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 a = 2 ft b = 6 ft c = 3 ft d = 4 ft Solution: F Ry = F 3 − F 2 − F 1 F Ry = 3 75 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 = 1 25 lb a = 2 ft F 2 = 350 ... writing from the publisher Engineering Mechanics - Statics 2 F = FRx + FRy Chapter 4 2 F = 922 lb ⎛ FRy ⎞ ⎟ ⎝ FRx ⎠ θ = atan ⎜ θ = 77 .5 deg F Ry x = −F2 a − F 3 ( a + b) − F 4 c + M x = − F2 ( a) + F 3 ( a + b) + F 4 c − M x = 3 .55 6 ft F Ry Problem 4-1 15 Replace the loading on the frame by a single resultant force Specify where the force acts , measured from end A Given: F 1 = 450 N a = 2m F 2 = 300 N... Engineering Mechanics - Statics Chapter 4 f ⎛ ⎞ ⎜ 2 2 ⎟ F1 a + F sin ( θ ) ( a + b − d) + F2 ( a + b) = M ⎝ e +f ⎠ ⎛F⎞ ⎜ θ ⎟ = Find ( F , θ , d) ⎜ ⎟ ⎝d⎠ θ = 57 .52 9 deg F = 2.608 kN d = 2.636 m Problem 4-1 20 Replace the loading on the frame by a single resultant force Specify where its line of action intersects member AB, measured from A Given: F 1 = 50 0 N a = 3m b = 2m F 2 = 300 N c = 1m F 3 = 250 ... 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... Engineering Mechanics - Statics Chapter 4 F 3 = 180 N a = 1. 25 m b = 0 .5 m c = 0. 75 m Solution: ⎛0 ⎞ ⎛ 0 ⎟ ⎛ 0 ⎟ ⎞ ⎜ ⎞ ⎜ ⎟ ⎜ FR = ⎜ 0 ⎟ + ⎜ 0 ⎟ + ⎜ 0 ⎟ ⎜ F 1 ⎟ ⎜ −F2 ⎟ ⎜ −F 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 0 ⎞ ⎜ 0 ⎟N FR = ⎜ ⎟ ⎝ −210 ⎠ ⎞ ⎞ ⎛a + c⎞ ⎛ 0 ⎞ ⎛a⎞ ⎛ 0 ⎟ ⎛a⎞ ⎛ 0 ⎟ ⎜ b ⎟ × ⎜ 0 ⎟ + ⎜b⎟ × ⎜ 0 + ⎜0⎟ × ⎜ 0 MO = ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ F 1 ⎟ ⎝ 0 ⎠ ⎜ −F 2 ⎟ ⎝ 0 ⎠ ⎜ −F 3 ⎟ ⎝ 0 ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ − 15 ⎞ ⎜ ⎟ MO = 2 25. .. publisher Engineering Mechanics - Statics Chapter 4 b = 0.8 m c = 0. 25 m d = 0.7 m e = 0.3 m f = 0.3 m g = 0 .5 m h = 0. 25 m P = 60 N Q = 40 N Solution: Initial Guess F = 1N MC = 1 N⋅ m Given ⎛ − M C ⎞ ⎡ − P ( c + h) ⎤ ⎡ 0 ⎜ ⎟ ⎢ ⎥+⎢ 0 ⎜ 0 ⎟=⎢ 0 ⎥ ⎢ ⎜ 0 ⎟ ⎣ 0 ⎦ ⎣ −F ( e + ⎝ ⎠ ⎛ F ⎞ ⎜ ⎟ = Find ( F , MC) ⎝ MC ⎠ ⎤ ⎛ a ⎞ ⎛ −Q ⎞ ⎥ + ⎜b⎟ × ⎜ 0 ⎟ ⎥ ⎜ ⎟ ⎜ ⎟ f) ⎦ ⎝ 0 ⎠ ⎝ 0 ⎠ MC = 30 N⋅ m F = 53 .3 N Problem 4-1 36 The... without permission in writing from the publisher Engineering Mechanics - Statics Given Chapter 4 ⎛ −e ⎞ F + F cos ( θ ) = 0 ⎜ 2 2⎟ 1 ⎝ e +f ⎠ ⎛ − f ⎞ F − F sin ( θ ) − F = −F 2 R ⎜ 2 2⎟ 1 ⎝ e +f ⎠ f ⎛ ⎞ ⎜ 2 2 ⎟ F1 a + F sin ( θ ) ( a + b − d) + F2 ( a + b) = M ⎝ e +f ⎠ ⎛F⎞ ⎜ θ ⎟ = Find ( F , θ , d) ⎜ ⎟ ⎝d⎠ θ = 71 .56 5 deg F = 4.427 kN d = 3 .52 4 m Problem 4-1 19 Determine the magnitude and direction θ of... 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 = 150 0 N⋅ m φ = 30deg Solution: F Rx = F 1 cos ( θ ) − F 3 sin ( φ ) F Rx = −1 25 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... from the publisher Engineering Mechanics - Statics y = F 1 cos ( θ ) b + F3 ( c) FRx Chapter 4 y = 4.617 ft (Below point B) Problem 4-1 24 Replace the force system acting on the frame by an equivalent resultant force and couple moment acting at point A Given: F 1 = 35 lb a = 2 ft F 2 = 20 lb b = 4 ft F 3 = 25 lb c = 3 ft θ = 30 deg d = 2 ft Solution: F Rx = F 1 sin ( θ ) + F3 F Rx = 42 .5 lb F Ry = F 1... reproduced, in any form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 4 acting at point P Express the results in Cartesian vector form Given: a = 120 mm b = 800 mm ⎛ 50 ⎞ ⎜ 80 ⎟ N Ft = ⎜ ⎟ ⎝ − 158 ⎠ ⎛ −6 ⎞ ⎜ ⎟ Mt = 4 N⋅ m ⎜ ⎟ ⎝2⎠ ⎛ −20 ⎞ ⎜ 60 ⎟ N Fh = ⎜ ⎟ ⎝ − 250 ⎠ ⎛ −20 ⎞ ⎜ 8 ⎟ N⋅ m Mh = ⎜ ⎟ ⎝ 3 ⎠ Solution: FR = Ft + Fh ⎛ −70 ⎞ ⎜ ⎟ F R = 140 N . Engineering Mechanics - Statics Chapter 4 α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos M M ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 155 .496 114 .50 4 90 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= Problem 4-9 6 Express the. moment? Given: F 1 25 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 4-9 7 If the couple. 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 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ +