y SOLUTIONweb.eng.fiu.edu/leonel/EGM3503/4_5-4_8.pdf · 2020. 9. 18. · y 300 mm 200 mm 200 mm 300 mm 300 mm F F SOLUTION It is easiest to find the couple moment of F by taking the

283

© 2016 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.

Ans:Mx = 15.0 lb # ftMy = 4.00 lb # ftMz = 36.0 lb # ft

*4–56.

Determine the magnitude of the moments of the force Fabout the x, y, and z axes. Solve the problem (a) using aCartesian vector approach and (b) using a scalar approach.

4 ft

3 ft

2 ft

y

z

C

A

B

F {4i 12j 3k} lb

x

SOLUTIONa) Vector Analysis

Position Vector:

Moment of Force F About x,y, and z Axes: The unit vectors along x, y, and z axes arei, j, and k respectively. Applying Eq. 4–11, we have

Ans.

Ans.

Ans.

b) Scalar Analysis

Ans.

Ans.

Ans.Mz = ©Mz ; Mz = -4(3) + 12(4) = 36.0 lb # ft

My = ©My ; My = -4(2) + 3(4) = 4.00 lb # ft

Mx = ©Mx ; Mx = 12(2) - 3(3) = 15.0 lb # ft

= 0 - 0 + 1[4(12) - (4)(3)] = 36.0 lb # ft

= 30 0 14 3 -24 12 -3

3

Mz = k # (rAB * F)

= 0 - 1[4(-3) - (4)(-2)] + 0 = 4.00 lb # ft

= 30 1 04 3 -24 12 -3

3

My = j # (rAB * F)

= 1[3(-3) - (12)(-2)] - 0 + 0 = 15.0 lb # ft

= 31 0 04 3 -24 12 -3

3

Mx = i # (rAB * F)

rAB = {(4 - 0) i + (3 - 0)j + (-2 - 0)k} ft = {4i + 3j - 2k} ft

288

© 2016 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.

Ans:MAB = 136 N # m

Solution

uAB ={3.5i + 0.5j}2(3.5)2 + (0.5)2

uAB = {0.9899i + 0.1414j}

MAB = uAB# (rAD * F) = †

0.9899 0.1414 02.5 0 450 -20 -80

†

MAB = 136 N # m Ans.

4–61.

Determine the magnitude of the moment of the force F = {50i - 20j - 80k} N about the base line AB of the tripod.

x

y

C A

D

B

F

z

0.5 m

2.5 m

1 m

2 m

1.5 m

2 m

4 m

301

© 2016 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.

Ans:MC = 22.5 N # mb

4–74.

The man tries to open the valve by applying the couple forcesof F = 75 N to the wheel. Determine the couple moment produced.

SOLUTIONa

b Ans. = -22.5 N # m = 22.5 N # m +Mc = ©M; Mc = -75(0.15 + 0.15)

150 mm 150 mm

F

F

302

© 2016 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.

Ans:F = 83.3 N

4–75.

If the valve can be opened with a couple moment of , determinethe required magnitude of each couple force which must be appliedto the wheel.

SOLUTIONa

Ans. F = 83.3 N+Mc = ©M; -25 = -F(0.15 + 0.15)

25 N # m 150 mm 150 mm

F

F

304

© 2016 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.

Ans:(MR)C = 240 lb # ft d

4–77.

Two couples act on the beam as shown. If F = 150 lb, determinethe resultant couple moment.

SOLUTION150 lb couple is resolved into their horizontal and vertical components as shown inFig. a

a

d Ans. = 240 lb # ft

+ (MR)c = 150 a45b (1.5) + 150 a3

5b (4) - 200(1.5)

200 lb

200 lb

1.5 ft

–F

435

F 435

4 ft

307

© 2016 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.

Ans:MC = 126 lb # ft

*4–80.

SOLUTION

(a)

Ans.

