457
SOLUTION
7
140
5–69.
Determine the tension in each cable and the components of
reaction at D needed to support the load.
C
z
B
x
y
3 m
2 m
6 m
30
A
D
459
5–70.
The stiff-leg derrick used on ships is supported by a ball-and-
socket joint at D and two cables BA and BC. The cables are
attached to a smooth collar ring at B, which allows rotation
of the derrick about z axis. If the derrick supports a crate
having a mass of 200 kg, determine the tension in the cables
and the x, y, z components of reaction at D.
z
y
x
ΣFx=0;
B
D
7.5 m
6 m
6 m
2 m
C
1 m
SOLUTION
2
7
6
9
ΣF
y
=0;
Dy–
3
7
TBA –
3
9
TBC =0
ΣFz=0;
Dz–
6
7
TBA –
6
9
TBC –200(9.81) =0
ΣMx=0;
3
7
TBA(6) +
3
9
TBC(6) –200(9.81)(4) =0
ΣM
y
=0;
2
7
TBA(6) –
6
9
TBC(6) +200(9.81)(1) =0
TBA =2.00 kN
Ans.
TBC =1.35 kN
Ans.
Dx=0.327 kN
Ans.
D
y
=1.31 kN
Ans.
Dz=4.58 kN
Ans.
Ans:
TBA =2.00 kN
TBC =1.35 kN
Dx=0.327 kN
D
y
=1.31 kN
Dz=4.58 kN
© 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
A(0, 0, 0),
B(6, 0, 0),
C(0, –2, 3) m,
D(0, 2, 3) m
and
E(3, 0, 0) m
respectively.
FBC =FBC
a
rBC
rBC
b
=FBC
£
(0 –6)i+(–2–0)j+(3 –0)k
2
(0 –6)2+(–2–0)2+(3 –0)2
§
=–6
7 FBCi –2
7 FBC
j+3
7 FBCk
FBD =FBD
a
rBD
rBD
b
=FBD
£
(0 –6)i+(2 –0)j+(3 –0)k
2
(0 –6)2+(2 –0)2+(3 –0)2
§
=–6
7 FBDi +2
7 FBD
j+3
7 FBDk
F
A
=A
x
i+A
y
j+A
z
k
F={–600k} N
rAB ={6i} m
rAE ={3i} m
Equations of Equilibrium. Referring to the FBD of the rod shown in Fig. a, the force
equation of equilibrium gives
ΣF=0;
FBC +FBD +FA+F=0
5–71.
Determine the components of reaction at the ball-and-socket
joint A and the tension in each cable necessary for
equilibrium of the rod.
z
x
A
D
E
C
3 m
3 m
2 m
2 m
B
© 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.
Equating i, j and k components,
–
6
7
FBC –
6
7
FBD +Ax=0 (1)
2
7
FBD –
2
7
FBC +Ay=0 (2)
3
7
FBC +
3
7
FBD +Az–600 =0 (3)
The moment equation of equilibrium gives
ΣMA=0;
rAE *F+rAB *(FBC +FBD )=0
†
i j k
3 0 0
0 0 –600
†
+
5
i j k
6 0 0
–6
7(FBC +FBD)2
7
(FBD –FBC)3
7
(FBC +FBD)
5
=0
c
1800 –
18
7
(FBC +FBD)
d
j+
12
7
(FBD –FBC)k=0
Equating j and k components,
1800 –
18
7
(FBC +FBD)=0 (4)
12
7
(FBD –FBC)=0 (5)
Solving Eqs. (1) to (5),
FBD =FBC =350 N
Ans.
Ax=600 N
Ans.
A
y
=0
Ans.
Az=300 N
Ans.
5–71. Continued
462
*5–72.
Determine the components of reaction at the ball-and-
socket joint A and the tension in the supporting cables DB
and DC.
SOLUTION
FDC =FDC
a
rDC
rDC
b
=FDC
£
(0 –1)i+(1.5 –0)j+(3 –1)k
2
(0 –1)2+(1.5 –0)2+(3 –1)2
§
=–
1
17.25
FCDi+
1.5
17.25
FDC
j+
2
17.25
FDC
k
FDB =FDB
a
rDB
rDB
b
=FDB
£
(0 –1)i+(–1.5 –0)j+(3 –1)k
2
(0 –1)2+(–1.5 –0)2+(3 –1)2
§
=–
1
17.25
FDBi+
1.5
17.25
FDB
j+
2
17.25
FDBk
F
A
=A
x
i+A
y
j+A
z
k
F={–2400k} N
rAD =(1 –0)i+(1 –0)k={i+k} m
rF={4i} m
y
800 N/m
1 m
1.5 m
3 m
1.5 m
1.5 m
3 m
B
z
C
A
D
466
5–75.
Member AB is supported by a cable BC and at A by a
square rod which fits loosely through the square hole in the
collar fixed to the member as shown. Determine the
components of reaction at A and the tension in the cable
needed to hold the rod in equilibrium.
1.5 m
400 N
200 N
3 m
C
z
y
A
SOLUTION
7
7
467
468
*5–76.
The member is supported by a pin at A and cable BC.
Determine the components of reaction at these supports if
the cylinder has a mass of 40 kg.
SOLUTION
FCB =FCB
ar
CB
rCB
b
=FCB
c(0 –3)i+(–0.5 –1)j+(1 –0)k
2
(0
–
3)2
+
(
–
0.5
–
1)2
+
(1
–
0)2
d
=–
6
7
FCBi–
3
7
FCBj+
2
7
FCBk
W={–40(9.81)k} N ={–392.4k} N
.
F
A
=A
x
i+A
y
j+A
z
k
MA
=
(
MA
)
x
i
+
(
MA
)
zk
rAC ={3i+j} m
rAD ={3i–j} m
Equations of Equilibrium. Referring to the FBD of the assembly shown in Fig. a. the
force equation of equilibrium gives
ΣF=0;
FCB +W+FA=0;
a
–
6
7
FCB +Ax
b
i+
a
–
3
7
FCB +Ay
b
j+
a2
7
FCB +Az–392.4
b
k=0
Equating i, j and k components
–
6
7
FCB +Ax=0 (1)
–
3
7
FCB +Ay=0 (2)
2
7
FCB +Az–392.4 =0 (3)
The moment equation of equilibrium gives
ΣMA=0;
rAC *FCB +rAD *W+MA=0
5i j k
3 1 0
–6
7FCB –3
7FCB
2
7FCB
5
+
†i j k
3–1 0
0 0 –392.4
†
+
(
MA
)
x i+
(
MA
)
Z k=0
c2
7
FCB +392.4 +
(
MA
)
x
d
i+
a
–
6
7
FCB +1177.2
b
j+
c
–
9
7
FCB +
6
7
FCB +
(
MA
)
z
d
k=0
0.5 m
z
A
B
D
1 m
1 m
y