Dynamics 14th edition by r c hibbeler chapter 09

Locate the centroid x of the shaded area.Solution Area And Moment Arm.. Locate the centroid of the area.SOLUTION Area: Integrating the area of the differential element gives Centroid: Th

- 2p 3

300a43pb

x= Lx'dL

LdL

= L

2p 3

- 2p 3

300 cos u (300du)

L

2p 3

- 2p 3

300d u

y' = 300 sin u

x' = 300 cos u

dL = 300 d u

Locate the center of mass of the homogeneous rod bent

into the shape of a circular arc

Determine the location (x, y) of the centroid of the wire.

SOLUTION

Length and Moment Arm: The length of the differential element is dL =

2dx2 + dy2 = ¢B1 + ady dx b2≤ dx and its centroid is ∼y = y = x2 Here, dy

L

2 ft -2 ft21 + 4x2 dx

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

A1+ ady dx b2d dx and its centroid is x~ = x Here dy dx = 2x

Perform the integration

1 m

0 Ax2 +

1

4 dx = c xAx2+ 14 + 14 ln ax + Ax2 + 14 b d

A1+ ady dx b2d dx and its centroid is y~ = y Here dy dx = 2x

Perform the integration,

1 m

0 Ax2 +

1

4 dx = c xAx2+ 14 + 14 ln ax + Ax2 + 14 b d

Determine the distance y to the center of gravity of the

Locate the centroid of the area.y

SOLUTION

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

Centroid: Due to symmetry

2m

-2m¢1 - 14x2≤dx

x= 0

y' = y2 = 1

Differential Element:The area element parallel to the x axis shown shaded in Fig a

will be considered The area of the element is

Centroid: The centroid of the element is located at and

Area: Integrating,

y' = yx

3ah

=

a22h¢y22≤`h

Locate the centroid of the shaded area

Solution

Area And Moment Arm The area of the differential element shown shaded in

Fig. a is dA = ydx = a cos pL x dx and its centroid is at y~ = y2 = a2 cos p

L>2 -L>2 a cos p

L x dx

= L

L>2 -L>2

a2

4 acos 2pL x + 1b dxL

Locate the centroid x of the shaded area.

Solution

Area And Moment Arm The area of the differential element shown shaded in Fig. a

is dA = x dy and its centroid is at x~ = 12 x Here, x = 2y1 >2

Centroid Perform the integration

Locate the centroid y of the shaded area.

Solution

Area And Moment Arm The area of the differential element shown shaded in Fig. a

is dA = x dy and its centroid is at ∼y = y Here, x = 2y1 >2

Centroid Perform the integration

Locate the centroid of the area.

Locate the centroid of the shaded area.

Locate the centroid of the shaded area Solve the problem by

evaluating the integrals using Simpson’s rule

x

y = 0.5e x2 y

x

1 m

Ans:

x = 0.398 m

Locate the centroid of the shaded area Solve the problem by

evaluating the integrals using Simpson’s rule

x

1 m

Ans:

y = 1.00 m

Locate the centroid of the area.

SOLUTION

Area: Integrating the area of the differential element gives

Centroid: The centroid of the element is located at Applying

Bhx - h1xn +12

an1n + 12R

0 a

x a

0 a

Locate the centroid x of the shaded area.

Solution

Area And Moment Arm The area of the differential element shown shaded

in  Fig.  a is dA = y dx = (4 - x1 >2)2 dx = (x - 8x1 >2 + 16)dx and its centroid is

Locate the centroid y of the shaded area.

Solution

Area And Moment Arm The area of the differential element shown shaded in

Fig.  a is dA = y dx = (4- x12)2 dx = (x - 8x1 >2 + 16)dx and its centroid is at

Locate the centroid x of the shaded area.

Solution

Area And Moment Arm The area of the differential element shown shaded in Fig. a

is dA = y dx = a -h

a2 x2 + hbdx and its centroid is at x~ = x.

Centroid Perform the integration,

a2 x2 + hbdx and its centroid is at y~ = y2 = 12a -h a2 x2 + hb.

