9–1.
SOLUTION
Locate t
h
e center of mass of t
h
e
h
omogeneous ro
d
b
ent
into the shape of a circular arc.
y
x
30
300 mm
880
9–2.
Determine the location (x, y) of the centroid of the wire.
SOLUTION
Applying Eq. 9–7 and performing the integration, we have
y
2 ft
4 ft
Ans:
881
9–3.
Locate the center of gravity
x
of the homogeneous rod. If
the rod has a weight per unit length of 100 N
>
m, determine
the vertical reaction at A and the x and y components of
reaction at the pin B.
SOLUTION
Length And Moment Arm. The length of the differential element is
Centroid.
A
B
x
1 m
1 m
y x2
y
882
*9–4.
Locate the center of gravity
y
of the homogeneous rod.
SOLUTION
Perform the integration,
Centroid.
Ans:
A
B
x
1 m
1 m
y x2
y
9–5.
Determine the distance
y
to the center of gravity of the
homogeneous rod.
SOLUTION
Centroid. Applying Eq. 9–7,
y
y 2x3
x
2 m
1 m
Ans:
9–6.
Locate the centroid of the area.y
SOLUTION
Centroid: Due to symmetry
y
x
2m
1m
y1– x
2
1
–
4
9–7.
SOLUTION
Differential Element:The area element parallel to the xaxis shown shaded in Fig. a
will be considered. The area of the element is
Determine the area and the centroid of the parabolic area.
x
x
h
a
y x2
h
––
a2
y
886
*9–8.
Locate the centroid of the shaded area.
SOLUTION
Area And Moment Arm. The area of the differential element shown shaded in
Due to Symmetry,
Ans:
y
x
L
a
y a cos L
px
2
L
2
9–9.
Locate the centroid
x
of the shaded area.
SOLUTION
Area And Moment Arm. The area of the differential element shown shaded in Fig.a
Centroid. Perform the integration
y
x
4 m
4 m
x2
y
1
4
888
9–10.
Locate the centroid
y
of the shaded area.
SOLUTION
Area And Moment Arm. The area of the differential element shown shaded in Fig.a
Centroid. Perform the integration
y
x
4 m
4 m
x2
y
1
4
9–11.
Locate the centroid of the area.
SOLUTION
x
y
y
x2
h
—
b2
890
*9–12.
y
y
x2
h
—
b2
Locate the centroid of the shaded area.
SOLUTION
y
Ans:
9–13.
Locate the centroid of the shaded area.
SOLUTION
x
y
x
4 m
y 4 x2
1
––
16
892
9–14.
Locate the centroid of the shaded area.
SOLUTION
y
y
4 m
y 4 x2
1
––
Ans:
893
9–15.
SOLUTION
Locate the centroid of the shaded area. Solve the problem by
evaluating the integrals using Simpson’s rule.
x
y= 0.5e
x2
y
Ans:
*9–16.
Locate the centroid of the shaded area.Solve the problem by
evaluating the integrals using Simpson’s rule.
y
SOLUTION
y= 0.5e
x2
y
Ans:
9–17.
y
y x
4 in.
2
––
3
Locate the centroid of the area.
SOLUTION
Area: Integrating the area of the differential element gives
y
Ans:
9–18.
Locate the centroid of the area.
SOLUTION
x
y
y h xn
h
h
—
an
897
9–19.
y
y h xn
h
h
—
an
Locate the centroid of the area.
SOLUTION
y
Ans:
898
*9–20.
SOLUTION
L
ocate the centroid of the shaded area.
y
y
h
yx
n
h
––
a
n
Ans: