎟ = acos ⎜ ⎟ ⎝ F1 ⎠ ⎜γ ⎟ ⎝ 1 F 2v ⎛0⎞ ⎜ ⎟ = F2 0 ⎜ ⎟ ⎝ 1 ⎠ ⎛ 2 ⎞ ⎜ ⎟ ⎛ F2v... publisher Engineering Mechanics - Statics + ↑ F Ry = ΣF y; Chapter 2 F R( cos ( θ 1 ) ) = F 1 cos ( θ 1 + θ ) + F2 + c ⎛ ⎞ ⎜ 2 2 ⎟ F3 ⎝ c +d ⎠ ⎛ F1 ⎞ ⎜ ⎟ = Find ( F1 , θ ) ⎝θ ⎠ F 1 = 869 N θ = 21 . 3 deg Problem 2- 4 9 Determine the magnitude and direction, measured counterclockwise from the positive x' axis, of the resultant force acting on the bracket Given: F 1 = 300 N F 2 = 20 0 N F 3 = 18 0 N θ 1 = 60... Engineering Mechanics - Statics Chapter 2 Problem 2- 5 5 Determine the magnitude and orientation, measured clockwise from the positive x axis, of the resultant force of the three forces acting on the bracket Given: F 1 = 80 lb F 2 = 15 0 lb F 3 = 52lb θ = 55 deg φ = 25 deg c = 12 m d = 5m Solution: ⎛ ⎞ + F cos ( θ + φ ) 2 ⎝ c +d ⎠ F Rx = F 1 + F 3 ⎜ ⎛ d 2 F Rx = 12 6.05 lb 2 ⎞ ⎟ − F2 sin ( θ + φ ) 2 2 . publisher. Engineering Mechanics - Statics Chapter 2 F 1 20 0 N= F 2 15 0 N= θ 45 deg= φ 30 deg= Solution: F 1x F 1 sin θ () = F 1x 14 1.4 N= F 1y F 1 cos θ () = F 1y 14 1.4 N= F 2x F 2 − cos φ () = F 2x 12 9.9−. writing from the publisher. Engineering Mechanics - Statics Chapter 2 F R 2 F 1 2 F 2 2 + F 2 + 2FF 1 sin θ () − 2F 2 Fcos θ () −= 2F R dF R dF 2F 2F 1 sin θ () − 2F 2 cos θ () −= If F is a minimum,. 3= F 2 400 lb= d 4= F 3 300 lb= e 3= F 4 300 lb= f 4= Solution: F 1x F 1 −= F 1x 20 0− lb= F 1y 0lb= F 2x F 2 d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F 2x 320 lb= F 2y F 2 − c c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F 2y 24 0−