Centroid Perform the integration,

The plate has a thickness of 0.25 ft and a specific weight of

Determine the location of its center of

gravity Also, find the tension in each of the cords used to

support it

SOLUTION

Area and Moment Arm: Here, The area of the differential

Evaluating the integrals, we have

Centroid: Applying Eq 9–6, we have

TC1162 - 192013.202 = 0 TC= 384 lb ©My = 0;

B

z

C y

Locate the centroid x of the shaded area.

Solution

Area And Moment Arm The area of the differential element shown shaded in Fig. a

is dA = y dx = 14 x2 dx and its centroid is at x~ = x.

Centroid Perform the integration

x = 1A x~ dA

4 ft 0

x a14 x2 dxbL

Locate the centroid y of the shaded area.

Solution

Area And Moment Arm The area of the differential element shown shaded in Fig. a

is dA = y dx = 14 x2 dx and its centroid is located at y~ = y2 = 12a14 x2b = 18 x2

Centroid Perform the integration,

y = 1A y~ dA

4 ft 0

Locate the centroid x of the shaded area.

Solution

Area And Moment Arm Here, y2 = x and y1 = 1001 x2 Thus the area of the

differential element shown shaded in Fig a is dA = (y2 - y1) dx = (x - 1001 x2)dx

and its centroid is at x~ = x.

Centroid Perform the integration

Locate the centroid y of the shaded area.

Solution

Area And Moment Arm Here, x2= 10y 1>2 and x1 = y Thus, the area of the

differential element shown shaded in Fig a is dA = (x2- x1) dy = (10y1 >2 - y)dy

and its centroid is at y~ = y.

Centroid Perform the integration,

Locate the centroid x of the shaded area.

Solution

Area And Moment Arm Here x1 = h a y and x2 = aa - b h b y + b Thus the area

of the differential element is dA = (x2 - x1) dy = c aa - b

Ans:

x = 13 (a + b)

Locate the centroid y of the shaded area.

Solution

Area And Moment Arm Here, x1= a h y and x2= aa - b h b y + b Thus the area

of the differential element is dA = (x2 - x1) dy = c aa - b

h b y + b - a h y d dy =

ab - b h y bdy and its centroid is at y~ = y.

Centroid Perform the integration,

2 bh

y

x h

Ans:

y = h3

Locate the centroid of the shaded area

SOLUTION

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

and its centroid are

a -a2cos x

a b`

0 pa

a

y  a sin x

Ans:

x = p2 a

Locate the centroid of the shaded area.

SOLUTION

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

and its centroid are

The steel plate is 0.3 m thick and has a density of

Determine the location of its center of mass

Also compute the reactions at the pin and roller support

Area And Moment Arm Here, y2 = h- a h n x n and y1 = h- h a x Thus, the

area of the differential element shown shaded in Fig a is dA = (y2 - y1) dx

= c h - a h n x n - ah - h a x b d dx = a h a x - a h n x n bdx and its centroid is x~ = x.

Centroid Perform the integration

x a

Locate the centroid y of the shaded area.

Solution

Area And Moment Arm Here, y2 = h- a h n x n and y1 = h- h a x Thus, the

area of the differential element shown shaded in Fig a is dA = (y2 - y1) dx

= c h - a h n x n - ah - h a x b d dx = a h a x - a h n x n bdx and its centroid is at

y

x a

Locate the centroid x of the circular sector.

Solution

Area And Moment Arm The area of the differential element shown in Fig a is

dA = 12 r2 du and its centroid is at x~ = 23 r cos u.

Centroid Perform the integration

a12 r2 ub `

-a a

Locate the center of gravity of the volume The material is

Volume and Moment Arm: The volume of the thin disk differential element is

and its centroid

Centroid: Due to symmetry about z axis

SOLUTION

Volume and Moment Arm: The volume of the thin disk differential element is

Centroid: Applying Eq 9–3 and performing the integration, we have

Ans.

=4p y

dV = pz2dy = p14y2dy

Ans:

y = 2.67 m

Volume and Moment Arm: From the geometry,

The volume of the thin disk differential element is

and its centroid

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

Ans.

