C3 Examples

Problem 9-1

Determine the distance xc to the center of mass of the homogeneous rod bent into the shapeshown. If the rod has a mass per unit length , determine the reactions at the fixed support O.Units Used :

g 9.81m

s2:=

Given:

0.5 kgm

:=

a 1m:=

b 1m:=

Solution:

Length and Moment Arm: The length of differential element is

dL dx2 dy2+= 1 dydx

2+

dx=

and its centroid is xc = x. Here,dydx

3 b2 a1.5

x=

Performing the integrations we have

L

0

a

x19 b2x4 a3

+

d:= L 1.440 m=

xc1L

0

a

xx 19 b2x4 a3

+

d:= xc 0.546 m=

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Equations of Equilibrium:

+ Fx = 0; Ox 0=

+ Fy = 0; Oy g L 0= Oy g L:= Oy 7.062 N=MO = 0; M0 g Lxc 0= M0 g L xc:= M0 3.854 N m=

Problem 9-2

Determine the location (xc, yc) of the centroid of the wire.

Given:

a 2 ft:=b 4 ft:=

Solution:

Length and Moment Arm: Thelength of differential element is

dL dx2 dy2+= 1 dydx

2+ dx=

dL 12 b x

a2

2+=

Performing the integrations

L

a

a

x12 b x

a2

2+

d:= L 9.29 ft=

xc1L

a

a

xx 12 b x

a2

2+

d:=

yc1L

a

a

xbxa

2 1 2 b x

a2

2+

d:=

xc

yc

0.00

1.82

ft=




Problem 9-3

Locate the center of mass of the homogeneous rod bent into the shape of a circular arc.

Given:

r 300mm:= 30deg:=

Solution:

dL r d=xc r cos ( )=yc r sin ( )=

xc

2

2

+r cos ( ) r

d

2

2

+r

d

:= xc 124mm=

yc 0= Bysymmetry( )




Problem 9-5

Determine the distance xc to the center of gravity of the homogeneous rod bent into the parabolicshape. If the rod has a weight per unit length determine the reactions at the fixed support O.Given:

0.5 lbft

:=

a 1 ft:=

b 0.5 ft:=

Solution:

dL dx2 dy2+= 1 dydx

2+=

dydx

2. b xa2

=

L

0

a

x12. b x

a2

2+

d:= L 1.148 ft=

xc1L

0

a

xx 12. b x

a2

2+

d

:= xc 0.531 ft=

+ Fx = 0; Ox 0=

+ Fy = 0; Oy L 0= Oy L:= Oy 0.574 lb=MO = 0; MO L xc 0= MO L xc:= MO 0.305 lb ft=




Problem 9-7

Locate the centroid of the parabolic area.

Solution: ah

b2=

dA x dy=

xcx2

=

yc y=

A

0

h

ybyh

d= A23

h b=

xc3

2 h b0

h

y12

byh

2

d= xc38

b=

xc38

b=

yc3

2 h b0

h

yy b yh

d= yc35

h=

yc35

h=




Problem 9-8

Locate the centroid (xc, yc) of the shaded area.

Given:

a 2m:=b 1m:=

Solution:

Area and Moment Arm: The area of the differential element is

dA ydx= b 1xa

2

dx=

and its centroid is

ycy2

=12

b 1 xa

2

=

Centroid: Due to symmetry

xc 0=

yca

a

x12

b 1xa

2

b 1

xa

2

d

a

a

xb 1xa

2

d

:= yc25

m=




Problem 9-9

Locate the centroid xc of the shaded area.

Solution:

dA y dx=

xc x=

ycy2

=

xc0

b

xxh

b2x2

d

0

b

xhx2

b2

d

= xc34

b=

xc34

b=

yc0

b

x12

h

b2x2

2

d

0

b

xh

b2x2

d

= yc310

h=

yc310

h=




Problem 9-10


Solution:

Area and Moment Area:The area of the differential element is

dA y dx= 2k x x2

2a

=

and its centroid isx1 x=

Centroid: Applying Eq.9-6 and performing the integration,we have

xc0

a

xx 2 k x x2

2a

d

0

a

x2 k x x2

2a

d

= xc58

a= xc58

a=




Problem *9-12

Locate the centroid of the shaded area.

