Engineering Mechanics - Statics Chapter α = acos ⎛ ⎜ d − a⎞ α = 76.9 deg β = acos ⎛ ⎜ e − b⎞ β = 142 deg γ = acos ⎛ ⎜ f − c⎞ γ = 124 deg ⎟ ⎝ r ⎠ ⎟ ⎝ r ⎠ ⎟ ⎝ r ⎠ Problem 2-81 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⎞ r = ⎜3⎟ m ⎜ ⎟ ⎝6⎠ ⎛α ⎞ ⎜ ⎟ ⎛ r ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ r ⎠ ⎜γ ⎟ ⎝ ⎠ ⎛ α ⎞ ⎛ 73.4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β ⎟ = ⎜ 64.6 ⎟ deg ⎜ γ ⎟ ⎝ 31.0 ⎠ ⎝ ⎠ Problem 2-82 Express the position vector r in Cartesian vector form; then determine its magnitude and coordinate direction angles Given: a = 4m 81 © 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 b = 8m c = 3m d = 4m Solution: ⎛ −c ⎞ r = ⎜ −d − b ⎟ ⎜ ⎟ ⎝ a ⎠ ⎛α ⎞ ⎜ ⎟ ⎛ r ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ r ⎠ ⎜γ ⎟ ⎝ ⎠ ⎛ −3 ⎞ r = ⎜ −12 ⎟ m ⎜ ⎟ ⎝ ⎠ r = 13 m ⎛ α ⎞ ⎛ 103.3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β ⎟ = ⎜ 157.4 ⎟ deg ⎜ γ ⎟ ⎝ 72.1 ⎠ ⎝ ⎠ Problem 2-83 Express the position vector r in Cartesian vector form; then determine its magnitude and coordinate direction angles Given: a = ft b = ft c = ft θ = 30 deg φ = 20 deg Solution: ⎛ −c cos ( φ ) sin ( θ ) ⎞ ⎜ ⎟ r = ⎜ a − c cos ( φ ) cos ( θ ) ⎟ ⎜ b + c sin ( φ ) ⎟ ⎝ ⎠ ⎛ −2.35 ⎞ r = ⎜ 3.93 ⎟ ft ⎜ ⎟ ⎝ 3.71 ⎠ r = 5.89 ft 82 © 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 ⎛α ⎞ ⎜ ⎟ ⎛ r ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ r ⎠ ⎜γ ⎟ ⎝ ⎠ ⎛ α ⎞ ⎛ 113.5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β ⎟ = ⎜ 48.2 ⎟ deg ⎜ γ ⎟ ⎝ 51 ⎠ ⎝ ⎠ Problem 2-84 Determine the length of the connecting rod AB by first formulating a Cartesian position vector from A to B and then determining its magnitude Given: b = 16 in a = in α = 30 deg Solution: r = ⎛ a sin ( α ) + b ⎞ ⎜ ⎟ ⎝ −a cos ( α ) ⎠ r= ⎛ 18.5 ⎞ ⎜ ⎟ in ⎝ −4.3 ⎠ r = 19 in Problem 2-85 Determine the length of member AB of the truss by first establishing a Cartesian position vector from A to B and then determining its magnitude Given: a = 1.2 m b = 0.8 m c = 0.3 m d = 1.5 m θ = 40 deg 83 © 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 Solution: ⎛ c + d cot ( θ ) ⎞ ⎜ ⎟ ⎝ d−a ⎠ r = r= ⎛ 2.09 ⎞ ⎜ ⎟m ⎝ 0.3 ⎠ r = 2.11 m Problem 2- 86 The positions of point A on the