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© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-1 Determine the distance xc to the center of mass of the homogeneous rod bent into the shape shown. If the rod has a mass per unit length ρ, determine the reactions at the fixed support O. Units Used : m

g := 9.81

2

s Given:

kg

ρ := 0.5

m

a := 1m b := 1m

Solution: Length and Moment Arm: The length of differential element is

2

2

dx + dy =

dL =

2⎤ ⎡ ⎢ 1 + ⎛⎜ dy ⎞ ⎥ ⋅ dx ⎣ ⎝ dx ⎠ ⎦

and its centroid is xc = x. Here,

dy dx

=

3⋅ b 2⋅ a

1.5

⋅ x

Performing the integrations we have ⌠ ⎮ L := ⎮ ⎮ ⌡

a 2

1+

9⋅ b x 3

L = 1.440 m

dx

4⋅ a

0 a

⌠ 1 ⎮ xc := ⋅ ⎮ x L ⎮ ⌡ 0

2

1+

9⋅ b x 4⋅ a

3

dx

xc = 0.546 m

Equations of Equilibrium:

+ Σ Fx = 0; →

+

↑Σ Fy = 0; ΣMO = 0;

Ox = 0

Oy − ρ ⋅ g L = 0

Oy := ρ ⋅ g⋅ L

Oy = 7.062 N

M0 − ρ ⋅ g L xc = 0

M0 := ρ ⋅ g⋅ L⋅ xc

M0 = 3.854 N⋅ m

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-2 Determine the location (x c, yc) of the centroid of the wire. Given: a := 2⋅ ft b := 4⋅ ft Solution: Length and Moment Arm: The length of differential element is 2

2

⎛ dy ⎞ ⋅ dx 1+⎜ ⎝ dx ⎠

2

dL =

dx + dy =

dL =

1+

⎛ 2⋅ b ⋅ x ⎞ ⎜ 2 ⎝ a ⎠

2

Performing the integrations a ⌠ 2 ⎮ 2⋅ b⋅ x ⎞ ⎛ L := ⎮ 1+⎜ dx 2 ⎮ a ⎝ ⎠ ⌡

L = 9.29 ft

−a

⌠ 1 ⎮ xc := ⋅ ⎮ L ⎮ ⌡

a

⌠ 1 ⎮ yc := ⋅ ⎮ L ⎮ ⌡

a

2

x 1+

−a

⎛ x⎞ ⎝ a⎠

b⋅ ⎜

−a

⎛⎜ xc ⎞ ⎛ 0.00 ⎞ =⎜ ft ⎜ yc ⎝ ⎠ ⎝ 1.82 ⎠

⎛ 2⋅ b⋅ x ⎞ dx ⎜ 2 ⎝ a ⎠

2

2

1+

⎛ 2⋅ b⋅ x ⎞ dx ⎜ 2 ⎝ a ⎠

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-3 Locate the center of mass of the homogeneous rod bent into the shape of a circular arc. Given: r := 300mm θ := 30deg Solution: dL = r⋅ dθ xc = r⋅ cos ( θ ) yc = r⋅ sin ( θ )

π

+θ

⌠2 ⎮ r⋅ cos ( θ ) r dθ ⎮ − π ⌡ −θ xc :=

2

π

+θ

⌠2 ⎮ r dθ ⎮ ⌡− π − θ 2

xc = 124 mm

yc = 0

( By symmetry )

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-5 Determine the distance xc to the center of gravity of the homogeneous rod bent into the parabolic shape. If the rod has a weight per unit length γ determine the reactions at the fixed support O. Given: lb

γ := 0.5

ft

a := 1⋅ ft b := 0.5 ⋅ ft

Solution:

2

dy dx

=

2

dx + dy =

dL =

1+

⎛ dy ⎞ ⎜ ⎝ dx ⎠

2

2. ⋅ b⋅ x 2

a

⌠ ⎮ L := ⎮ ⎮ ⌡

a 2

⎛ 2. ⋅ b⋅ x ⎞ dx 1+⎜ 2 ⎝ a ⎠

L = 1.148 ft

0

⎤ ⎡⌠a 2 ⎥ ⎢ ⎮ 1 ⎛ 2.⋅ b⋅ x ⎞ dx xc := ⎢⎮ x 1 + ⎜ ⎥ 2 L ⎢⎮ a ⎝ ⎠ ⎥ ⌡ ⎣0 ⎦ + Σ Fx = 0; →

+

↑Σ Fy = 0; ΣMO = 0;

xc = 0.531 ft

Ox = 0

Oy − γ ⋅ L = 0 MO − γ ⋅ L⋅ xc = 0

Oy := γ ⋅ L MO := γ ⋅ L⋅ xc

Oy = 0.574 lb MO = 0.305 lb⋅ ft

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-7 Locate the centroid of the parabolic area. Solution:

a=

h b

2

dA = x⋅ dy x xc = 2 yc = y h

⌠ ⎮ A= b⋅ ⎮ ⌡0

y h

dy → A =

2 ⋅ h⋅ b 3

h

⌠ ⎮ 3 ⎮ 1 ⎛ xc = ⋅ ⋅ ⎜ b⋅ 2⋅ h ⋅ b ⎮ 2 ⎝ ⌡0

y⎞ h⎠

2

dy → xc =

3 xc = ⋅ b 8

h

⌠ ⎮ ⋅ yc = y⋅ b ⋅ 2⋅ h ⋅ b ⎮ ⌡0 3

3 yc = ⋅ h 5

y h

dy → yc =

3 ⋅h 5

3 ⋅b 8

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-8 Locate the centroid (x c, yc) of the shaded area.

Given: a := 2m b := 1m Solution:

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

⎡ ⎛ x ⎞ 2⎤ ⎥ dx dA = ydx = b ⎢1 − ⎜ ⎣ ⎝ a⎠ ⎦ and its centroid is y 1 ⎡ ⎛ x⎞ yc = = ⋅ b⋅ ⎢1 − ⎜ 2 2 ⎣ ⎝ a⎠

2⎤

⎥ ⎦

Centroid: Due to symmetry

xc = 0

yc :=

⌠ ⎮ ⎮ ⌡

a

−a

1 2

⎡

b⎢1 −

⎣

⌠ ⎮ ⎮ ⌡

a

−a

2 2 ⎛ x ⎞ ⎥⎤ b⎡⎢1 − ⎛ x ⎞ ⎥⎤ dx ⎜ ⎜ ⎝ a⎠ ⎦ ⎣ ⎝ a⎠ ⎦

⎡ ⎛ x ⎞ 2⎤ ⎥ dx b⋅ ⎢1 − ⎜ ⎣ ⎝ a⎠ ⎦

2 yc = m 5

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-9 Locate the centroid xc of the shaded area. Solution: dA = y⋅ dx xc = x y yc = 2 b

⌠ h 2 ⎮ x ⋅ x dx 2 ⎮ b ⌡ xc =

0

b

⌠ 2 ⎮ x h ⋅ dx ⎮ 2 ⎮ b ⌡0

→ xc =

3 ⋅b 4

3 xc = ⋅ b 4

⌠ ⎮ ⎮ ⎮ ⌡ yc =

b 2

1 ⎛ h 2⎞ ⋅x dx 2⎜ 2 ⎝b ⎠

0

⌠ ⎮ ⎮ ⌡

→ yc =

b

0

3 yc = ⋅h 10

h b

2

2

⋅ x dx

3 ⋅h 10

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-10 Locate the centroid xc of the shaded area.

Solution:

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

⎛

x

⎝

2a ⎠

dA = y⋅ dx = 2k⋅ ⎜ x −

and its centroid is

2⎞

x1 = x

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

a

⌠ ⎛ x2 ⎞ ⎮ ⎮ x⋅ 2⋅ k⋅ ⎜ x − 2a dx ⎝ ⎠ ⌡ xc =

0

a

⌠ ⎛ x2 ⎞ ⎮ ⎜x − 2 ⋅ k ⋅ dx ⎮ 2a ⎠ ⎝ ⌡ 0

→ xc =

5 ⋅a 8

5 xc = ⋅ a 8

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem *9-12 Locate the centroid of the shaded area.

Solution: dA = ydx

xc = x

y yc = 2

L

⌠ ⎛ πx ⎞ dx → A = 2 ⋅ L⋅ a A = ⎮ a⋅ sin ⎜ ⎮ π ⎝L⎠ ⌡0 L

⌠ ⎛ πx ⎞ dx → x = 1 ⋅ L xc = ⋅ ⎮ x⋅ a⋅ sin ⎜ c 2 2⋅ L⋅ a ⎮ ⎝L⎠ ⌡0 L ⌠ 2 π ⎮ 1 ⎛ ⎛ π ⋅ x ⎞⎞ dx → y = 1 ⋅ π ⋅ a yc = ⋅⎮ ⋅ ⎜ a⋅ sin ⎜ c 8 2 ⎝ 2⋅ L⋅ a ⎝ L ⎠⎠ ⌡0 π

L

0

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem *9-16 Locate the centroid of the shaded area bounded by the parabola and the line y = a

Solution: x xc = 2

dA = x⋅ dy a

⌠ A= ⎮ ⌡0

( 2) 2

xc =

yc =

2⋅ a 3 2⋅ a

2

2⋅ a A= 3

2 a a⋅ y dy → A = ⋅ a 3

⎡ 1 ⎢⌠ ⋅ ⎮ 2 2 ⎢⌡ 0

a

3

yc = y 3

⎣

a

(

⎤ 3 ⎥ a⋅ y) dy⎥ → xc = ⋅ a 8 ⎦ 2

⌠ ⋅ ⎮ y a⋅ y dy → yc = 2 ⌡ 0

3 5⋅ a

3 xc = a 8 5

2 2) ( 4

⋅ a

3 yc = ⋅ a 5

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-17 Locate the centroid of the quarter elliptical area.

Solution:

dA = y⋅ dx

xc = x

y yc = 2

a

⌠ 2 ⎮ ⎛ x ⎞ dx A = ⎮ b⋅ 1 − ⎜ ⎮ ⎝ a⎠ ⌡0

A=

π ⋅ a⋅ b 4

a

⌠ 2 ⎮ 4 ⎛ x ⎞ dx → x = 4 ⋅ a xc = ⋅ ⎮ x⋅ b ⋅ 1 − ⎜ c π ⋅ a⋅ b ⎮ 3⋅ π ⎝ a⎠ ⌡0 a

⌠ 2 ⎮ 2⎤ ⎡ 4 1 x ⎛ ⎞ ⎥ dx → y = 4 ⋅ b yc = ⋅⎮ ⋅ ⎢b⋅ 1 − ⎜ c π ⋅ a⋅ b ⎮ 2 ⎣ 3⋅ π ⎝ a⎠ ⎦ ⌡0

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-21 Locate the centroid yc of the shaded area. Solve the problem by evaluating the integrals using Simpson's rule. Given: a := 2 ft 1

b :=

a

2

5

+ 2⋅ a

3

Solution:

a

⌠ ⎮ ⎮ A := ⎮ ⌡0

1 5⎞ ⎛ ⎜ 2 3 ⎝b − x + 2⋅ x ⎠ dx

a

⌠ ⎛ ⎮ 1 ⎮ 1 ⎜ yc := ⋅ ⎮ ⋅ ⎝b + A ⌡ 2 0

A = 2.177

1 5⎞ ⎛ ⎜ 2 3 2 3 x + 2⋅ x ⎠ ⋅ ⎝ b − x + 2⋅ x ⎠ dx 1

2

ft

5⎞

yc = 2.040

ft

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-23 Locate the centroid xc of the shaded area. Solution: Area and Moment Arm : Here,

⎛y⎞ x2 = a⋅ ⎜ ⎝ b⎠

a⋅ y

x1 =

b

2

The area of the differential element is 2 ⎡ a⋅ y y⎞ ⎤ ⎛ ⎥ ⋅ dy dA = ⎢ − a⋅ ⎜ ⎣b ⎝ b⎠ ⎦

and its centroid is 1 ⎡ a⋅ y ⎛y⎞ xc = ⋅ ⎢ + a⋅ ⎜ 2 ⎣b ⎝ b⎠

2⎤

⎥ ⎦

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

⌠ ⎮ A= ⎮ ⌡0

2 ⎡ a⋅ y 1 y ⎤ ⎢ − a⋅ ⎛⎜ ⎞ ⎥ dy → A = ⋅ a⋅ b 6 ⎣b ⎝ b⎠ ⎦

b

⌠ 2 2 2 6 ⎮ 1 ⎡ a⋅ y y ⎞ ⎤ ⎡ a⋅ y y⎞ ⎤ ⎛ ⎛ ⎥ ⎢ − a⋅ ⎜ ⎥ dy → xc = ⋅ a ⋅⎮ xc = ⋅⎢ + a⋅ ⎜ 5 2 ⎣b a⋅ b ⎝ b ⎠ ⎦⎣ b ⎝ b⎠ ⎦ ⌡0 Given:

a := 1ft

b := 2ft

2⋅ a xc := 5

xc = 0.40 ft

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-24 Locate the centroid yc of the shaded area. Solution: Area and Moment Arm : Here,

⎛y⎞ x2 = a⋅ ⎜ ⎝ b⎠

a⋅ y

x1 =

b

2

The area of the differential element is 2 ⎡ a⋅ y y⎞ ⎤ ⎛ ⎥ ⋅ dy dA = ⎢ − a⋅ ⎜ ⎣b ⎝ b⎠ ⎦

and its centroid is 1 ⎡ a⋅ y ⎛y⎞ xc = ⋅ ⎢ + a⋅ ⎜ 2 ⎣b ⎝ b⎠

2⎤

⎥ ⎦

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

⌠ ⎮ A= ⎮ ⌡0

2 ⎡ a⋅ y 1 y ⎤ ⎢ − a⋅ ⎛⎜ ⎞ ⎥ dy → A = ⋅ a⋅ b 6 ⎣b ⎝ b⎠ ⎦

b

⌠ 2 ⎡ a⋅ y 6 ⎮ y⎞ ⎤ 1 ⎛ ⎥ dy → yc = ⋅ b yc = ⋅ ⎮ y⋅ ⎢ − a⋅ ⎜ 2 a⋅ b b ⎝ b⎠ ⎦ ⌡0 ⎣ Given:

a := 1ft

b := 2ft

b yc := 2

yc = 1.00 ft

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-27 Locate the centroid xc of the shaded area.

Given :

a := 2in

Solution :

y=

b := 0.5in

a⋅ b x

a

xc :=

⌠ ⎮ x⋅ ⎛⎜ a⋅ b ⎞ dx ⎮ ⎝ x ⎠ ⌡ b

a

⌠ a⋅ b ⎮ dx ⎮ x ⌡b

xc = 1.08 in

Area and Moment Arm :The area of the differential element is dA = y⋅ dx =

1

x

⋅ dx

and its centroid is

xc = x

Centroid :Applying Eq. 9-6 and performing the integration, we have ⌠ ⎮ x dA ⎮ c ⌡ xc =

⌠ ⎮ ⎮ 1 dA ⌡ a

⌠ ⎮ x⋅ ⎛⎜ a⋅ b ⎞ dx ⎮ ⎝ x ⎠ ⌡

⌠ ⎮ x dA = b ⎮ c a ⌠ a⋅ b ⌡ ⎞ dx ⎮ ⎛ ⎜ ⎮ ⎝ x ⎠ ⌡b

xc = 1.08 in

a

xc :=

⌠ ⎮ x⋅ ⎛⎜ a⋅ b ⎞ dx ⎮ ⎝ x ⎠ ⌡ b

a

⌠ ⎮ ⎮ ⌡b

⎛ a⋅ b ⎞ dx ⎜ ⎝ x ⎠

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-32 Locate the centroid of the ellipsoid of revolution.

Solution :

2

dV = π ⋅ z ⋅ dy

⎛ y2 ⎞ z = a ⋅ ⎜1 − ⎜ b2 ⎝ ⎠ 2

2

b

⌠ 2 ⎮ y ⎞ 2 2⎛ 2 ⎜ V = ⎮ π⋅a ⋅ 1 − dy → V = ⋅ b⋅ π ⋅ a ⎜ b2 3 ⎮ ⎝ ⎠ ⌡ 0

b

⌠ 2 ⎮ 3 y ⎞ 3 2⎛ ⎜ yc = ⋅ ⎮ yπ ⋅ a ⋅ 1 − dy → yc = ⋅ b 2 ⎜ b2 8 2⋅ b ⋅ a ⋅ π ⎮ ⎝ ⎠ ⌡ 0 By symmetry

3⋅ b yc = 8 xc = zc = 0

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-33 Locate the center of gravity of the volume. The material is homogeneous. Given: a := 2m b := 2m 2

⎛ y⎞ = z ⎜ b ⎝ a⎠

2

2 z

y = a ⋅

b

Solution: b

zc :=

⌠ 2 a ⋅z ⎮ zπ ⋅ dz ⎮ b ⌡ 0

b

⌠ 2 a ⋅z ⎮ ⎮ π ⋅ b dz ⌡ 0

zc = 1.33 m

xc = yc = 0

by symmetry

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-36 Locate the centroid of the quarter-cone.

Solution : r=

a h

⋅ ( h − z)

zc = z

xc = yc =

4⋅ r 3⋅ π

h

⌠ 2 1 2 ⎮ π ⎡a ⎤ V= ⎮ ⋅ ⎢ ⋅ ( h − z)⎥ dz → V = ⋅ h⋅ π ⋅ a 4 ⎣h 12 ⎦ ⌡0 h

⌠ 2 12 ⎮ π ⎡a 1 ⎤ zc = ⋅ ⎮ z ⋅ ⎢ ⋅ ( h − z)⎥ dz → zc = ⋅ h 2 4 ⎣h 4 ⎦ h⋅ a ⋅ π ⌡ 0

⎡⌠h ⎤ 2 ⎥ ⎢⎮ 4 a 12 π ⎡a 1 ⎡ ⎤ ⎤ xc = ⋅ ⎢⎮ ⋅ ⎢ ⋅ ( h − z)⎥ ⋅ ⎢ ⋅ ( h − z)⎥ dz⎥ → xc = ⋅ a 2 π ⎦ 4 ⎣h ⎦ ⎥ h⋅ a ⋅ π ⎢⌡ 3⋅ π ⎣ h ⎣0 ⎦

xc = yc =

a π

h zc = 4

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-39 Locate the centroid yc of the paraboloid. Given: a := 4m b := 4m Solution: 2

⎛ z⎞ = y ⎜ a ⎝ b⎠

y

z = b⋅

2

a

2 y

dV = π ⋅ z ⋅ dy = π ⋅ b ⋅

and its centroid

a

yc = y

a

yc :=

⌠ ⎮ y⋅ π ⋅ b2⋅ y dy ⎮ a ⌡ 0

a

⌠ ⎮ π ⋅ b2⋅ y dy ⎮ a ⌡0

yc = 2.67 m

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-41 Locate the centroid zc of the frustum of the right-circular cone.

Solution : Volume and Moment Arm : From the geometry, y−r R−r y=

=

h−z h

( r − R) ⋅ z + R⋅ h h

The volume of thin disk differential element is 2

⎡( r − R) ⋅ z + R⋅ h⎤ ⋅ dz dV = π ⋅ y ⋅ dz = π ⋅ ⎢ ⎥ h ⎣ ⎦ 2

h

⌠ 2 ( r − R) ⋅ z + R⋅ h⎤ ⎮ ⎡ ⎥ dz ⎮ z π⋅⎢ h ⎣ ⎦ ⌡ zc =

0

h

⌠ 2 ⎮ ⎡( r − R) ⋅ z + R⋅ h⎤ dz π ⋅ ⎢ ⎥ ⎮ h ⎣ ⎦ ⌡ 0

1

→ zc =

4

2 2

⋅π⋅h ⋅r + 1 3

3

⋅ h⋅ r ⋅

1 6

2

⋅ π ⋅ h ⋅ r⋅ R +

π r−R

−

1 3

3

1

2

12

⋅ R ⋅ h⋅

2

⋅π⋅h ⋅R π

r−R

2

This can be simplified to give

zc =

2

R + 2⋅ r⋅ R + 3⋅ r

(

2

)

2

4⋅ R + r⋅ R + r

⋅h

π

dV =

h

2

⋅ ⎡⎣( r − R) ⋅ z + 2⋅ R⋅ h⋅ ( r − R) ⋅ z + R ⋅ h ⎤⎦ ⋅ dz 2 2

2

2

and its centroid z' = z Centroid :Applying Eq. 9-5 and performing the integration,we have h

⌠ ⎮ ⎮ z' dV ⌡ z' = V = ⌠ ⎮ 1 dV ⎮ ⌡ V

⌠ π ⎮ z⋅ ⎡ ⋅ ⎡⎣( r − R) 2⋅ z2 + 2⋅ R⋅ h⋅ ( r − R) ⋅ z + r2⋅ h2⎤⎦⎤ dz ⎢ ⎥ 2 ⎮ h ⎣ ⎦ ⌡0 h

⌠ π 2 2 2 2 ⎮ ⋅ ⎡⎣( r − R) ⋅ z + 2⋅ R⋅ h⋅ ( r − R) ⋅ z + R ⋅ h ⎤⎦ dz 2 ⎮ h ⌡0

4⎞ 2 ⎞⎤ ⎡ ⎛ z3 ⎞ 2 ⎛z 2 2 ⎛z ⎢ ⎜ ⎜ ⎜ ⎥ ⋅ ( r − R) ⋅ + 2⋅ R⋅ h⋅ ( r − R) ⋅ + R ⋅h ⋅ 2 ⎣ 4⎠ 3⎠ 2 ⎠⎦ ⎝ ⎝ ⎝ h

π

z' =

π h

2

⎡

⎛ z3 ⎞

⋅ ⎢( r − R) ⋅ ⎜

⎣

2

⎝ 3⎠

⎛ z2 ⎞

+ 2⋅ R⋅ h⋅ ( r − R) ⋅ ⎜

⎝ 2⎠

2

0

⎤

h

⎦

0

+ R ⋅ h ⋅ ( z)⎥ 2

h

2

z' =

2

R + 3⋅ r + 2⋅ r⋅ R

(

2

2

)

4⋅ R + r + r⋅ R

⋅h

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-44 Locate the center of gravity G of the five particles with respect to the origin O. g := 9.81 Given:

m 2

s

a := 2m

m1 := 1kg

b := 1m

m2 := 10kg

c := 1m

m3 := 2kg

d := 3m

m4 := 5kg

e := 2m

m5 := 6kg Solution:

xc =

xc :=

Σxc⋅ W ΣW −m1⋅ g⋅ ( a + b + c) − m2⋅ g⋅ ( b + c) − m3⋅ g⋅ c + m4⋅ g⋅ d + m5⋅ g⋅ ( d + e )

xc = 0.792 m

(m1 + m2 + m3 + m4 + m5)⋅ g

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

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 :

a⋅ xc :=

a 2

+ c⋅

2

⎛ ⎝

+ ( b − d) ⋅ c + ( a − c) ⋅ ⎜ c +

a − c⎞ 2

⎠

a + b + c + ( b − d) + ( a − c)

a⋅ b + b⋅ yc :=

c

b 2

+ ( b − d) ⋅

b−d 2

xc = 34.38 mm

+ ( a − c) ⋅ ( b − d)

a + b + c + ( b − d) + ( a − c)

yc = 85.83 mm

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-57 Determine the location yc of the centroidal axis xcxc of the beam's cross-sectional area. Neglect the size of the corner welds at A and B for the calculation. Given : r := 50mm t := 15mm a := 150mm b := 15mm c := 150mm

Solution : b⋅ c⋅ yc :=

b 2

⎛ ⎝

+ a⋅ t⋅ ⎜ b +

a⎞ 2⎠

2

+ π ⋅ r ⋅ ( b + a + r) 2

b ⋅ c + a⋅ t + π ⋅ r

yc = 154.44 mm

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

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 :

a⋅ b⋅ xc :=

a⋅ b⋅ yc :=

b 2

a 2

+

+

1

⎛ ⎝

c⎞

2 ⋅ ( b + c) ⋅ d⋅ ⋅ ( b + c) 2 3⎠ 2 3 1 1 a⋅ b + ⋅ c⋅ a + ⋅ ( b + c) ⋅ d 2 2

1 2

a⋅ b +

⋅ a⋅ c⋅ ⎜ b +

⋅ a⋅ c⋅ 1 2

a 3

⋅ c⋅ a +

− 1 2

1 2

+

1

⋅ ( b + c) ⋅ d⋅

⋅ ( b + c) ⋅ d

xc = 4.62 in

d 3

yc = 1.00 in

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-62 Determine the location xc of the centroid C of the shaded area which is part of a circle having a radius r.

Solution : 1 2 ⎛ 2⋅ r ⎞ 1 ⎞ ⎛2 ΣxcA = ⋅ r ⋅ α ⋅ ⎜ ⋅ sin ( α ) − ⋅ r⋅ sin ( α ) ⋅ r⋅ cos ( α ) ⋅ ⎜ ⋅ r⋅ cos ( α ) 2 ⎝3 ⎠ ⎝ 3⋅ α ⎠ 2 3

3

3

r r r 2 3 ΣxcA = ⋅ sin ( α ) − ⋅ sin ( α ) ⋅ cos ( α ) = ⋅ sin ( α ) 3 3 3 ΣA =

xc =

1 2 1 1 2⎛ sin ( 2⋅ α ) ⎞ ⋅ r ⋅ α − ⋅ r⋅ sin ( α ) ⋅ r⋅ cos ( α ) = ⋅ r ⋅ ⎜ α − 2 2 2 2 ⎝ ⎠ ΣxcA ΣA 3

r xc =

3

2

3

⋅ sin ( α )

sin ( 2⋅ α ) ⎞ 1 2⎛ ⋅ r ⋅ ⎜α − 2 2 ⎝ ⎠

=

3

3

⋅ r⋅ sin ( α )

α−

sin ( 2⋅ α ) 2

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-69 Determine the distance h to which a hole of diameter d must be bored into the base of the cone so that the center of mass of the resulting shape is located at zc. The material has a density ρ. Given: d := 100mm zc := 115mm ρ := 8

mg m

3

a := 150mm b := 500mm

Solutions: 1 zc =

3

2

⋅ π ⋅ a ⋅ b⋅

b 4

2

⎛ d ⎞ ⋅ h⋅ h 2 ⎝2⎠

− π⋅⎜

2

⎛ d⎞ ⋅h ⋅π⋅a ⋅b − π⋅⎜ 3 ⎝2⎠ 1

2

Choosing the positive root,

h :=

1 2 2 2 2 2 zc⋅ d + ⋅ 9⋅ zc ⋅ d − 24⋅ zc⋅ a ⋅ b + 6⋅ a ⋅ b 3

h = 323 mm

d

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-71 The sheet metal part has the dimensions shown. Determine the location (xc, yc, zc) of its centroid. Given: a := 3in b := 4in c := 6in

Solution:

−a⋅ b⋅ xc :=

2

a⋅ b +

1

a

+

a⋅ b⋅ yc :=

2

2

2

a⋅ c⋅

a⋅ b +

1 2

xc = −1.14 in

⋅ a⋅ c

1

a⋅ b +

−1 zc :=

b

⋅ a⋅ c⋅

2 1 2

⋅ a⋅ c

2⋅ a 3

yc = 1.71 in

c 3

⋅ a⋅ c

zc = −0.857 in

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-76 Locate the center of gravity of the two-block assembly. The specific weights of the materials A and B are γA and γB, respectively. Given: lb

γ A := 150

3

ft

lb

γ B := 400

3

ft a := 2in b := 6in c := 2in d := 6in e := 6in Solution WA := γ A⋅

xc :=

yc :=

zc :=

WB⋅

a⋅ b ⋅ d 2

⎛ b⎞ + WA⋅ ⎜ c + 2 ⎝ 3⎠ c

WB + WA WB⋅

a + WA⋅ 2 2

xc = 1.47 in

e

WB + WA WB⋅

WB := γ B⋅ c⋅ d⋅ e

d + WA⋅ 3 2

yc = 2.68 in

d

WB + WA

zc = 2.84 in

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-79 Locate the centroid zc of the top made from a hemisphere and a cone. Given: r := 24mm h := 120mm

Solution: π zc :=

3

2

⋅ r ⋅ h⋅

3⋅ h 4

+

2 3

3

⎛ ⎝

⋅ π ⋅ r ⋅ ⎜h +

2 π 2 3 ⋅r ⋅h + π⋅r 3 3

3⋅ r ⎞ 8

⎠

zc = 101.14 mm

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-83 The assembly consists of a wooden dowel rod of length L and a tight-fitting steel collar. Determine the distance xc to its center of gravity if the specific weights of the materials are γw and γst.The radii of the dowel and collar are shown. Given: L := 20in γ w := 150

lb 3

ft γ st := 490

lb 3

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

2

xc :=

γ w⋅ π ⋅ r1 ⋅ L⋅

2 2 ⎛ b⎞ + γ st⋅ π ⋅ ⎛ r2 − r1 ⎞ ⋅ b⋅ ⎜ a + ⎝ ⎠ ⎝ 2⎠ 2

L

2 2 γ w⋅ π ⋅ r1 ⋅ L + γ stπ ⋅ ⎛ r2 − r1 ⎞ ⋅ b ⎝ ⎠ 2

xc = 8.22 in

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-85 The anchor ring is made of steel having specific weight γst. Determine the surface area of the ring. The cross section is circular as shown. Given γ st := 490

lb 3

ft a := 4in b := 8in

Solution:

⎛ a + b − a ⎞ ⋅ 2⋅ π ⋅ b − a 4 4 ⎠ ⎝2

A := 2⋅ π ⋅ ⎜

A = 118 in

2

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-101 A V-belt has as inner radius r and a cross-sectional area as shown. Determine the volume of material used in making the V-belt. Given : r := 6in a := 0.25in b := 0.5in θ := 30deg

Solution : Volume : Applying the theorem of Pappus and Guldinus, Eq.9-12

⎛θ⎞ ⎝2⎠

h := a⋅ cot ⎜

⎡ ⎣

⎛ ⎝

V := 2⋅ π ⋅ ⎢b⋅ h⋅ ⎜ r +

h⎞ 2⎠

+ 2⋅

⎛ 2 ⎝

a⋅ h

⋅ ⎜r +

2⋅ h ⎞⎤ 3

⎥ ⎠⎦

V = 28.66 in

3

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-102 The full circular aluminum housing is used in an automotive brake system. The cross section is shown in the figure. Determine its weight if aluminum has a specific weight γ. Given : γ := 169

lb 3

ft

a := 2.00in b := 0.25in

d := 4.00in e := 3.25in f := 0.25in

c := 0.15in Solution : Volume : Applying the theorem of Pappus and Guldinus, Eq.9-12, with r :=

a −b 2

⎡ ⎣

V := 2⋅ π ⋅ ⎢( r + c) ⋅ f⋅

W := γ ⋅ V

r+c ⎛ c ⎞ + b⋅ ( d − e − f) ⋅ ⎛ r + b ⎞⎤ + c⋅ e⋅ ⎜ r + ⎜ ⎥ 2 ⎝ 2⎠ ⎝ 2 ⎠⎦

W = 0.377 lb

V = 3.85 in

3

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-124 A circular V-belt has an inner radius r and a cross-sectional area as shown. Determine the volume of material required to make the belt. Given r := 600mm a := 25mm b := 50mm c := 75mm Solution

⎡⎛ ⎣⎝

V := 2⋅ π ⋅ ⎢⎜ r +

c ⎞ ⎛ 1⎞ ⎛ c ⎞ ⋅ b⋅ c⎤ ⋅ 2⋅ ⎜ ⋅ a⋅ c + ⎜ r + ⎥ 3⎠ ⎝ 2⎠ ⎝ 2⎠ ⎦

V = 22.4 × 10

−3

m

3

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Distributed Forces

Sample Problem 5/1 Centroid of a circular arc.

Locate the centroid of a circular arc as shown in

the figure.

r

α α

Solution. Choosing the axis of symmetry as the x-axis makes y 0. A differential 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 [Lx

x dL]

(2r)x

y

(r cos ) r d

C

r cos θ dθ

2rx 2r2 sin r sin Ans. For a semicircular arc 2 , which gives x 2r/. By symmetry we see immediately that this result also applies to the quarter-circular arc when the measurement is made as shown. x

r dθ

α

θ x

α r

Helpful Hint

It should be perfectly evident that polar coordinates are preferable to rectan-

C

gular coordinates to express the length of a circular arc.

C 2r/π r

Sample Problem 5/2

r

y

Determine the distance h from the base of a triangle of altitude h to the centroid of its area.

Centroid of a triangular area.

dy h

Solution.

x

The x-axis is taken to coincide with the base. A differential strip of

area dA x dy is chosen. By similar triangles x/(h y) b/h. Applying the sec-

y

ond of Eqs. 5/5a gives [Ay and

y dA] c

x

bh y 2

b h

0

y

b(h y) bh2 dy 6 h y

h 3

Helpful Hint Ans.

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

We save one integration here by using the first-order element of area. Recognize that dA must be expressed in terms of the integration variable y; hence, x ƒ( y) is required.

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Article 5/3

245

Centroids of Lines, Areas, and Volumes

Sample Problem 5/3 Centroid of the area of a circular sector.

r

Locate the centroid of the area

of a circular sector with respect to its vertex.

α α

Solution I. The x-axis is chosen as the axis of symmetry, and y is therefore automatically zero. We may cover the area by moving an element in the form of a partial circular ring, as shown in the figure, from the center to the outer periphery. The radius of the ring is r0 and its thickness is dr0, so that its area is dA 2r0 dr0. The x-coordinate to the centroid of the element from Sample Problem 5/1 is xc r0 sin /, where r0 replaces r in the formula. Thus, the first of Eqs. 5/5a gives [Ax

x dA] c

2 (r2)x 2

r

0

y r

α

x

r0 sin α xc = ——––– α

2

2 r sin 3

dr0

r0

α

r0 sin (2r0 dr0)

r2x 3r 3 sin x

C

Ans.

Solution I

Helpful Hints

Solution II.

The area may also be covered by swinging a triangle of differential area 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 terms are neglected. From Sample Problem 5/2 the centroid of the triangular element of area is two-thirds of its altitude from its vertex, so that the x-coordinate to the centroid of the element is xc 32 r cos . Applying the first of Eqs. 5/5a gives [Ax

x dA] c

(r2)x

(23 r cos )(12 r2 d)

and as before

guish between the variable r0 and the constant r. Be careful not to use r0 as the centroidal coordinate for the element.

y xc = 2 – r cos θ 3

r2x 23r3 sin 2 r sin x 3

Note carefully that we must distin-

Ans.

For a semicircular area 2 , which gives x 4r/3. By symmetry we see immediately that this result also applies to the quarter-circular area where the measurement 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 I began. On the other hand, integration with respect to r0 initially would give the triangular element with which Solution II began.

dθ

α

θ

α

x

r

Solution II

4r/3π C C r

r

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Distributed Forces

Sample Problem 5/4

y x = ky3

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

– x

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

[Ax

x dA]

x

c

a

0

y dx

– y

a

xy dx

3ab 3a2b x 7 4

4

x 7a

y b

Ans.

x = ky3

In the solution for y from the second of Eqs. 5/5a, the coordinate to the centroid of the rectangular element is yc y/2, where y is the height of the strip governed by the equation of the curve x ky3. Thus, the moment principle becomes

y dA]

3ab y 4

c

x

a

0

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

[Ay

b

C

Solution I.

y y yc = – 2

2y y dx a

x

0

x

a

dx

Substituting y b(x/a)1/3 and integrating give 3ab2 3ab y 4 10

y 52 b

Ans. y

Solution II.

The horizontal element of area shown in the lower figure may be employed in place of the vertical element. The x-coordinate to the centroid of the 1 rectangular element is seen to be xc x 2(a x) (a x)/2, which is simply the average of the coordinates a and x of the ends of the strip. Hence, [Ax

x dA] c

x

b

0

a 2 x(a x) dy

b

a+x xc = –––– 2 x = ky3 dy

b

(a x) dy

0

x

a–x

y

The value of y is found from [Ay

y dA] c

y

b

0

(a x) dy

a

b

0

x

y(a x) dy

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

Helpful Hint

Note that xc x for the vertical element.

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Article 5/3

Centroids of Lines, Areas, and Volumes

247

Sample Problem 5/5 Hemispherical volume.

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

Solution I. With the axes chosen as shown in the figure, x z 0 by symmetry. The most convenient element is a circular slice of thickness dy parallel to the 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 r2 y2. The volume of the elemental slice becomes

z

y2 + z2 = r2

z

dV (r2 y2) dy

y r

The second of Eqs. 5/6a requires [V y

y dV] c

y

r

0

(r2

y2)

dy

dy

r

y(r2

0

y2)

dy

where yc y. Integrating gives

Solution I 3 y 8r

2 3 r y 14r4 3

Ans. z

Solution II.

Alternatively we may use for our differential element a cylindrical shell of length y, radius z, and thickness dz, as shown in the lower figure. By expanding the radius of the shell from zero to r, we cover the entire volume. By symmetry 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 from 2 the equation of the circle gives y r2 z2. Using the value of 3 r3 computed in Solution I for the volume of the hemisphere and substituting in the second of Eqs. 5/6a give us [V y

y dV] c

2

r

2

y dz

yc = y/2

z y r

r 2 z (2zr z ) dz r (r z z ) dz 4

(3r3)y

yc = y

x

2

2

x

2

0

r

2

3

4

Solution II

0

3

y 8r

Ans.

z

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

dθ

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

r dθ

θ

y

Solution III

Helpful Hint

Can you identify the higher-order element of volume which is omitted from the expression for dV?

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Distributed Forces

Sample Problem 5/6

y 12″

Locate the centroid of the shaded area.

4″

Solution.

The composite area is divided into the four elementary shapes shown in the lower figure. The centroid locations of all these shapes may be obtained from Table D/3. Note that the areas of the “holes” (parts 3 and 4) are taken as negative in the following table:

4″ 3″ 3″

A in.2

x in.

y in.

1 2 3 4

120 30 14.14 8

6 14 6 12

5 10/3 1.273 4

TOTALS

127.9

PART

xA in.3

yA in.3

720 420 84.8 96

600 100 18 32

959

650

2″ 2″ 2″

1

x

5″

4

2

3

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

X ΣAx ΣA

X

959 7.50 in. 127.9

Ans.

Y ΣAy ΣA

Y

650 5.08 in. 127.9

Ans.

Sample Problem 5/7

6 5 A, m2

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

4 3 2

Solution.

The body is divided into five sections. For each section, the average area, volume, and centroid location are determined and entered in the following table:

INTERVAL

Aav m2

Volume V m3

x m

Vx m4

0–0.2 0.2–0.4 0.4–0.6 0.6–0.8 0.8–1.0

3 4.5 5.2 5.2 4.5

0.6 0.90 1.04 1.04 0.90

0.1 0.3 0.5 0.7 0.9

0.060 0.270 0.520 0.728 0.810

TOTALS

4.48

1 0 0

0.2

0.4

0.6

0.8

x, m

2.388 Helpful Hint

X ΣΣVVx

X

2.388 0.533 m 4.48

Ans.

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

1.0

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Article 5/4

Composite Bodies and Figures; Approximations

Sample Problem 5/8

259

z

Locate the center of mass of the bracket-and-shaft combination. The vertical face is made from sheet metal which has a mass of 25 kg/m2. The material of the horizontal base has a mass of 40 kg/m2, and the steel shaft has a density of 7.83 Mg/m3.

50 150 x 40

Solution.

The composite body may be considered to be composed of the five elements shown in the lower portion of the illustration. The triangular part will be taken as a negative mass. For the reference axes indicated it is clear by symmetry 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 explanation. For Part 1 we have from Sample Problem 5/3 z

150

75

y

50

50

25 100

150

Dimensions in millimeters

4(50) 4r 21.2 mm 3 3

1

For Part 3 we see from Sample Problem 5/2 that the centroid of the triangular mass is one-third of its altitude above its base. Measurement from the coordinate axes becomes z [150 25 13 (75)] 100 mm

1 2 3 4 5 TOTALS

m kg

y mm

z mm

my kg m

mz kg mm

0.098 0.562 0.094 0.600 1.476

0 0 0 50.0 75.0

21.2 75.0 100.0 150.0 0

0 0 0 30.0 110.7

2.08 42.19 9.38 90.00 0

140.7

120.73

2.642

5 3

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

PART

2

Equations 5/7 are now applied and the results are

Y Σmy Σm

Y

140.7 53.3 mm 2.642

Ans.

Z Σmz Σm

Z

120.73 45.7 mm 2.642

Ans.

4

c05.qxd

1/26/06

268

1:30 PM

Chapter 5

Page 268

Distributed Forces

Sample Problem 5/9

z R

Determine the volume V and surface area A of the complete torus of circular cross section. a

Solution.

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

V rA 2(R)(a2) 22Ra2

Ans.

Similarly, using Eq. 5/8a gives

z

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

–r = R

Ans.

a

Helpful Hint

We note that the angle of revolution is 2 for the complete ring. This common but special-case result is given by Eq. 5/9.

Sample Problem 5/10

z

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

60 mm x

Solution.

With the angle of revolution 180, Eq. 5/9a gives V rA [30 13(60)][12(60)(60)] 2.83(105) mm3

Ans.

30 mm

60 mm

The mass of the body is then z

kg

1m m V 7830 3 [2.83(10 ) mm ] 1000 mm m 2.21 kg

5

3

3

Ans.

60 mm

C r– 30 mm

60 mm

Helpful Hint

Note that must be in radians.

Lihat lebih banyak...
Problem 9-1 Determine the distance xc to the center of mass of the homogeneous rod bent into the shape shown. If the rod has a mass per unit length ρ, determine the reactions at the fixed support O. Units Used : m

g := 9.81

2

s Given:

kg

ρ := 0.5

m

a := 1m b := 1m

Solution: Length and Moment Arm: The length of differential element is

2

2

dx + dy =

dL =

2⎤ ⎡ ⎢ 1 + ⎛⎜ dy ⎞ ⎥ ⋅ dx ⎣ ⎝ dx ⎠ ⎦

and its centroid is xc = x. Here,

dy dx

=

3⋅ b 2⋅ a

1.5

⋅ x

Performing the integrations we have ⌠ ⎮ L := ⎮ ⎮ ⌡

a 2

1+

9⋅ b x 3

L = 1.440 m

dx

4⋅ a

0 a

⌠ 1 ⎮ xc := ⋅ ⎮ x L ⎮ ⌡ 0

2

1+

9⋅ b x 4⋅ a

3

dx

xc = 0.546 m

Equations of Equilibrium:

+ Σ Fx = 0; →

+

↑Σ Fy = 0; ΣMO = 0;

Ox = 0

Oy − ρ ⋅ g L = 0

Oy := ρ ⋅ g⋅ L

Oy = 7.062 N

M0 − ρ ⋅ g L xc = 0

M0 := ρ ⋅ g⋅ L⋅ xc

M0 = 3.854 N⋅ m

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-2 Determine the location (x c, yc) of the centroid of the wire. Given: a := 2⋅ ft b := 4⋅ ft Solution: Length and Moment Arm: The length of differential element is 2

2

⎛ dy ⎞ ⋅ dx 1+⎜ ⎝ dx ⎠

2

dL =

dx + dy =

dL =

1+

⎛ 2⋅ b ⋅ x ⎞ ⎜ 2 ⎝ a ⎠

2

Performing the integrations a ⌠ 2 ⎮ 2⋅ b⋅ x ⎞ ⎛ L := ⎮ 1+⎜ dx 2 ⎮ a ⎝ ⎠ ⌡

L = 9.29 ft

−a

⌠ 1 ⎮ xc := ⋅ ⎮ L ⎮ ⌡

a

⌠ 1 ⎮ yc := ⋅ ⎮ L ⎮ ⌡

a

2

x 1+

−a

⎛ x⎞ ⎝ a⎠

b⋅ ⎜

−a

⎛⎜ xc ⎞ ⎛ 0.00 ⎞ =⎜ ft ⎜ yc ⎝ ⎠ ⎝ 1.82 ⎠

⎛ 2⋅ b⋅ x ⎞ dx ⎜ 2 ⎝ a ⎠

2

2

1+

⎛ 2⋅ b⋅ x ⎞ dx ⎜ 2 ⎝ a ⎠

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-3 Locate the center of mass of the homogeneous rod bent into the shape of a circular arc. Given: r := 300mm θ := 30deg Solution: dL = r⋅ dθ xc = r⋅ cos ( θ ) yc = r⋅ sin ( θ )

π

+θ

⌠2 ⎮ r⋅ cos ( θ ) r dθ ⎮ − π ⌡ −θ xc :=

2

π

+θ

⌠2 ⎮ r dθ ⎮ ⌡− π − θ 2

xc = 124 mm

yc = 0

( By symmetry )

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-5 Determine the distance xc to the center of gravity of the homogeneous rod bent into the parabolic shape. If the rod has a weight per unit length γ determine the reactions at the fixed support O. Given: lb

γ := 0.5

ft

a := 1⋅ ft b := 0.5 ⋅ ft

Solution:

2

dy dx

=

2

dx + dy =

dL =

1+

⎛ dy ⎞ ⎜ ⎝ dx ⎠

2

2. ⋅ b⋅ x 2

a

⌠ ⎮ L := ⎮ ⎮ ⌡

a 2

⎛ 2. ⋅ b⋅ x ⎞ dx 1+⎜ 2 ⎝ a ⎠

L = 1.148 ft

0

⎤ ⎡⌠a 2 ⎥ ⎢ ⎮ 1 ⎛ 2.⋅ b⋅ x ⎞ dx xc := ⎢⎮ x 1 + ⎜ ⎥ 2 L ⎢⎮ a ⎝ ⎠ ⎥ ⌡ ⎣0 ⎦ + Σ Fx = 0; →

+

↑Σ Fy = 0; ΣMO = 0;

xc = 0.531 ft

Ox = 0

Oy − γ ⋅ L = 0 MO − γ ⋅ L⋅ xc = 0

Oy := γ ⋅ L MO := γ ⋅ L⋅ xc

Oy = 0.574 lb MO = 0.305 lb⋅ ft

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-7 Locate the centroid of the parabolic area. Solution:

a=

h b

2

dA = x⋅ dy x xc = 2 yc = y h

⌠ ⎮ A= b⋅ ⎮ ⌡0

y h

dy → A =

2 ⋅ h⋅ b 3

h

⌠ ⎮ 3 ⎮ 1 ⎛ xc = ⋅ ⋅ ⎜ b⋅ 2⋅ h ⋅ b ⎮ 2 ⎝ ⌡0

y⎞ h⎠

2

dy → xc =

3 xc = ⋅ b 8

h

⌠ ⎮ ⋅ yc = y⋅ b ⋅ 2⋅ h ⋅ b ⎮ ⌡0 3

3 yc = ⋅ h 5

y h

dy → yc =

3 ⋅h 5

3 ⋅b 8

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Problem 9-8 Locate the centroid (x c, yc) of the shaded area.

Given: a := 2m b := 1m Solution:

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

⎡ ⎛ x ⎞ 2⎤ ⎥ dx dA = ydx = b ⎢1 − ⎜ ⎣ ⎝ a⎠ ⎦ and its centroid is y 1 ⎡ ⎛ x⎞ yc = = ⋅ b⋅ ⎢1 − ⎜ 2 2 ⎣ ⎝ a⎠

2⎤

⎥ ⎦

Centroid: Due to symmetry

xc = 0

yc :=

⌠ ⎮ ⎮ ⌡

a

−a

1 2

⎡

b⎢1 −

⎣

⌠ ⎮ ⎮ ⌡

a

−a

2 2 ⎛ x ⎞ ⎥⎤ b⎡⎢1 − ⎛ x ⎞ ⎥⎤ dx ⎜ ⎜ ⎝ a⎠ ⎦ ⎣ ⎝ a⎠ ⎦

⎡ ⎛ x ⎞ 2⎤ ⎥ dx b⋅ ⎢1 − ⎜ ⎣ ⎝ a⎠ ⎦

2 yc = m 5

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Problem 9-9 Locate the centroid xc of the shaded area. Solution: dA = y⋅ dx xc = x y yc = 2 b

⌠ h 2 ⎮ x ⋅ x dx 2 ⎮ b ⌡ xc =

0

b

⌠ 2 ⎮ x h ⋅ dx ⎮ 2 ⎮ b ⌡0

→ xc =

3 ⋅b 4

3 xc = ⋅ b 4

⌠ ⎮ ⎮ ⎮ ⌡ yc =

b 2

1 ⎛ h 2⎞ ⋅x dx 2⎜ 2 ⎝b ⎠

0

⌠ ⎮ ⎮ ⌡

→ yc =

b

0

3 yc = ⋅h 10

h b

2

2

⋅ x dx

3 ⋅h 10

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-10 Locate the centroid xc of the shaded area.

Solution:

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

⎛

x

⎝

2a ⎠

dA = y⋅ dx = 2k⋅ ⎜ x −

and its centroid is

2⎞

x1 = x

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

a

⌠ ⎛ x2 ⎞ ⎮ ⎮ x⋅ 2⋅ k⋅ ⎜ x − 2a dx ⎝ ⎠ ⌡ xc =

0

a

⌠ ⎛ x2 ⎞ ⎮ ⎜x − 2 ⋅ k ⋅ dx ⎮ 2a ⎠ ⎝ ⌡ 0

→ xc =

5 ⋅a 8

5 xc = ⋅ a 8

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Problem *9-12 Locate the centroid of the shaded area.

Solution: dA = ydx

xc = x

y yc = 2

L

⌠ ⎛ πx ⎞ dx → A = 2 ⋅ L⋅ a A = ⎮ a⋅ sin ⎜ ⎮ π ⎝L⎠ ⌡0 L

⌠ ⎛ πx ⎞ dx → x = 1 ⋅ L xc = ⋅ ⎮ x⋅ a⋅ sin ⎜ c 2 2⋅ L⋅ a ⎮ ⎝L⎠ ⌡0 L ⌠ 2 π ⎮ 1 ⎛ ⎛ π ⋅ x ⎞⎞ dx → y = 1 ⋅ π ⋅ a yc = ⋅⎮ ⋅ ⎜ a⋅ sin ⎜ c 8 2 ⎝ 2⋅ L⋅ a ⎝ L ⎠⎠ ⌡0 π

L

0

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem *9-16 Locate the centroid of the shaded area bounded by the parabola and the line y = a

Solution: x xc = 2

dA = x⋅ dy a

⌠ A= ⎮ ⌡0

( 2) 2

xc =

yc =

2⋅ a 3 2⋅ a

2

2⋅ a A= 3

2 a a⋅ y dy → A = ⋅ a 3

⎡ 1 ⎢⌠ ⋅ ⎮ 2 2 ⎢⌡ 0

a

3

yc = y 3

⎣

a

(

⎤ 3 ⎥ a⋅ y) dy⎥ → xc = ⋅ a 8 ⎦ 2

⌠ ⋅ ⎮ y a⋅ y dy → yc = 2 ⌡ 0

3 5⋅ a

3 xc = a 8 5

2 2) ( 4

⋅ a

3 yc = ⋅ a 5

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Problem 9-17 Locate the centroid of the quarter elliptical area.

Solution:

dA = y⋅ dx

xc = x

y yc = 2

a

⌠ 2 ⎮ ⎛ x ⎞ dx A = ⎮ b⋅ 1 − ⎜ ⎮ ⎝ a⎠ ⌡0

A=

π ⋅ a⋅ b 4

a

⌠ 2 ⎮ 4 ⎛ x ⎞ dx → x = 4 ⋅ a xc = ⋅ ⎮ x⋅ b ⋅ 1 − ⎜ c π ⋅ a⋅ b ⎮ 3⋅ π ⎝ a⎠ ⌡0 a

⌠ 2 ⎮ 2⎤ ⎡ 4 1 x ⎛ ⎞ ⎥ dx → y = 4 ⋅ b yc = ⋅⎮ ⋅ ⎢b⋅ 1 − ⎜ c π ⋅ a⋅ b ⎮ 2 ⎣ 3⋅ π ⎝ a⎠ ⎦ ⌡0

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Problem 9-21 Locate the centroid yc of the shaded area. Solve the problem by evaluating the integrals using Simpson's rule. Given: a := 2 ft 1

b :=

a

2

5

+ 2⋅ a

3

Solution:

a

⌠ ⎮ ⎮ A := ⎮ ⌡0

1 5⎞ ⎛ ⎜ 2 3 ⎝b − x + 2⋅ x ⎠ dx

a

⌠ ⎛ ⎮ 1 ⎮ 1 ⎜ yc := ⋅ ⎮ ⋅ ⎝b + A ⌡ 2 0

A = 2.177

1 5⎞ ⎛ ⎜ 2 3 2 3 x + 2⋅ x ⎠ ⋅ ⎝ b − x + 2⋅ x ⎠ dx 1

2

ft

5⎞

yc = 2.040

ft

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-23 Locate the centroid xc of the shaded area. Solution: Area and Moment Arm : Here,

⎛y⎞ x2 = a⋅ ⎜ ⎝ b⎠

a⋅ y

x1 =

b

2

The area of the differential element is 2 ⎡ a⋅ y y⎞ ⎤ ⎛ ⎥ ⋅ dy dA = ⎢ − a⋅ ⎜ ⎣b ⎝ b⎠ ⎦

and its centroid is 1 ⎡ a⋅ y ⎛y⎞ xc = ⋅ ⎢ + a⋅ ⎜ 2 ⎣b ⎝ b⎠

2⎤

⎥ ⎦

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

⌠ ⎮ A= ⎮ ⌡0

2 ⎡ a⋅ y 1 y ⎤ ⎢ − a⋅ ⎛⎜ ⎞ ⎥ dy → A = ⋅ a⋅ b 6 ⎣b ⎝ b⎠ ⎦

b

⌠ 2 2 2 6 ⎮ 1 ⎡ a⋅ y y ⎞ ⎤ ⎡ a⋅ y y⎞ ⎤ ⎛ ⎛ ⎥ ⎢ − a⋅ ⎜ ⎥ dy → xc = ⋅ a ⋅⎮ xc = ⋅⎢ + a⋅ ⎜ 5 2 ⎣b a⋅ b ⎝ b ⎠ ⎦⎣ b ⎝ b⎠ ⎦ ⌡0 Given:

a := 1ft

b := 2ft

2⋅ a xc := 5

xc = 0.40 ft

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-24 Locate the centroid yc of the shaded area. Solution: Area and Moment Arm : Here,

⎛y⎞ x2 = a⋅ ⎜ ⎝ b⎠

a⋅ y

x1 =

b

2

The area of the differential element is 2 ⎡ a⋅ y y⎞ ⎤ ⎛ ⎥ ⋅ dy dA = ⎢ − a⋅ ⎜ ⎣b ⎝ b⎠ ⎦

and its centroid is 1 ⎡ a⋅ y ⎛y⎞ xc = ⋅ ⎢ + a⋅ ⎜ 2 ⎣b ⎝ b⎠

2⎤

⎥ ⎦

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

⌠ ⎮ A= ⎮ ⌡0

2 ⎡ a⋅ y 1 y ⎤ ⎢ − a⋅ ⎛⎜ ⎞ ⎥ dy → A = ⋅ a⋅ b 6 ⎣b ⎝ b⎠ ⎦

b

⌠ 2 ⎡ a⋅ y 6 ⎮ y⎞ ⎤ 1 ⎛ ⎥ dy → yc = ⋅ b yc = ⋅ ⎮ y⋅ ⎢ − a⋅ ⎜ 2 a⋅ b b ⎝ b⎠ ⎦ ⌡0 ⎣ Given:

a := 1ft

b := 2ft

b yc := 2

yc = 1.00 ft

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-27 Locate the centroid xc of the shaded area.

Given :

a := 2in

Solution :

y=

b := 0.5in

a⋅ b x

a

xc :=

⌠ ⎮ x⋅ ⎛⎜ a⋅ b ⎞ dx ⎮ ⎝ x ⎠ ⌡ b

a

⌠ a⋅ b ⎮ dx ⎮ x ⌡b

xc = 1.08 in

Area and Moment Arm :The area of the differential element is dA = y⋅ dx =

1

x

⋅ dx

and its centroid is

xc = x

Centroid :Applying Eq. 9-6 and performing the integration, we have ⌠ ⎮ x dA ⎮ c ⌡ xc =

⌠ ⎮ ⎮ 1 dA ⌡ a

⌠ ⎮ x⋅ ⎛⎜ a⋅ b ⎞ dx ⎮ ⎝ x ⎠ ⌡

⌠ ⎮ x dA = b ⎮ c a ⌠ a⋅ b ⌡ ⎞ dx ⎮ ⎛ ⎜ ⎮ ⎝ x ⎠ ⌡b

xc = 1.08 in

a

xc :=

⌠ ⎮ x⋅ ⎛⎜ a⋅ b ⎞ dx ⎮ ⎝ x ⎠ ⌡ b

a

⌠ ⎮ ⎮ ⌡b

⎛ a⋅ b ⎞ dx ⎜ ⎝ x ⎠

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-32 Locate the centroid of the ellipsoid of revolution.

Solution :

2

dV = π ⋅ z ⋅ dy

⎛ y2 ⎞ z = a ⋅ ⎜1 − ⎜ b2 ⎝ ⎠ 2

2

b

⌠ 2 ⎮ y ⎞ 2 2⎛ 2 ⎜ V = ⎮ π⋅a ⋅ 1 − dy → V = ⋅ b⋅ π ⋅ a ⎜ b2 3 ⎮ ⎝ ⎠ ⌡ 0

b

⌠ 2 ⎮ 3 y ⎞ 3 2⎛ ⎜ yc = ⋅ ⎮ yπ ⋅ a ⋅ 1 − dy → yc = ⋅ b 2 ⎜ b2 8 2⋅ b ⋅ a ⋅ π ⎮ ⎝ ⎠ ⌡ 0 By symmetry

3⋅ b yc = 8 xc = zc = 0

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Problem 9-33 Locate the center of gravity of the volume. The material is homogeneous. Given: a := 2m b := 2m 2

⎛ y⎞ = z ⎜ b ⎝ a⎠

2

2 z

y = a ⋅

b

Solution: b

zc :=

⌠ 2 a ⋅z ⎮ zπ ⋅ dz ⎮ b ⌡ 0

b

⌠ 2 a ⋅z ⎮ ⎮ π ⋅ b dz ⌡ 0

zc = 1.33 m

xc = yc = 0

by symmetry

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-36 Locate the centroid of the quarter-cone.

Solution : r=

a h

⋅ ( h − z)

zc = z

xc = yc =

4⋅ r 3⋅ π

h

⌠ 2 1 2 ⎮ π ⎡a ⎤ V= ⎮ ⋅ ⎢ ⋅ ( h − z)⎥ dz → V = ⋅ h⋅ π ⋅ a 4 ⎣h 12 ⎦ ⌡0 h

⌠ 2 12 ⎮ π ⎡a 1 ⎤ zc = ⋅ ⎮ z ⋅ ⎢ ⋅ ( h − z)⎥ dz → zc = ⋅ h 2 4 ⎣h 4 ⎦ h⋅ a ⋅ π ⌡ 0

⎡⌠h ⎤ 2 ⎥ ⎢⎮ 4 a 12 π ⎡a 1 ⎡ ⎤ ⎤ xc = ⋅ ⎢⎮ ⋅ ⎢ ⋅ ( h − z)⎥ ⋅ ⎢ ⋅ ( h − z)⎥ dz⎥ → xc = ⋅ a 2 π ⎦ 4 ⎣h ⎦ ⎥ h⋅ a ⋅ π ⎢⌡ 3⋅ π ⎣ h ⎣0 ⎦

xc = yc =

a π

h zc = 4

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Problem 9-39 Locate the centroid yc of the paraboloid. Given: a := 4m b := 4m Solution: 2

⎛ z⎞ = y ⎜ a ⎝ b⎠

y

z = b⋅

2

a

2 y

dV = π ⋅ z ⋅ dy = π ⋅ b ⋅

and its centroid

a

yc = y

a

yc :=

⌠ ⎮ y⋅ π ⋅ b2⋅ y dy ⎮ a ⌡ 0

a

⌠ ⎮ π ⋅ b2⋅ y dy ⎮ a ⌡0

yc = 2.67 m

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-41 Locate the centroid zc of the frustum of the right-circular cone.

Solution : Volume and Moment Arm : From the geometry, y−r R−r y=

=

h−z h

( r − R) ⋅ z + R⋅ h h

The volume of thin disk differential element is 2

⎡( r − R) ⋅ z + R⋅ h⎤ ⋅ dz dV = π ⋅ y ⋅ dz = π ⋅ ⎢ ⎥ h ⎣ ⎦ 2

h

⌠ 2 ( r − R) ⋅ z + R⋅ h⎤ ⎮ ⎡ ⎥ dz ⎮ z π⋅⎢ h ⎣ ⎦ ⌡ zc =

0

h

⌠ 2 ⎮ ⎡( r − R) ⋅ z + R⋅ h⎤ dz π ⋅ ⎢ ⎥ ⎮ h ⎣ ⎦ ⌡ 0

1

→ zc =

4

2 2

⋅π⋅h ⋅r + 1 3

3

⋅ h⋅ r ⋅

1 6

2

⋅ π ⋅ h ⋅ r⋅ R +

π r−R

−

1 3

3

1

2

12

⋅ R ⋅ h⋅

2

⋅π⋅h ⋅R π

r−R

2

This can be simplified to give

zc =

2

R + 2⋅ r⋅ R + 3⋅ r

(

2

)

2

4⋅ R + r⋅ R + r

⋅h

π

dV =

h

2

⋅ ⎡⎣( r − R) ⋅ z + 2⋅ R⋅ h⋅ ( r − R) ⋅ z + R ⋅ h ⎤⎦ ⋅ dz 2 2

2

2

and its centroid z' = z Centroid :Applying Eq. 9-5 and performing the integration,we have h

⌠ ⎮ ⎮ z' dV ⌡ z' = V = ⌠ ⎮ 1 dV ⎮ ⌡ V

⌠ π ⎮ z⋅ ⎡ ⋅ ⎡⎣( r − R) 2⋅ z2 + 2⋅ R⋅ h⋅ ( r − R) ⋅ z + r2⋅ h2⎤⎦⎤ dz ⎢ ⎥ 2 ⎮ h ⎣ ⎦ ⌡0 h

⌠ π 2 2 2 2 ⎮ ⋅ ⎡⎣( r − R) ⋅ z + 2⋅ R⋅ h⋅ ( r − R) ⋅ z + R ⋅ h ⎤⎦ dz 2 ⎮ h ⌡0

4⎞ 2 ⎞⎤ ⎡ ⎛ z3 ⎞ 2 ⎛z 2 2 ⎛z ⎢ ⎜ ⎜ ⎜ ⎥ ⋅ ( r − R) ⋅ + 2⋅ R⋅ h⋅ ( r − R) ⋅ + R ⋅h ⋅ 2 ⎣ 4⎠ 3⎠ 2 ⎠⎦ ⎝ ⎝ ⎝ h

π

z' =

π h

2

⎡

⎛ z3 ⎞

⋅ ⎢( r − R) ⋅ ⎜

⎣

2

⎝ 3⎠

⎛ z2 ⎞

+ 2⋅ R⋅ h⋅ ( r − R) ⋅ ⎜

⎝ 2⎠

2

0

⎤

h

⎦

0

+ R ⋅ h ⋅ ( z)⎥ 2

h

2

z' =

2

R + 3⋅ r + 2⋅ r⋅ R

(

2

2

)

4⋅ R + r + r⋅ R

⋅h

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-44 Locate the center of gravity G of the five particles with respect to the origin O. g := 9.81 Given:

m 2

s

a := 2m

m1 := 1kg

b := 1m

m2 := 10kg

c := 1m

m3 := 2kg

d := 3m

m4 := 5kg

e := 2m

m5 := 6kg Solution:

xc =

xc :=

Σxc⋅ W ΣW −m1⋅ g⋅ ( a + b + c) − m2⋅ g⋅ ( b + c) − m3⋅ g⋅ c + m4⋅ g⋅ d + m5⋅ g⋅ ( d + e )

xc = 0.792 m

(m1 + m2 + m3 + m4 + m5)⋅ g

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

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 :

a⋅ xc :=

a 2

+ c⋅

2

⎛ ⎝

+ ( b − d) ⋅ c + ( a − c) ⋅ ⎜ c +

a − c⎞ 2

⎠

a + b + c + ( b − d) + ( a − c)

a⋅ b + b⋅ yc :=

c

b 2

+ ( b − d) ⋅

b−d 2

xc = 34.38 mm

+ ( a − c) ⋅ ( b − d)

a + b + c + ( b − d) + ( a − c)

yc = 85.83 mm

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-57 Determine the location yc of the centroidal axis xcxc of the beam's cross-sectional area. Neglect the size of the corner welds at A and B for the calculation. Given : r := 50mm t := 15mm a := 150mm b := 15mm c := 150mm

Solution : b⋅ c⋅ yc :=

b 2

⎛ ⎝

+ a⋅ t⋅ ⎜ b +

a⎞ 2⎠

2

+ π ⋅ r ⋅ ( b + a + r) 2

b ⋅ c + a⋅ t + π ⋅ r

yc = 154.44 mm

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

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 :

a⋅ b⋅ xc :=

a⋅ b⋅ yc :=

b 2

a 2

+

+

1

⎛ ⎝

c⎞

2 ⋅ ( b + c) ⋅ d⋅ ⋅ ( b + c) 2 3⎠ 2 3 1 1 a⋅ b + ⋅ c⋅ a + ⋅ ( b + c) ⋅ d 2 2

1 2

a⋅ b +

⋅ a⋅ c⋅ ⎜ b +

⋅ a⋅ c⋅ 1 2

a 3

⋅ c⋅ a +

− 1 2

1 2

+

1

⋅ ( b + c) ⋅ d⋅

⋅ ( b + c) ⋅ d

xc = 4.62 in

d 3

yc = 1.00 in

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-62 Determine the location xc of the centroid C of the shaded area which is part of a circle having a radius r.

Solution : 1 2 ⎛ 2⋅ r ⎞ 1 ⎞ ⎛2 ΣxcA = ⋅ r ⋅ α ⋅ ⎜ ⋅ sin ( α ) − ⋅ r⋅ sin ( α ) ⋅ r⋅ cos ( α ) ⋅ ⎜ ⋅ r⋅ cos ( α ) 2 ⎝3 ⎠ ⎝ 3⋅ α ⎠ 2 3

3

3

r r r 2 3 ΣxcA = ⋅ sin ( α ) − ⋅ sin ( α ) ⋅ cos ( α ) = ⋅ sin ( α ) 3 3 3 ΣA =

xc =

1 2 1 1 2⎛ sin ( 2⋅ α ) ⎞ ⋅ r ⋅ α − ⋅ r⋅ sin ( α ) ⋅ r⋅ cos ( α ) = ⋅ r ⋅ ⎜ α − 2 2 2 2 ⎝ ⎠ ΣxcA ΣA 3

r xc =

3

2

3

⋅ sin ( α )

sin ( 2⋅ α ) ⎞ 1 2⎛ ⋅ r ⋅ ⎜α − 2 2 ⎝ ⎠

=

3

3

⋅ r⋅ sin ( α )

α−

sin ( 2⋅ α ) 2

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-69 Determine the distance h to which a hole of diameter d must be bored into the base of the cone so that the center of mass of the resulting shape is located at zc. The material has a density ρ. Given: d := 100mm zc := 115mm ρ := 8

mg m

3

a := 150mm b := 500mm

Solutions: 1 zc =

3

2

⋅ π ⋅ a ⋅ b⋅

b 4

2

⎛ d ⎞ ⋅ h⋅ h 2 ⎝2⎠

− π⋅⎜

2

⎛ d⎞ ⋅h ⋅π⋅a ⋅b − π⋅⎜ 3 ⎝2⎠ 1

2

Choosing the positive root,

h :=

1 2 2 2 2 2 zc⋅ d + ⋅ 9⋅ zc ⋅ d − 24⋅ zc⋅ a ⋅ b + 6⋅ a ⋅ b 3

h = 323 mm

d

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-71 The sheet metal part has the dimensions shown. Determine the location (xc, yc, zc) of its centroid. Given: a := 3in b := 4in c := 6in

Solution:

−a⋅ b⋅ xc :=

2

a⋅ b +

1

a

+

a⋅ b⋅ yc :=

2

2

2

a⋅ c⋅

a⋅ b +

1 2

xc = −1.14 in

⋅ a⋅ c

1

a⋅ b +

−1 zc :=

b

⋅ a⋅ c⋅

2 1 2

⋅ a⋅ c

2⋅ a 3

yc = 1.71 in

c 3

⋅ a⋅ c

zc = −0.857 in

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-76 Locate the center of gravity of the two-block assembly. The specific weights of the materials A and B are γA and γB, respectively. Given: lb

γ A := 150

3

ft

lb

γ B := 400

3

ft a := 2in b := 6in c := 2in d := 6in e := 6in Solution WA := γ A⋅

xc :=

yc :=

zc :=

WB⋅

a⋅ b ⋅ d 2

⎛ b⎞ + WA⋅ ⎜ c + 2 ⎝ 3⎠ c

WB + WA WB⋅

a + WA⋅ 2 2

xc = 1.47 in

e

WB + WA WB⋅

WB := γ B⋅ c⋅ d⋅ e

d + WA⋅ 3 2

yc = 2.68 in

d

WB + WA

zc = 2.84 in

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-79 Locate the centroid zc of the top made from a hemisphere and a cone. Given: r := 24mm h := 120mm

Solution: π zc :=

3

2

⋅ r ⋅ h⋅

3⋅ h 4

+

2 3

3

⎛ ⎝

⋅ π ⋅ r ⋅ ⎜h +

2 π 2 3 ⋅r ⋅h + π⋅r 3 3

3⋅ r ⎞ 8

⎠

zc = 101.14 mm

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-83 The assembly consists of a wooden dowel rod of length L and a tight-fitting steel collar. Determine the distance xc to its center of gravity if the specific weights of the materials are γw and γst.The radii of the dowel and collar are shown. Given: L := 20in γ w := 150

lb 3

ft γ st := 490

lb 3

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

2

xc :=

γ w⋅ π ⋅ r1 ⋅ L⋅

2 2 ⎛ b⎞ + γ st⋅ π ⋅ ⎛ r2 − r1 ⎞ ⋅ b⋅ ⎜ a + ⎝ ⎠ ⎝ 2⎠ 2

L

2 2 γ w⋅ π ⋅ r1 ⋅ L + γ stπ ⋅ ⎛ r2 − r1 ⎞ ⋅ b ⎝ ⎠ 2

xc = 8.22 in

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-85 The anchor ring is made of steel having specific weight γst. Determine the surface area of the ring. The cross section is circular as shown. Given γ st := 490

lb 3

ft a := 4in b := 8in

Solution:

⎛ a + b − a ⎞ ⋅ 2⋅ π ⋅ b − a 4 4 ⎠ ⎝2

A := 2⋅ π ⋅ ⎜

A = 118 in

2

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-101 A V-belt has as inner radius r and a cross-sectional area as shown. Determine the volume of material used in making the V-belt. Given : r := 6in a := 0.25in b := 0.5in θ := 30deg

Solution : Volume : Applying the theorem of Pappus and Guldinus, Eq.9-12

⎛θ⎞ ⎝2⎠

h := a⋅ cot ⎜

⎡ ⎣

⎛ ⎝

V := 2⋅ π ⋅ ⎢b⋅ h⋅ ⎜ r +

h⎞ 2⎠

+ 2⋅

⎛ 2 ⎝

a⋅ h

⋅ ⎜r +

2⋅ h ⎞⎤ 3

⎥ ⎠⎦

V = 28.66 in

3

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-102 The full circular aluminum housing is used in an automotive brake system. The cross section is shown in the figure. Determine its weight if aluminum has a specific weight γ. Given : γ := 169

lb 3

ft

a := 2.00in b := 0.25in

d := 4.00in e := 3.25in f := 0.25in

c := 0.15in Solution : Volume : Applying the theorem of Pappus and Guldinus, Eq.9-12, with r :=

a −b 2

⎡ ⎣

V := 2⋅ π ⋅ ⎢( r + c) ⋅ f⋅

W := γ ⋅ V

r+c ⎛ c ⎞ + b⋅ ( d − e − f) ⋅ ⎛ r + b ⎞⎤ + c⋅ e⋅ ⎜ r + ⎜ ⎥ 2 ⎝ 2⎠ ⎝ 2 ⎠⎦

W = 0.377 lb

V = 3.85 in

3

© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. For the exclusive use of adopters of the Hibbeler series of books.

Problem 9-124 A circular V-belt has an inner radius r and a cross-sectional area as shown. Determine the volume of material required to make the belt. Given r := 600mm a := 25mm b := 50mm c := 75mm Solution

⎡⎛ ⎣⎝

V := 2⋅ π ⋅ ⎢⎜ r +

c ⎞ ⎛ 1⎞ ⎛ c ⎞ ⋅ b⋅ c⎤ ⋅ 2⋅ ⎜ ⋅ a⋅ c + ⎜ r + ⎥ 3⎠ ⎝ 2⎠ ⎝ 2⎠ ⎦

V = 22.4 × 10

−3

m

3

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Distributed Forces

Sample Problem 5/1 Centroid of a circular arc.

Locate the centroid of a circular arc as shown in

the figure.

r

α α

Solution. Choosing the axis of symmetry as the x-axis makes y 0. A differential 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 [Lx

x dL]

(2r)x

y

(r cos ) r d

C

r cos θ dθ

2rx 2r2 sin r sin Ans. For a semicircular arc 2 , which gives x 2r/. By symmetry we see immediately that this result also applies to the quarter-circular arc when the measurement is made as shown. x

r dθ

α

θ x

α r

Helpful Hint

It should be perfectly evident that polar coordinates are preferable to rectan-

C

gular coordinates to express the length of a circular arc.

C 2r/π r

Sample Problem 5/2

r

y

Determine the distance h from the base of a triangle of altitude h to the centroid of its area.

Centroid of a triangular area.

dy h

Solution.

x

The x-axis is taken to coincide with the base. A differential strip of

area dA x dy is chosen. By similar triangles x/(h y) b/h. Applying the sec-

y

ond of Eqs. 5/5a gives [Ay and

y dA] c

x

bh y 2

b h

0

y

b(h y) bh2 dy 6 h y

h 3

Helpful Hint Ans.

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

We save one integration here by using the first-order element of area. Recognize that dA must be expressed in terms of the integration variable y; hence, x ƒ( y) is required.

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245

Centroids of Lines, Areas, and Volumes

Sample Problem 5/3 Centroid of the area of a circular sector.

r

Locate the centroid of the area

of a circular sector with respect to its vertex.

α α

Solution I. The x-axis is chosen as the axis of symmetry, and y is therefore automatically zero. We may cover the area by moving an element in the form of a partial circular ring, as shown in the figure, from the center to the outer periphery. The radius of the ring is r0 and its thickness is dr0, so that its area is dA 2r0 dr0. The x-coordinate to the centroid of the element from Sample Problem 5/1 is xc r0 sin /, where r0 replaces r in the formula. Thus, the first of Eqs. 5/5a gives [Ax

x dA] c

2 (r2)x 2

r

0

y r

α

x

r0 sin α xc = ——––– α

2

2 r sin 3

dr0

r0

α

r0 sin (2r0 dr0)

r2x 3r 3 sin x

C

Ans.

Solution I

Helpful Hints

Solution II.

The area may also be covered by swinging a triangle of differential area 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 terms are neglected. From Sample Problem 5/2 the centroid of the triangular element of area is two-thirds of its altitude from its vertex, so that the x-coordinate to the centroid of the element is xc 32 r cos . Applying the first of Eqs. 5/5a gives [Ax

x dA] c

(r2)x

(23 r cos )(12 r2 d)

and as before

guish between the variable r0 and the constant r. Be careful not to use r0 as the centroidal coordinate for the element.

y xc = 2 – r cos θ 3

r2x 23r3 sin 2 r sin x 3

Note carefully that we must distin-

Ans.

For a semicircular area 2 , which gives x 4r/3. By symmetry we see immediately that this result also applies to the quarter-circular area where the measurement 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 I began. On the other hand, integration with respect to r0 initially would give the triangular element with which Solution II began.

dθ

α

θ

α

x

r

Solution II

4r/3π C C r

r

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Distributed Forces

Sample Problem 5/4

y x = ky3

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

– x

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

[Ax

x dA]

x

c

a

0

y dx

– y

a

xy dx

3ab 3a2b x 7 4

4

x 7a

y b

Ans.

x = ky3

In the solution for y from the second of Eqs. 5/5a, the coordinate to the centroid of the rectangular element is yc y/2, where y is the height of the strip governed by the equation of the curve x ky3. Thus, the moment principle becomes

y dA]

3ab y 4

c

x

a

0

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

[Ay

b

C

Solution I.

y y yc = – 2

2y y dx a

x

0

x

a

dx

Substituting y b(x/a)1/3 and integrating give 3ab2 3ab y 4 10

y 52 b

Ans. y

Solution II.

The horizontal element of area shown in the lower figure may be employed in place of the vertical element. The x-coordinate to the centroid of the 1 rectangular element is seen to be xc x 2(a x) (a x)/2, which is simply the average of the coordinates a and x of the ends of the strip. Hence, [Ax

x dA] c

x

b

0

a 2 x(a x) dy

b

a+x xc = –––– 2 x = ky3 dy

b

(a x) dy

0

x

a–x

y

The value of y is found from [Ay

y dA] c

y

b

0

(a x) dy

a

b

0

x

y(a x) dy

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

Helpful Hint

Note that xc x for the vertical element.

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247

Sample Problem 5/5 Hemispherical volume.

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

Solution I. With the axes chosen as shown in the figure, x z 0 by symmetry. The most convenient element is a circular slice of thickness dy parallel to the 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 r2 y2. The volume of the elemental slice becomes

z

y2 + z2 = r2

z

dV (r2 y2) dy

y r

The second of Eqs. 5/6a requires [V y

y dV] c

y

r

0

(r2

y2)

dy

dy

r

y(r2

0

y2)

dy

where yc y. Integrating gives

Solution I 3 y 8r

2 3 r y 14r4 3

Ans. z

Solution II.

Alternatively we may use for our differential element a cylindrical shell of length y, radius z, and thickness dz, as shown in the lower figure. By expanding the radius of the shell from zero to r, we cover the entire volume. By symmetry 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 from 2 the equation of the circle gives y r2 z2. Using the value of 3 r3 computed in Solution I for the volume of the hemisphere and substituting in the second of Eqs. 5/6a give us [V y

y dV] c

2

r

2

y dz

yc = y/2

z y r

r 2 z (2zr z ) dz r (r z z ) dz 4

(3r3)y

yc = y

x

2

2

x

2

0

r

2

3

4

Solution II

0

3

y 8r

Ans.

z

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

dθ

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

r dθ

θ

y

Solution III

Helpful Hint

Can you identify the higher-order element of volume which is omitted from the expression for dV?

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Sample Problem 5/6

y 12″

Locate the centroid of the shaded area.

4″

Solution.

The composite area is divided into the four elementary shapes shown in the lower figure. The centroid locations of all these shapes may be obtained from Table D/3. Note that the areas of the “holes” (parts 3 and 4) are taken as negative in the following table:

4″ 3″ 3″

A in.2

x in.

y in.

1 2 3 4

120 30 14.14 8

6 14 6 12

5 10/3 1.273 4

TOTALS

127.9

PART

xA in.3

yA in.3

720 420 84.8 96

600 100 18 32

959

650

2″ 2″ 2″

1

x

5″

4

2

3

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

X ΣAx ΣA

X

959 7.50 in. 127.9

Ans.

Y ΣAy ΣA

Y

650 5.08 in. 127.9

Ans.

Sample Problem 5/7

6 5 A, m2

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

4 3 2

Solution.

The body is divided into five sections. For each section, the average area, volume, and centroid location are determined and entered in the following table:

INTERVAL

Aav m2

Volume V m3

x m

Vx m4

0–0.2 0.2–0.4 0.4–0.6 0.6–0.8 0.8–1.0

3 4.5 5.2 5.2 4.5

0.6 0.90 1.04 1.04 0.90

0.1 0.3 0.5 0.7 0.9

0.060 0.270 0.520 0.728 0.810

TOTALS

4.48

1 0 0

0.2

0.4

0.6

0.8

x, m

2.388 Helpful Hint

X ΣΣVVx

X

2.388 0.533 m 4.48

Ans.

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

1.0

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Composite Bodies and Figures; Approximations

Sample Problem 5/8

259

z

Locate the center of mass of the bracket-and-shaft combination. The vertical face is made from sheet metal which has a mass of 25 kg/m2. The material of the horizontal base has a mass of 40 kg/m2, and the steel shaft has a density of 7.83 Mg/m3.

50 150 x 40

Solution.

The composite body may be considered to be composed of the five elements shown in the lower portion of the illustration. The triangular part will be taken as a negative mass. For the reference axes indicated it is clear by symmetry 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 explanation. For Part 1 we have from Sample Problem 5/3 z

150

75

y

50

50

25 100

150

Dimensions in millimeters

4(50) 4r 21.2 mm 3 3

1

For Part 3 we see from Sample Problem 5/2 that the centroid of the triangular mass is one-third of its altitude above its base. Measurement from the coordinate axes becomes z [150 25 13 (75)] 100 mm

1 2 3 4 5 TOTALS

m kg

y mm

z mm

my kg m

mz kg mm

0.098 0.562 0.094 0.600 1.476

0 0 0 50.0 75.0

21.2 75.0 100.0 150.0 0

0 0 0 30.0 110.7

2.08 42.19 9.38 90.00 0

140.7

120.73

2.642

5 3

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

PART

2

Equations 5/7 are now applied and the results are

Y Σmy Σm

Y

140.7 53.3 mm 2.642

Ans.

Z Σmz Σm

Z

120.73 45.7 mm 2.642

Ans.

4

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Sample Problem 5/9

z R

Determine the volume V and surface area A of the complete torus of circular cross section. a

Solution.

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

V rA 2(R)(a2) 22Ra2

Ans.

Similarly, using Eq. 5/8a gives

z

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

–r = R

Ans.

a

Helpful Hint

We note that the angle of revolution is 2 for the complete ring. This common but special-case result is given by Eq. 5/9.

Sample Problem 5/10

z

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

60 mm x

Solution.

With the angle of revolution 180, Eq. 5/9a gives V rA [30 13(60)][12(60)(60)] 2.83(105) mm3

Ans.

30 mm

60 mm

The mass of the body is then z

kg

1m m V 7830 3 [2.83(10 ) mm ] 1000 mm m 2.21 kg

5

3

3

Ans.

60 mm

C r– 30 mm

60 mm

Helpful Hint

Note that must be in radians.

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