= 4 RR22+ 3r2 + 2rR+ r2 + rR h

=

p

h2B1r - R22¢z44≤ + 2Rh1r - R2¢z33≤ + R2h2¢z22≤ R`

0 h

p

h2 1r - R22 z3

3 + 2Rh1r - R2 z22 + R2h21z2 `

0 h

z = LVz'dV

y x

Differential Element: The thin disk element shown shaded in Fig a will be considered The

volume of the element is

Centroid: The centroid of the element is located at We have

=

a3h12

pa2h12

= ap

= pa2

4 h2aa h3p b =3 a123h

= p4 ha22

4a3phah4- 3h2 +4 h4 - h4 b4

pa2h12

Ans:

z = h4

x = y = pa

The hemisphere of radius r is made from a stack of very thin

plates such that the density varies with height

where k is a constant Determine its mass and the distance

to the center of mass G.

SOLUTION

Mass and Moment Arm: The density of the material is The mass of the thin

centroid Evaluating the integrals, we have

Locate the centroid z of the volume.

Solution

Volume And Moment Arm The volume of the thin disk differential element shown

shaded in Fig a is dV = py2 dz = p(0.5z)dz and its centroid is at z~ = z.

Centroid Perform the integration

Locate the center of gravity of the solid.

SOLUTION

Differential Element: The thin disk element shown shaded in Fig a will be

considered The volume of the element is

Centroid: The centroid of the element is located at We have

Volume And Moment Arm The volume of the thin disk differential element shown

shaded in Fig a is dV = pz2 dy = pa1001 y2b2 dy = 10000p y4 dy and its centroid is

Locate the centroid of the spherical segment.z

a

a 2

z (a2 - z2)dz

pL

a

a 2

z 2 a2 y2

Ans:

z = 0.675a

Determine the location of the centroid for the

tetrahedron Hint: Use a triangular “plate” element parallel

to the x–y plane and of thickness dz.

The truss is made from five members, each having a length

of 4 m and a mass of If the mass of the gusset plates

at the joints and the thickness of the members can be

neglected, determine the distance d to where the hoisting

cable must be attached, so that the truss does not tip

(rotate) when it is lifted

Determine the location (x, y, z) of the centroid of the

homogeneous rod

Solution

Centroid Referring to Fig a, the length of the segments and the locations of their

respective centroids are tabulated below

A rack is made from roll-formed sheet steel and has the

cross section shown Determine the location of the

centroid of the cross section The dimensions are indicated

at the center thickness of each segment

Locate the centroid ( ) of the metal cross section Neglect

the thickness of the material and slight bends at the corners

yx,

The steel and aluminum plate assembly is bolted together

and fastened to the wall Each plate has a constant width in

the z direction of 200 mm and thickness of 20 mm If the

density of A and B is and for C,

determine the location of the center ofmass Neglect the size of the bolts

Ans:

x = 179 mm

Locate the center of gravity of the streetlight Neglect

the thickness of each segment The mass per unit length of

each segment is as follows:

Determine the location of the centroidal axis of the

beam’s cross-sectional area Neglect the size of the corner

welds at A and B for the calculation.

x xy

50 mm

A C

-Ans:

y = 154 mm

Locate the centroid (x, y) of the shaded area.

Solution

Centroid Referring to Fiq a, the areas of the segments and the locations of their

respective centroids are tabulated below

Segment A(in 2 ) x∼(in.) y∼(in.) x∼A(in 3 ) y∼A(in 3 )

Thus,

x = Σx~A ΣA = 72.0 in.126 in.23 = 0.5714 in = 0.571 in Ans.

y = Σy~A ΣA = -72.0 in.126 in.23 = -0.5714 in = -0.571 in Ans.

Centroid The locations of the centroids measuring from the x axis for segments

1 and 2 are indicated in Fig a Thus

y = Σy~A ΣA = 300(120)(600)120(600) + 120(240)(120)+ 240(120)

= 248.57 mm = 249 mm

Ans:

y = 249 mm

Determine the location y of the centroid C of the beam

having the cross-sectional area shown

Solution

Centroid The locations of the centroids measuring from the x axis for segments 1,

2 and 3 are indicated in Fig a Thus

y = Σy~ A ΣA = 7.5(15)(150)15(150)+ 90(150)(15) + 172.5(15)(100)+ 150(15) + 15(100)

A

C B

Ans:

y = 79.7 mm

Centroid Referring to Fig a, the areas of the segments and the locations of their respective

centroids are tabulated below

Segment A(in 2 ) x~(in.) y~(in.) x~A (in 3 ) y~A (in 3 )

x = Σx~A ΣA = -72.0 in.72.0 in.23 = -1.00 in Ans.

y = Σy~A ΣA = 333.0 in.3

Ans:

x = -1.00 in

y = 4.625 in.

Determine the location of the centroid of the beam’s

cross-sectional area Neglect the size of the corner welds at A and B

for the calculation

B y

Ans:

y = 85.9 mm

Centroid Referring to Fig a, the areas of the segments and the locations of their respective

centroids are tabulated below

Segment A(in 2 ) x~(in.) y~(in.) x~A (in 3 ) y~A (in 3 )

x = Σx~A ΣA = -22.50 in.3

18.9978 in.2 = -1.1843 in = -1.18 in Ans.

y = Σy~A ΣA = 26.33 in.

Determine the location (x, y) of the centroid C of the area.

Solution

Centroid Referring to Fig a, the areas of the segments and the locations of their respective

centroids are tabulated below

Segment A(in 2 ) x~(in.) y~(in.) x~A (in 3 ) y~A (in 3 )

6.1079 in.2 = 1.5656 in = 1.57 in Ans.

y = Σy~A ΣA = 9.5625 in.3

6.1079 in.2 = 1.5656 in = 1.57 in Ans.

Determine the location of the centroid C for a beam having

the cross-sectional area shown The beam is symmetric with

respect to the y axis.

Ans:

y = 2 in.

Locate the centroid of the cross-sectional area of the

beam constructed from a channel and a plate Assume all

corners are square and neglect the size of the weld at A.

A triangular plate made of homogeneous material has a

constant thickness which is very small If it is folded over as

shown, determine the location of the plate’s center of

A triangular plate made of homogeneous material has a

constant thickness which is very small If it is folded over as

shown, determine the location of the plate’s center of

Locate the center of mass of the forked lever, which

is made from a homogeneous material and has the

Determine the location of the centroid C of the shaded

area which is part of a circle having a radius

Using symmetry, to simplify, consider just the top half:

r.x

= r33sin a - r33sin a cos2a

©x'A = 12r2aa3a2r sin ab - 12(r sin a) (r cos a)a23r cos ab

y

x C

A toy skyrocket consists of a solid conical top,

a hollow cylinder, and astick having a circular cross section,

Determine the length of the stick, x, so that the center of

gravity G of the skyrocket is located along line aa.

Centroid : The area and the centroid for segments 1 and 2 are

Listed in a tabular form, we have

Determine the location ( ) of the center of gravity of the

three-wheeler The location of the center of gravity of each

component and its weight are tabulated in the figure If the

three-wheeler is symmetrical with respect to the x–y plane,

determine the normal reaction each of its wheels exerts on

32SOLUTION

NB = 72.1 lb

21NB214.52 - 23112.812 = 0+ ©MA = 0;

Locate the center of mass of the homogeneous

SOLUTION

Centroid: Since the block is made of a homogeneous material, the center of mass of

the block coincides with the centroid of its volume The centroid of each composite

segment is shown in Fig a.

Ans:

x = 120 mm

y = 305 mm

z = 73.4 mm

The sheet metal part has the dimensions shown Determine

the location 1x, y, z2of its centroid

SOLUTION

Ans.

Ans.

Ans.

z = ©z'A©A = -1821 = -0.857 in

y = ©y'A©A = 3621 = 1.71 in

Since the material is homogeneous, the center of gravity coincides with the centroid

See solution to Prob 9-74

Ans.

u = tan-1a0.857 b =1.14 53.1°

The sheet metal part has a weight per unit area of

and is supported by the smooth rod and at C If the cord is

cut, the part will rotate about the y axis until it reaches

equilibrium Determine the equilibrium angle of tilt,

measured downward from the negative x axis, that AD

makes with the -xaxis