*5–76. Continued
470
5–77.
The member is supported by a square rod which fits loosely
through the smooth square hole of the attached collar at A
and by a roller at B. Determine the components of reaction
at these supports when the member is subjected to the
loading shown.
y
z
xA
B1 m 2 m
2 m
C
SOLUTION
471
5–78.
The bent rod is supported at A, B, and Cby smooth journal
bearings. Compute the x, y, zcomponents of reaction at the
bearings if the rod is subjected to forces and
lies in the y–z plane.The bearings are in
proper alignment and exert only force reactions on the rod.
F
1
F
2=250 lb.
F
1=300 lb
SOLUTION
={–212.1j–212.1k}lb
F1=(–300 cos 45°j–300 sin 45°k)
z
5ft
45
45
F
1
C
4ft
1ft
A
B
5–79.
SOLUTION
F2=(F2cos 45° sin 30°i+F2cos 45° cos 30°j–F2sin 45°k)
={–212.1j–212.1k}lb
F1=(–300 cos 45°j–300 sin 45°k)
The bent rod is supported at A, B, and Cby smooth journal
bearings. Determine the magnitude of which will cause
the reaction at the bearing Cto be equal to zero. The
bearings are in proper alignment and exert only force
reactions on the rod. Set F
1=300 lb.
Cy
F2
z
5ft
45
45
F
1
C
4ft
1ft
A
B
474
*5–80.
The bar AB is supported by two smooth collars. AtA the
connection is with a ball-and-socket joint and at B it is a
rigid attachment. If a 50-lb load is applied to the bar,
determine the x, y, z components of reaction at A and B.z
y
Ax+Bx=0
A
C
B
D
F
6 ft
4 ft
6 ft
SOLUTION
B
y
=–50 lb
Ans.
Az+Bz=0
(2)
M
B
z
=0
Ans.
M
B
x+50(6) =0
M
Bx
=–300 lb #ft
Ans.
B
C
D
=–9i+3j
B
C
D
=–0.94868i+0.316228j
Require
FB#uCD =0
(Bxi–50j+Bzk)#(–0.94868i+0.316228j)=0
–0.94868Bx–50(0.316228) =0
Bx=–16.667 =–16.7 lb
Ans.
From Eq. (1);
Ax=16.7 lb
Ans.
Require
MB#uCD =0
(–300i+M
By
j)#(–0.94868i+0.316228j)=0
300(0.94868) +M
By
(0.316228) =0
M
By
=–900 lb #ft
Ans.
Ans:
B
y
=–50 lb
M
Bz
=0
M
B
x
=–300 lb #ft
Bx=–16.7 lb
Ax=16.7 lb
475
5–81.
The rod has a weight of 6 lb
>
ft. If it is supported by a ball-
and-socket joint at C and a journal bearing at D, determine
the x, y, z components of reaction at these supports and the
moment M that must be applied along the axis of the rod to
hold it in the position shown.
z
y
A
D
C
M
0.5 ft
1 ft
B1 ft
x
60
45
SOLUTION
ΣFx=0;
Cx+Dx–15 sin 45°=0
(1)
ΣF
y
=0;
C
y
+D
y
=0
(2)
ΣFz=0;
Cz–15 cos 45°=0
Cz=10.6 lb
Ans.
ΣMx=0;
–3 cos 45°(0.25 sin 60°)–D
y
(2) =0
D
y
=–0.230 lb
Ans.
From Eq. (2);
C
y
=0.230 lb
Ans.
ΣM
y
=0;
–(12 sin 45°)(1) –(3 sin 45°)(1) +(3 cos 45°)(0.25 cos 60°)
+Dx(2) =0
Dx=5.17 lb
Ans.
From Eq. (1);
Cx=5.44 lb
Ans.
ΣMz=0;
–M+(3 sin 45°)(0.25 sin 60°)=0
M=0.459 lb #ft
Ans.
Ans:
Cz=10.6 lb
D
y
=–0.230 lb
C
y
=0.230 lb
Dx=5.17 lb
Cx=5.44 lb
M=0.459 lb #ft
476
5–82.
The sign has a mass of 100 kg with center of mass at G.
Determine the x,y,zcomponents of reaction at the ball-and-
socket joint Aand the tension in wires BC and BD.
SOLUTION
Applying the forces equation of equilibrium, we have
Equating i,j, and kcomponents, we have
(1)
(2)
(3)
In order to write the moment equation of equilibrium about point A,the position
vectors and must be determined first.
r
AB =
{2j} m
rAG ={1j} m
rAB
rAG
Az+1
3F
BD +2
3F
BC –981 =0
Ay–2
3F
BD –2
3F
BC =0
Ax–2
3F
BD +1
3F
BC =0
aAx–2
3 F
BD +1
3 F
BC bi+aAy–2
3F
BD –2
3F
BC bj+aAz+1
3F
BD +2
3F
BC –981bk=0
(Axi+Ayj+Azk)+a–2
3F
BDi–2
3F
BDj+1
3F
BDkb+a1
3F
BCi–2
3F
BCj+2
3F
BCkb+(–981 k)=0
©F=0; FA+FBD +FBC +W=0
FBC =F
BCuBC =F
BC ≥(1 –0)i+(0 –2)j+(2 –0)k
2(1 –0)2+(0 –2)2+(2 –0)2¥=a1
3F
BCi–2
3F
BCj+2
3F
BCkb
FBD =F
BDuBD =F
BD ≥(–2–0)i+(0 –2)j+(1 –0)k
BDi–2
BDj+1
BDkb
z
D
C
1 m
2 m
1 m
5–82. Continued
5–83.
Both pulleys are fixed to the shaft and as the shaft turns
with constant angular velocity, the power of pulley Ais
transmitted to pulley B. Determine the horizontal tension T
in the belt on pulley Band the x, y, zcomponents of
reaction at the journal bearing Cand thrust bearing Dif
The bearings are in proper alignment and exert only
force reactions on the shaft.
u=0°.
SOLUTION
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.Dz=58.0 N
Dz+87.0 –80 –65 =0©Fz=0;
Dy=79.2 N
Dy+28.8 –50 –58.0 =0©Fy=0;
Dx=0©Fx=0;
Cy=28.8 N
150 +58.0210.22–Cy10.752=0©Mz=0;
Cz=87.0 N
165 +80210.452–Cz10.752=0©My=0;
T=58.0 N
300 mm
250 mm
150 mm
80 mm
200 mm
50 N
z
y
A
B
C
D
x
u
*5–84.
Both pulleys are fixed to the shaft and as the shaft turns
with constant angular velocity, the power of pulley Ais
transmitted to pulley B. Determine the horizontal tension T
in the belt on pulley Band the x, y, zcomponents of
reaction at the journal bearing Cand thrust bearing Dif
The bearings are in proper alignment and exert
only force reactions on the shaft.
u=45°.
SOLUTION
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.D
z=
32.1 N
Dz+77.57 +50 sin 45° –80 –65 =0©Fz=0;
Dy=68.5 N
Dy+24.89 –50 cos 45° –58.0 =0©Fy=0;
Dx=0©Fx=0;
Cy=24.89 N =24.9 N
58.010.22+50 cos 45°10.22–Cy10.752=0©Mz=0;
Cz=77.57 N =77.6 N
165 +80210.452–50 sin 45°10.22–Cz10.752=0©My=0;
T=58.0 N
300 mm
250 mm
150 mm
80 mm
200 mm
50 N
z
y
A
B
C
D
x
u
5–85.
Mem
b
er AB
i
s supporte
d
b
y a ca
bl
e BC an
d
at A
b
y a square
rod which fits loosely through the square hole at
the end joint of the member as shown. Determine the
components of reaction at Aand the tension in the cable
needed to hold the 800-lb cylinder in equilibrium.
SOLUTION
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.©Mz=0; (MA)z=0
©My=0; (MA)y=0
(MA)x=4.80 kip #ft
©Mx=0; (MA)x–800(6) =0
©Fz=0; Az=800 lb
©Fy=0; Ay=0
FBC =0
©Fx=0; FBC a3
7b=0
FBC =FBC a3
7i–6
7j+2
7kb
B
3ft
6ft
2ft
C
z
x
y
A