PROBLEM 5.18
Determine the x coordinate of the centroid of the trapezoid shown in
terms of h1, h2, and a.
SOLUTION
Area 1:
A
x
x
A
1 1
1
2ha 1
3a 2
1
1
6ha
PROBLEM 5.19
For the semiannular area of Prob. 5.12, determine the ratio r1 to r2 so that the centroid
of the area is located at x = 2
1
2r and y = 0.
SOLUTION
First, determine the location of the centroid.
22
4
3
x
r
x
A
PROBLEM 5.19 (Continued)
Now substituting for 2
1
2
Xr in equation (1) gives.
22
221 2121
12
(r)
22 3
rrr rrr
(1)
Dividing through by 2 and 2/3,r
PROBLEM 5.20
A composite beam is constructed by bolting
four plates to four 60 60 12-mm angles as
shown. The bolts are equally spaced along the
beam, and the beam supports a vertical load.
As proved in mechanics of materials, the
shearing forces exerted on the bolts at A and B
are proportional to the first moments with
respect to the centroidal x-axis of the red-
shaded areas shown, respectively, in parts a
and b of the figure. Knowing that the force
exerted on the bolt at A is 280 N, determine
the force exerted on the bolt at B.
SOLUTION
PROBLEM 5.21
The horizontal x-axis is drawn through the centroid C of the area shown,
and it divides the area into two component areas A
1
and A
2
. Determine the
first moment of each component area with respect to the x-axis, and
explain the results obtained.
SOLUTION
PROBLEM 5.22
The horizontal x-axis is drawn through the centroid C of the area shown,
and it divides the area into two component areas A
1
and A
2
. Determine the
first moment of each component area with respect to the x-axis, and
explain the results obtained.
SOLUTION
Dimensions in mm
PROBLEM 5.23
The first moment of the shaded area with respect to the x-axis is denoted by
Q
x
. (a) Express Q
x
in terms of b, c, and the distance y from the base of the
shaded area to the x-axis. (b) For what value of y is
x
Q
maximum, and what is
that maximum value?
SOLUTION
Shaded area:
()
x
Abcy
QyA
PROBLEM 5.24
A thin, homogeneous wire is bent to form the perimeter of the figure indicated.
Locate the center of gravity of the wire figure thus formed.
SOLUTION
,mmL
,mmx
,mmy
2
,mmxL
2
,mmyL
1 45 49.5 0 2227.5 0
PROBLEM 5.25
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
L, mm
,mmx
,mmy
2
,mmxL
2
,mmyL
1
22
24 12 26.833 12 322
2 32 24 28 768 896
PROBLEM 5.26
A thin, homogeneous wire is bent to form the perimeter of the figure indicated.
Locate the center of gravity of the wire figure thus formed.
SOLUTION
L, in. ,in.x ,in.y
2
,inxL
2
,inyL
1 4 0 2 0 8
PROBLEM 5.27
A thin, homogeneous wire is bent to form the perimeter of the
figure indicated. Locate the center of gravity of the wire figure
thus formed.
SOLUTION
First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of
the corresponding line.
6
2(38 in.)Y
PROBLEM 5.28
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
for
which the wire is in equilibrium for the indicated position.
SOLUTION
First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through A.
Further, because the wire is homogeneous, its center of gravity will coincide with the centroid of the
corresponding line. Thus,
PROBLEM 5.29
The frame for a sign is fabricated from thin, flat steel bar stock of mass
per unit length 4.73 kg/m. The frame is supported by a pin at C and by a
cable AB. Determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
First note that because the frame is fabricated from uniform bar stock, its center of gravity will coincide
with the centroid of the corresponding line.
PROBLEM 5.29 (Continued)
Equilibrium then requires
(a)
3
0: (1.55 m) (0.54247 m)(214.75 N) 0
5
CBA
MT
or 125.264 N
BA
T or 125.3 N
BA
T
PROBLEM 5.30
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.
SOLUTION
First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through C.
Further, because the wire is homogeneous, the center of gravity of the wire will coincide with the
centroid of the corresponding line. Thus,
PROBLEM 5.31
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
PROBLEM 5.32
Determine the distance h for which the centroid of the shaded
area is as far above line BB′ as possible when (a) k 0.10,
(b) k 0.80.
SOLUTION
PROBLEM 5.32 (Continued)
Then
22
2(2)4()()
2
11
aaka
hk
ak
k
Note that only the negative root is acceptable since .ha
Then
PROBLEM 5.33
Knowing that the distance h has been selected to maximize
the distance
y
from line BB′ to the centroid of the shaded
area, show that 2/3.yh
SOLUTION
See solution to Problem 5.32 for analysis leading to the following equations:
22
3( )
akh
Yakh
(1)
PROBLEM 5.34
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and h.
SOLUTION
We have
h
yx
a
and
()
1
dA h y dx
x
hdx
a