(b) a

Ans.MC = 126 lb # ft

+ MC = -45

(80)(3) +45

(80)(7) + 50 cos 30°(2) - 50 cos 30°(5)

MC = {126k} lb # ft

= 3i j k3 0 0

-50 sin 30° -50 cos 30° 0

3 + 3i j k0 4 0

-45(80) -3

5(80) 0

3

MC = ©(r * F)

Two couples act on the frame. If determine theresultant couple moment. Compute the result by resolvingeach force into x and y components and (a) finding themoment of each couple (Eq. 4–13) and (b) summing themoments of all the force components about point A.

d = 4 ft,2 ft

B

A

y

1 ft

3 ft50 lb

80 lb

50 lb30�

30�

5

43

80 lb

3 ft

d

x

5

43

317

© 2016 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.

Ans:MC = 45.1 N # m

4–90.

SOLUTION

Ans.MC = 2(37.5)2 + (-25)2 = 45.1 N # mMC = {37.5i - 25j} N # mMC = (0.2i + 0.3j) * (125 k)

MC = rAB * (125 k)

Express the moment of the couple acting on the pipe inCartesian vector form. What is the magnitude of the couplemoment? Take F = 125 N.

z

O

x

y

A

B

–F

F

600 mm

200 mm

150 mm

150 mm

320

© 2016 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.

Ans:F = 98.1 N

4–93.

If the magnitude of the couple moment acting on the pipeassembly is , determine the magnitude of the coupleforces applied to each wrench.The pipe assembly lies in thex–y plane.

50 N # m

x

z

y

300 mm

200 mm

200 mm300 mm

300 mm

F

F

SOLUTIONIt is easiest to find the couple moment of F by taking the moment of either F or –Fabout point A or B, respectively, Fig. a. Here the position vectors and mustbe determined first.

The force vectors F and –F can be written asand

Thus, the couple moment of F can be determined from

The magnitude of is given by

Since is required to equal ,

Ans.F = 98.1 N

50 = 0.5099F

50 N # mMc

Mc = 2Mx2 + My

2 + Mz2 = 2(0.5F)2 + (0.1F)2 + 02 = 0.5099F

Mc

Mc = rAB * F = 3i j k

0.1 0.5 00 0 F

3 = 0.5Fi - 0.1Fj

-F = [-Fk]NF = {Fk} N

rBA = (0.2 - 0.3)i + (0.3 - 0.8)j + (0 - 0)k = [-0.1i - 0.5j] m

rAB = (0.3 - 0.2)i + (0.8 - 0.3)j + (0 - 0)k = [0.1i + 0.5j] m

rBArAB

324

© 2016 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.

Ans:FR = 365 N

u = 70.8° d(MR)O = 2364 N # m (counterclockwise)

SolutionEquivalent Resultant Force And Couple Moment At O.

+S (FR)x = ΣFx; (FR)x = 600 cos 60° - 455 a1213

b = -120 N = 120 N d

+ c (FR)y = ΣFy; (FR)y = 455 a 513

b - 600 sin 60° = -344.62 N = 344.62 NT

As indicated in Fig. a

FR = 2 (FR)x2 + (FR)y

2 = 21202 + 344.622 = 364.91 N = 365 N Ans.

And

u = tan-1 c (FR)y

(FR)xd = tan-1 a344.62

120b = 70.80° = 70.8° d Ans.

Also,

a+(MR)O = ΣMO; (MR)O = 455 a1213

b(2) + 600 cos 60° (0.75) + 600 sin 60° (2.5)

= 2364.04 N # m

= 2364 N # m (counterclockwise) Ans.

4–97.

Replace the force system by an equivalent resultant force and couple moment at point O.

y

xO

600 N

60�

2.5 m 2 m

0.75 m0.75 m

1 m

512

13

455 N

P

326

© 2016 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.

Ans:FR = 5.93 kNu = 77.8° dMRA

= 34.8 kN # m b

4–99.

SOLUTION

Thus,

Ans.

and

Ans.

a

Ans.= -34.8 kN # m = 34.8 kN # m (Clockwise)

+MRA= ©MA ; MRA

= -2.5a35b(2) - 1.5 cos 30°(6) - 3(8)

u = tan - 1 ¢FRy

FRx

≤ = tan - 1a5.7991.25

b = 77.8° d

FR = 2F2Rx

+ F2Ry

= 21.252 + 5.7992 = 5.93 kN

= -5.799 kN = 5.799 kN T

+ cFRy= ©Fy ; FRy

= -1.5 cos 30° - 2.5a35b - 3

= -1.25 kN = 1.25 kN ;

:+ FRx= ©Fx ; FRx

= 1.5 sin 30° - 2.5a45b

Replace the force system acting on the beam by anequivalent force and couple moment at point A. 2.5 kN 1.5 kN

3 kN

A B

4 m

34

5

2 m 2 m

30

328

© 2016 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.

Ans:FR = 294 Nu = 40.1° dMRO = 39.6 N # m b

Solution

d+ FRx = ΣFx ; FRx = 450 sin 30° = 225.0

+ T FRy = ΣFy ; FRy = 450 cos 30° - 200 = 189.7

FR = 2(225)2 + (189.7)2 = 294 N Ans.

u = tan-1 a189.7225

b = 40.1° d Ans.

c+MRO = ΣMO ; MRO = 450 cos 30° (1.5) - 450 (sin 30°)(0.2) - 200 (3.5) + 200

MRO = 39.6 N # m b Ans.

4–101.

Replace the loading system acting on the beam by an equivalent resultant force and couple moment at point O. 30�

y

x

450 N

O

200 N

0.2 m 200 N � m

2 m1.5 m 1.5 m

335

© 2016 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.

Ans:FR = {-200i + 700j - 600k} N

(MR)O = {-1200i + 450j + 1450k} N # m

SolutionPosition And Force Vectors.

r1 = {2j} m r2 = {1.5i + 3.5j} r3 = {1.5i + 2j} m

F1 = {-300k} N F2 = {200j} N F3 = {-200i + 500j - 300k} N

Equivalent Resultant Force And Couple Moment At Point O.

FR = ΣF; FR = F1 + F2 + F3

= (-300k) + 200j + (-200i + 500j - 300k)

= {-200i + 700j - 600k} N Ans.

(MR)O = ΣMO; (MR)O = r1 * F1 + r2 * F2 + r3 * F3

= 3 i j k0 2 00 0 -300

3 + 3 i j k1.5 3.5 00 200 0

3 + 3 i j k1.5 2 0

-200 500 -300

3 = (-600i) + (300k) + (-600i + 450j + 1150k)

= {-1200i + 450j + 1450k} N # m Ans.

*4–108.

Replace the force system by an equivalent resultant force  and couple moment at point O. Take F3 = {-200i + 500j - 300k} N.

y

O

z

x2 m

F2 = 200 N

F1 = 300 N

1.5 m

1.5 m

F3

344

© 2016 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.

Ans:F = 1302 Nu = 84.5° dx = 7.36 m

4–117.

Replace the loading acting on the beam by a single resultantforce. Specify where the force acts, measured from end A.

SOLUTION

Ans.

Ans.

c

Ans.x = 7.36 m

+ MRA = ©MA ; 1296(x) = 450 sin 60°(2) + 300(6) + 700 cos 30°(9) + 1500

du = tan-1a1296125b = 84.5°

F = 2(-125)2 + (-1296)2 = 1302 N

+ cFRy = ©Fy ; FRy = -450 sin 60° - 700 cos 30° - 300 = -1296 N = 1296 N T

:+ FRx = ©Fx ; FRx = 450 cos 60° - 700 sin 30° = -125 N = 125 N ;

2 m

300 N 30

60

1500 N m4 m 3 m

450 N700 N

AB

354

© 2016 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.

Ans:FR = 197 lbu = 42.6°ad = 5.24 ft

4–125.

Replace the force and couple system acting on the frame byan equivalent resultant force and specify where theresultant’s line of action intersects member AB, measuredfrom A.

SOLUTION

Ans.

Ans.u = tan -1 a133.3145

b = 42.6°

FR = 2(145)2 + (133.3)2 = 197 lb + cFRy = ©Fy ; FRy = 50 cos 30° + 150a3

5b = 133.3 lb

:+ FRx = ©Fx ; FRx = 150a45b + 50 sin 30° = 145 lb

3 ft30

4 ft

35

4

2 ft

150 lb

50 lb

500 lb ft

C B

A

a

Ans.d = 5.24 ft

+MRA = ©MA ; 145 d = 150 a45b (2) - 50 cos 30° (3) + 50 sin 30° (6) + 500