PROBLEM 5.21
The homogeneous wire ABCD is bent as shown and is attached to
a hinge at C. Determine the length L for which portion BCD of the
wire is horizontal.
PROBLEM 5.22
The homogeneous wire ABCD is bent as shown and is attached to
a hinge at C. Determine the length L for which portion AB of the
wire is horizontal.
SOLUTION
I II III
80 100W wW wWLw= = =
0: (80 )(32) (100 )(14) ( )(0.4 ) 0
C
M w w Lw L
Σ= + − =
29900L=
99.5 mmL=
consent of McGraw–Hill Education.
PROBLEM 5.23
A uniform circular rod of weight 8 lb and radius 10 in. is attached to a pin at C and
to the cable AB. Determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
For quarter circle,
2r
r
π
=
(a)
2
0: 0
C
r
M W Tr
π
Σ = −=
22
(8 lb)TW
ππ
= =
5.09 lbT=
(b)
0: 0 5.09 lb 0
xx x
F TC CΣ= −= −=
5.09 lb
x
=C
0: 0 8 lb 0
yy y
F CW CΣ= −= − =
8 lb
y
=C
9.48 lb=C
57.5°
consent of McGraw–Hill Education.
PROBLEM 5.24
The homogeneous wire ABC is bent into a semicircular arc and a straight
section as shown and is attached to a hinge at A. Determine the value of
θ
for
which the wire is in equilibrium for the indicated position.
SOLUTION
PROBLEM 5.25
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and h.
SOLUTION
h
PROBLEM 5.26
SOLUTION
3
(1 )y h kx= −
For
, 0.x ay= =
3
0 (1 )h ka= −
3
1
ka
∴=
3
3
1x
yh a
= −
1
,2
EL EL
x x y y dA ydx= = =
34
33
00 0
3
14
4
a
aa
xx
A dA ydx h dx h x ah
aa
===−=−=
∫∫ ∫
4 25 2
33
00 0
3
2 10
5
a
aa
EL
x xx
x dA xydx h x dx h a h
aa
= = − =−=
∫∫∫
3 2 36
23 36
00 0
11 2
11
22 2
aa a
EL
x h xx
y dA y ydx h dx dx
a aa
= = − = −+
∫∫ ∫ ∫
2 47 2
36
0
9
2 28
27
a
h xx
x ah
aa
= −+ =
2
33
:4 10
EL
xA x dA x ah a h
= =
∫
2
5
xa=
2
39
:4 28
EL
yA y dA y ah ah
= =
∫
3
7
yh=
consent of McGraw–Hill Education.
PROBLEM 5.27
Determine by direct integration the centroid of the area shown.
Express your answer in terms of a and h.
SOLUTION
At
( , ),ah
2
1
:y h ka=
or
2
h
ka
=
2
:y h ma=
or
h
ma
=
Now
12
1()
2
EL
EL
xx
y yy
=
= +
and
2
21 2
2
2
()
()
hh
dA y y dx x x dx
aa
hax x dx
a
=−=−
= −
Then
2 23
22
00
11
() 23 6
a
ah ha
A dA ax x dx x x ah
aa
== −= − =
∫∫
and
( )
2 34 2
22
00
22
12 21 21
11
() 3 4 12
11
( )[( ) ]
22
a
a
EL
EL
h ha
x dA x ax x dx x x a h
aa
y dA y y y y dx y y dx
= −= − =
=+ −= −
∫∫
∫∫ ∫
consent of McGraw–Hill Education.
SOLUTION Continued
22
24
24
0
22
35
40
2
1
2
11
235
1
15
a
a
hh
x x dx
aa
ha
xx
a
ah
= −
= −
=
∫
2
11
:6 12
EL
xA x dA x ah a h
= =
∫
1
2
xa=
2
11
:6 15
EL
yA y dA y ah ah
= =
∫
2
5
yh=
PROBLEM 5.28
Determine by direct integration the centroid of the area shown. Express your
answer in terms of a and h.
SOLUTION
For the element (EL) shown
At
33
, : or h
x a y h h ka k a
= = = =
Then
1/3
1/3
a
xy
h
=
Now
1/3
1/3
1/3
1/3
11
22
EL
EL
a
dA xdy y dy
ha
xx y
h
yy
= =
= =
=
Then
( )
1/3 4/3
1/3 1/3
00
33
44
h
h
aa
A dA y dy y ah
hh
= = = =
∫∫
and
1/3 1/3 5/3 2
1/3 1/3 2/3
00
1/3 7/3 2
1/3 1/3
00
1 13 3
2 2 5 10
33
77
h
h
EL
h
h
EL
aa a
x dA y y dy y a h
hh h
aa
y dA y y dy y ah
hh
= = =
= = =
∫∫
∫∫
Hence
2
33
: 4 10
EL
xA x dA x ah a h
= =
∫
2
5
xa
=
2
33
: 47
EL
yA y dA y ah ah
= =
∫
4
7
yh=
PROBLEM 5.29
SOLUTION
First note that symmetry implies
0y=
EL
dA adx
xx
=
=
1()
2
2cos
3
EL
dA a ad
xa
φ
φ
=
=
Then
2
0
22
0
1
2
[ ] [ ] (1 )
2
a
a
A dA adx a d
a
ax a
a
a
a
a
φ
φa
−
= = −
=−=−
∫∫ ∫
and
2
0
23
0
3
21
( ) cos
32
1[sin ]
23
12
sin
23
a
EL
a
x dA x adx a a d
x
aa
a
a
a
aa
φφ
φ
a
−
−
= −
= −
= −
∫∫ ∫
23
12
: [ (1 )] sin
23
EL
xA x dA x a a
aa
= −= −
∫
or
3 4sin
6(1 )
xa
a
a
−
=−
consent of McGraw–Hill Education.
PROBLEM 5.22
The homogeneous wire ABCD is bent as shown and is attached to
a hinge at C. Determine the length L for which portion AB of the
wire is horizontal.
SOLUTION
I II III
80 100W wW wWLw= = =
0: (80 )(32) (100 )(14) ( )(0.4 ) 0
C
M w w Lw L
Σ= + − =
29900L=
99.5 mmL=
consent of McGraw–Hill Education.
PROBLEM 5.23
A uniform circular rod of weight 8 lb and radius 10 in. is attached to a pin at C and
to the cable AB. Determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
For quarter circle,
2r
r
π
=
(a)
2
0: 0
C
r
M W Tr
π
Σ = −=
22
(8 lb)TW
ππ
= =
5.09 lbT=
(b)
0: 0 5.09 lb 0
xx x
F TC CΣ= −= −=
5.09 lb
x
=C
0: 0 8 lb 0
yy y
F CW CΣ= −= − =
8 lb
y
=C
9.48 lb=C
57.5°
consent of McGraw–Hill Education.
PROBLEM 5.24
The homogeneous wire ABC is bent into a semicircular arc and a straight
section as shown and is attached to a hinge at A. Determine the value of
θ
for
which the wire is in equilibrium for the indicated position.
SOLUTION
PROBLEM 5.25
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and h.
SOLUTION
h
PROBLEM 5.26
SOLUTION
3
(1 )y h kx= −
For
, 0.x ay= =
3
0 (1 )h ka= −
3
1
ka
∴=
3
3
1x
yh a
= −
1
,2
EL EL
x x y y dA ydx= = =
34
33
00 0
3
14
4
a
aa
xx
A dA ydx h dx h x ah
aa
===−=−=
∫∫ ∫
4 25 2
33
00 0
3
2 10
5
a
aa
EL
x xx
x dA xydx h x dx h a h
aa
= = − =−=
∫∫∫
3 2 36
23 36
00 0
11 2
11
22 2
aa a
EL
x h xx
y dA y ydx h dx dx
a aa
= = − = −+
∫∫ ∫ ∫
2 47 2
36
0
9
2 28
27
a
h xx
x ah
aa
= −+ =
2
33
:4 10
EL
xA x dA x ah a h
= =
∫
2
5
xa=
2
39
:4 28
EL
yA y dA y ah ah
= =
∫
3
7
yh=
consent of McGraw–Hill Education.
PROBLEM 5.27
Determine by direct integration the centroid of the area shown.
Express your answer in terms of a and h.
SOLUTION
At
( , ),ah
2
1
:y h ka=
or
2
h
ka
=
2
:y h ma=
or
h
ma
=
Now
12
1()
2
EL
EL
xx
y yy
=
= +
and
2
21 2
2
2
()
()
hh
dA y y dx x x dx
aa
hax x dx
a
=−=−
= −
Then
2 23
22
00
11
() 23 6
a
ah ha
A dA ax x dx x x ah
aa
== −= − =
∫∫
and
( )
2 34 2
22
00
22
12 21 21
11
() 3 4 12
11
( )[( ) ]
22
a
a
EL
EL
h ha
x dA x ax x dx x x a h
aa
y dA y y y y dx y y dx
= −= − =
=+ −= −
∫∫
∫∫ ∫
consent of McGraw–Hill Education.
SOLUTION Continued
22
24
24
0
22
35
40
2
1
2
11
235
1
15
a
a
hh
x x dx
aa
ha
xx
a
ah
= −
= −
=
∫
2
11
:6 12
EL
xA x dA x ah a h
= =
∫
1
2
xa=
2
11
:6 15
EL
yA y dA y ah ah
= =
∫
2
5
yh=
PROBLEM 5.28
Determine by direct integration the centroid of the area shown. Express your
answer in terms of a and h.
SOLUTION
For the element (EL) shown
At
33
, : or h
x a y h h ka k a
= = = =
Then
1/3
1/3
a
xy
h
=
Now
1/3
1/3
1/3
1/3
11
22
EL
EL
a
dA xdy y dy
ha
xx y
h
yy
= =
= =
=
Then
( )
1/3 4/3
1/3 1/3
00
33
44
h
h
aa
A dA y dy y ah
hh
= = = =
∫∫
and
1/3 1/3 5/3 2
1/3 1/3 2/3
00
1/3 7/3 2
1/3 1/3
00
1 13 3
2 2 5 10
33
77
h
h
EL
h
h
EL
aa a
x dA y y dy y a h
hh h
aa
y dA y y dy y ah
hh
= = =
= = =
∫∫
∫∫
Hence
2
33
: 4 10
EL
xA x dA x ah a h
= =
∫
2
5
xa
=
2
33
: 47
EL
yA y dA y ah ah
= =
∫
4
7
yh=
PROBLEM 5.29
SOLUTION
First note that symmetry implies
0y=
EL
dA adx
xx
=
=
1()
2
2cos
3
EL
dA a ad
xa
φ
φ
=
=
Then
2
0
22
0
1
2
[ ] [ ] (1 )
2
a
a
A dA adx a d
a
ax a
a
a
a
a
φ
φa
−
= = −
=−=−
∫∫ ∫
and
2
0
23
0
3
21
( ) cos
32
1[sin ]
23
12
sin
23
a
EL
a
x dA x adx a a d
x
aa
a
a
a
aa
φφ
φ
a
−
−
= −
= −
= −
∫∫ ∫
23
12
: [ (1 )] sin
23
EL
xA x dA x a a
aa
= −= −
∫
or
3 4sin
6(1 )
xa
a
a
−
=−
consent of McGraw–Hill Education.