Solution:

dA ydx= xc x= ycy2

=

A

0

L

xa sinxL

d= A2 L a=

xc

2 L a0

L

xx a sin xL

d= xc12

L=

yc

2 L a0

L

x12

a sin xL

2

d= yc18

a=

L

0




Problem *9-16

Locate the centroid of the shaded area bounded by the parabola and the line y = a

Solution:

dA x dy= xcx2

= yc y=

A0

aya y d= A

23

a2( )32

a= A 2 a2

3=

xc3

2 a212

0

a

ya y( )2 d

= xc

38

a= xc38

a=

yc3

2 a2 0

ayy a y d= yc

3

5 a4a2( )

52

= yc35

a=




Problem 9-17

Locate the centroid of the quarter elliptical area.

Solution: dA y dx= xc x= ycy2

=

A

0

a

xb 1xa

2

d= A a b

4=

xc4

a b0

a

xx b 1 xa

2

d= xc4

3 a=

yc4

a b0

a

x12

b 1xa

2

2

d= yc4

3 b=




Problem 9-21

Locate the centroid yc of the shaded area. Solve the problem by evaluating the integrals usingSimpson's rule.

Given:

a 2:= ft

b a

12 2 a

53+:=

Solution:

A0

a

xb x

12 2 x

53+

d:= A 2.177= ft

2

yc1A

0

a

x12

b x

12 2 x

53++

b x

12 2 x

53+

d:= yc 2.040= ft




Problem 9-23


Solution:

Area and Moment Arm : Here,

x1a yb

= x2 ayb

2=

The area of the differential element is

dAa yb

ayb

2

dy=

and its centroid is

xc12

a yb

ayb

2+

=

Centroid : Applying Eq.9-6 and performing theintegration, we have

A

0

b

ya yb

ayb

2

d= A16

a b=

xc6

a b0

b

y12

a yb

ayb

2+

a yb

ayb

2

d= xc25

a=

Given: a 1ft:= b 2ft:= xc2 a5

:= xc 0.40 ft=




Problem 9-24

Locate the centroid yc of the shaded area.

Solution:

Area and Moment Arm : Here,

x1a yb

= x2 ayb

2=

The area of the differential element is

dAa yb

ayb

2

dy=

and its centroid is

xc12

a yb

ayb

2+

=

Centroid : Applying Eq.9-6 and performing theintegration, we have

A

0

b

ya yb

ayb

2

d= A16

a b=

yc6

a b0

b

yya yb

ayb

2

d= yc12

b=

Given: a 1ft:= b 2ft:= ycb2

:= yc 1.00 ft=




Problem 9-27


Given : a 2in:= b 0.5in:=

Solution : ya bx

=

xcb

a

xxa bx

d

b

a

xa bx

d

:= xc 1.08 in=




Area and Moment Arm :The area of the differential element is

dA y dx= 1x

dx= and its centroid is xc x=

Centroid :Applying Eq. 9-6 and performing the integration, we have

xc

Axc

d

A1

d

=

Axc

db

a

xxa bx

d

b

a

xa bx

d

= xcb

a

xxa bx

d

b

a

xa bx

d

:=

xc 1.08 in=

Problem 9-32

Locate the centroid of the ellipsoid of revolution.

Solution : dV z2 dy= z2 a2 1 y2

b2

=

V

0

b

y a2 1 y2

b2

d= V23

b a2=

yc3

2 b a2 0

b

yy a2 1 y2

b2

d= yc38

b= yc3 b8

=

By symmetry xc zc= 0=




Problem 9-33

Locate the center of gravity of the volume. The material is homogeneous.

Given:

a 2m:=b 2m:=

ya

2 zb

= y2 a2zb

=

Solution:

zc0

b

zz a2 zb

d

0

b

z a2 zb

d

:= zc 1.33 m= xc yc= 0= by symmetry




Problem 9-36

Locate the centroid of the quarter-cone.

Solution :

rah

h z( )= zc z= xc yc=4 r3 =

V

0

h

z4

ah

h z( )

2

d= V112

h a2=

zc12

h a2 0

h

zz4

ah

h z( )

2

d= zc14

h=

xc12

h a2 0

h

z4

3 ah

h z( )4

ah

h z( )

2

d

= xc

1 a=

xc yc=a= zc

h4

=




Problem 9-39

Locate the centroid yc of the paraboloid.

Given:

a 4m:=b 4m:=

Solution:

zb

2 ya

= z bya

=

dV z2 dy= b2 ya

=

and its centroid yc y=

yc0

a

yy b2 ya

d

0

a

y b2 ya

d

:= yc 2.67 m=




Problem 9-41

Locate the centroid zc of the frustum of theright-circular cone.

Solution :

Volume and Moment Arm :From the geometry,

y rR r

h zh

=

yr R( ) z R h+

h=

The volume of thin disk differential element is

dV y2 dz= r R( ) z R h+h

2 dz=

zc0

h

zz r R( ) z R h+h

2

d

0

h

z r R( ) z R h+h

2

d

= zc

14 h2 r2 1

6 h2 r R+ 1

12 h2 R2+

13

h r3

r R13

R3 h

r R=




This can be simplified to give zcR2 2 r R+ 3 r2+4 R2 r R+ r2+( ) h=

dVh2

r R( )2 z2 2 R h r R( ) z+ R2 h2+ dz=

and its centroid

z' z=

Centroid :Applying Eq. 9-5 and performing the integration,we have

z'

Vz'

d

V1

d

=0

h

zzh2

r R( )2 z2 2 R h r R( ) z+ r2 h2+

d

0

h

zh2

r R( )2 z2 2 R h r R( ) z+ R2 h2+

d

=V

V

h0

z'

h2

r R( )2 z4

4

2 R h r R( )

z3

3

+ R

2 h2 z2

2

+

h2

r R( )2 z3

3

2 R h r R( )

z2

2

+ R

2 h2 z( )+

=

h0

z'R2 3 r2+ 2 r R+4 R2 r2+ r R+( ) h=

Problem 9-44

Locate the center of gravity G of the five particles with respect to the origin O.

g 9.81m

s2:=

Given:

a 2m:= m1 1kg:=b 1m:= m2 10kg:=c 1m:=

m3 2kg:=d 3m:=

m4 5kg:=e 2m:=m5 6kg:=

Solution:

xcxc WW=

xcm1 g a b+ c+( ) m2 g b c+( ) m3 g c m4 g d+ m5 g d e+( )+

m1 m2+ m3+ m4+ m5+( ) g:=xc 0.792 m=




Problem 9-46

Locate the centroid (xc,yc) of the uniform wire bent in the shape shown.

Given :

a 100mm:=b 150mm:=c 50mm:=d 20mm:=

Solution :

xc

aa2

c c2

+ b d( ) c+ a c( ) c a c2

++

a b+ c+ b d( )+ a c( )+:= xc 34.38 mm=

yc

a b b b2

+ b d( ) b d2

+ a c( ) b d( )+a b+ c+ b d( )+ a c( )+:= yc 85.83 mm=




Problem 9-57

Determine the location yc of the centroidal axis xcxc of the beam's cross-sectional area. Neglect thesize of the corner welds at A and B for the calculation.

Given :

r 50mm:=t 15mm:=a 150mm:=b 15mm:=c 150mm:=

Solution :

yc

b c b2

a t b a2

++ r

2 b a+ r+( )+

b c a t+ r2+:= yc 154.44 mm=




Problem 9-58

Determine the location (xc,yc) of the centroid C of the area.

Given :

a 6in:=b 6in:=c 3in:=d 6in:=

Solution :

xc

a b b2

12

a c b c3

++

12

b c+( ) d 23

b c+( )+

a b 12

c a+ 12

b c+( ) d+:= xc 4.62 in=

yc

a b a2

12

a c a3

+ 12

b c+( ) d d3

a b 12

c a+ 12

b c+( ) d+:= yc 1.00 in=




Problem 9-62

Determine the location xc of the centroid C of the shaded area which is part of a circle having aradius r.

Solution :

xcA12

r2 2 r3 sin ( )

12

r sin ( ) r cos ( ) 23

r cos ( )=

xcAr3

3sin ( ) r

3

3sin ( ) cos2 ( )= r

3

3sin ( )3=

A 12

r2 12

r sin ( ) r cos ( )= 12

r2 sin 2 ( )2

=

xcxcAA=

xc

r3

3sin3 ( )

12

r2 sin 2 ( )2

=

23

r sin3 ( )

sin 2 ( )2

=




Problem 9-69

Determine the distance h to which a hole of diameter d must be bored into the base of the cone sothat the center of mass of the resulting shape is located at zc. The material has a density .

Given:

d 100mm:=zc 115mm:=

8 mgm3

:=

a 150mm:=b 500mm:=

Solutions:

zc

13

a2 b b4

d2

2 h h

2

13

a2 b d2

2 h

=

Choosing the positive root,

hzc d

13

9 zc2 d2 24 zc a2 b 6 a2 b2++

d:=

h 323 mm=




Problem 9-71

The sheet metal part has the dimensions shown. Determine the location (xc, yc, zc) ofits centroid.

Given:

a 3in:=

b 4in:=

c 6in:=

Solution:

xc

a b b2

a b 12

a c+:= xc 1.14 in=

yc

a b a2

12

a c 2 a3

+

a b 12

a c+:= yc 1.71 in=

zc

12

a c c3

a b 12

a c+:= zc 0.857 in=




Problem 9-76

Locate the center of gravity of the two-block assembly. The specific weights of the materials Aand B are A and B, respectively.

Given:

A 150lb

ft3:=

B 400lb

ft3:=

a 2in:=b 6in:=c 2in:=d 6in:=e 6in:=

Solution

WA Aa b d

2:= WB B c d e:=

xc

WBc2

WA cb3

++

WB WA+:= xc 1.47 in=

yc

WBe2

WAa2

+WB WA+

:= yc 2.68 in=

zc

WBd2

WAd3

+WB WA+

:= zc 2.84 in=




Problem 9-79

Locate the centroid zc of the top made from a hemisphere and a cone.

Given:

r 24mm:=h 120mm:=

Solution:

zc

3

r2 h 3 h4

23

r3 h 3 r8

++

3

r2 h 23 r3+

:= zc 101.14 mm=




Problem 9-83

The assembly consists of a wooden dowel rod of length L and a tight-fitting steel collar. Determinethe distance xc to its center of gravity if the specific weights of the materials are w and st.The radiiof the dowel and collar are shown.

Given:

L 20in:=w 150

lb

ft3:=

st 490lb

ft3:=

a 5in:=b 5in:=r1 1in:=r2 2in:=

Solution:

xc

w r12 LL2

st r22 r12 b a b2+

+

w r12 L st r22 r12 b+:= xc 8.22 in=




Problem 9-85

The anchor ring is made of steel having specific weight st. Determine the surface area of the ring.The cross section is circular as shown.

Given

st 490lb

ft3:=

a 4in:=

b 8in:=

Solution:

A 2 a2

b a4

+ 2

b a4

:= A 118 in2=




Problem 9-101

A V-belt has as inner radius r and a cross-sectional area as shown. Determine thevolume of material used in making the V-belt.

Given :

r 6in:=a 0.25in:=b 0.5in:= 30deg:=

Solution :

Volume : Applying the theorem ofPappus and Guldinus, Eq.9-12

h a cot2

:=

V 2 b h r h2

+ 2

a h2

r 2 h3

++

:= V 28.66 in

3=




Problem 9-102

The full circular aluminum housing is used in an automotive brake system. The cross section isshown in the figure. Determine its weight if aluminum has a specific weight .

Given :

169 lbft3

:= d 4.00in:=e 3.25in:=

a 2.00in:=f 0.25in:=

b 0.25in:=c 0.15in:=

Solution :

Volume : Applying the theorem of Pappusand Guldinus, Eq.9-12, with

ra2

b:=

V 2 r c+( ) f r c+2

c e r c2

++ b d e f( ) r

b2

++

:= V 3.85 in

3=

W V:= W 0.377 lb=




Problem 9-124

A circular V-belt has an inner radius r and a cross-sectional area as shown. Determine the volumeof material required to make the belt.

Given

r 600mm:=

a 25mm:=

b 50mm:=

c 75mm:=

Solution

V 2 r c3

+ 2

12

a c r

c2

+ b c+

:=

V 22.4 10 3 m3=




Sample Problem 5/1

Centroid of a circular arc. Locate the centroid of a circular arc as shown inthe figure.

Solution. Choosing the axis of symmetry as the x-axis makes 0. A differ-ential element of arc has the length dL r d expressed in polar coordinates,and the x-coordinate of the element is r cos .

Applying the first of Eqs. 5/4 and substituting L 2r give

Ans.

For a semicircular arc 2 , which gives 2r/. By symmetry we seeimmediately that this result also applies to the quarter-circular arc when themeasurement is made as shown.

Helpful Hint

It should be perfectly evident that polar coordinates are preferable to rectan-gular coordinates to express the length of a circular arc.

x

x r sin

2rx 2r2 sin

(2r)x

(r cos ) r d[Lx x dL]

y

244 Chapter 5 Distr ibuted Forces

r

C

r

C C

x

yr cos

r dd

rr

2r/

y

x

h

bx

y

dy

Helpful Hint

We save one integration here byusing the first-order element of area.Recognize that dA must be expressedin terms of the integration variabley; hence, x (y) is required.

Sample Problem 5/2

Centroid of a triangular area. Determine the distance from the base of atriangle of altitude h to the centroid of its area.

Solution. The x-axis is taken to coincide with the base. A differential strip ofarea dA x dy is chosen. By similar triangles x/(h y) b/h. Applying the sec-ond of Eqs. 5/5a gives

and Ans.

This same result holds with respect to either of the other two sides of thetriangle considered a new base with corresponding new altitude. Thus, the cen-troid lies at the intersection of the medians, since the distance of this point fromany side is one-third the altitude of the triangle with that side considered thebase.

y h3

bh2

y h0

y b(h y)

h dy bh

2

6[Ay yc dA]

h

c05.qxd 1/26/06 1:30 PM Page 244

Sample Problem 5/3

Centroid of the area of a circular sector. Locate the centroid of the areaof a circular sector with respect to its vertex.

Solution I. The x-axis is chosen as the axis of symmetry, and is thereforeautomatically zero. We may cover the area by moving an element in the form ofa partial circular ring, as shown in the figure, from the center to the outer pe-riphery. The radius of the ring is r0 and its thickness is dr0, so that its area isdA 2r0 dr0.

The x-coordinate to the centroid of the element from Sample Problem 5/1 isxc r0 sin /, where r0 replaces r in the formula. Thus, the first of Eqs. 5/5agives

Ans.

Solution II. The area may also be covered by swinging a triangle of differentialarea about the vertex and through the total angle of the sector. This triangle,shown in the illustration, has an area dA (r/2)(r d), where higher-order termsare neglected. From Sample Problem 5/2 the centroid of the triangular elementof area is two-thirds of its altitude from its vertex, so that the x-coordinate to thecentroid of the element is xc cos . Applying the first of Eqs. 5/5a gives

and as before Ans.

For a semicircular area 2 , which gives 4r/3. By symmetry we seeimmediately that this result also applies to the quarter-circular area where themeasurement is made as shown.

It should be noted that, if we had chosen a second-order element r0 dr0 d,one integration with respect to would yield the ring with which Solution Ibegan. On the other hand, integration with respect to r0 initially would give thetriangular element with which Solution II began.

x

x 23

r sin

r2x 23r3 sin

(r2)x

(23 r cos )(12 r

2 d)[Ax xc dA]

23 r

x 23

r sin

r2x 23r 3 sin

22

(r2)x r0

r0 sin (2r0 dr0)[Ax xc dA]

y

Art ic le 5/3 Centroids of L ines , Areas , and Volumes 245

C

r

r

y

dr0

xr0

r0 sin

xc =

Solution I

Helpful Hints

Note carefully that we must distin-guish between the variable r0 andthe constant r.

Be careful not to use r0 as the cen-troidal coordinate for the element.

x

r

y

d

CC

rr

4r/3

xc = r cos 23

Solution II

c05.qxd 1/26/06 1:30 PM Page 245

Sample Problem 5/4

Locate the centroid of the area under the curve x ky3 from x 0 to x a.

