328
329
SOLUTION
4–102.
Replace the loading system acting on the post by an
equivalent resultant force and couple moment at point A.
3 m
500 N
30
60
1500 N m
5 m 2 m
650 N
300 N
B
A
330
SOLUTION
4–103.
Replace the loading system acting on the post by an
equivalent resultant force and couple moment at point B.
3 m
500 N
30
60
1500 N m
5 m 2 m
650 N
300 N
B
A
331
332
SOLUTION
4–105.
Replace the force system acting on the frame by an
equivalent resultant force and couple moment acting at
point A.
1 m
0.5 m
500 N
300 N
A
30
© 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.
SOLUTION
FR=F1+F2={–1i–2j–5k} kN
Ans.
MRO =r1*F1+r2*F2
=
†i j k
4–0.15 0.25
–4 2 –3†
+
†i j k
4 0.15 0.25
3–4–2†
=(–0.05i+11j+7.4k)+(0.7i+8.75j–16.45k)
=(0.65i+19.75j–9.05k)
MRO ={0.650i+19.75j–9.05k} kN #m
Ans.
4–106.
The forces
F1={–4i+2j–3k} kN
and F2
={3i–4j–
2k} kN
act on the end of the beam. Replace these forces by
an equivalent force and couple moment acting at point O.
y
z
x
F1150 mm
150 mm F2
250 mm
O
4 m
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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
SOLUTION
F3={–200i+500j–300k} N
*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
F1 = 300 N
1.5 m
F3
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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={6i+5j–5k} kN
(MR)O={2.5i–7j} kN #m
4–109.
Replace the loading by an equivalent resultant force and
couple moment at point O.
SOLUTION
0.8 m
0.5 m
0.7 m
x
y
O
z
F = {–2 i + 5 j – 3 k} kN
F {8 i – 2 k} kN
1
2
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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={44.5 i+53.1 j+40 k} N
MRA ={–5.39 i+13.1 j+11.4 k} N #m
SOLUTION
FR=ΣF ;
FR=80 cos 30° sin 40° i+80 cos 30° cos 40° j–80 sin 30° k
=44.53 i+53.07 j–40 k
={44.5 i+53.1 j–40 k} N
Ans.
MRA =ΣMA ;
MRA =
†i j k
0.55 0.4 –0.2
44.53 53.07 –40 †
={–5.39 i+13.1 j+11.4 k} N #m
Ans.
4–110.
Replace the force of F
=
80 N acting on the pipe assembly
by an equivalent resultant force and couple moment at
point A.
[
400 mm
y
300 mm
200 mm
250 mm
200 mm
40
B
A
z
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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=5–40j–40k6 N
MRA =5–12j+12k6 N #m
4–111.
T
h
e
b
e
l
tpass
i
ng over t
h
epu
ll
ey
i
ssu
bj
ecte
d
to forces an
d
each having amagnitude of 40 N. acts in the direction.
Replace these forces by an equivalent force and couple
moment at point A.Express the result in Cartesian vector form.
Set so that acts in the direction.–jF2
u=0°
–kF1
F2,
F
1
y
z
300 mm
r80 mm
A
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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=5–28.3j–68.3k6 N
MRA =5–20.5j+8.49k6 N #m
*4–112.
T
h
e
b
e
l
t pass
i
ng over t
h
e pu
ll
ey
i
s su
bj
ecte
d
to two forces
and each having a magnitude of 40 N. acts in the
direction. Replace these forces by an equivalent force and
couple moment at point A. Express the result in Cartesian
vector form. Take u=45°.
–k
F1
F2,
F
1
Ans.
Ans.
Also,
Ans.M
RA
=
{
–20.5
j
+8.49k
}
N#m
MRAz=8.49 N #m
MRAz=28.28(0.3)
MRAz=©MAz
MRAy=-20.5 N #m
MRAy=-28.28(0.3) –40(0.3)
MRAy=©MAy
MRAx=0
MRAx=28.28(0.0566) +28.28(0.0566) –40(0.08)
MRAx=©MAx
MRA ={–20.5j+8.49k}N#m
=3ijk
–0.3 0.08 0
00–40 3+3ij k
–0.3 –0.0566 0.0566
0–40 cos 45° –40 sin 45° 3
MRA =(rAF1*F1)+(rAF2*F2)
={–0.3i–0.0566j+0.0566k}m
rAF2=-0.3i–0.08 sin 45°j+0.08 cos 45°k
rAF1={–0.3i+0.08j}m
FR={–28.3j–68.3k}N
=-40 cos 45°j+(–40 – 40 sin 45°)k
FR=F1+F2
y
z
300 mm
r80 mm
A
F
1
F
2
© 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=798
lb
67.9°
d
x=7.43
ft
4–115.
Replace the three forces acting on the shaft by a single
resultant force. Specify where the force acts, measured from
end A.
SOLUTION
+FRx
300 b=67.9°Ans.
c+MRA =ΣMA; 740(x)=500 a3
5b(5)+200(8) +260a12
13 b(10)
740(x)=5500
x=7.43 ft Ans.
5 ft 3 ft 2 ft 4 ft
200 lb
3
4
5
500 lb
260 lb
5
12 13
BA
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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=798
lb
u=67.9°
d
x=6.57
ft
*4–116.
Replace the three forces acting on the shaft by a single
resultant force. Specify where the force acts, measured from
end B.
SOLUTION
+ΣFRx
300 b=67.9° Ans.
a+MRB =ΣMB; 740(x)=500
a3
5b(9)+200(6)+260
a12
13 b(4)
x=6.57 ft Ans.
5 ft 3 ft 2 ft 4 ft
200 lb
3
4
5
500 lb
260 lb
5
12 13
BA
344
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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
N
u=84.5°
d
x=7.36
m
4–117.
Replace the loading acting on the beam byasingle resultant
force.Specify where the force acts,measured from end A.
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–1a1296
125 b=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
:
2 m
300 N 30
60
1500 Nm
4 m 3 m
450 N
700 N
A
B
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
SOLUTION
4–119.
Replace the loading on the frame by a single resultant force.
Specify where its line of action intersects a vertical line
along member AB, measured from A.
1.5 m
0.5 m 0.5 m
200 N200 N
400 N
600 N
B
C
347
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*4–120.
Replace the loading on the frame by a single resultant force.
Specify where its line of action intersects a vertical line
along member AB, measured from A.
B
y
0.5 m
1 m
0.5 m
400 N
600 N
54
3
900 N
1.5 m
5
3