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Engineering Mechanics - Statics Chapter 4 Solution: F R F C − F B − F A − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = F R 108 lb= Guesses y 1ft= z 1ft= M 1lbft⋅= Given M F R F R 0 y z ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F R ×+ 0 a 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F C − 0 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ × 0 a b ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 F B − 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+= M y z ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Find My, z,()= M 624− lb ft⋅= y z ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0.414 8.69 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ft= Problem 4-139 The loading on the bookshelf is distributed as shown. Determine the magnitude of the equivalent resultant location, measured from point O. Given: w 1 2 lb ft = w 2 3.5 lb ft = a 2.75 ft= b 4ft= c 1.5 ft= Solution: Guesses R 1lb= d 1ft= Given w 1 bw 2 c+ R= w 1 ba b 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ w 2 c c 2 b+ a− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − d− R= R d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find Rd,()= R 13.25 lb= d 0.34 ft= 316 © 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 4 Replace the loading by an equivalent resultant force and couple moment acting at point A. Units Used: kN 10 3 N= Given: w 1 600 N m = w 2 600 N m = a 2.5 m= b 2.5 m= Solution: F R w 1 aw 2 b−= F R 0N= M RA w 1 a ab+ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = M RA 3.75 kN m⋅= Problem 4-141 Replace the loading by an equivalent force and couple moment acting at point O. Units Used: kN 10 3 N= Given: w 6 kN m = F 15 kN= M 500 kN m⋅= a 7.5 m= b 4.5 m= 317 Problem 4-140 © 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 4 Solution: F R 1 2 wa b+()F+= F R 51.0 kN= M R M− 1 2 wa ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 3 a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 1 2 wb ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a b 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − Fa b+()−= M R 914− kN m⋅= Problem 4-142 Replace the loading by a single resultant force, and specify the location of the force on the beam measured from point O. Units Used: kN 10 3 N= Given: w 6 kN m = F 15 kN= M 500 kN m⋅= a 7.5 m= b 4.5 m= Solution: Initial Guesses: F R 1kN= d 1m= Given F R 1 2 wa b+()F+= F R − dM− 1 2 wa ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 3 a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 1 2 wb ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a b 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − Fa b+()−= F R d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find F R d,()= F R 51 kN= d 17.922 m= Problem 4-143 The column is used to support the floor which exerts a force P on the top of the column. The effect of soil pressure along its side is distributed as shown. Replace this loading by an 318 © 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 4 pg p gy equivalent resultant force and specify where it acts along the column, measured from its base A. Units Used: kip 10 3 lb= Given: P 3000 lb= w 1 80 lb ft = w 2 200 lb ft = h 9ft= Solution: F Rx w 1 h 1 2 w 2 w 1 −()h+= F Rx 1260 lb= F Ry P= F R F Rx 2 P 2 += F R 3.25 kip= θ atan P F Rx ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 67.2 deg= F Rx y 1 2 w 2 w 1 −()h h 3 w 1 h h 2 += y 1 6 h 2 w 2 2 w 1 + F Rx = y 3.86 ft= Problem 4-144 Replace the loading by an equivalent force and couple moment at point O. Units Used: kN 10 3 N= Given: w 1 15 kN m = w 2 5 kN m = 319 © 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 4 d 9m= Solution: F R 1 2 w 1 w 2 +()d= F R 90 kN= M RO w 2 d d 2 1 2 w 1 w 2 −()d d 3 += M RO 338 kN m⋅= Problem 4-145 Replace the distributed loading by an equivalent resultant force, and specify its location on the beam, measured from the pin at C. Units Used: kip 10 3 lb= Given: w 800 lb ft = a 15 ft= b 15 ft= θ 30 deg= Solution: F R wa wb 2 += F R 18 kip= F R xwa a 2 wb 2 a b 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ += x wa a 2 wb 2 a b 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + F R = x 11.7 ft= Problem 4-146 The beam supports the distributed load caused by the sandbags. Determine the resultant force on the beam and specify its location measured from point A. 320 © 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 4 Units Used: kN 10 3 N= Given: w 1 1.5 kN m = a 3m= w 2 1 kN m = b 3m= w 3 2.5 kN m = c 1.5 m= Solution: F R w 1 aw 2 b+ w 3 c+= F R 11.25 kN= M A w 1 a a 2 w 2 ba b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + w 3 ca b+ c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ += M A 45.563 kN m⋅= d M A F R = d 4.05 m= Problem 4-147 Determine the length b of the triangular load and its position a on the beam such that the equivalent resultant force is zero and the resultant couple moment is M clockwise. Units Used: kN 10 3 N= Given: w 1 4 kN m = w 2 2.5 kN m = M 8kNm⋅= c 9m= Solution: Initial Guesses: a 1m= b 1m= Given 1− 2 w 1 b 1 2 w 2 c+ 0= 321 © 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 4 1 2 w 1 ba 2b 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 2 w 2 c 2c 3 − M−= a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find ab,()= a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1.539 5.625 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m= Problem 4-148 Replace the distributed loading by an equivalent resultant force and specify its location, measured from point A. Units Used: kN 10 3 N= Given: w 1 800 N m = w 2 200 N m = a 2m= b 3m= Solution: F R w 2 bw 1 a+ 1 2 w 1 w 2 −()b+= F R 3.10 kN= xF R w 1 a a 2 1 2 w 1 w 2 −()ba b 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + w 2 ba b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ += x w 1 a a 2 1 2 w 1 w 2 −()ba b 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + w 2 ba b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + F R = x 2.06 m= Problem 4-149 The distribution of soil loading on the bottom of a building slab is shown. Replace this loading by an equivalent resultant force and specify its location, measured from point O. Units Used: 322 © 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 4 kip 10 3 lb= Given: w 1 50 lb ft = w 2 300 lb ft = w 3 100 lb ft = a 12 ft= b 9ft= Solution: F R w 1 a 1 2 w 2 w 1 −()a+ 1 2 w 2 w 3 −()b+ w 3 b+= F R 3.9 kip= F R dw 1 a a 2 1 2 w 2 w 1 −()a 2a 3 + 1 2 w 2 w 3 −()ba b 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + w 3 ba b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ += d 3 w 3 ba 2 w 3 b 2 + w 1 a 2 + 2 a 2 w 2 + 3 bw 2 a+ w 2 b 2 + 6F R = d 11.3 ft= Problem 4-150 The beam is subjected to the distributed loading. Determine the length b of the uniform load and its position a on the beam such that the resultant force and couple moment acting on the beam are zero. Given: w 1 40 lb ft = c 10ft= d 6ft= w 2 60 lb ft = Solution: Initial Guesses: a 1ft= b 1ft= 323 © 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 4 1 2 w 2 dw 1 b− 0= 1 2 w 2 dc d 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ w 1 ba b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find ab,()= a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 9.75 4.5 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ft= Problem 4-151 Replace the loading by an equivalent resultant force and specify its location on the beam, measured from point B. Units Used: kip 10 3 lb= Given: w 1 800 lb ft = w 2 500 lb ft = a 12 ft= b 9ft= Solution: F R 1 2 aw 1 1 2 w 1 w 2 −()b+ w 2 b+= F R 10.65 kip= F R x 1 2 − aw 1 a 3 1 2 w 1 w 2 −()b b 3 + w 2 b b 2 += x 1 2 − aw 1 a 3 1 2 w 1 w 2 −()b b 3 + w 2 b b 2 + F R = x 0.479 ft= ( to the right of B ) Problem 4-152 Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member AB, measured from A. 324 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. Engineering Mechanics - Statics Chapter 4 Given: w 1 200 N m = w 2 100 N m = w 3 200 N m = a 5m= b 6m= Solution: F Rx w 3 − a= F Rx 1000− N= F Ry 1− 2 w 1 w 2 +()b= F Ry 900− N= y− F Rx w 3 a a 2 w 2 b b 2 − 1 2 w 1 w 2 −()b b 3 −= y w 3 a a 2 w 2 b b 2 − 1 2 w 1 w 2 −()b b 3 − F Rx − = y 0.1 m= Problem 4-153 Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member BC, measured from C. Units Used: kN 10 3 N= 325 © 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 4 Given: w 1 200 N m = w 2 100 N m = w 3 200 N m = a 5m= b 6m= Solution: F Rx w 3 − a= F Rx 1000− N= F Ry 1− 2 w 1 w 2 +()b= F Ry 900− N= x− F Ry w 3 − a a 2 w 2 b b 2 + 1 2 w 1 w 2 −()b 2b 3 += x w 3 − a a 2 w 2 b b 2 + 1 2 w 1 w 2 −()b 2b 3 + F Ry − = x 0.556 m= F Rx F Ry ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1.345 kN= Problem 4-154 Replace the loading by an equivalent resultant force and couple moment acting at point O. Units Used: kN 10 3 N= Given: w 1 7.5 kN m = 326 © 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 4 w 2 20 kN m = a 3m= b 3m= c 4.5 m= Solution: F R 1 2 w 2 w 1 −()cw 1 c+ w 1 b+ 1 2 w 1 a+= F R 95.6 kN= M Ro 1 2 − w 2 w 1 −()c c 3 w 1 c c 2 − w 1 bc b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 1 2 w 1 ab c+ a 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= M Ro 349− kN m⋅= Problem 4-155 Determine the equivalent resultant force and couple moment at point O. Units Used: kN 10 3 N= Given: a 3m= w O 3 kN m = wx() w O x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 = Solution: F R 0 a xwx() ⌠ ⎮ ⌡ d= F R 3kN= M O 0 a xwx()ax−() ⌠ ⎮ ⌡ d= M O 2.25 kN m⋅= 327 © 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 4 Problem 4-156 Wind has blown sand over a platform such that the intensity of the load can be approximated by the function ww 0 x d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 = . Simplify this distributed loading to an equivalent resultant force and specify the magnitude and location of the force, measured from A. Units Used: kN 10 3 N= Given: w 0 500 N m = d 10 m= wx() w 0 x d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 = Solution: F R 0 d xwx() ⌠ ⎮ ⌡ d= F R 1.25 kN= d 0 d xxw x() ⌠ ⎮ ⌡ d F R = d 8m= Problem 4-157 Determine the equivalent resultant force and its location, measured from point O. 328 © 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 4 Solution: F R 0 L xw 0 sin πx L ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⌠ ⎮ ⎮ ⌡ d= 2w 0 L π = d 0 L xxw 0 sin πx L ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⌠ ⎮ ⎮ ⌡ d F R = L 2 = Problem 4-158 Determine the equivalent resultant force acting on the bottom of the wing due to air pressure and specify where it acts, measured from point A. Given: a 3ft= k 86 lb ft 3 = wx() kx 2 = Solution: F R 0 a xwx() ⌠ ⎮ ⌡ d= F R 774 lb= x 0 a xxw x() ⌠ ⎮ ⌡ d F R = x 2.25 ft= Problem 4-159 Currently eighty-five percent of all neck injuries are caused by rear-end car collisions. To 329 © 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 4 ygy p j y alleviate this problem, an automobile seat restraint has been developed that provides additional pressure contact with the cranium. During dynamic tests the distribution of load on the cranium has been plotted and shown to be parabolic. Determine the equivalent resultant force and its location, measured from point A. Given: a 0.5 ft= w 0 12 lb ft = k 24 lb ft 3 = wx() w 0 kx 2 += Solution: F R 0 a xwx() ⌠ ⎮ ⌡ d= F R 7lb= x 0 a xxw x() ⌠ ⎮ ⌡ d F R = x 0.268 ft= Problem 4-160 Determine the equivalent resultant force of the distributed loading and its location, measured from point A. Evaluate the integrals using Simpson's rule. Units Used: kN 10 3 N= Given: c 1 5= c 2 16= a 3= 330 © 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 4 b 1= Solution: F R 0 ab+ xc 1 xc 2 x 2 ++ ⌠ ⎮ ⌡ d= F R 14.9= d 0 ab+ xxc 1 xc 2 x 2 ++ ⌠ ⎮ ⌡ d F R = d 2.27= Problem 4-161 Determine the coordinate direction angles of F, which is applied to the end A of the pipe assembly, so that the moment of F about O is zero. Given: F 20 lb= a 8in= b 6in= c 6in= d 10 in= Solution: Require M o = 0. This happens when force F is directed either towards or away from point O. r c ab+ d ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = u r r = u 0.329 0.768 0.549 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = If the force points away from O, then α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos u()= α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 70.774 39.794 56.714 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= If the force points towards O, then 331 © 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 4 α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos u−()= α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 109.226 140.206 123.286 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= Problem 4-162 Determine the moment of the force F about point O. The force has coordinate direction angles α, β, γ. Express the result as a Cartesian vector. Given: F 20 lb= a 8in= α 60 deg= b 6in= β 120 deg= c 6in= γ 45 deg= d 10 in= Solution: r c ab+ d ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F v F cos α() cos β() cos γ() ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = MrF v ×= M 297.99 15.147 200− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb in⋅= Problem 4-163 Replace the force at A by an equivalent resultant force and couple moment at point P. Express the results in Cartesian vector form. Units Used: 332 © 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 4 kN 10 3 N= Given: a 4m= b 6m= c 8m= d 4m= F 300− 200 500− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= Solution: F R F= F R 300− 200 500− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= M P a− c− b d ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F×= M P 3.8− 7.2− 0.6− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN m⋅= Problem 4-164 Determine the moment of the force F C about the door hinge at A. Express the result as a Cartesian vector. 333 © 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 4 Given: F 250 N= b 1m= c 2.5 m= d 1.5 m= e 0.5 m= θ 30 deg= Solution: r CB ce− bdcos θ()+ d− sin θ() ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r AB 0 b 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F v F r CB r CB = M A r AB F v ×= M A 59.7− 0.0 159.3− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Nm⋅= Problem 4-165 Determine the magnitude of the moment of the force F C about the hinged axis aa of the door. 334 © 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 4 Given: F 250 N= b 1m= c 2.5 m= d 1.5 m= e 0.5 m= θ 30 deg= Solution: r CB ce− bdcos θ()+ d− sin θ() ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r AB 0 b 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F v F r CB r CB = u a 1 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = M aa r AB F v ×()u a ⋅= M aa 59.7− Nm⋅= Problem 4-166 A force F 1 acts vertically downward on the Z-bracket. Determine the moment of this force about the bolt axis (z axis), which is directed at angle θ from the vertical. 335 © 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 4 Given: F 1 80 N= a 100 mm= b 300 mm= c 200 mm= θ 15 deg= Solution: r b− ac+ 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F F 1 sin θ() 0 cos θ()− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = k 0 0 1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = M z rF×()k= M z 6.212− Nm⋅= Problem 4-167 Replace the force F having acting at point A by an equivalent force and couple moment at point C. Units Used: kip 10 3 lb= Given: F 50 lb= a 10 ft= b 20 ft= c 15 ft= 336 © 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. dhanesh_h Mathcad - CombinedMathcads

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