Engineering Mechanics Dynamics 14th
Edition Hibbeler Solutions Manual
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17–1.
z
Determine the moment of inertia Iy for the slender rod. The
rod’s density r and cross-sectional area A are constant.
Express the result in terms of the rod’s total mass m.
y
l
SOLUTION
A
x
Iy =
LM
x 2 dm
l
=
=
L0
x 2 (r A dx)
1
r A l3
3
m = rAl
Thus,
Iy =
1
m l2
3
Ans.
Ans:
Iy =
791
1
m l2
3
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17–2.
z
The solid cylinder has an outer radius R, height h, and is
made from a material having a density that varies from its
center as r = k + ar2, where k and a are constants.
Determine the mass of the cylinder and its moment of
inertia about the z axis.
R
h
SOLUTION
Consider a shell element of radius r and mass
dm = r dV = r(2p r dr)h
R
m =
(k + ar2)(2p r dr)h
L0
m = 2p h(
aR4
kR2
+
)
2
4
m = p h R2(k +
aR2
)
2
Ans.
dI = r2 dm = r2(r)(2p r dr)h
R
Iz =
L0
r2(k + ar2)(2p r dr) h
R
Iz = 2ph
L0
Iz = 2ph[
Iz =
(k r3 + a r5) dr
k R4
aR6
+
]
4
6
2 aR2
p h R4
[k +
]
2
3
Ans.
Ans:
aR2
b
2
p h R4
2 aR2
Iz =
ck +
d
2
3
m = p h R2 ak +
792
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17–3.
y
Determine the moment of inertia of the thin ring about the
z axis. The ring has a mass m.
R
x
SOLUTION
2p
Iz =
L0
r A(R du)R2 = 2p r A R3
2p
m =
L0
r A R du = 2p r A R
Thus,
I z = m R2
Ans.
Ans:
Iz = mR2
793
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*17–4.
y
The paraboloid is formed by revolving the shaded area
around the x axis. Determine the radius of gyration kx. The
density of the material is r = 5 Mg>m3.
y2
50x
100 mm
x
SOLUTION
200 mm
dm = r p y2 dx = r p (50x) dx
Ix =
1 2
1
y dm =
2 L0
L2
= r pa
200
50 x {p r (50x)} dx
502 1 3 200
bc x d
2
3
0
= rp a
502
b(200)3
6
200
m =
L
dm =
L0
p r (50x) dx
200
1
= r p (50)c x2 d
2
0
= rp a
kx =
50
b (200)2
2
Ix
50
(200) = 57.7 mm
=
Am
A3
Ans.
Ans:
kx = 57.7 mm
794
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17–5.
y
Determine the radius of gyration kx of the body. The
specific weight of the material is g = 380 lb>ft3.
y3 = x
2 in.
x
8 in.
SOLUTION
dm = r dV = rp y2 dx
d Ix =
1
1
(dm) y2 = pry4 dx
2
2
8
Ix =
1
prx4/3 dx = 86.17r
L0 2
8
m =
kx =
L0
prx2/3 dx = 60.32r
86.17r
Ix
=
= 1.20 in.
Am
A 60.32r
Ans.
Ans:
kx = 1.20 in.
795
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17–6.
y
The sphere is formed by revolving the shaded area around
the x axis. Determine the moment of inertia Ix and express
the result in terms of the total mass m of the sphere. The
material has a constant density r.
x2 + y2 = r2
x
SOLUTION
dIx =
y2 dm
2
dm = r dV = r(py2 dx) = r p(r2 - x2) dx
dIx =
1
r p(r2 - x2)2 dx
2
r
Ix =
=
1
2
2 2
r p(r - x ) dx
2
L- r
8
pr r5
15
r
m =
=
L- r
2
2
Ix =
2
m r2
5
r p(r - x ) dx
4
r p r3
3
Thus,
Ans.
Ans:
Ix =
796
2
mr2
5
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17–7.
y
The frustum is formed by rotating the shaded area around
the x axis. Determine the moment of inertia Ix and express
the result in terms of the total mass m of the frustum. The
frustum has a constant density r.
y
–ba x
b
2b
b
x
z
SOLUTION
a
dm = r dV = rpy2 dx = rp A
2b2
b2 2
x + b2 B dx
x
+
a
a2
dIx =
1
1
dmy2 = rpy4 dx
2
2
dIx =
1
4 b4
6 b4
4b4
b4
rp A 4 x4 + 3 x3 + 2 x2 +
x + b4 B dx
a
2
a
a
a
a
Ix =
L
dIx =
=
1
4b4
6 b4
4 b4
b4
x + b4 B dx
rp
A 4 x4 + 3 x3 + 2 x2 +
a
2 L0 a
a
a
31
rpab4
10
a
m =
Ix =
Lm
dm = rp
L0
A 2 x2 +
b2
a
2b2
7
x + b2 B dx = rpab2
a
3
93 2
mb
70
Ans.
Ans:
Ix =
797
93 2
mb
70
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*17–8.
y
The hemisphere is formed by rotating the shaded area
around the y axis. Determine the moment of inertia Iy and
express the result in terms of the total mass m of the
hemisphere. The material has a constant density r.
x2
y2
r2
x
SOLUTION
r
m =
LV
r dV = r
= rp cr 2 y -
Iy =
=
L0
r
p x2 dy = rp
L0
(r2 - y2)dy
1 3 r
2
y d = rp r3
3
3
0
r
r
r
rp
1
(dm) x2 =
px4 dy =
(r2 - y2)2 dy
2 L0
2 L0
Lm 2
rp 4
y5 r
4rp 5
2
c r y - r2 y3 +
d =
r
2
3
5 0
15
Thus,
Iy =
2
m r2
5
Ans.
Ans:
Iy =
798
2
mr 2
5
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17–9.
Determine the moment of inertia of the homogeneous
triangular prism with respect to the y axis. Express the result
in terms of the mass m of the prism. Hint: For integration, use
thin plate elements parallel to the x–y plane and having a
thickness dz.
z
–h (x – a)
z = ––
a
h
SOLUTION
dV = bx dz = b(a)(1 -
z
) dz
h
x
b
a
y
x
dIy = dIy + (dm)[( )2 + z2]
2
=
x2
1
dm(x2) + dm( ) + dmz2
12
4
= dm(
x2
+ z2)
3
= [b(a)(1 -
a2
z
z
)dz](r)[ (1 - )2 + z2]
h
3
h
k
Iy = abr
a3 h - z 3
z
(
) + z2(1 - )]dz
h
h
L0 3
= abr[
=
[
3
1
1 1
1
a2
(h4 - h4 + h4 - h4) + ( h4 - h4 )]
2
4
h 3
4
3h3
1
abhr(a2 + h2)
12
m = rV =
1
abhr
2
Thus,
Iy =
m 2
(a + h2)
6
Ans.
Ans:
Iy =
799
m 2
( a + h2 )
6
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17–10.
The pendulum consists of a 4-kg circular plate and a
2-kg slender rod. Determine the radius of gyration of the
pendulum about an axis perpendicular to the page and
passing through point O.
O
2m
Solution
Using the parallel axis theorem by referring to Fig. a,
IO = Σ ( IG + md 2 )
= c
1m
1
1
(2) ( 22 ) + 2 ( 12 ) d + c (4) ( 0.52 ) + 4 ( 2.52 ) d
12
2
= 28.17 kg # m2
Thus, the radius of gyration is
kO =
28.17
IO
=
= 2.167 m = 2.17 m
Am
A4 + 2
Ans.
Ans:
kO = 2.17 m
800
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17–11.
The assembly is made of the slender rods that have a mass
per unit length of 3 kg>m. Determine the mass moment of
inertia of the assembly about an axis perpendicular to the
page and passing through point O.
O
0.4 m
0.8 m
0.4 m
Solution
Using the parallel axis theorem by referring to Fig. a,
IO = Σ ( IG + md 2 )
= e
1
33(1.2)4 ( 1.22 ) + 33(1.2)4 ( 0.22 ) f
12
+ e
1
33(0.4)4 ( 0.42 ) + 33(0.4)4 ( 0.82 ) f
12
= 1.36 kg # m2
Ans.
Ans:
IO = 1.36 kg # m2
801
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*17–12.
Determine the moment of inertia of the solid steel assembly
about the x axis. Steel has a specific weight of
gst = 490 lb>ft3.
0.25 ft
0.5 ft
2 ft
3 ft
SOLUTION
Ix =
1
3
3
m (0.5)2 +
m (0.5)2 m (0.25)2
2 1
10 2
10 3
1
3 1
3 1
490
a b p(0.5)2 (4)(0.5)2 a bp(0.25)2(2)(0.25)2 d a
b
= c p(0.5)2(3)(0.5)2 +
2
10 3
10 2
32.2
= 5.64 slug # ft2
Ans.
Ans:
Ix = 5.64 slug # ft 2
802
x
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17–13.
The wheel consists of a thin ring having a mass of 10 kg and
four spokes made from slender rods, each having a mass of
2 kg. Determine the wheel’s moment of inertia about an
axis perpendicular to the page and passing through point A.
500 mm
SOLUTION
A
IA = Io + md3
= c2c
1
(4)(1)2 d + 10(0.5)2 d + 18(0.5)2
12
= 7.67 kg # m2
Ans.
Ans:
IA = 7.67 kg # m2
803
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17–14.
If the large ring, small ring and each of the spokes weigh
100 lb, 15 lb, and 20 lb, respectively, determine the mass
moment of inertia of the wheel about an axis perpendicular
to the page and passing through point A.
4 ft
1 ft
O
SOLUTION
Composite Parts: The wheel can be subdivided into the segments shown in Fig. a.
The spokes which have a length of (4 - 1) = 3 ft and a center of mass located at a
3
distance of a1 + b ft = 2.5 ft from point O can be grouped as segment (2).
2
A
Mass Moment of Inertia: First, we will compute the mass moment of inertia of the
wheel about an axis perpendicular to the page and passing through point O.
IO = a
100
1 20
20
15
b (4 2) + 8c a
b (32) + a
b(2.52) d + a
b(12)
32.2
12 32.2
32.2
32.2
= 84.94 slug # ft2
The mass moment of inertia of the wheel about an axis perpendicular to the page
and passing through point A can be found using the parallel-axis theorem
100
20
15
IA = IO + md 2, where m =
+ 8a
b +
= 8.5404 slug and d = 4 ft.
32.2
32.2
32.2
Thus,
IA = 84.94 + 8.5404(42) = 221.58 slug # ft2 = 222 slug # ft2
Ans.
Ans:
IA = 222 slug # ft 2
804
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17–15.
Determine the moment of inertia about an axis perpendicular
to the page and passing through the pin at O. The thin plate
has a hole in its center. Its thickness is 50 mm, and the
material has a density r = 50 kg>m3.
O
150 mm
SOLUTION
IG =
1.40 m
1
1
C 50(1.4)(1.4)(0.05) D C (1.4)2 + (1.4)2 D - C 50(p)(0.15)2(0.05) D (0.15)2
12
2
1.40 m
= 1.5987 kg # m2
IO = IG + md2
m = 50(1.4)(1.4)(0.05) - 50(p)(0.15)2(0.05) = 4.7233 kg
IO = 1.5987 + 4.7233(1.4 sin 45°)2 = 6.23 kg # m2
Ans.
Ans:
IO = 6.23 kg # m2
805
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*17–16.
Determine the mass moment of inertia of the thin plate
about an axis perpendicular to the page and passing
through point O. The material has a mass per unit area of
20 kg>m2.
200 mm
O
200 mm
SOLUTION
Composite Parts: The plate can be subdivided into two segments as shown in Fig. a.
Since segment (2) is a hole, it should be considered as a negative part. The
perpendicular distances measured from the center of mass of each segment to the
point O are also indicated.
200 mm
Mass Moment of Inertia: The moment of inertia of segments (1) and (2) are computed
as m1 = p(0.2 2)(20) = 0.8p kg and m2 = (0.2)(0.2)(20) = 0.8 kg. The moment of
inertia of the plate about an axis perpendicular to the page and passing through point
O for each segment can be determined using the parallel-axis theorem.
IO = ©IG + md2
1
1
= c (0.8p)(0.22) + 0.8p(0.22) d - c (0.8)(0.22 + 0.22) + 0.8(0.22) d
2
12
= 0.113 kg # m2
Ans.
Ans:
IO = 0.113 kg # m2
806
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17–17.
Determine the location y of the center of mass G of the
assembly and then calculate the moment of inertia about
an axis perpendicular to the page and passing through G.
The block has a mass of 3 kg and the semicylinder has a
mass of 5 kg.
400 mm
300 mm
G
–y
Solution
200 mm
O
Moment inertia of the semicylinder about its center of mass:
(IG)cyc =
y =
4R 2
1
mR2 - m a b = 0.3199mR2
2
3p
Σ∼
ym
Σm
=
c 0.2 -
4(0.2)
d (5) + 0.35(3)
3p
= 0.2032 m = 0.203 m
5 + 3
IG = 0.3199(5)(0.2)2 + 5c 0.2032 - a0.2 -
4(0.2)
3p
2
bd +
Ans.
1
(3) ( 0.32 + 0.42 )
12
+ 3(0.35 - 0.2032)2
= 0.230 kg # m2
Ans.
Ans:
IG = 0.230 kg # m2
807
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17–18.
Determine the moment of inertia of the assembly about an
axis perpendicular to the page and passing through point O.
The block has a mass of 3 kg, and the semicylinder has a
mass of 5 kg.
400 mm
300 mm
G
–y
Solution
(IG)cyl =
200 mm
O
2
1
4R
mR2 - m a b = 0.3199 mR2
2
3p
IO = 0.3199(5)(0.2)2 + 5 a0.2 -
4(0.2)
3p
= 0.560 kg # m2
2
b +
1
(3) ( (0.3)2 + (0.4)2 ) + 3(0.350)2
12
Ans.
Also from the solution to Prob. 17–22,
IO = IG + md 2
= 0.230 + 8(0.2032)2
= 0.560 kg # m2
Ans.
Ans:
IO = 0.560 kg # m2
808
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17–19.
Determine the moment of inertia of the wheel about an
axis which is perpendicular to the page and passes through
the center of mass G. The material has a specific weight
g = 90 lb>ft 3.
0.25 ft
1 ft
G
0.25 ft
2 ft
0.5 ft
Solution
IG =
O
1 ft
1 90
1 90
c
(p)(2)2(0.25) d (2)2 + c
(p)(2.5)2(1) d (2.5)2
2 32.2
2 32.2
-
1 90
1 90
c
b(p)(0.25)2(0.25) d (0.25)2
(p)(2)2(1) d (2)2 - 4c a
2 32.2
2 32.2
- 4 ca
90
b(p)(0.25)2(0.25) d (1)2
32.2
= 118.25 = 118 slug # ft 2
Ans.
Ans:
IG = 118 slug # ft 2
809
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*17–20.
Determine the moment of inertia of the wheel about an axis
which is perpendicular to the page and passes through point O.
The material has a specific weight g = 90 lb>ft 3.
0.25 ft
1 ft
G
0.25 ft
2 ft
0.5 ft
Solution
m =
O
1 ft
90
3p(2)2(0.25) + p 5 (2.5)2(1) - (2)2(1)6 - 4p(0.25)2(0.25)4 = 27.99 slug
32.2
From the solution to Prob. 17–18,
IG = 118.25 slug # ft 2
IO = 118.25 + 27.99(2.5)2 = 293 slug # ft 2
Ans.
Ans:
IO = 293 slug # ft 2
810
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17–21.
The pendulum consists of the 3-kg slender rod and the
5-kg thin plate. Determine the location y of the center of
mass G of the pendulum; then calculate the moment of
inertia of the pendulum about an axis perpendicular to the
page and passing through G.
O
2m
G
SOLUTION
y =
y
0.5 m
1(3) + 2.25(5)
© ym
=
= 1.781 m = 1.78 m
©m
3 + 5
Ans.
1m
IG = ©IG + md 2
=
1
1
(3)(2)2 + 3(1.781 - 1)2 +
(5)(0.52 + 12) + 5(2.25 - 1.781)2
12
12
= 4.45 kg # m2
Ans.
Ans:
y = 1.78 m
IG = 4.45 kg # m2
811
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17–22.
20 mm
Determine the moment of inertia of the overhung crank
about the x axis. The material is steel having a destiny of
r = 7.85 Mg>m3.
30 mm
90 mm
50 mm
x
180 mm
20 mm
SOLUTION
mc = 7.85(10 ) A (0.05)p(0.01)
3
2
x¿
B = 0.1233 kg
30 mm
20 mm
mr = 7.85(103)((0.03)(0.180)(0.02)) = 0.8478 kg
50 mm
30 mm
1
Ix = 2 c (0.1233)(0.01)2 + (0.1233)(0.06)2 d
2
+ c
1
(0.8478) A (0.03)2 + (0.180)2 B d
12
= 0.00325 kg # m2 = 3.25 g # m2
Ans.
Ans:
Ix = 3.25 g # m2
812
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17–23.
20 mm
Determine the moment of inertia of the overhung crank
about the x¿ axis. The material is steel having a destiny of
r = 7.85 Mg>m3.
30 mm
90 mm
50 mm
x
180 mm
20 mm
SOLUTION
x¿
mc = 7.85 A 103 B A (0.05)p(0.01)2 B = 0.1233 kg
30 mm
mp = 7.85 A 103 B A (0.03)(0.180)(0.02) B = 0.8478 kg
20 mm
50 mm
30 mm
1
1
Ix¿ = c (0.1233)(0.01)2 d + c (0.1233)(0.02)2 + (0.1233)(0.120)2 d
2
2
+ c
1
(0.8478) A (0.03)2 + (0.180)2 B + (0.8478)(0.06)2 d
12
= 0.00719 kg # m2 = 7.19 g # m2
Ans.
Ans:
Ix′ = 7.19 g # m2
813
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*17–24.
The door has a weight of 200 lb and a center of gravity at G.
Determine how far the door moves in 2 s, starting from rest,
if a man pushes on it at C with a horizontal force F = 30 lb.
Also, find the vertical reactions at the rollers A and B.
A
F
SOLUTION
+ ©F = m(a ) ;
:
x
G x
6 ft
G
C
3 ft
200
)aG
30 = (
32.2
6 ft
B
12 ft
5 ft
aG = 4.83 ft>s2
a+©MA = ©(Mk)A;
NB(12) - 200(6) + 30(9) = (
200
)(4.83)(7)
32.2
Ans.
NB = 95.0 lb
+ c ©Fy = m(aG)y ;
NA + 95.0 - 200 = 0
Ans.
NA = 105 lb
+ )
(:
s = s0 + v0t +
s = 0 + 0 +
1 2
a t
2 G
1
(4.83)(2)2 = 9.66 ft
2
Ans.
Ans:
NB = 95.0 lb
NA = 105 lb
s = 9.66 ft
814
© 2016 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.
17–25.
The door has a weight of 200 lb and a center of gravity at G.
Determine the constant force F that must be applied to the
door to push it open 12 ft to the right in 5 s, starting from
rest. Also, find the vertical reactions at the rollers A and B.
A
F
SOLUTION
12 = 0 + 0 +
6 ft
G
C
3 ft
+ )s = s + v t + 1 a t2
(:
0
0
G
2
6 ft
B
12 ft
5 ft
1
aG(5)2
2
ac = 0.960 ft>s2
+ ©F = m(a ) ;
:
x
G x
F =
200
(0.960)
32.2
Ans.
F = 5.9627 lb = 5.96 lb
a + ©MA = ©(Mk)A ;
NB(12) - 200(6) + 5.9627(9) =
200
(0.960)(7)
32.2
Ans.
NB = 99.0 lb
+ c ©Fy = m(aG)y ;
NA + 99.0 - 200 = 0
Ans.
NA = 101 lb
Ans:
F = 5.96 lb
NB = 99.0 lb
NA = 101 lb
815
© 2016 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.
17–26.
The jet aircraft has a total mass of 22 Mg and a center of
mass at G. Initially at take-off the engines provide a thrust
2T = 4 kN and T¿ = 1.5 kN. Determine the acceleration of
the plane and the normal reactions on the nose wheel and
each of the two wing wheels located at B. Neglect the mass
of the wheels and, due to low velocity, neglect any lift
caused by the wings.
T¿
2.5 m
G
2T
2.3 m
1.2 m
B
3m
6m
A
SOLUTION
+ ©F = ma ;
:
x
x
+ c ©Fy = 0 ;
1.5 + 4 = 22aG
2By + Ay - 22(9.81) = 0
a + ©MB = ©(MK)B ;
4(2.3) - 1.5(2.5) - 22(9.81)(3) + Ay (9) = - 22aG(1.2)
Ay = 72.6 kN
Ans.
By = 71.6 kN
Ans.
aG = 0.250 m>s2
Ans.
Ans:
Ay = 72.6 kN
By = 71.6 kN
aG = 0.250 m>s2
816
© 2016 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.
17–27.
The sports car has a weight of 4500 lb and center of gravity
at G. If it starts from rest it causes the rear wheels to slip
as it accelerates. Determine how long it takes for it to reach
a speed of 10 ft/s. Also, what are the normal reactions at
each of the four wheels on the road? The coefficients of
static and kinetic friction at the road are ms = 0.5 and
mk = 0.3, respectively. Neglect the mass of the wheels.
G
2.5 ft
B
4 ft
2 ft
A
SOLUTION
a + ©MA = ©(Mk)A ; - 2NB(6) + 4500(2) =
- 4500
a (2.5)
32.2 G
+ ©F = m(a ) ;
:
x
G x
0.3(2NB) =
+ c ©Fy = m(aG)y ;
2NB + 2NA - 4500 = 0
4500
a
32.2 G
Solving,
NA = 1393 lb
Ans.
NB = 857 lb
Ans.
aG = 3.68 ft>s2
+ )
(:
v = v0 + act
10 = 0 + 3.68 t
Ans.
t = 2.72 s
Ans:
NA = 1393 lb
NB = 857 lb
t = 2.72 s
817
© 2016 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.
*17–28.
The assembly has a mass of 8 Mg and is hoisted using the
boom and pulley system. If the winch at B draws in the cable
with an acceleration of 2 m>s2, determine the compressive
force in the hydraulic cylinder needed to support the boom.
The boom has a mass of 2 Mg and mass center at G.
6m
2m
C
4m
Solution
B
sB + 2sL = l
1m
aB = - 2aL
2 = -2aL
G
60
A
D
2m
aL = - 1 m>s2
Assembly:
+ c ΣFy = may; 2T - 8 ( 103 ) (9.81) = 8 ( 103 ) (1)
T = 43.24 kN
Boom:
FCD(2) - 2 ( 103 ) (9.81)(6 cos 60°) - 2(43.24) ( 103 ) (12 cos 60°) = 0
a+ ΣMA = 0; Ans.
FCD = 289 kN
Ans:
FCD = 289 kN
818
© 2016 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.
17–29.
The assembly has a mass of 4 Mg and is hoisted using the
winch at B. Determine the greatest acceleration of the
assembly so that the compressive force in the hydraulic
cylinder supporting the boom does not exceed 180 kN. What
is the tension in the supporting cable? The boom has a mass
of 2 Mg and mass center at G.
6m
2m
B
Boom:
a + ΣMA = 0;
C
4m
Solution
180 ( 103 ) (2) - 2 ( 103 ) (9.81)(6 cos 60°) - 2T(12 cos 60°) = 0
1m
G
60
A
D
Ans.
T = 25 095 N = 25.1 kN
2m
Assembly:
+ c ΣFy = may ; 2(25 095) - 4 ( 103 ) (9.81) = 4 ( 103 ) a
a = 2.74 m>s2
Ans.
Ans:
a = 2.74 m>s2
T = 25.1 kN
819
© 2016 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.
17–30.
The uniform girder AB has a mass of 8 Mg. Determine the
internal axial, shear, and bending-moment loadings at the
center of the girder if a crane gives it an upward acceleration
of 3 m>s2.
3 m/s2
C
Solution
Girder:
+ c ΣFy = may ; 2T sin 60° - 8000(9.81) = 8000(3)
A
T = 59 166.86 N
60
4m
60
B
Segment:
+ ΣFx = max ; 59 166.86 cos 60° - N = 0
S
Ans.
N = 29.6 kN
+ c ΣFy = may ; 59 166.86 sin 60° - 4000(9.81) + V = 4000(3)
Ans.
V = 0
a+ ΣMC = Σ(Mk)C ; M + 4000(9.81)(1) - 59 166.86 sin 60°(2) = - 4000(3)(1)
M = 51.2 kN # m
Ans.
Ans:
N = 29.6 kN
V = 0
M = 51.2 kN # m
820
© 2016 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.
17–31.
z
A car having a weight of 4000 lb begins to skid and turn with
the brakes applied to all four wheels. If the coefficient of
kinetic friction between the wheels and the road is mk = 0.8,
determine the maximum critical height h of the center of
gravity G such that the car does not overturn. Tipping will
begin to occur after the car rotates 90° from its original
direction of motion and, as shown in the figure, undergoes
translation while skidding. Hint: Draw a free-body diagram
of the car viewed from the front. When tipping occurs, the
normal reactions of the wheels on the right side (or passenger
side) are zero.
2.5 ft
x
h
2.5 ft
G
y
Solution
NA represents the reaction for both the front and rear wheels on the left side.
+ ΣF = m(a ) ; 0.8N = 4000 a
d
A
x
G x
32.2 G
+ c ΣFy = m(aG)y ; NA - 4000 = 0
a + ΣMA = Σ(Mk)A; 4000(2.5) =
4000
(a )(h)
32.2 G
Solving,
NA = 4000 lb
aG = 25.76 ft>s2
Ans.
h = 3.12 ft
Ans:
h = 3.12 ft
821
© 2016 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.
*17–32.
A force of P = 300 N is applied to the 60-kg cart. Determine
the reactions at both the wheels at A and both the wheels
at B. Also, what is the acceleration of the cart? The mass
center of the cart is at G.
P
30
G
0.4 m
0.3 m
B
A
Solution
0.3 m
Equations of Motions. Referring to the FBD of the cart, Fig. a,
0.2 m
0.08 m
+ ΣF = m(a ) ; 300 cos 30° = 60a
d
x
G x
a = 4.3301 m>s2 = 4.33 m>s2 d Ans.
+c ΣFy = m(aG)y ; NA + NB + 300 sin 30° - 60(9.81) = 60(0)
(1)
a+ΣMG = 0; NB(0.2) - NA(0.3) + 300 cos 30°(0.1)
- 300 sin 30°(0.38) = 0
(2)
Solving Eqs. (1) and (2),
NA = 113.40 N = 113 N
Ans.
NB = 325.20 N = 325 N
Ans.
Ans:
a = 4.33 m>s2 d
NA = 113 N
NB = 325 N
822
© 2016 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.
17–33.
Determine the largest force P that can be applied to the
60-kg cart, without causing one of the wheel reactions,
either at A or at B, to be zero. Also, what is the acceleration
of the cart? The mass center of the cart is at G.
P
30
G
0.4 m
0.3 m
B
A
Solution
Equations of Motions. Since (0.38 m) tan 30° = 0.22 m 7 0.1 m, the line of action
of P passes below G. Therefore, P tends to rotate the cart clockwise. The wheels at A
will leave the ground before those at B. Then, it is required that NA = 0. Referring,
to the FBD of the cart, Fig. a
+ c ΣFy = m(aG)y; NB + P sin 30° - 60(9.81) = 60(0)
(1)
a+ ΣMG = 0; P cos 30°(0.1) - P sin 30°(0.38) + NB(0.2) = 0
(2)
0.3 m
0.2 m
0.08 m
Solving Eqs. (1) and (2)
Ans.
P = 578.77 N = 579 N
NB = 299.22 N
Ans:
P = 579 N
823
© 2016 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.
17–34.
The trailer with its load has a mass of 150 kg and a center of
mass at G. If it is subjected to a horizontal force of
P = 600 N, determine the trailer’s acceleration and the
normal force on the pair of wheels at A and at B. The
wheels are free to roll and have negligible mass.
G
P
1.25 m
0.25 m
SOLUTION
0.75 m
Equations of Motion: Writing the force equation of motion along the x axis,
+ ©F = m(a ) ;
:
x
G x
600 = 150a
a = 4 m>s2 :
B
0.25 m
A
600 N
0.5 m
1.25 m
Ans.
Using this result to write the moment equation about point A,
a + ©MA = (Mk)A ;
150(9.81)(1.25) - 600(0.5) - NB(2) = -150(4)(1.25)
Ans.
NB = 1144.69 N = 1.14 kN
Using this result to write the force equation of motion along the y axis,
+ c ©Fy = m(a G)y ;
NA + 1144.69 - 150(9.81) = 150(0)
Ans.
NA = 326.81 N = 327 N
Ans:
a = 4 m>s2 S
NB = 1.14 kN
NA = 327 N
824
© 2016 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.
17–35.
The desk has a weight of 75 lb and a center of gravity at G.
Determine its initial acceleration if a man pushes on it with
a force F = 60 lb. The coefficient of kinetic friction at A
and B is mk = 0.2.
F
1 ft
30
G
2 ft
A
Solution
+ ΣFx = max; 60 cos 30° - 0.2NA - 0.2NB = 75 aG
S
32.2
B
2 ft
2 ft
+ c ΣFy = may; NA + NB - 75 - 60 sin 30° = 0
a+ΣMG = 0; 60 sin 30°(2) - 60 cos 30°(1) - NA(2) + NB(2) - 0.2NA(2)
- 0.2NB(2) = 0
Solving,
aG = 13.3 ft>s2
Ans.
NA = 44.0 lb
NB = 61.0 lb
Ans:
aG = 13.3 ft>s2
825
© 2016 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.
*17–36.
The desk has a weight of 75 lb and a center of gravity at G.
Determine the initial acceleration of a desk when the man
applies enough force F to overcome the static friction at A
and B. Also, find the vertical reactions on each of the
two legs at A and at B. The coefficients of static and kinetic
friction at A and B are ms = 0.5 and mk = 0.2, respectively.
F
1 ft
30
G
2 ft
A
Solution
B
2 ft
Force required to start desk moving;
2 ft
+ ΣFx = 0; F cos 30° - 0.5NA - 0.5NB = 0
S
+ c ΣFy = 0; NA + NB - F sin 30° - 75 = 0
Solving for F by eliminating NA + NB,
F = 60.874 lb
Desk starts to slide.
+ ΣFx = m(aG)x; 60.874 cos 30° - 0.2NA - 0.2NB = 75 aG
S
32.2
+ c ΣFy = m(aG)y; NA + NB - 60.874 sin 30° - 75 = 0
Solving for aG by eliminating NA + NB,
aG = 13.58 = 13.6 ft>s2
Ans.
a+ ΣMA = Σ(MK)A; NB(4) - 75(2) - 60.874 cos 30°(3) =
-75
(13.58)(2)
32.2
NB = 61.2 lb
So that
NA = 44.2 lb
For each leg,
′
NA
=
44.2
= 22.1 lb
2
Ans.
NB′ =
61.2
= 30.6 lb
2
Ans.
Ans:
aG = 13.6 ft>s2
′
NA
= 22.1 lb
NB′ = 30.6 lb
826
© 2016 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.
17–37.
The 150-kg uniform crate rests on the 10-kg cart. Determine
the maximum force P that can be applied to the handle
without causing the crate to tip on the cart. Slipping does
not occur.
0.5 m
P
1m
Solution
Equation of Motion. Tipping will occur about edge A. Referring to the FBD and
kinetic diagram of the crate, Fig. a,
a+ ΣMA = Σ(MK)A; 150(9.81)(0.25) = (150a)(0.5)
a = 4.905 m>s2
Using the result of a and refer to the FBD of the crate and cart, Fig. b,
+ ΣF = m(a ) P = (150 + 10)(4.905) = 784.8 N = 785 N
d
x
G x
Ans.
Ans:
P = 785 N
827
© 2016 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.
17–38.
The 150-kg uniform crate rests on the 10-kg cart. Determine
the maximum force P that can be applied to the handle
without causing the crate to slip or tip on the cart. The
coefficient of static friction between the crate and cart is
ms = 0.2.
0.5 m
P
1m
Solution
Equation of Motion. Assuming that the crate slips before it tips, then Ff = ms N = 0.2 N.
Referring to the FBD and kinetic diagram of the crate, Fig. a
+ c ΣFy = may; N - 150 (9.81) = 150 (0) N = 1471.5 N
+ ΣF = m(a ) ; 0.2(1471.5) = 150 a a = 1.962 m>s2
d
x
G x
a+ΣMA = (Mk)A; 150(9.81)(x) = 150(1.962)(0.5)
x = 0.1 m
Since x = 0.1 m 6 0.25 m, the crate indeed slips before it tips. Using the result of a
and refer to the FBD of the crate and cart, Fig. b,
+ ΣFx = m(aG)x; P = (150 + 10)(1.962) = 313.92 N = 314 N
d
Ans.
Ans:
P = 314 N
828
© 2016 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.
17–39.
The bar has a weight per length w and is supported by the
smooth collar. If it is released from rest, determine the
internal normal force, shear force, and bending moment in
the bar as a function of x.
30
x
Solution
Entire bar:
ΣFx′ = m(aG)x′; wl cos 30° =
wl
(a )
g G
aG = g cos 30°
Segment:
+ ΣF = m(a ) ; N = (wx cos 30°) sin 30° = 0.433wx
d
x
G x
Ans.
+ T ΣFy = m(aG)y; wx - V = wx cos 30°(cos 30°)
Ans.
V = 0.25wx
x
x
a+ ΣMS = Σ(Mk)S; wxa b - M = wx cos 30°(cos 30°)a b
2
2
M = 0.125wx2
Ans.
Ans:
N = 0.433wx
V = 0.25wx
M = 0.125wx2
829
© 2016 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.
*17–40.
The smooth 180-lb pipe has a length of 20 ft and a negligible
diameter. It is carried on a truck as shown. Determine the
maximum acceleration which the truck can have without
causing the normal reaction at A to be zero. Also determine
the horizontal and vertical components of force which the
truck exerts on the pipe at B.
20 ft
A
5 ft
B
Solution
12 ft
+ ΣF = ma ; B = 180 a
S
x
x
x
32.2 T
+ c ΣFy = 0; By - 180 = 0
c+ΣMB = Σ(Mk)B; 180(10)a
Solving,
180
12
5
aT (10)a b
b =
13
32.2
13
Ans.
Bx = 432 lb
Ans.
By = 180 lb
2
Ans.
aT = 77.3 ft>s Ans:
Bx = 432 lb
By = 180 lb
aT = 77.3 ft>s2
830
© 2016 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.
17–41.
The smooth 180-lb pipe has a length of 20 ft and a negligible
diameter. It is carried on a truck as shown. If the truck
accelerates at a = 5 ft>s2, determine the normal reaction at
A and the horizontal and vertical components of force
which the truck exerts on the pipe at B.
20 ft
A
5 ft
B
Solution
12 ft
+ ΣF = ma ; B - N a 5 b = 180 (5)
S
x
A
x
x
13
32.2
+ c ΣFy = 0 ; By - 180 + NAa
12
b = 0
13
a+ΣMB = Σ(Mk)B; - 180(10)a
Solving,
12
180
5
(5)(10)a b
b + NA(13) = 13
32.2
13
Bx = 73.9 lb
Ans.
By = 69.7 lb
Ans.
NA = 120 lb
Ans.
Ans:
Bx = 73.9 lb
By = 69.7 lb
NA = 120 lb
831
© 2016 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.
17–42.
The uniform crate has a mass of 50 kg and rests on the cart
having an inclined surface. Determine the smallest
acceleration that will cause the crate either to tip or slip
relative to the cart. What is the magnitude of this
acceleration? The coefficient of static friction between the
crate and the cart is ms = 0.5.
0.6 m
F
1m
SOLUTION
15
Equations of Motion: Assume that the crate slips, then Ff = ms N = 0.5N.
a + ©MA = ©(Mk)A ;
50(9.81) cos 15°(x) - 50(9.81) sin 15°(0.5)
= 50a cos 15°(0.5) + 50a sin 15°(x)
(1)
+Q©Fy¿ = m(aG)y¿ ;
N - 50(9.81) cos 15° = -50a sin 15°
(2)
R + ©Fx¿ = m(aG)x¿ ;
50(9.81) sin 15° - 0.5N = -50a cos 15°
(3)
Solving Eqs. (1), (2), and (3) yields
N = 447.81 N
x = 0.250 m
a = 2.01 m>s2
Ans.
Since x 6 0.3 m , then crate will not tip. Thus, the crate slips.
Ans.
Ans:
a = 2.01 m>s2
The crate slips.
832
© 2016 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.
17–43.
1 ft
Determine the acceleration of the 150-lb cabinet and the
normal reaction under the legs A and B if P = 35 lb. The
coefficients of static and kinetic friction between the
cabinet and the plane are ms = 0.2 and mk = 0.15,
respectively. The cabinet’s center of gravity is located at G.
P
G
4 ft
SOLUTION
Equations of Equilibrium: The free-body diagram of the cabinet under the static
condition is shown in Fig. a, where P is the unknown minimum force needed to move
the cabinet. We will assume that the cabinet slides before it tips. Then,
FA = msNA = 0.2NA and FB = msNB = 0.2NB.
+ ©F = 0;
:
x
P - 0.2NA - 0.2NB = 0
(1)
+ c ©Fy = 0;
NA + NB - 150 = 0
(2)
+ ©MA = 0;
NB(2) - 150(1) - P(4) = 0
(3)
1 ft
3.5 ft
A
B
Solving Eqs. (1), (2), and (3) yields
P = 30 lb
NA = 15 lb
NB = 135 lb
Since P 6 35 lb and NA is positive, the cabinet will slide.
Equations of Motion: Since the cabinet is in motion, FA = mkNA = 0.15NA and
FB = mkNB = 0.15NB. Referring to the free-body diagram of the cabinet shown in
Fig. b,
+ ©F = m(a ) ;
:
x
G x
+ ©F = m(a ) ;
:
x
G x
+ ©MG = 0;
35 - 0.15NA - 0.15NB = a
150
ba
32.2
(4)
(5)
NA + NB - 150 = 0
NB(1) - 0.15NB(3.5) - 0.15NA(3.5) - NA(1) - 35(0.5) = 0
(6)
Solving Eqs. (4), (5), and (6) yields
a = 2.68 ft>s2
NA = 26.9 lb
Ans.
Ans.
NB = 123 lb
Ans:
a = 2.68 ft>s2
NA = 26.9 lb
NB = 123 lb
833
© 2016 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.
*17–44.
A
The uniform bar of mass m is pin connected to the collar,
which slides along the smooth horizontal rod. If the collar is
given a constant acceleration of a, determine the bar’s
inclination angle u. Neglect the collar’s mass.
L
a
u
SOLUTION
Equations of Motion: Writing the moment equation of motion about point A,
+ ©MA = (Mk)A;
mg sin u a
L
L
b = ma cos u a b
2
2
a
u = tan-1 a b
g
Ans.
Ans:
a
u = tan - 1 a b
g
834
© 2016 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.
17–45.
The drop gate at the end of the trailer has a mass
of 1.25 Mg and mass center at G. If it is supported by the
cable AB and hinge at C, determine the tension in the cable
when the truck begins to accelerate at 5 m>s2. Also, what
are the horizontal and vertical components of reaction at
the hinge C?
B
30
$
G
C
1m
1.5 m
45
Solution
a+ ΣMC = Σ(Mk)C; T sin 30°(2.5) - 12 262.5(1.5 cos 45°) = 1250(5)(1.5 sin 45°)
Ans.
T = 15 708.4 N = 15.7 kN
+ ΣF = m(a ) ; - C + 15 708.4 cos 15° = 1250(5)
d
x
x
G x
Ans.
Cx = 8.92 kN
+ c ΣFy = m(aG)y; Cy - 12 262.5 - 15 708.4 sin 15° = 0
Ans.
Cy = 16.3 kN
Ans:
T = 15.7 kN
Cx = 8.92 kN
Cy = 16.3 kN
835
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open one into another, giving a fine perspective, and they lead,
through a dozen different doorways, on to a splendid, white-tiled
verandah which runs out to the bank of the Pásig River. There is a
picturesque, moss-covered river landing on the verandah below.
There are about twenty rooms on the one floor, all of them good
sized and some of them enormous, and it took a great many servants
to keep the place in order. The floors were all of beautiful hardwoods
and it required a permanent force of six muchachos to keep them in
a proper state of polish. The Filipino method of polishing floors is
interesting. Your muchacho ties either banana leaves or some sort of
bags on his bare feet, then he skates up and down, up and down,
until the floors get so slick that he himself can hardly stand up on
them. It is easy to imagine that six boys skating together in the
spaciousness of the Palace might cut fancy figures and have a
delightful time generally, if they thought they were unobserved.
Filipinos of the muchacho class always play like children, no matter
what they are doing, and they have to be treated like children.
The Palace furniture, which must have been very fine in Spanish
days, was of red narra, or Philippine mahogany, handsomely carved
and displaying on every piece the Spanish coat-of-arms. But during
the changing Spanish régimes some one with a bizarre taste had
covered all the beautiful wood with a heavy coat of black paint. The
effect was depressingly sombre to me.
The porcelain, however, or what was left of it, was unusually good.
The Spanish coat-of-arms in beautiful colours was reproduced on
each plate against a background of a dark blue canopy. I must say
there were quite as many reminders of Spanish authority as I could
wish for and I frequently felt that some noble Don might walk in at
any moment and catch me living in his house.
But, it didn’t take us long to get settled down in our new domain,
and I soon ceased to regret the sea breezes and the salt baths of
Malate. Malacañan enjoyed a clean sweep of air from the river and
our open verandah was in many ways an improvement on the
gaudily glazed one that we had gradually become accustomed to in
the other house. The Malacañan verandah, being much of it roofless,
was of little use in the daytime, but on clear evenings it was the most
delightful spot I have ever seen. I began to love the tropical nights
and to feel that I never before had known what nights can be like.
The stars were so large and hung so low that they looked almost like
raised silver figures on a dark blue field. And when the moon shone—
but why try to write about tropical moonlight? The wonderful
sunsets and the moonlit nights have tied more American hearts to
Manila and the Philippines than all the country’s other charms
combined. And they are both indescribable.
TWO VIEWS OF MALACAÑAN
PALACE. THE FIRST PICTURE SHOWS
THE WIDE, ROOFLESS VERANDA
OVER THE PASIG RIVER
When I lived in Malate and could look out across the open, whitecapped bay to far-away Mt. Meriveles, I sometimes forgot I was in
the Tropics. But at Malacañan when we gazed down on the lowlapping Pásig, glinting in the starlight, and across the rice fields on
the other side where swaying lanterns twinkled from beneath the
outline of thatched roofs, there was little to remind us that we were
Americans or that we had ever felt any air less soothing than the soft
breeze which rustled the bamboo plumes along the bank.
Our household was in every way much enlarged on our change of
residence and circumstances. There were eight or nine muchachos in
the house, two extra Chinese helpers in the kitchen, and the staff of
coachmen and gardeners increased on even a larger scale. Our stable
of ponies multiplied to sixteen, and even then there were too few for
our various needs. It is difficult for the dweller in the Temperate
Zone to realise how small an amount of work the native of the
Tropics, either man or beast, is capable of.
We thought at first that the salary attached to the office of
Governor of the Philippines was quite splendid, but we soon gave up
any idea we might have had of saving a little of it for a rainy day. Our
rainy day was upon us. It rained official obligations which we had to
meet. The mere cost of lighting Malacañan was enough to keep a
modest family in comfort. I don’t know about conditions at the
Palace now, but I imagine they have not changed much, and I do
know that Manila is a more expensive place in which to live than it
was in my time. And yet there is serious talk of reducing the salary of
the Governor-General. It seems a pity. This would place the office in
a class with Ambassadorships which nobody but rich men can
accept. The present salary, with nice management and a not too
ambitious programme, will just about cover expenses, but I feel sorry
for the wife of the Governor who must try to do what is expected of
her on less.
My cook, who had been quite independent of me at Malate,
became at Malacañan wholly unapproachable. I don’t know why, but
so it was. He occupied quarters opening on one of the courts below
and connected with the dining-room by an outside staircase up
which I was never able to inveigle him. I had to deliver my orders
from the top of the stairs and when he had listened to just as much as
he cared to hear he would disappear through the kitchen door, and
no amount of calling would bring him back. As the kitchen was an
ante-chamber to a sort of Chinese catacombs, extending over a good
part of the basement, I never ventured to follow him and I had to
swallow my wrath as best I could.
But he was a jewel despite his eccentricities. He could produce the
most elaborate and varied buffet suppers I ever saw and I never knew
a cook who could make such a wonderful variety of cakes and fruit
tarts and cream-puffs. He took a real delight in their construction,
and for two days before a reception he would spend all his time
filling every pan in the house with patisseries elaborately iced in
every imaginable colour.
I began at once to give an afternoon reception every week and if it
hadn’t been for my disagreeable, but capable, old Ah Sing I should
have been in a constant turmoil of engagements with caterers and
confectioners. As it was, I never had to give an order, really.
“Reception Wednesday, Ah Sing,” was all that was necessary, and
except for a glance now and then to see that the muchachos were
giving the floors and the furniture a little extra polish on Wednesday
morning, the only preparations I had to make for receiving two
thousand people were to put on an embroidered muslin gown and
compose myself.
These afternoon receptions were public, our only form of
invitation being an “At Home” notice in the newspapers, and
considering the unsettled state of Manila society in those days, it is
really remarkable that we had so few unwelcome guests. There were
a great many derelicts and generally disreputable people, both
American and European, trying to better their fortunes or add to the
excitement in our agitated community, but we suffered no
unpleasant consequences from our open hospitality, though every
Wednesday the Palace was thronged and every Wednesday many
new faces appeared. Army and Navy people, civilians of every
occupation and many foreigners—Germans and British mostly—
came nearly always. I remember especially the first instalment of
American school teachers. They were, for the most part, a fine lot of
men and women who had come out with high hopes and ideals and
an enthusiastic desire to pass them on. There were some pretty girls
among them and a number of very clever looking men. I believe they
used to enjoy my parties as much as anybody in Manila. They were
homesick, no doubt, especially the girls, and I suppose the sight of so
many friendly American faces cheered them up.
The Filipinos had to have a little coaxing before they began to avail
themselves very freely of our general invitation. But by asking many
of them personally and persistently to “be sure and come
Wednesday” we prevailed on a good number to believe they were
really wanted; and after a little while there began to be as many
brown faces as white among our guests.
Speaking of school teachers reminds me that it was just about this
time that our minds were relieved of all anxiety with regard to Bob’s
and Helen’s education. My husband had wanted to send our tenyear-old son back across the Pacific and the United States, all by
himself, to his Uncle Horace’s school in Connecticut, and I had
opposed the idea with all my might without being able to offer a
satisfactory substitute plan. But now a school for American children
was opened and they were as well taught as they would have been at
home. Moreover, Bob and Helen found a large number of congenial
companions, and I don’t think I ever saw a happier set of boys and
girls. They lived out of doors and did everything that children usually
do, but their most conspicuous performance was on the Luneta in
the evenings, where they would race around the drive on their little
ponies, six abreast, or play games all over the grass plots which were
then, and always have been, maintained chiefly for the benefit of
children, both brown and white.
My husband’s change in title and station made very little
difference in the character of his duties, but it gave him increased
authority in the performance of them. The onerous necessity for
submitting legislation to an executive whose point of view was
different from that of the Commission came to an end, and he was
able to see that such laws as the Commission passed were put in
operation without delay. Under General Chaffee the feeling on the
part of the Army against the encroachments of civil government gave
way, slowly but surely, to an attitude of, at least, friendly toleration.
It was as if they said: “Well, let them alone; we know they are wrong;
but they must learn by experience, and, after all, they mean well.”
General Chaffee and General MacArthur were two quite different
types of men. General Chaffee was less precise, less analytical.
General MacArthur had always been given to regarding everything in
its “psychological” aspect and, indeed, “psychological” was a word so
frequently on his lips that it became widely popular. General Chaffee
was impetuous; he was much less formal than his predecessor both
in thought and manner, and Mr. Taft found co-operation with him
much less difficult. He made no secret of his conviction, which was
shared by most of the Army, that civil government was being
established prematurely, but he was not unreasonable about it.
He refused at first to listen to the proposition for the establishment
of a native Constabulary. This had been the Commission’s pet project
ever since they had been in the Islands, and it was a great
disappointment to them to find that the opposition which they had
encountered in the former administration was to be continued.
What they wanted was a force of several thousand Filipinos,
trained and commanded by American Army officers, either from the
regular Army or from the volunteers. The same thing had been done
with success by the British in India and the Straits Settlements, by
the Dutch in Java and by our own General Davis in Porto Rico, and
as the insurrectionary force had dwindled to a few bands and to
scattered groups of murderers and ladrones, so acknowledged by
everybody, there was no reason why a native constabulary should not
be employed to clear these out.
This plan was among the first things submitted to General Chaffee,
but he was evidently not impressed. “Pin them down with a bayonet
for at least ten years” was a favourite expression of Army sentiment
which sometimes made the Commissioners’ explanations to the
natives rather difficult.
General Wright, on behalf of the Commission, called on General
Chaffee and was much surprised to learn that he had not even read
the Constabulary bill which had been passed some time before and
held up pending the hoped for opportunity to carry it into effect.
When General Wright explained the purport of the measure General
Chaffee said,
“I am opposed to the whole business. It seems to me that you are
trying to introduce something to take the place of my Army.”
“Why, so we are,” said General Wright. “We are trying to create a
civil police force to do the police work which we understood the
Army was anxious to be relieved of. You have announced your
purpose to concentrate the Army in the interest of economy, and to
let our civil governments stand alone to see what is in them and we
consider it necessary to have a constabulary, or some such force, to
take care of the lawless characters that are sure to be in the country
after four years of war, and especially in a country where the natives
take naturally to ladronism. The Municipal police as now organised
are not able to meet all the requirements in this regard.”
“There you are,” said General Chaffee, “you give your whole case
away.”
“I have no case to give away,” replied General Wright. “We are
trying to put our provincial governments on a basis where they will
require nothing but the moral force of the military arm, and actually
to preserve law and order through the civil arm. The people desire
peace, but they also desire protection and we intend through the civil
government to give it to them.”
The Commissioner then suggested the names of some Army
officers whose peculiar tact in handling Filipinos had marked them
as the best available men for organising and training native soldiers,
but General Chaffee was not inclined to detail them for the work, so
General Wright returned to the Commission quite cast down and
communicated to his colleagues the feeling that they were to have a
continuance of the same difficulties with which they were required to
contend under the former administration.
