Engineering Mechanics - Statics Episode 1 Part 9 pdf

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 0.479ft= ( 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 π x L ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⌠ ⎮ ⎮ ⌡ d= 2w 0 L π = d 0 L xxw 0 sin π x 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 774lb= x 0 a xxw x() ⌠ ⎮ ⌡ d F R = x 2.25ft= Problem 4-159 Currentl y ei g ht y -five p ercent of all neck in j uries are caused b y 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 0.268ft= 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. [...]... writing from the publisher Engineering Mechanics - Statics NA = 1 lb Chapter 5 NB = 1 lb Given NB cos ( θ 1 − 90 deg) − NA cos ( θ 2 ) = 0 NA sin ( θ 2 ) − NB sin ( θ 1 − 90 deg) − W = 0 ⎛ NA ⎞ ⎜ ⎟ = Find ( NA , NB) ⎝ NB ⎠ ⎛ NA ⎞ ⎛ 19 .3 ⎞ ⎜ ⎟=⎜ ⎟ lb ⎝ NB ⎠ ⎝ 14 .1 ⎠ Problem 5 -1 2 Determine the magnitude of the resultant force acting at pin A of the handpunch Given: F = 8 lb a = 1. 5 ft b = 0.2 ft c = 2 ft... θ 1 = 20 deg c = 1m θ 2 = 30 deg g = 9. 81 m 2 s Solution: Guesses F = 1 kN NA = 1 kN NB = 1 kN Given NA + NB + F sin ( θ 2 ) − M g cos ( θ 1 ) = 0 −F cos ( θ 2 ) + M g sin ( θ 1 ) = 0 F cos ( θ 2 ) a − F sin ( θ 2 ) b − M g cos ( θ 1 ) c − M g sin ( θ 1 ) e + NB( c + d) = 0 ⎛F ⎞ ⎜ ⎟ ⎜ NA ⎟ = Find ( F , NA , NB) ⎜ NB ⎟ ⎝ ⎠ ⎛ F ⎞ ⎛ 19 .37 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ NA ⎟ = ⎜ 13 .05 ⎟ kN ⎜ NB ⎟ ⎝ 23.36 ⎠ ⎝ ⎠ Problem 5 -1 7... writing from the publisher Engineering Mechanics - Statics Chapter 5 Problem 5 -1 0 Draw the free-body diagram of the beam, which is pin-connected at A and rocker-supported at B Given: F = 500 N M = 800 N⋅ m a = 8m b = 4m c = 5m Solution: Problem 5 -1 1 The sphere of weight W rests between the smooth inclined planes Determine the reaactions at the supports Given: W = 10 lb θ 1 = 10 5 deg θ 2 = 45 deg Solution:... from the publisher Engineering Mechanics - Statics FA = Chapter 5 2 Ax + Ay 2 F A = 10 lb Problem 5 -1 3 The C-bracket is supported at A, B, and C by rollers Determine the reactions at the supports Given: a = 3 ft b = 4 ft θ 1 = 30 deg θ 2 = 20 deg F = 200 lb Solution: Initial Guesses: NA = 1 lb NB = 1 lb NC = 1 lb Given NA a − F b = 0 NB sin ( θ 1 ) − NC sin ( θ 2 ) = 0 NB cos ( θ 1 ) + NC cos ( θ 2.. .Engineering Mechanics - Statics Chapter 4 b = 1 Solution: a+ b ⌠ FR = ⎮ ⌡0 2 c1 x + c2 + x dx x c1 x + F R = 14 .9 c2 + x dx a+ b d = ⌠ ⎮ ⌡0 2 d = 2.27 FR Problem 4 -1 61 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 = 8 in b = 6 in c = 6 in d = 10 in Solution: Require... the publisher Engineering Mechanics - Statics Chapter 4 Given: F 1 = 80 N a = 10 0 mm b = 300 mm c = 200 mm θ = 15 deg Solution: ⎛ −b ⎞ ⎜ ⎟ r = a+c ⎜ ⎟ ⎝ 0 ⎠ ⎛ sin ( θ ) ⎞ ⎜ 0 ⎟ F = F1 ⎜ ⎟ ⎝ −cos ( θ ) ⎠ ⎛0⎞ ⎜ ⎟ k = 0 ⎜ ⎟ 1 Mz = ( r × F ) k Mz = −6. 212 N⋅ m Problem 4 -1 67 Replace the force F having acting at point A by an equivalent force and couple moment at point C Units Used: 3 kip = 10 lb Given:... 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: rCB ⎛ c−e ⎞ ⎜ ⎟ = b + d cos ( θ ) ⎜ ⎟ ⎝ −d sin ( θ ) ⎠ Maa = ( rAB × Fv) ⋅ ua rAB ⎛0⎞ ⎜ ⎟ = b ⎜ ⎟ ⎝0⎠ Fv = F rCB rCB 1 ⎜ ⎟ ua = 0 ⎜ ⎟ ⎝0⎠ Maa = − 59. 7 N⋅ m Problem 4 -1 66 A force F1 acts vertically downward on the Z-bracket Determine the moment of this... publisher Engineering Mechanics - Statics ⎛α ⎞ ⎜ ⎟ ⎜ β ⎟ = acos ( −u) ⎜γ ⎟ ⎝ ⎠ Chapter 4 ⎛ α ⎞ ⎛ 10 9. 226 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β ⎟ = ⎜ 14 0.206 ⎟ deg ⎜ γ ⎟ ⎝ 12 3.286 ⎠ ⎝ ⎠ Problem 4 -1 62 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 = 8 in α = 60 deg b = 6 in β = 12 0 deg c = 6 in γ = 45 deg d = 10 in Solution:... writing from the publisher Engineering Mechanics - Statics Chapter 5 Problem 5 -1 Draw the free-body diagram of the sphere of weight W resting between the smooth inclined planes Explain the significance of each force on the diagram Given: W = 10 lb θ 1 = 10 5 deg θ 2 = 45 deg Solution: NA, NB force of plane on sphere W force of gravity on sphere Problem 5-2 Draw the free-body diagram of the hand punch,... without permission in writing from the publisher Engineering Mechanics - Statics Chapter 5 Given: F = 20 lb a = 1 in b = 6 in Solution: Initial Guesses: Ax = 1 lb Ay = 1 lb NB = 1 lb Given − Ax + NB = 0 Ay − F = 0 ⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ = Find ( Ax , Ay , NB) ⎜N ⎟ ⎝ B⎠ −F ( a + b) + Ax a = 0 ⎛ Ax ⎞ ⎛ 14 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 20 ⎟ lb ⎜ N ⎟ ⎝ 14 0 ⎠ ⎝ B⎠ Problem 5 -1 6 The automobile is being towed at constant velocity . 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 +. 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 −. 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 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ +