Engineering Mechanics - Statics Episode 1 Part 8 docx

Engineering Mechanics - Statics Chapter 4 α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos M M ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 155.496 114.504 90 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= Problem 4-96 Express the moment of the couple acting on the pipe in Cartesian vector form. What is the magnitude of the couple moment? Given: F 125 N= a 150 mm= b 150 mm= c 200 mm= d 600 mm= Solution: M c ab+ 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 0 F ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ×= M 37.5 25− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Nm⋅= M 45.1 N m⋅= Problem 4-97 If the couple moment acting on the pipe has a magnitude M, determine the magnitude F of the forces applied to the wrenches. 281 © 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: M 300 N m⋅= a 150 mm= b 150 mm= c 200 mm= d 600 mm= Solution: Initial guess: F 1N= Given c ab+ 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 0 F ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ × M= F Find F()= F 832.1 N= Problem 4-98 Replace the force at A by an equivalent force and couple moment at point O. Given: F 375 N= a 2m= b 4m= c 2m= d 1m= θ 30 deg= 282 © 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 F v F sin θ () cos θ () − 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F v 187.5 324.76− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= M O a− b 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F v ×= M O 0 0 100.481− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Nm⋅= Problem 4-99 Replace the force at A by an equivalent force and couple moment at point P . Given: F 375 N= a 2m= b 4m= c 2m= d 1m= θ 30 deg= Solution: F v F sin θ () cos θ () − 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = F v 187.5 324.76− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= M P a− c− bd− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F v ×= M P 0 0 736.538 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Nm⋅= Problem 4-100 Replace the force system by an equivalent resultant force and couple moment at point O. 283 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 4 F 1 60 lb= a 2ft= F 2 85 lb= b 3ft= F 3 25 lb= c 6ft= θ 45 deg= d 4ft= e 3= f 4= Solution: F 0 F 1 − 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ F 2 e 2 f 2 + e− f− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ + F 3 cos θ () sin θ () 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ += F 33.322− 110.322− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= F 115.245 lb= M O c− 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 F 1 − 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ × 0 a 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 e 2 f 2 + e− f− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+ d b 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 3 cos θ () sin θ () 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+= M O 0 0 480 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb ft⋅= M O 480lb ft⋅= Problem 4-101 Replace the force system by an equivalent resultant force and couple moment at point P. 284 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: F 1 60 lb= a 2ft= F 2 85 lb= b 3ft= F 3 25 lb= c 6ft= θ 45 deg= d 4ft= e 3= f 4= Solution: F 0 F 1 − 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ F 2 e 2 f 2 + e− f− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ + F 3 cos θ () sin θ () 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ += F 33.322− 110.322− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= F 115.245 lb= M P c− d− 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 F 1 − 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ × d− a 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 e 2 f 2 + e− f− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+ 0 b 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 3 cos θ () sin θ () 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+= M P 0 0 921 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb ft⋅= M P 921lb ft⋅= Problem 4-102 Replace the force system by an equivalent force and couple moment at point O. Units Used: kip 10 3 lb= Given: F 1 430 lb= F 2 260 lb= a 2ft= e 5ft= 285 © 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 8ft= f 12= c 3ft= g 5= da= θ 60 deg= Solution: F R F 1 sin θ () − cos θ () − 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 g 2 f 2 + g f 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ += F R 272− 25 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= F R 274lb= M O d− b 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 1 sin θ () − cos θ () − 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ × e 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 g 2 f 2 + g f 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+= M O 0 0 4.609 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kip ft⋅= Problem 4-103 Replace the force system by an equivalent force and couple moment at point P. Units Used: kip 10 3 lb= Given: F 1 430 lb= F 2 260 lb= a 2ft= e 5ft= b 8ft= f 12= c 3ft= g 5= da= θ 60 deg= Solution: F R F 1 sin θ () − cos θ () − 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 g 2 f 2 + g f 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ += F R 272− 25 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= F R 274lb= 286 © 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 M P 0 bc+ 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 1 sin θ () − cos θ () − 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ × de+ c 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 g 2 f 2 + g f 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+= M P 0 0 5.476 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kip ft⋅= Problem 4-104 Replace the loading system acting on the post by an equivalent resultant force and couple moment at point O. Given: F 1 30 lb= a 1ft= d 3= F 2 40 lb= b 3ft= e 4= F 3 60 lb= c 2ft= Solution: F R F 1 0 1− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 1 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ + F 3 d 2 e 2 + d− e− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ += F R 4 78− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= F R 78.1 lb= M O 0 ab+ c+ 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 1 0 1− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ × 0 c 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 1 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+ 0 bc+ 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 3 d 2 e 2 + d− e− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+= M O 0 0 100 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb ft⋅= Problem 4-105 Replace the loading system acting on the post by an equivalent resultant force and couple moment at point P. 287 © 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 30 lb= F 2 40 lb= F 3 60 lb= a 1ft= b 3ft= c 2ft= d 3= e 4= Solution: F R F 1 0 1− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 1 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ + F 3 d 2 e 2 + d− e− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ += F R 4 78− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= F R 78.1 lb= M P 0 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ft ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ F 1 0 1− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ × 0 a− b− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 1 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+ 0 a− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 3 d 2 e 2 + d− e− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+= M P 0 0 124 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb ft⋅= Problem 4-106 Replace the force and couple system by an equivalent force and couple moment at point O. Units Used: kN 10 3 N= 288 © 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: M 8kNm= θ 60 deg= a 3m= f 12= b 3m= g 5= c 4m= F 1 6kN= d 4m= F 2 4kN= e 5m= Solution: F R F 1 f 2 g 2 + g f 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 cos θ () − sin θ () − 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ += F R 0.308 2.074 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= F R 2.097 kN= M O 0 0 M ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ c− e− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 1 f 2 g 2 + g f 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+ 0 d− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 cos θ () − sin θ () − 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+= M O 0 0 10.615− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN m⋅= Problem 4-107 Replace the force and couple system by an equivalent force and couple moment at point P. Units Used: kN 10 3 N= 289 © 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: M 8kNm⋅= θ 60 deg= a 3m= f 12= b 3m= g 5= c 4m= F 1 6kN= d 4m= F 2 4kN= e 5m= Solution: F R F 1 f 2 g 2 + g f 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 cos θ () − sin θ () − 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ += F R 0.308 2.074 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= F R 2.097 kN= M P 0 0 M ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ c− b− e− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 1 f 2 g 2 + g f 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+ b− d− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 cos θ () − sin θ () − 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×+= M P 0 0 16.838− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN m⋅= Problem 4-108 Replace the force system by a single force resultant and specify its point of application, measured along the x axis from point O. Given: F 1 125 lb= F 2 350 lb= F 3 850 lb= 290 © 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 = 450 N a = 2m F 2 = 300 N b = 4m F 3 = 700 N c = 3m θ = 60 deg M = 15 00 N⋅ m φ = 30deg Solution: F Rx = F 1 cos ( θ ) − F 3 sin ( φ ) F Rx = 12 5 N F Ry = −F 1 sin ( θ ) − F3 cos ( φ ) − F2 F Ry = 1. 296 × 10 N F = 2 FRx + FRy 3 3 2 F = 1. 302 × 10 N ⎛ FRy ⎞ ⎟ ⎝ FRx ⎠ θ 1 = atan ⎜ θ 1 = 84 .5 deg F Ry x = F1 sin ( θ ) b − F3 cos ( φ ) c − M x = F1... publisher Engineering Mechanics - Statics Chapter 4 Given: F 1 = 40 N F 2 = 40 N θ = 0 deg r = 80 mm a = 300 mm θ = 45 deg Solution: F 1v ⎛0⎞ = F1 ⎜ 0 ⎟ ⎜ ⎟ ⎝ 1 ⎠ F R = F1v + F2v ⎛ 0 ⎞ F R = ⎜ − 28. 28 ⎟ N ⎜ ⎟ ⎝ − 68. 28 ⎠ F 2v ⎛ 0 ⎞ = F2 ⎜ −cos ( θ ) ⎟ ⎜ ⎟ ⎝ −sin ( θ ) ⎠ ⎛ −a ⎞ r1 = ⎜ r ⎟ ⎜ ⎟ ⎝0⎠ ⎛ −a ⎞ r2 = ⎜ −r sin ( θ ) ⎟ ⎜ ⎟ ⎝ r cos ( θ ) ⎠ MA = r1 × F1v + r2 × F2v ⎛ 0 ⎞ MA = ⎜ −20.49 ⎟ N⋅ m ⎜ ⎟ ⎝ 8. 49... = 15 00 N⋅ m φ = 30 deg Solution: F Rx = F 1 cos ( θ ) − F 3 sin ( φ ) F Rx = 12 5 N F Ry = −F 1 sin ( θ ) − F3 cos ( φ ) − F2 F Ry = 1. 296 × 10 N F = 2 FRx + FRy 3 3 2 F = 1. 302 × 10 N ⎛ FRy ⎞ ⎟ ⎝ FRx ⎠ θ 1 = atan ⎜ θ 1 = 84 .5 deg F Ry( x) = −F1 sin ( θ ) a − F2 ( a + b) − F3 cos ( φ ) ( a + b + c) − M x = −F1 sin ( θ ) a − F2 ( a + b) − F3 cos ( φ ) ( a + b + c) − M F Ry x = 7.36 m Problem 4 -1 16 ... in writing from the publisher Engineering Mechanics - Statics Chapter 4 Units Used: 3 kN = 10 N Given: 8 ⎜ ⎟ F = 6 kN ⎜ ⎟ 8 a = 3m b = 3m ⎛ −20 ⎞ ⎜ ⎟ M = −70 kN⋅ m ⎜ ⎟ ⎝ 20 ⎠ c = 4m e = 5m f = 6m g = 5m d = 6m Solution: ⎛−f⎞ ⎜ ⎟ MR = M + e × F ⎜ ⎟ ⎝g⎠ FR = F 8 ⎜ ⎟ F R = 6 kN ⎜ ⎟ 8 ⎛ 10 ⎞ ⎜ ⎟ MR = 18 kN⋅ m ⎜ ⎟ ⎝ −56 ⎠ Problem 4 -1 26 Replace the force and couple-moment system by an equivalent... publisher Engineering Mechanics - Statics Chapter 4 p y Cartesian vector form q p p p Given: F 1 = 40 N r = 80 mm F 2 = 40 N a = 300 mm θ = 0 deg Solution: F 1v F 2v ⎛0⎞ = F1 ⎜ 0 ⎟ ⎜ ⎟ ⎝ 1 ⎠ ⎛ 0 ⎞ = F2 ⎜ −cos ( θ ) ⎟ ⎜ ⎟ ⎝ −sin ( θ ) ⎠ ⎛ −a ⎞ r2 = ⎜ −r sin ( θ ) ⎟ ⎜ ⎟ ⎝ r cos ( θ ) ⎠ ⎛ −a ⎞ r1 = ⎜ r ⎟ ⎜ ⎟ ⎝0⎠ F R = F1v + F2v MA = r1 × F1v + r2 × F2v ⎛ 0 ⎞ F R = ⎜ −40 ⎟ N ⎜ ⎟ ⎝ −40 ⎠ ⎛ 0 ⎞ MA = ⎜ 12 ⎟... means, without permission in writing from the publisher Engineering Mechanics - Statics F 1v = ⎛b⎞ ⎜ −a ⎟ ⎟ 2 2⎜ b +a ⎝ 0 ⎠ F1 Chapter 4 F 2v ⎛ 0 ⎞ ⎜ ⎟ = ⎜ −F 2 ⎟ ⎜ 0 ⎟ ⎝ ⎠ F 3v = M = 1 lb⋅ ft R x = 1 lb ⎛ −b ⎞ ⎜a⎟ ⎟ 2 2⎜ b +a ⎝ 0 ⎠ F3 Place the wrench in the x - z plane x = 1ft z = 1ft Guesses R y = 1 lb R z = 1 lb ⎛ Rx ⎞ ⎜ ⎟ Given ⎜ Ry ⎟ = F1v + F2v + F3v ⎜R ⎟ ⎝ z⎠ ⎛ Rx ⎞ ⎛ 0 ⎞ ⎞ ⎛ x ⎞ ⎛ Rx ⎟ ⎛0⎞... Engineering Mechanics - Statics Chapter 4 Units Used: 3 kN = 10 N Given: F 1 = 20 kN a = 3m F 2 = 50 kN b = 8m F 3 = 20 kN c = 2m F 4 = 50 kN d = 6m e = 4m Solution: FR = F1 + F2 + F3 + F4 F R = 14 0 kN F R x = F2 e + F1 ( d + e) + F 2 ( d + e) x = 2 F2 e + F1 d + F1 e + F2 d x = 6.43 m FR −F R y = −F 2 a − F3 ( a + b) − F2 ( a + b + c) y = 2 F2 a + F3 a + F3 b + F2 b + F2 c FR y = 7.29 m Problem 4 -1 35... writing from the publisher Engineering Mechanics - Statics Chapter 4 Units Used: 3 kN = 10 N Given: F 1 = 30 kN a = 3m F 2 = 40 kN b = 8m F 3 = 20 kN c = 2m F 4 = 50 kN d = 6m e = 4m Solution: + ↑ FR = ΣFx; FR = F1 + F2 + F3 + F4 F R = 14 0 kN MRx = ΣMx; −F R( y) = −( F4 ) ( a) − y = ⎡( F1 ) ( a + b)⎤ − ⎡( F2 ) ( a + b + c)⎤ ⎣ ⎦ ⎣ ⎦ F4 a + F1 a + F1 b + F2 a + F2 b + F2 c FR y = 7 .14 m MRy = ΣMy; (FR)x =.. .Engineering Mechanics - Statics Chapter 4 a = 2 ft b = 6 ft c = 3 ft d = 4 ft Solution: F Ry = F 3 − F 2 − F 1 F Ry = 375 lb F Ry x = F3 ( b + c) − F 2 ( b) + F1 ( a) x = F3 ( b + c) − F 2 ( b) + F1 a FRy x = 15 .5 ft Problem 4 -1 09 Replace the force system by a single force resultant and specify its point of application, measured along the x axis from point P Given: F 1 = 12 5 lb a = 2 ft... publisher Engineering Mechanics - Statics Chapter 4 Problem 4 -1 10 The forces and couple moments which are exerted on the toe and heel plates of a snow ski are F t, Mt, and F h, Mh, respectively Replace this system by an equivalent force and couple moment acting at point O Express the results in Cartesian vector form Given: a = 12 0 mm b = 80 0 mm Solution: ⎛ −50 ⎞ F t = ⎜ 80 ⎟ N ⎜ ⎟ ⎝ 15 8 ⎠ ⎛ −20 ⎞ . deg= Solution: F Rx F 1 cos θ () F 3 sin φ () −= F Rx 12 5− N= F Ry F 1 − sin θ () F 3 cos φ () − F 2 −= F Ry 1. 296− 10 3 × N= FF Rx 2 F Ry 2 += F 1. 302 10 3 × N= θ 1 atan F Ry F Rx ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 1 84 .5 deg= F Ry x(). N= F Ry F 1 − sin θ () F 3 cos φ () − F 2 −= F Ry 1. 296− 10 3 × N= FF Rx 2 F Ry 2 += F 1. 302 10 3 × N= θ 1 atan F Ry F Rx ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 1 84 .5 deg= F Ry xF 1 sin θ () bF 3 cos φ () c− M−= x F 1 sin θ () bF 3 cos φ () c−. writing from the publisher. Engineering Mechanics - Statics Chapter 4 acting at point P. Express the results in Cartesian vector form. Given: a 12 0 mm= b 80 0 mm= F t 50− 80 15 8 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= M t 6− 4 2 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Nm⋅= F h 20− 60 250− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= M h 20− 8 3 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Nm⋅= Solution: F R F t F h +=