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308
© 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
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
30
5
4
3
d
5
© 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=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
20 lb
1 ft
C
1.5 ft
3 ft
2 ft
1 ft
© 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.
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
2ft
150 lb ft
1ft
45
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
2ft
150 lb ft
1ft
45
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
© 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=64.0
lb #ft
a=94.7°
b
=13.2°
4–85.
SOLUTION
Ans.
Ans.
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
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 Nm
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
© 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={-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)
© 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=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)
© 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.
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
© 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=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
© 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.
*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
F
SOLUTION
Mc=40.8
N#m
a=11.3°
© 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 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
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
© 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={-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
0.2 m
0.2 m
0.2 m
0.3 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
0.2 m
0.3 m
0.3 m
30
–F4 [ 150 k] N
324
325
326
