Engineering Mechanics - Statics Episode 1 Part 4 ppsx

Engineering Mechanics - Statics Chapter 2 Given: F 100 lb= a 2ft= b 8ft= c 6ft= d 4ft= e 2ft= Solution: r CD c− b e ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = u CD r CD r CD = r CB c− d− e ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = u CB r CB r CB = F BC Fu CD () u CB ⋅= F BC 10.5 lb= Problem 2-129 Determine the angle θ between pipe segments BA and BC. Given: F 100 lb= a 3ft= b 8ft= c 6ft= d 4ft= e 2ft= Solution: r BC c d e− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r BA a− 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = θ acos r BC r BA ⋅ r BC r BA ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 143.3 deg= 121 © 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 2-130 Determine the angles θ and φ made between the axes OA of the flag pole and AB and AC, respectively, of each cable. Given: F B 55 N= c 2m= F c 40 N= d 4m= a 6m= e 4m= b 1.5 m= f 3m= Solution: r AO 0 e− f− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r AB b e− af− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r AC c− e− df− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = θ acos r AB r AO ⋅ r AB r AO ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 74.4 deg= φ acos r AC r AO ⋅ r AC r AO ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = φ 55.4 deg= Problem 2-131 Determine the magnitude and coordinate direction angles of F 3 so that resultant of the three forces acts along the positive y axis and has magnitude F R . 122 © 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 R 600 lb= F 1 180 lb= F 2 300 lb= φ 40 deg= θ 30 deg= Solution: The initial guesses : F 3 100 lb= β 30 deg= α 10 deg= γ 60 deg= Given F Rx = Σ F x ; F 1 − F 2 cos θ () sin φ () + F 3 cos α () + 0= F Ry = Σ F y ; F 2 cos θ () cos φ () F 3 cos β () + F R = F Rz = Σ F z ; F 2 − sin θ () F 3 cos γ () + 0= cos α () 2 cos β () 2 + cos γ () 2 + 1= Solving: F 3 α β γ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find F 3 α , β , γ , () = α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 88.3 20.6 69.5 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= F 3 428.3 lb= 123 © 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 Determine the magnitude and coordinate direction angles of F 3 so that resultant of the three forces is zero. Given: F 1 180 lb= F 2 300 lb= φ 40 deg= θ 30 deg= Solution: The initial guesses : α 10 deg= β 30 deg= γ 60 deg= F 3 100 lb= Given F Rx = Σ F x ; F 1 − F 2 cos θ () sin φ () + F 3 cos α () + 0= F Ry = Σ F y ; F 2 cos θ () cos φ () F 3 cos β () + 0= F Rz = Σ F z ; F 2 − sin θ () F 3 cos γ () + 0= cos α () 2 cos β () 2 + cos γ () 2 + 1= Solving: F 3 α β γ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find F 3 α , β , γ , () = α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 87 142.9 53.1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= F 3 249.6 lb= 124 Problem 2-132 © 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 2-133 Resolve the force F into two components, one acting parallel and the other acting perpendicular to the u axis. Given: F 600 lb= θ 1 60 deg= θ 2 20 deg= Solution: F perpendicular F cos θ 1 θ 2 − () = F perpendicular 460lb= F parallel F sin θ 1 θ 2 − () = F parallel 386lb= Problem 2-134 The force F has a magnitude F and acts at the midpoint C of the thin rod. Express the force as a Cartesian vector. 125 © 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 80 lb= a 2ft= b 3ft= c 6ft= Solution: r CO b− 2 a 2 c− 2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = F v F r CO r CO = F v 34.3− 22.9 68.6− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= Problem 2-135 Determine the magnitude and direction of the resultant F R = F 1 + F 2 + F 3 of the three forces by first finding the resultant F ' = F 1 + F 3 and then forming F R = F ' + F 2 . Specify its direction measured counterclockwise from the positive x axis. 126 © 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 80 N= F 2 75 N= F 3 50 N= θ 1 30 deg= θ 2 30 deg= θ 3 45 deg= Solution: F 1v F 1 sin θ 1 () − cos θ 1 () ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F 2v F 2 cos θ 2 θ 3 + () sin θ 2 θ 3 + () ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F 3v F 3 cos θ 3 () sin θ 3 () ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F' F 1v F 3v += F' 4.6− 104.6 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= i 1 0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = j 0 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F R F' F 2v += F R 14.8 177.1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= F R 177.7 N= θ atan F R j F R i ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 85.2 deg= Problem 2-136 The leg is held in position by the quadriceps AB, which is attached to the pelvis at A. If the force exerted on this muscle by the pelvis is F , in the direction shown, determine the stabilizing force component acting along the positive y axis and the supporting force component acting along the negative x axis. Given: F 85 N= 127 © 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 1 55 deg= θ 2 45 deg= Solution: F x F cos θ 1 θ 2 − () = F x 83.7 N= F y F sin θ 1 θ 2 − () = F y 14.8 N= Problem 2-137 Determine the magnitudes of the projected components of the force F in the direction of the cables AB and AC . Given: F 60 12 40− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= a 3m= b 1.5 m= c 1m= d 0.75 m= e 1m= 128 © 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 Find the unit vectors, then use the dot product r AB a− d− e ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r AB 3− 0.8− 1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m= u AB r AB r AB = u AB 0.9− 0.2− 0.3 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r AC a− c b ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r AC 3− 1 1.5 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m= u AC r AC r AC = u AC 0.9− 0.3 0.4 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F AB Fu AB = F AB 78.5− 19.6− 26.2 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= F AC Fu AC = F AC 72.9− 24.3 36.4 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= Problem 2-138 Determine the magnitude and coordinate direction angles of F 3 so that resultant of the three forces is zero. Given: F 1 180 lb= φ 40 deg= F 2 300 lb= θ 30 deg= Solution: The initial guesses : α 10 deg= β 30 deg= γ 60 deg= F 3 100 lb= Given F Rx = Σ F x ; F 1 − F 2 cos θ () sin φ () + F 3 cos α () + 0= F Ry = Σ F y ; F 2 cos θ () cos φ () F 3 cos β () + 0= 129 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 2 F Rz = Σ F z ; F 2 − sin θ () F 3 cos γ () + 0= cos α () 2 cos β () 2 + cos γ () 2 + 1= Solving: F 3 α β γ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find F 3 α , β , γ , () = α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 87 142.9 53.1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= F 3 249.6 lb= Problem 2-139 Determine the angles θ and φ so that the resultant force is directed along the positive x axis and has magnitude F R. . Given: F 1 30 lb= F 2 30 lb= F R 20 lb= Solution: Initial Guesses: θ 20 deg= φ 20 deg= Given F 1 sin φ () F 2 sin θ () = F R 2 F 1 2 F 2 2 + 2 F 1 F 2 cos 180 deg θ − φ − () −= 130 © 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. [...]... from the publisher Engineering Mechanics - Statics Chapter 3 Problem 3 -1 Determine the magnitudes of F1 and F2 so that the particle is in equilibrium Given: F = 500 N θ 1 = 45 deg θ 2 = 30deg Solution: Initial Guesses F 1 = 1N F 2 = 1N Given + → ΣFx = 0; F 1 cos ( θ 1 ) + F2 cos ( θ 2 ) − F = 0 + F 1 sin ( θ 1 ) − F 2 sin ( θ 2 ) = 0 ↑ ΣFy = 0; ⎛ F1 ⎞ ⎜ ⎟ = Find ( F1 , F2) ⎝ F2 ⎠ ⎛ F1 ⎞ ⎛ 259 ⎞ ⎜ ⎟=⎜... without permission in writing from the publisher Engineering Mechanics - Statics Chapter 3 Given: F = 12 kN F 1 = 8 kN F 2 = 5 kN θ 1 = 45 deg Solution: Initial Guesses T = 1kN θ = 10 deg Given + → ΣFx = 0; F 1 − T cos ( θ ) + F 2 sin ( θ 1 ) = 0 + −T sin ( θ ) − F2 cos ( θ 1 ) + F = 0 ↑ ΣFy = 0; ⎛T⎞ ⎜ ⎟ = Find ( T , θ ) ⎝θ⎠ T = 14 . 31 kN θ = 36.27 deg Problem 3-7 Determine the maximum weight of the engine.. .Engineering Mechanics - Statics Chapter 2 ⎛θ⎞ ⎜ ⎟ = Find ( θ , φ ) ⎝φ⎠ θ = 70.5 deg φ = 70.5 deg Problem 2 -1 40 Determine the magnitude of the resultant force and its direction measured counterclockwise from the x axis Given: F 1 = 300 lb F 2 = 200 lb θ 1 = 40 deg θ 2 = 10 0 deg Solution: F Rx = F 1 cos ( 18 0 deg − θ 2 ) + F2 cos ( θ 1 ) F Rx = 205.3 lb F Ry = F 1 sin ( 18 0 deg − θ 2 )... publisher Engineering Mechanics - Statics Chapter 3 Problem 3-5 The members of a truss are connected to the gusset plate If the forces are concurrent at point O, determine the magnitudes of F and T for equilibrium Units Used: 3 kN = 10 N Given: F 1 = 8 kN F 2 = 5 kN θ 1 = 45 deg θ = 30 deg Solution: + → ΣFx = 0; −T cos ( θ ) + F 1 + F 2 sin ( θ 1 ) = 0 T = F1 + F2 sin ( θ 1 ) cos ( θ ) T = 13 .3 kN +... publisher Engineering Mechanics - Statics + ↑ ΣFy = 0; Chapter 3 F + F2 cos ( θ 2 ) + F 1 sin ( θ 1 ) − F 3 sin ( θ ) = 0 ⎛F⎞ ⎜ ⎟ = Find ( F , θ ) ⎝θ ⎠ F = 28.25 N θ = 53.02 deg Problem 3 -4 Determine the magnitude and angle θ of F so that the particle is in equilibrium Units Used: 3 kN = 10 N Given: F 1 = 4. 5 kN F 2 = 7.5 kN F 3 = 2.25 kN α = 60 deg φ = 30 deg Solution: Guesses: F = 1 kN θ = 1 Given... max ( θ 1 , θ 2 ) θ 2 = 11 . 54 deg TAB2 = 244 9 lb θ = 11 . 54 deg 13 9 © 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. .. from the publisher Engineering Mechanics - Statics Chapter 3 Problem 3-2 0 The sack has weight W and is supported by the six cords tied together as shown Determine the tension in each cord and the angle θ for equilibrium Cord BC is horizontal Given: W = 15 lb θ 1 = 30 deg θ 2 = 45 deg θ 3 = 60 deg Solution: Guesses TBE = 1 lb TAB = 1 lb θ = 20deg TBC = 1 lb TAC = 1 lb TCD = 1 lb TAH = 1 lb Given At H:... permission in writing from the publisher Engineering Mechanics - Statics Chapter 3 ⎛ TBE ⎞ ⎜ ⎟ ⎜ TAB ⎟ ⎜ TBC ⎟ ⎜ ⎟ ⎜ TAC ⎟ = Find ( TBE , TAB , TBC , TAC , TCD , TAH , θ ) ⎜T ⎟ ⎜ CD ⎟ ⎜ TAH ⎟ ⎜ ⎟ ⎝ θ ⎠ ⎛ TBE ⎞ ⎜ ⎟ ⎛ 10 .98 ⎞ ⎜ ⎟ TAB ⎟ ⎜ ⎜ 7.76 ⎟ ⎜ ⎟ ⎜ TBC ⎟ = ⎜ 4. 02 ⎟ lb ⎜ TAC ⎟ ⎜ 10 .98 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ 13 .45 ⎟ ⎜ TCD ⎟ ⎜ TAH ⎟ ⎜ 15 .00 ⎟ ⎝ ⎠ ⎝ ⎠ θ = 45 .00 deg Problem 3-2 1 Each cord can sustain a maximum tension... publisher Engineering Mechanics - Statics Chapter 3 Solution: F = 1kN The initial guesses: θ = 30deg Given Equations of equilibrium: + → Σ Fx = 0; + ↑ Σ F y = 0; ⎛ −d ⎞ F + F cos ( θ ) = 0 ⎜ 2 2⎟ 1 ⎝ c +d ⎠ c ⎛ ⎞ ⎜ 2 2 ⎟ F1 − F2 − F sin ( θ ) = 0 ⎝ c +d ⎠ ⎛F⎞ ⎜ ⎟ = Find ( F , θ ) ⎝θ ⎠ F = 4. 94 kN θ = 31. 8 deg Problem 3-3 Determine the magnitude of F and the orientation θ of the force F 3 so that the particle... writing from the publisher Engineering Mechanics - Statics Chapter 3 Problem 3 -1 4 The unstretched length of spring AB is δ If the block is held in the equilibrium position shown, determine the mass of the block at D Given: δ = 2m a = 3m b = 3m c = 4m kAB = 30 N kAC = 20 N kAD = 40 N g = 9. 81 m m m m 2 s Solution: F AB = kAB ( 2 2 a +c −δ ) The initial guesses: mD = 1 kg F AC = 1 N Given + → Σ Fx = 0; . the publisher. Engineering Mechanics - Statics Chapter 2 θ 1 55 deg= θ 2 45 deg= Solution: F x F cos θ 1 θ 2 − () = F x 83.7 N= F y F sin θ 1 θ 2 − () = F y 14 .8 N= Problem 2 -1 37 Determine the. publisher. Engineering Mechanics - Statics Chapter 3 Given: F 12 kN= F 1 8kN= F 2 5kN= θ 1 45 deg= Solution: Initial Guesses T 1kN= θ 10 deg= Given + → Σ F x = 0; F 1 Tcos θ () − F 2 sin θ 1 () +. from the publisher. Engineering Mechanics - Statics Chapter 2 Given: F 1 80 N= F 2 75 N= F 3 50 N= θ 1 30 deg= θ 2 30 deg= θ 3 45 deg= Solution: F 1v F 1 sin θ 1 () − cos θ 1 () ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ =