978-0133915426 Chapter 4 Part 5

308

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

d

4–81.

Two couples act on the frame.If determine the

resultant couple moment. Compute the result by resolving

each force into xand ycomponents and (a) finding the

moment of each couple (Eq. 4–13) and (b) summing the

moments of all the force components about point B.

d=4ft,

SOLUTION

2ft

B

y

1ft

3ft

50 lb

80 lb

50 lb

30

5

4

3

d

5

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=78.1 lb #ft

SOLUTION

rCB ={–3i–2.5j} ft

MC=rCB *F

=

†i j k

–3–2.5 0

0 0 20 †

MC={–50i+60j} lb #ft

Ans.

M

C=2

(

–

50)2

+

(60)2

=

78.1 lb

#

ft Ans.

4–82.

Express the moment of the couple acting on the pipe

assembly in Cartesian vector form. What is the magnitude

of the couple moment?

z

y

x

B

A

20 lb

1 ft

C

1.5 ft

3 ft

2 ft

1 ft

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4–83.

If ,,and ,

determine the magnitude and coordinate direction angles

of the resultant couple moment.

M3=120 lb

#

ftM2=90 lb

#

ftM1=180 lb

#

ft

SOLUTION

effect.Thus, the couple moments M1,M2,M3, and M4acting on the gear deducer can

be simplified, as shown in Fig. a. Expressing each couple moment in Cartesian

vector form,

M2=[–90i]lb #ft

M1=[180j]lb #ft

x

z

2ft

150 lb ft

1ft

45

M

1

M

2

M

3

M4=150[cos 45° sin 45°i–cos 45° cos 45°j–sin 45°k]=[75i–75j–106.07k]lb #ft

M3=M3u=120C(2 –0)i+(–2–0)j+(1 +0)k

2(2 –0)2+(–2–0)2+(1 –0)2S=[80i–80j+40k]lb #ft

*4–84.

M2=-M2i

Determine the magnitudes of couple moments

so that the resultant couple moment is zero.M1,M2, and M3

x

z

2ft

150 lb ft

1ft

45

M

3

M4=150[cos 45° sin 45°i–cos 45° cos 45°j–sin 45°k]=[75i–75j–106.07k]lb #ft

M3=M3u=M3C(2 –0)i+(–2–0)j+(1 +0)k

2(2 –0)2+(–2–0)2+(1 –0)2S=2

3M3i–2

3M3j+1

3M3k

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Ans:

MR=64.0

lb #ft

a=94.7°

b

=13.2°

4–85.

SOLUTION

Ans.

Ans.g=cos–1a–13.681

63.990 b=102°

b=cos–1a62.288

63.990 b=13.2°

a=cos–1a–5.272

63.990 b=94.7°

MR=2(–5.272)2+(62.288)2+(–13.681)2=63.990 =64.0 lb #ft

MR=M1+M2=-5.272 i+62.288 j–13.681 k

=-15 i+25.981 j

M2=-30 sin 30° i +30 cos 30° j

=9.728 i+36.307 j–13.681 k

M1=40 cos 20° sin 15° i +40cos 20° cos 15° j–40 sin 20° k

T

h

e gears are su

bj

ecte

d

to t

h

e coup

l

e moments s

h

own.

Determine the magnitude and coordinate direction angles

of the resultant couple moment.

z

x

y

M

1

40 lb ft

M

2

30 lb ft

20

30

15

=102°

313

Ans:

M2=424

N#m

M3=300

N#m

4–86.

Determine the required magnitude of the couple moments

and so that the resultant couple moment is zero.M3

M2

M

3

M

2

45

M

1

300 Nm

SOLUTION

Fig. a. Since the resultant of ,,and is required to be zero,

Ans.

Ans.M3=300 N #m

(MR)x=©Mx;0=424.26 cos 45° –M3

M2=424.26 N #m=424 N #m

(MR)y=©My;0=M2sin 45° –300

M3

M2

M1

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Ans:

MC={–65.0i–37.5k} N #m

*4–88.

Express the moment of the couple acting on the frame in

Cartesian vector form. The forces are applied perpendicular

to the frame.What is the magnitude of the couple moment?

Take .F=50 N

F

x

y

z

O

1.5 m

3m

30

F

SOLUTION

Ans.={–65.0i–37.5k}N#m

MC=-75(cos 30° i+cos 60° k)

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=15.4

N

4–89.

In order to turn over the frame,a couple moment is applied

as shown. If the component of this couple moment along

the xaxis is , determine the magnitude

Fof the couple forces.

Mx=5–20i6N#m

F

x

y

z

O

1.5 m

3m

30

F

SOLUTION

Thus

Ans.F=15.4 N

20 =F(1.5) cos 30°

MC=F(1.5)

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

MC=45.1

N#m

4–90.

SOLUTION

MC={37.5i–25j}N#m

MC=(0.2i+0.3j)*(125 k)

MC=rAB *(125 k)

Express the moment of the couple acting on the pipe in

Cartesian vector form. What is the magnitude of the couple

moment? Take F=125 N.

z

O

x

y

A

B

–F

F

600 mm

200mm

150 mm

150mm

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=832

N

4–91.

If the couple moment acting on the pipe has a magnitude of

determine the magnitude Fof the forces applied

to the wrenches.

300 N #m,

SOLUTION

Ans.F=832 N

300 =0.3606 F

MC=F2(0.2F)2+(–0.3F)

=0.3606 F

={0.2Fi–0.3Fj}N#m

=(0.2i+0.3j)*(Fk)

MC=rAB *(Fk)

z

O

x

y

A

B

–F

F

600 mm

200mm

150 mm

150mm

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*4–92.

If , determine the magnitude and coordinate

direction angles of the couple moment. The pipe assembly

lies in the x–y plane.

F=80 N

x

z

300 mm

200 mm

300 mm

F

SOLUTION

Mc=40.8

N#m

a=11.3°

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 pipe

assembly is , determine the magnitude of the couple

forces applied to each wrench.The pipe assembly lies in the

x–y plane.

50 N #m

x

z

300 mm

200 mm

300 mm

F

SOLUTION

be determined first.

The force vectors Fand –Fcan be written as

and

Thus, the couple moment of Fcan 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+My2+Mz2=2(0.5F)2+(0.1F)2+02=0.5099F

Mc

Mc=rAB *F=3ijk

0.1 0.5 0

00F3=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

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={–2i+20j+17k} kN #m

MC=26.3 kN #m

4–94.

Express the moment of the couple acting on the rod in

Cartesian vector form. What is the magnitude of the couple

moment?

SOLUTION

1 m

2 m

1 m 3 m

x

y

A

z

F  { 4i  3j  4k} kN

4–95.

If , and , determine the

magnitude and coordinate direction angles of the resultant

couple moment.

SOLUTION

F

3=80 NF

2=120 NF

1=100 N

– F1

y

x

z

0.2 m

0.3 m

30

F1

–F4 [ 150 k] N

F4 [150 k] N

*4–96.

Determine the required magnitude of

F

1,

F

2,and

F

3

so that the resultant couple moment is

.

SOLUTION

[50 i–45 j–20 k] N #m

(Mc)R=

z

0.2 m

0.3 m

30

–F4 [ 150 k] N

324

325

326