1026 Solutions Manual
which yields
NxD4:98 in:(3)
Problem 7.48
The assembly shown is made of a semicircular plate having weight per area of
0:125 lb=in:2
and uniform rods having weight per length of
0:05 lb=in
. Deter-
mine the coordinates of the center of gravity.
2.3in:/2D1:767 lb 04
3 .3in:/D1:273 in:
2.0:5 lb=in./ .5in:/D0:25 lb 2:5 in: 0
Problem 7.49
The object shown consists of a rectangular solid, a plate, and a quarter-circular
wire, with masses as follows: solid:
0:0005 g=mm3
, plate:
0:01 g=mm2
, wire:
0:05 g=mm. Determine the coordinates of the center of mass.
NxDPQximi
Pmi
;
D.5mm/ .0:240 g/C10 mm
3.0:300 g/C0
(1)
D.3mm/ .0:240 g/C0C.3:820 mm/ .0:4710 g/
(3)
D.4mm/ .0:240 g/C.10 mm/ .0:300 g/C.11:82 mm/ .0:471 g/
(5)
Problem 7.50
A sign frame is constructed of metal pipe having a mass per length of
4kg=m
.
Portion
CDE
is semicircular. The frame is supported by a pin at
B
and a
weightless cable
CF
. Determine the coordinates of the center of mass for the
frame, the reactions at B, and the force in the cable.
1.4kg=m/ .2m/D8kg 0
2.4kg=m/ .2m/D8kg 2m
Using the information in the table, the xlocation of the center of mass is
Problem 7.51
In Example 7.9 on p. 456, the center of gravity of a cone with variable specific weight is found using
integration with a thin disk mass element. Discuss why, for this example, a thin disk mass element is
considerably more convenient than a thin shell mass element.
Note: Concept problems are about explanations, not computations.
Solution
Problem 7.52
The object shown consists of a metal outside housing with density
0
and a
hole filled with plastic having density
0=2
. Use integration to determine the
mass of the solid and the xposition of the center of mass.
Problem 7.53
The object shown consists of a metal outside housing with density
0
and a
hole filled with plastic having density
0=2
. Use integration to determine the
mass of the solid and the xposition of the center of mass.
Solution
Problem 7.54
For the object indicated:
(a)
Fully set up the integral, including the limits of integration, that will yield
the center of mass of the object.
(b)
Evaluate the integral determined in Part (a), using computer software
such as Mathematica or Maple.
The solid hemisphere of radius
r
shown has density
0=2
for
0xr=2
and 0for r=2 xr.
Problem 7.55
For the object indicated:
(a)
Fully set up the integral, including the limits of integration, that will yield
the center of mass of the object.
(b)
Evaluate the integral determined in Part (a), using computer software
such as Mathematica or Maple.
The hollow cone shown is constructed of a material with uniform density
0
and has wall thickness that varies linearly from
2t0
at
xD0
to
t0
at
xDL
.
Assume t0is much smaller than Land R.
Problem 7.56
For the object indicated:
(a)
Fully set up the integral, including the limits of integration, that will yield the
center of mass of the object.
(b)
Evaluate the integral determined in Part (a), using computer software such as
Mathematica or Maple.
A solid of revolution is formed by revolving the area shown 360
ı
about the
x
axis. The material has uniform density 0.
Problem 7.57
For the object indicated:
(a)
Fully set up the integral, including the limits of integration, that will yield the
center of mass of the object.
(b)
Evaluate the integral determined in Part (a), using computer software such as
Mathematica or Maple.
A solid of revolution is formed by revolving the area shown 360
ı
about the
y
axis. The material has uniform density 0.
Problem 7.58
For the object indicated:
(a)
Fully set up the integral, including the limits of integration, that will yield
the center of gravity of the object.
(b)
Evaluate the integral determined in Part (a), using computer software
such as Mathematica or Maple.
The solid shown has a cone-shaped cavity. The material has uniform specific
weight 0.
D0“Rx
L1=32
R
2
x
L2#dx;
(2)
Problem 7.59
For the object indicated:
(a)
Fully set up the integral, including the limits of integration, that will yield the
center of gravity of the object.
(b)
Evaluate the integral determined in Part (a), using computer software such as
Mathematica or Maple.
A solid of revolution is formed by revolving the area shown 360
ı
about the
x
axis.
The material has uniform specific weight 0.
D0h1Cx22x2idx:
(2)
The center of gravity for
dw
is at
QxDx
. The
x
location of the center of gravity
Problem 7.60
For the object indicated:
(a)
Fully set up the integral, including the limits of integration, that will yield the
center of gravity of the object.
(b)
Evaluate the integral determined in Part (a), using computer software such as
Mathematica or Maple.
A solid of revolution is formed by revolving the area shown 360
ı
about the
y
axis.
The material has uniform specific weight 0.
Problem 7.61
The cross section of a rubber V belt is shown. If the belt has circular shape
about the axis of revolution with an inside radius of
6in:
, determine the volume
of material in the belt and the surface area of the belt.
2.0:25 in:/ .0:5 in:/D0:0625 in:2;(1)
3.0:5 in:/;Qr2D6in:C1
2.0:5 in:/(3)
Noting that the contributions from shapes 1 and 3 are the same, we employ the Pappus-Guldinus theorem,
doubling the first term, to obtain the volume of the belt as
i
which yields
VD12:8 in:3:(5)
To determine the surface area of the belt, the Pappus-Guldinus theorem will be used with the composite
lines shown in the second figure above, where
i
(8)
Problem 7.62
A pharmaceutical company’s design for a medicine capsule consists of hemispherical
ends and a cylindrical body. Determine the volume and outside surface area of the
capsule.
Solution
The distances from the axis of revolution to the centroid of each region is
Noting that the contributions from shapes 1 and 3 are the same, we employ the Pappus-Guldinus theorem,
doubling the first term, to obtain the surface area of the capsule as
Problem 7.63
A solid is produced by rotating a triangular area 360
ı
about the vertical axis of
revolution shown. Determine the volume and surface area of the solid.
Solution
To determine the volume of the solid, the Pappus-Guldinus theorem will be used where
the generating area Atfor the solid of revolution is
3.5mm/D10:67 mm:(2)
Employing the Pappus-Guldinus theorem, the volume of the object is
The distances from the axis of revolution to the centroid of each line segment is
Using the Pappus-Guldinus theorem, the surface area of the solid is
Problem 7.64
Repeat Prob. 7.63 if the solid is produced by rotating the triangular area about
the horizontal axis shown.
2.5mm/ .10 mm/D25 mm2;(1)
The distance from the axis of revolution to the centroid of the generating area is
2.10 mm/D13 mm;Qr2D8mm;Qr3D8mm C1
2.10 mm/D13 mm:(6)
Using the Pappus-Guldinus theorem, the surface area of the solid is