building and point B on the antenna have been measured relative to the electronic distance meter (EDM) at O Determine the distance between A and B Hint: Formulate a position vector directed from A to B; then determine its magnitude Given: a = 460 m b = 653 m α = 60 deg β = 55 deg θ = 30 deg φ = 40 deg Solution: rOA ⎛ −a cos ( φ ) sin ( θ ) ⎟ ⎞ ⎜ = ⎜ a cos ( φ ) cos ( θ ) ⎟ ⎜ ⎟ a sin ( φ ) ⎝ ⎠ rOB ⎛ −b cos ( β ) sin ( α ) ⎟ ⎞ ⎜ = ⎜ −b cos ( β ) cos ( α ) ⎟ ⎜ ⎟ b sin ( β ) ⎝ ⎠ rAB = rOB − rOA ⎛ −148.2 ⎞ ⎜ ⎟ rAB = −492.4 m ⎜ ⎟ ⎝ 239.2 ⎠ rAB = 567.2 m Problem 2-87 Determine the lengths of cords ACB and CO The knot at C is located midway between A and B 84 © 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 Given: a = ft b = ft c = ft Solution: rAB ⎛c ⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝ −a ⎠ rAC = rOA ⎛0⎞ ⎜ ⎟ = ⎜ ⎟ ⎝a⎠ rAB rAB = 7.8 ft rOC = rOA + rAC rOC = 3.91 ft Problem 2-88 Determine the length of the crankshaft AB by first formulating a Cartesian position vector from A to B and then determining its magnitude 85 © 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 Given: a = 400 b = 125 θ = 25 deg Solution: rAB ⎡a + b sin ( θ ) ⎤ ⎢ ⎥ = −( b cos ( θ ) ) mm ⎢ ⎥ ⎣ ⎦ rAB = 467 mm Problem 2-89 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 © 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 Solution: rAD ⎛ −c ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ d ⎟ = ⎜ ⎟ ⎜e b⎟ ⎜ − ⎟ ⎝2 2⎠ ⎛ ⎜ ⎜ = ⎜ ⎜ ⎜ ⎜a + ⎝ rCD rAD ⎛ −1 ⎞ ⎜ ⎟m = ⎜ ⎟ ⎝ −0.5 ⎠ rAD = 1.5 m rBD = −rAD rBD = 1.5 m ⎞ ⎟ ⎟ d ⎟ ⎟ b e⎟ − ⎟ 2⎠ c rCD ⎛1⎞ ⎜ ⎟ = m ⎜ ⎟ ⎝1⎠ rCD = 1.7 m Problem 2-90 Express force F as a Cartesian vector; then determine its coordinate direction angles Given: F = 600 lb c = ft a = 1.5 ft φ = 60 deg b = ft Solution: r = bi + ( a + c sin ( φ ) ) j + ( − c cos ( φ ) ) k r = b + ( a + c sin ( φ ) ) + ( c cos ( φ ) ) 2 r=2m 87 © 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 ⎡ −( c cos ( φ ) )⎤ ⎥ r ⎣ ⎦ b d = F r ⎛ a + c sin ( φ ) ⎞ e = F⎜ ⎟ r ⎝ ⎠ f = F⎢ d = 452 lb e = 370 lb f = −136 lb F = ( di + ej + fk) lb d α = acos ⎛ ⎞ ⎜ ⎟ α = 41.1 deg e β = acos ⎛ ⎞ ⎜ ⎟ β = 51.9 deg f γ = acos ⎛ ⎞ ⎜ ⎟ γ = 103 deg ⎝F⎠ ⎝ F⎠ ⎝ F⎠ Problem 2-91 Express force F as a Cartesian vector; then determine its coordinate direction angles Given: a = 1.5 ft b = ft c = ft θ = 60 deg F = 600 lb Solution: b ⎛ ⎞ ⎜ a + c sin ( θ ) ⎟ r = ⎜ ⎟ ⎝ −c cos ( θ ) ⎠ F = F ⎛ 452 ⎞ ⎜ ⎟ F = 370 lb ⎜ ⎟ ⎝ −136 ⎠ r r ⎛α ⎞ ⎜ ⎟ ⎛ F ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ F ⎠ ⎜γ ⎟ ⎝ ⎠ ⎛ α ⎞ ⎛ 41.1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β ⎟ = ⎜ 51.9 ⎟ deg ⎜ γ ⎟ ⎝ 103.1 ⎠ ⎝ ⎠ 88 © 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 Problem 2-92 Determine the magnitude and coordinate direction angles of the resultant force acting at point A Given: F = 150 N F = 200 N a = 1.5 m b = 4m c = 3m d = 2m e = 3m θ = 60 deg Solution: Define the position vectors and then the forces rAB ⎛ e cos ( θ ) ⎞ = ⎜ a + e sin ( θ ) ⎟ ⎜ ⎟ −b ⎝ ⎠ rAC ⎛ c ⎞ = ⎜a − d⎟ ⎜ ⎟ ⎝ −b ⎠ F 1v = F1 F 2v = F2 ⎛ 38 ⎞ F 1v = ⎜ 103.8 ⎟ N ⎜ ⎟ ⎝ −101.4 ⎠ rAB rAB ⎛ 119.4 ⎞ F 2v = ⎜ −19.9 ⎟ N ⎜ ⎟ ⎝ −159.2 ⎠ rAC rAC Add the forces and find the magnitude of the resultant F R = F1v + F2v ⎛ 157.4 ⎞ F R = ⎜ 83.9 ⎟ N ⎜ ⎟ ⎝ −260.6 ⎠ F R = 316 N Find the direction cosine angles ⎛α ⎞ ⎜ ⎟ ⎛ FR ⎞ ⎜ β ⎟ = acos ⎜ ⎟ ⎝ FR ⎠ ⎜γ ⎟ ⎝ ⎠ ⎛ α ⎞ ⎛ 60.1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β ⎟ = ⎜ 74.6 ⎟ deg ⎜ γ ⎟ ⎝ 145.6 ⎠ ⎝ ⎠ 89 © 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 Problem 2-93 The plate is suspended using the three cables which exert the forces shown Express each force as a Cartesian vector Given: F BA = 350 lb F CA = 500 lb F DA = 400 lb a = ft b = ft c = ft d = 14 ft e = ft f = ft g = ft Solution: rBA ⎛ −e − g ⎞ ⎜ ⎟ = a+b ⎜ ⎟ ⎝ d ⎠ rCA ⎛f⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝d⎠ rDA ⎛ −g ⎞ ⎜ ⎟ = −c ⎜ ⎟ ⎝d⎠ F BAv = FBA F CAv = FCA F DAv = FDA rBA rBA rCA rCA rDA rDA F BAv ⎛ −109.2 ⎞ ⎜ 131 ⎟ lb = ⎜ ⎟ ⎝ 305.7 ⎠ ⎛ 102.5 ⎞ ⎜ ⎟ F CAv = 102.5 lb ⎜ ⎟ ⎝ 478.5 ⎠ ⎛ −52.1 ⎞ ⎜ ⎟ F DAv = −156.2 lb ⎜ ⎟ ⎝ 364.5 ⎠ 90 © 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 rBA ⎛0⎞ = ⎜ −e ⎟ m ⎜ ⎟ ⎝ −c ⎠ rCB ⎛ −b ⎞ = ⎜ −d ⎟ m ⎜ ⎟ ⎝c ⎠ Chapter rBC ⎛b⎞ = ⎜ d ⎟ ft ⎜ ⎟ ⎝ −c ⎠ θ = acos ⎜ rCD ⎛ −b ⎞ = ⎜ ⎟ ft ⎜ ⎟ ⎝0⎠ φ = acos ⎜ ⎛ rBA⋅ rBC ⎞ ⎟ ⎝ rBA rBC ⎠ θ = 74.0 deg ⎛ rCB⋅ rCD ⎞ ⎟ ⎝ rCB rCD ⎠ φ = 33.9 deg ⎛ rAC⋅ rAB ⎞ ⎟ ⎝ rAC rAB ⎠ θ = 64.6 deg Problem 2-112 Determine the angle θ between the two cords Given: a = 3m b = 2m c = 6m d = 3m e = 4m Solution: rAC ⎛b⎞ = ⎜ a ⎟ ft ⎜ ⎟ ⎝c⎠ rAB ⎛0⎞ = ⎜ −d ⎟ ft ⎜ ⎟ ⎝e ⎠ θ = acos ⎜ Problem 2-113 Determine the angle θ between the two cables Given: a = ft b = 10 ft 106 © 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 c = ft d = 10 ft e = ft f = ft F AB = 12 lb Solution: rAC ⎛a − f⎞ = ⎜ −c ⎟ ft ⎜ ⎟ ⎝ b ⎠ rAB ⎛ −f ⎞ = ⎜ d − c ⎟ ft ⎜ ⎟ ⎝ e ⎠ ⎛ rAC⋅ rAB ⎞ ⎟ ⎝ rAC rAB ⎠ θ = acos ⎜ θ = 82.9 deg Problem 2-114 Determine the projected component of the force F acting in the direction of cable AC Express the result as a Cartesian vector Given: F = 12 lb 107 © 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 a = ft b = 10 ft c = ft d = 10 ft e = ft f = ft Solution: rAC ⎛a − f⎞ ⎜ −c ⎟ m = ⎜ ⎟ ⎝ b ⎠ rAB ⎛ −f ⎞ ⎜ ⎟ = d−c ⎜ ⎟ ⎝ e ⎠ F AC = ( F AB⋅ uAC) uAC uAC = rAC rAC F AB = F F AC rAB rAB uAC ⎛ 0.2 ⎞ ⎜ ⎟ = −0.6 ⎜ ⎟ ⎝ 0.8 ⎠ ⎛ −9.6 ⎞ ⎜ ⎟ F AB = 3.2 lb ⎜ ⎟ ⎝ 6.4 ⎠ ⎛ 0.229 ⎞ ⎜ ⎟ = −0.916 lb ⎜ ⎟ ⎝ 1.145 ⎠ Problem 2-115 Determine the components of F that act along rod AC and perpendicular to it Point B is located at the midpoint of the rod 108 © 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 Given: F = 600 N c = m a = 4m d = 3m b = 6m e = 4m Solution: Find the force vector and the unit vector uAC rBD ⎛c + d ⎞ ⎜ 2⎟ ⎜ ⎟ ⎜b − e ⎟ = ⎜ 2⎟ ⎜ −a ⎟ ⎜ ⎟ ⎝ ⎠ rAC ⎛ −d ⎞ ⎜ ⎟ = e ⎜ ⎟ ⎝ −a ⎠ ⎛ 5.5 ⎞ ⎜ ⎟m rBD = ⎜ ⎟ ⎝ −2 ⎠ rAC Fv = F ⎛ −3 ⎞ ⎜ ⎟ = m ⎜ ⎟ ⎝ −4 ⎠ uAC = rBD rBD rAC rAC ⎛ 465.5 ⎞ ⎜ ⎟ F v = 338.6 N ⎜ ⎟ ⎝ −169.3 ⎠ uAC ⎛ −0.5 ⎞ ⎜ ⎟ = 0.6 ⎜ ⎟ ⎝ −0.6 ⎠ Now find the component parallel to AC F parallel = Fv ⋅ uAC F parallel = 99.1 N The perpendicular component is now found F perpendicular = Fv ⋅ F v − F parallel F perpendicular = 591.8 N Problem 2-116 Determine the components of F that act along rod AC and perpendicular to it Point B is located a distance f along the rod from end C 109 © 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 Given: F = 600 N c = 4m a = 4m d = 3m b = 6m e = 4m f = 3m Solution: f r = 2 d +e +a Find the force vector and the unit vector uAC rBD ⎡c + d( − r) ⎤ ⎢ ⎥ = b − e( − r) ⎢ ⎥ ⎣ −a r ⎦ Fv = F rAC ⎛ 5.5944 ⎞ ⎜ ⎟ rBD = 3.8741 m ⎜ ⎟ ⎝ −1.8741 ⎠ ⎛ 475.6 ⎞ ⎜ ⎟ F v = 329.3 N ⎜ ⎟ ⎝ −159.3 ⎠ rBD rBD ⎛ −d ⎞ ⎜ ⎟ = e ⎜ ⎟ ⎝ −a ⎠ rAC ⎛ −3 ⎞ ⎜ ⎟ = m ⎜ ⎟ ⎝ −4 ⎠ uAC = rAC rAC uAC ⎛ −0.5 ⎞ ⎜ ⎟ = 0.6 ⎜ ⎟ ⎝ −0.6 ⎠ Now find the component parallel to AC F parallel = Fv ⋅ uAC F parallel = 82.4 N The perpendicular component is now found F perpendicular = Fv ⋅ F v − F parallel F perpendicular = 594.3 N Problem 2-117 Determine the magnitude of the projected component of the length of cord OA along the Oa axis 110 © 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 Given: a = 10 ft b = ft c = 15 ft d = ft θ = 45 deg θ = 60 deg Solution: rOA ⎛ cos ( θ 1) cos ( θ ) ⎞ ⎜ ⎟ = d ⎜ cos ( θ ) sin ( θ ) ⎟ ⎜ ⎟ sin ( θ ) ⎝ ⎠ rOa = rOA ⋅ uOa rOa ⎛c ⎞ = ⎜ a ⎟ ⎜ ⎟ ⎝ −b ⎠ uOa = rOa rOa rOa = 2.1 ft 111 © 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 Problem 2-118 Force F acts at the end of the pipe Determine the magnitudes of the components F1 and F which are directed along the pipe's axis and perpendicular to it Given: a = ft ⎛ ⎞ ⎜ ⎟ lb F = ⎜ ⎟ ⎝ −40 ⎠ b = ft c = ft Solution: ⎛b⎞ ⎜ ⎟ r = a ⎜ ⎟ ⎝ −c ⎠ u = F1 = F⋅ u F2 = F⋅ F − F1 r r F = 18.3 lb F = 35.6 lb Problem 2-119 Determine the projected component of the force F acting along the axis AB of the pipe 112 © 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 Given: F = 80 N a = 4m b = 3m c = 12 m d = 2m e = 6m Solution: Find the force and the unit vector ⎛ −e ⎞ ⎜ ⎟ rA = −a − b ⎜ ⎟ ⎝ d−c ⎠ rAB ⎛ −e ⎞ ⎜ ⎟ = −b ⎜ ⎟ ⎝d⎠ ⎛ −6 ⎞ ⎜ ⎟ rA = −7 m ⎜ ⎟ ⎝ −10 ⎠ ⎛ −6 ⎞ ⎜ ⎟ rAB = −3 m ⎜ ⎟ ⎝2⎠ Fv = F uAB = rA rA rAB rAB ⎛ −35.3 ⎞ ⎜ ⎟ F v = −41.2 N ⎜ ⎟ ⎝ −58.8 ⎠ ⎛ −0.9 ⎞ ⎜ ⎟ uAB = −0.4 ⎜ ⎟ ⎝ 0.3 ⎠ Now find the projection using the Dot product F AB = F v ⋅ uAB F AB = 31.1 N Problem 2-120 Determine the angles θ and φ between the axis OA of the pole and each cable, AB and AC Given: F = 50 N 113 © 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 F = 35 N a = 1m b = 3m c = 2m d = 5m e = 4m f = 6m g = 4m Solution: rAO ⎛0⎞ ⎜ ⎟ = −g ⎜ ⎟ ⎝−f ⎠ rAB ⎛e⎞ ⎜ ⎟ = a ⎜ ⎟ ⎝−f⎠ rAC ⎛ rAO⋅ rAB ⎞ ⎟ ⎝ rAO rAB ⎠ θ = 52.4 deg ⎛ rAO⋅ rAC ⎞ ⎟ ⎝ rAO rAC ⎠ ⎛ −c ⎞ ⎜ ⎟ = a+b ⎜ ⎟ ⎝ −f ⎠ φ = 68.2 deg θ = acos ⎜ φ = acos ⎜ Problem 2-121 The two cables exert the forces shown on the pole Determine the magnitude of the projected component of each force acting along the axis OA of the pole Given: F = 50 N F = 35 N a = 1m b = 3m c = 2m 114 © 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 d = 5m e = 4m f = 6m g = 4m Solution: rAB ⎛e⎞ ⎜ ⎟ = a ⎜ ⎟ ⎝−f⎠ F 1v = F1 F 2v = F2 rAC rAC rAC rAB rAB ⎛ −c ⎞ ⎜ ⎟ = a+b ⎜ ⎟ ⎝ −f ⎠ rAO ⎛0⎞ ⎜ ⎟ = −g ⎜ ⎟ ⎝−f ⎠ F 1AO = F 1v⋅ uAO rAO rAO F 1AO = 18.5 N F 2AO = F 2v⋅ uAO uAO = F 2AO = 21.3 N Problem 2-122 Force F is applied to the handle of the wrench Determine the angle θ between the tail of the force and the handle AB 115 © 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 Given: a = 300 mm b = 500 mm F = 80 N θ = 30 deg θ = 45 deg Solution: ⎛ −cos ( θ 1) sin ( θ 2) ⎞ ⎜ ⎟ F v = F ⎜ cos ( θ ) cos ( θ ) ⎟ ⎜ ⎟ sin ( θ ) ⎝ ⎠ ⎛ Fv⋅ uab ⎞ ⎟ ⎝ F ⎠ θ = acos ⎜ uab ⎛0⎞ ⎜ ⎟ = −1 ⎜ ⎟ ⎝0⎠ θ = 127.8 deg Problem 2-123 Two cables exert forces on the pipe Determine the magnitude of the projected component of F along the line of action of F2 Given: F = 30 lb β = 30 deg F = 25 lb γ = 60 deg α = 30 deg ε = 60 deg Solution: We first need to find the third angle ( > 90 deg) that locates force F Initial Guess: φ = 120 deg 116 © 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 Given cos ( ε ) + cos ( γ ) + cos ( φ ) = 2 φ = Find ( φ ) φ = 135 deg Find the force F 1v and the unit vector u2 F 1v ⎛ cos ( α ) sin ( β ) ⎞ ⎜ ⎟ = F1 ⎜ cos ( α ) cos ( β ) ⎟ ⎜ −sin ( α ) ⎟ ⎝ ⎠ ⎛ cos ( φ ) ⎞ ⎜ ⎟ u2 = ⎜ cos ( ε ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠ Now find the projection ⎛ 13 ⎞ F 1v = ⎜ 22.5 ⎟ lb ⎜ ⎟ ⎝ −15 ⎠ ⎛ −0.7 ⎞ u2 = ⎜ 0.5 ⎟ ⎜ ⎟ ⎝ 0.5 ⎠ F 12 = F1v⋅ u2 F 12 = 5.4 lb Problem 2-124 Determine the angle θ between the two cables attached to the pipe Given: F = 30 lb β = 30 deg F = 25 lb γ = 60 deg α = 30 deg ε = 60 deg Solution: We first need to find the third angle ( > 90 deg) that locates force F Initial Guesses: φ = 120 deg 117 © 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 Given cos ( ε ) + cos ( γ ) + cos ( φ ) = 2 φ = Find ( φ ) φ = 135 deg Find the unit vectors u1 and u2 ⎛ cos ( α ) sin ( β ) ⎞ ⎜ ⎟ u1 = ⎜ cos ( α ) cos ( β ) ⎟ ⎜ −sin ( α ) ⎟ ⎝ ⎠ ⎛ 0.4 ⎞ u1 = ⎜ 0.8 ⎟ ⎜ ⎟ ⎝ −0.5 ⎠ ⎛ cos ( φ ) ⎞ ⎜ ⎟ u2 = ⎜ cos ( ε ) ⎟ ⎜ cos ( γ ) ⎟ ⎝ ⎠ ⎛ −0.7 ⎞ u2 = ⎜ 0.5 ⎟ ⎜ ⎟ ⎝ 0.5 ⎠ Find the angle using the dot product θ = acos ( u1 ⋅ u2 ) θ = 100.4 deg Problem 2-125 Determine the angle θ between the two cables Given: a = 7.5 ft b = ft c = ft d = ft e = ft f = ft F = 60 lb F = 30 lb 118 © 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 Solution: rAC ⎛ −d − b ⎞ = ⎜ −c − f ⎟ ⎜ ⎟ ⎝ a−e ⎠ rAB ⎛ −d = ⎜ −c − ⎜ ⎝ −e ⎛ rAC⋅ rAB ⎞ ⎟ ⎝ rAC rAB ⎠ θ = acos ⎜ ⎞ f⎟ ⎟ ⎠ θ = 59.2 deg Problem 2-126 Determine the projection of the force F along cable AB Determine the projection of the force F along cable AC Given: a = 7.5 ft b = ft c = ft d = ft e = ft f = ft F = 60 lb F = 30 lb Solution: rAC ⎛ −d − b ⎞ = ⎜ −c − f ⎟ ⎜ ⎟ ⎝ a−e ⎠ rAB ⎛ −d = ⎜ −c − ⎜ ⎝ −e ⎞ ⎟ f ⎟ ⎠ uAC = rAC rAC F 1v = F1 uAC F 1AB = F1v⋅ uAB F 2AC = F2v⋅ uAC rAB rAB F 1AB = 30.8 lb F 2v = F2 uAB uAB = F 2AC = 15.4 lb 119 © 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 Problem 2-127 Determine the angle θ between the edges of the sheet-metal bracket Given: a = 50 mm b = 300 mm c = 250 mm d = 400 mm Solution: Find the unit vectors and use the dot product ⎛d⎞ ⎜ ⎟ r1 = ⎜ ⎟ ⎝c⎠ u1 = ⎛a⎞ ⎜ ⎟ r2 = b ⎜ ⎟ ⎝0⎠ u2 = r1 r1 r2 r2 ⎛ 0.848 ⎞ ⎛ 0.164 ⎞ ⎜ 0.000 ⎟ u = ⎜ 0.986 ⎟ u1 = ⎜ ⎟ ⎜ ⎟ ⎝ 0.530 ⎠ ⎝ 0.000 ⎠ θ = acos ( u1 ⋅ u2 ) θ = 82 deg Problem 2-128 Determine the magnitude of the projected component of the force F acting along the axis BC of the pipe 120 © 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 ... ⎜d ⎟ ⎜ ⎟ ⎝ −e ⎠ F 1v = F1 F 2v = F2 F 3v = F3 ⎛ 282.4 ⎞ F 1v = ⎜ − 31 7 .6 ⎟ N ⎜ ⎟ ⎝ −4 23. 5 ⎠ rDC rDC ⎛ 230 .8 ⎞ F 2v = ⎜ 17 3 .1 ⎟ N ⎜ ⎟ ⎝ −277 ⎠ rDA rDA ⎛ ? ?19 1.5 ⎞ F 3v = ⎜ 12 7.7 ⎟ N ⎜ ⎟ ⎝ −766.2... ⎛ 38 ⎞ F 1v = ⎜ 10 3. 8 ⎟ N ⎜ ⎟ ⎝ ? ?10 1.4 ⎠ rAB rAB ⎛ 11 9.4 ⎞ F 2v = ⎜ ? ?19 .9 ⎟ N ⎜ ⎟ ⎝ ? ?15 9.2 ⎠ rAC rAC Add the forces and find the magnitude of the resultant F R = F1v + F2v ⎛ 15 7.4 ⎞ F R = ⎜ 83. 9... FDA rBA rBA rCA rCA rDA rDA F BAv ⎛ ? ?10 9.2 ⎞ ⎜ 13 1 ⎟ lb = ⎜ ⎟ ⎝ 30 5.7 ⎠ ⎛ 10 2.5 ⎞ ⎜ ⎟ F CAv = 10 2.5 lb ⎜ ⎟ ⎝ 478.5 ⎠ ⎛ −52 .1 ⎞ ⎜ ⎟ F DAv = ? ?15 6.2 lb ⎜ ⎟ ⎝ 36 4.5 ⎠ 90 © 2007 R C Hibbeler Published