Solution I. A vertical element of area dA y dx is chosen as shown in the fig-ure. The x-coordinate of the centroid is found from the first of Eqs. 5/5a. Thus,

Substituting y (x/k)1/3 and k a/b3 and integrating give

Ans.

In the solution for from the second of Eqs. 5/5a, the coordinate to thecentroid of the rectangular element is yc y/2, where y is the height of the stripgoverned by the equation of the curve x ky3. Thus, the moment principle be-comes

Substituting y b(x/a)1/3 and integrating give

Ans.

Solution II. The horizontal element of area shown in the lower figure may beemployed in place of the vertical element. The x-coordinate to the centroid of therectangular element is seen to be xc x (a x)/2, which is simplythe average of the coordinates a and x of the ends of the strip. Hence,

The value of is found from

where yc y for the horizontal strip. The evaluation of these integrals will checkthe previous results for and y.x

y b0

(a x) dy b0

y(a x) dy[Ay yc dA]y

x b0

(a x) dy b0

a x2 (a x) dy[Ax xc dA]

12(a x)

3ab4

y 3ab2

10 y 25 b

3ab4

y a0

y2y dx[Ay yc dA]

y

3ab4

x 3a2b

7 x 47a

x a0

y dx a0

xy dx[Ax xc dA]


y

x = ky3

a

b

x

x

y

C

y

x = ky3

x a

y

b

xdx

yc =y2

y

x = ky3

x

a

y

b

x

a x

dy

xc =a + x

2

Helpful Hint

Note that xc x for the verticalelement.

c05.qxd 1/26/06 1:30 PM Page 246

Sample Problem 5/5

Hemispherical volume. Locate the centroid of the volume of a hemisphereof radius r with respect to its base.

Solution I. With the axes chosen as shown in the figure, 0 by symme-try. The most convenient element is a circular slice of thickness dy parallel tothe x-z plane. Since the hemisphere intersects the y-z plane in the circle y2 z2 r2, the radius of the circular slice is z The volume of the elementalslice becomes

The second of Eqs. 5/6a requires

where yc y. Integrating gives

Ans.

Solution II. Alternatively we may use for our differential element a cylindricalshell of length y, radius z, and thickness dz, as shown in the lower figure. By ex-panding the radius of the shell from zero to r, we cover the entire volume. Bysymmetry the centroid of the elemental shell lies at its center, so that yc y/2.The volume of the element is dV (2z dz)(y). Expressing y in terms of z fromthe equation of the circle gives y Using the value of computedin Solution I for the volume of the hemisphere and substituting in the second ofEqs. 5/6a give us

Ans.

Solutions I and II are of comparable use since each involves an element ofsimple shape and requires integration with respect to one variable only.

Solution III. As an alternative, we could use the angle as our variable withlimits of 0 and /2. The radius of either element would become r sin , whereasthe thickness of the slice in Solution I would be dy (r d) sin and that of theshell in Solution II would be dz (r d) cos . The length of the shell would be y r cos .

y 38 r

r0

(r2z z3) dz r4

4

(23r3)y r

0 r2 z2

2 (2zr2 z2) dz[V y yc dV]

23 r

3r2 z2.

23 r

3 y 14r

4 y 38 r

y r0

(r2 y2) dy r0

y(r2 y2) dy[V y yc dV]

dV (r2 y2) dy

r2 y2.

zx

Art ic le 5/3 Centroids of L ines , Areas , and Volumes 247

z

y2 + z2 = r2

yc = y

y

xdy

dz

r

z

z

yc = y/2

y

y

r

y

z

r

x

r d

d

Solution I

Solution II

Solution III

z

Helpful Hint

Can you identify the higher-order el-ement of volume which is omittedfrom the expression for dV?

c05.qxd 1/26/06 1:30 PM Page 247

Sample Problem 5/6

Locate the centroid of the shaded area.

Solution. The composite area is divided into the four elementary shapesshown in the lower figure. The centroid locations of all these shapes may be ob-tained from Table D/3. Note that the areas of the holes (parts 3 and 4) aretaken as negative in the following table:

APART in.2 in. in. in.3 in.3

1 120 6 5 720 6002 30 14 10/3 420 1003 14.14 6 1.273 84.8 184 8 12 4 96 32

TOTALS 127.9 959 650

The area counterparts to Eqs. 5/7 are now applied and yield

Ans.

Ans.

Sample Problem 5/7

Approximate the x-coordinate of the volume centroid of a body whose lengthis 1 m and whose cross-sectional area varies with x as shown in the figure.

Solution. The body is divided into five sections. For each section, the averagearea, volume, and centroid location are determined and entered in the followingtable:

Aav Volume VINTERVAL m2 m3 m m4

00.2 3 0.6 0.1 0.0600.20.4 4.5 0.90 0.3 0.2700.40.6 5.2 1.04 0.5 0.5200.60.8 5.2 1.04 0.7 0.7280.81.0 4.5 0.90 0.9 0.810

TOTALS 4.48 2.388

Ans.X 2.3884.48

0.533 mX V xV

V xx

Y 650127.9

5.08 in.Y AyA

X 959127.9

7.50 in.X AxA

yAxAyx


Helpful Hint

Note that the shape of the body as afunction of y and z does not affect X.

x

y

12

3

3 2 2

2

5

4

4

14

2

3

00

1

2

3

4

5

6

0.2 0.4x, m

A, m

2

0.6 0.8 1.0

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Sample Problem 5/8

Locate the center of mass of the bracket-and-shaft combination. The verti-cal face is made from sheet metal which has a mass of 25 kg/m2. The material ofthe horizontal base has a mass of 40 kg/m2, and the steel shaft has a density of7.83 Mg/m3.

Solution. The composite body may be considered to be composed of the five el-ements shown in the lower portion of the illustration. The triangular part will betaken as a negative mass. For the reference axes indicated it is clear by symme-try that the x-coordinate of the center of mass is zero.

The mass m of each part is easily calculated and should need no further ex-planation. For Part 1 we have from Sample Problem 5/3

For Part 3 we see from Sample Problem 5/2 that the centroid of the triangularmass is one-third of its altitude above its base. Measurement from the coordinateaxes becomes

The y- and z-coordinates to the mass centers of the remaining parts should be ev-ident by inspection. The terms involved in applying Eqs. 5/7 are best handled inthe form of a table as follows:

mPART kg mm mm

1 0.098 0 21.2 0 2.082 0.562 0 75.0 0 42.193 0.094 0 100.0 0 9.384 0.600 50.0 150.0 30.0 90.005 1.476 75.0 0 110.7 0

TOTALS 2.642 140.7 120.73

Equations 5/7 are now applied and the results are

Ans.

Ans. Z 120.732.642

45.7 mmZ mzm

Y 140.72.642

53.3 mmY mym

kg mmkg mmzmyzy

z [150 25 13 (75)] 100 mm

z 4r3

4(50)

3 21.2 mm

Art ic le 5/4 Composite Bodies and F igures; Approximat ions 259

y

x

z

40

5050

50

25

100

150 75

150

Dimensions in millimeters

150

1

2

3

4

5

c05.qxd 1/26/06 1:30 PM Page 259

Sample Problem 5/9

Determine the volume V and surface area A of the complete torus of circularcross section.

Solution. The torus can be generated by revolving the circular area of radius athrough 360 about the z-axis. With the use of Eq. 5/9a, we have

Ans.

Similarly, using Eq. 5/8a gives

Ans.A rL 2(R)(2a) 42Ra

V rA 2(R)(a2) 22Ra2


Helpful Hint

We note that the angle of revolu-tion is 2 for the complete ring. Thiscommon but special-case result isgiven by Eq. 5/9.

Sample Problem 5/10

Calculate the volume V of the solid generated by revolving the 60-mm right-triangular area through 180 about the z-axis. If this body were constructed ofsteel, what would be its mass m?

Solution. With the angle of revolution 180, Eq. 5/9a gives

Ans.

The mass of the body is then

Ans. 2.21 kg

m V 7830 kgm3[2.83(105) mm3] 1 m1000 mm3

V rA [30 13(60)][12(60)(60)] 2.83(10

5) mm3

Helpful Hint

Note that must be in radians.

Rz

a

60mm

60mm

30mm

x

z

60mm

60mm

30mm

z

rCC

z

r = R

aa

c05.qxd 1/26/06 1:30 PM Page 268

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