But a peacemaker came along in the person of General Corbin. He
spent some time with General Chaffee and then came to Malacañan
to visit us. He made a hurried, but quite extensive trip through the
Islands and gave the whole situation pretty thorough inspection.
After he left, a change was found to have come over the spirit of
affairs, and it was thought that he had managed to make clear to
everybody concerned that, while there was a military arm and a civil
arm of the government in the Philippines, they represented a single
American purpose and that that purpose had been expressed by the
administration at Washington when the Commission was sent out to
do the work it was then engaged upon.
After that General Chaffee seems not only to have been amenable
to reason, but to have been imbued with a spirit of cordiality and
helpfulness which was most gratifying to the long-harassed
Commission. To facilitate co-operation, a private telephone was
installed between the offices of Mr. Taft and the Commanding
General, and it seemed to me that my husband suddenly lost some of
the lines of worry which had begun to appear in his face.
The Constabulary, as everybody knows, was eventually established
and perhaps no finer body of men, organised for such a purpose,
exists. It took a long time to get them enlisted, equipped and
properly drilled, but to-day they are a force which every man and
woman in the Philippines, of whatever nationality, colour, creed or
occupation, regards with peculiar satisfaction. They include corps
enlisted from nearly every tribe in the Islands, not excepting the
Moros and the Igorrotes. The Moro constabulario is distinguishable
from the Christian in that he wears a jaunty red fez with his smart
khaki uniform instead of the regulation cap, while the Igorrote
refuses trousers and contents himself with the cap, the tight jacket,
the cartridge belt and a bright “G-string.” To the Ifugao Igorrote
uniform is added a distinguishing spiral of brass which the natty
soldier wears just below the knee. It is difficult to imagine anything
more extraordinary than a “crack” company of these magnificent
barelegged Ifugaos going through dress-parade drill under the sharp
commands of an American officer. The Constabulary Band of eightyodd pieces, under the direction of Captain Loving, an American
negro from the Boston Conservatory of Music, is well known in
America and is generally considered one of the really great bands of
the world. All its members are Filipinos.
Press clippings and some correspondence which I have before me
remind me that even at this period there began to manifest itself in
the Taft family, and otherwheres, a mild interest in the possibility
that my husband might become President of the United States. Mr.
Taft himself treated all such “far-fetched speculation” with the
derision which he thought it deserved, but to me it did not seem at all
unreasonable. We received first a copy of the Boston Herald
containing two marked articles in parallel columns, one of which,
headed by a picture of Mr. Taft, stated that in Washington there had
been serious suggestion of his name as a Presidential candidate and
the other giving a sympathetic account of an anti-imperialistic
meeting at Faneuil Hall. We thought the two articles as “news items”
hardly warranted juxtaposition, and it seemed to us the editor was
indulging a sort of sardonic sense of humour when he placed them
so. Not that my husband was an “imperialist,” but that he was
generally so considered. Indeed, he was the most active antiimperialist of them all. He was doing the work of carrying out a
thoroughly anti-imperialistic policy, but he recognised the difference
between abandoning the Philippines to a certain unhappy fate and
guiding them to substantial independence founded on selfdependence. It took a long time to get the shouters from the
housetops to accept this interpretation of our national obligation, but
there was reassurance in the fact that where our honour is involved
Americanism can always be trusted to rise above purely partisan
politics.
Mr. Taft’s mother, who took an active and very intelligent interest
in her son’s work and who sent him letters by nearly every mail
which were filled with entertaining and accurate comment on
Philippine affairs, took the suggestion of his being a Presidential
possibility quite seriously. And she did not at all approve of it.
Having seen a number of press notices about it she sat down and
wrote him a long letter in which she discussed with measured
arguments the wisdom of his keeping out of politics. At that time the
idea appealed to nothing in him except his sense of humour. He
wrote to his brother Charles: “To me such a discussion has for its
chief feature the element of humour. The idea that a man who has
issued injunctions against labour unions, almost by the bushel, who
has sent at least ten or a dozen violent labour agitators to jail, and
who is known as one of the worst judges for the maintenance of
government by injunction, could ever be a successful candidate on a
Presidential ticket, strikes me as intensely ludicrous; and had I the
slightest ambition in that direction I hope that my good sense would
bid me to suppress it. But, more than this, the horrors of a modern
Presidential campaign and the political troubles of the successful
candidate for President, rob the office of the slightest attraction for
me. I have but one ambition, and if that cannot be satisfied I am
content to return to the practice of the law with reasonable assurance
that if my health holds out I can make a living, and make Nellie and
the children more comfortable than I could if I went to Washington.”
This letter is dated August 27, 1901, and was written on a Spanish
steamer which the Commission had taken from Aparri, on the north
coast of Luzon, after they finished the last of the long trips they had
to make for the purpose of organising civil government in the
provinces.
It was just after they returned from this trip; just when things were
at their brightest; when everything seemed to be developing so
rapidly and our hopes were running high, that we were shaken by the
appalling news of the attack on President McKinley. We had kept
luncheon waiting for Mr. Taft until it seemed useless to wait any
longer and we were at table when he came in. He looked so white and
stunned and helpless that I was frightened before he could speak.
Then he said, “The President has been shot.”
I suppose that throughout the United States the emotions of
horror and grief were beyond expression, but I cannot help thinking
that to the Americans in the Philippines the shock came with more
overwhelming force than to any one else. Mr. McKinley was our chief
in a very special sense. He was the director of our endeavours and
the father of our destinies. It was he who had sent the civil officials
out there and it was on the strength of his never failing support that
we had relied in all our troubles. It might, indeed, have been Mr.
Root in whose mind the great schemes for the development of the
islands and their peoples had been conceived, but Mr. Root exercised
his authority through the wise endorsement of the President and it
was to the President that we looked for sanction or criticism of every
move that was made. Then, too, the extraordinary sweetness of his
nature inspired in every one with whom he came in close contact a
strong personal affection, and we had reason to feel this more than
most people. Truly, it was as if the foundations of our world had
crumbled under us.
But he was not dead; and on the fact that he was strong and clean
we began to build hopes. Yet the hush which fell upon the
community on the day that he was shot was not broken until a couple
of days before he died when we received word that he was
recovering. We were so far away that we could not believe anybody
would send us such a cable unless it were founded on a practical
certainty, and our “Thank God!” was sufficiently fervent to dispel all
the gloom that had enveloped us. Then came the cable announcing
his death. I need not dwell on that.
Mr. Taft and Mr. Roosevelt knew each other very well. They had
been in Washington together years before, Mr. Taft as Solicitor
General, Mr. Roosevelt as Civil Service Commissioner, and they had
corresponded with some frequency since we had been in Manila. So,
in so far as the work in the Philippines was concerned, my husband
knew where the new President’s sympathies were and he had no
fears on that score. At the same time he was most anxious to have
Mr. Root continued as Secretary of War in order that there might not
be any delay or radical change in carrying out the plans which had
been adopted and put in operation under his direction. All activities
suffered a sort of paralysis from the crushing blow of the President’s
assassination, but the press of routine work continued. We were very
much interested in learning that a great many Filipinos, clever
politicians as they are, thought that after Mr. McKinley’s death Mr.
Bryan would become President, and that, after all, they would get
immediate independence.
Then came the awful tragedy of Balangiga. It happened only a few
days after the President died, while our nerves were still taut, and
filled us all with unspeakable horror intensified by the first actual
fear we had felt since we had been in the Philippine Islands.
Company “C” of the 9th Infantry, stationed at the town of Balangiga
on the island of Samar, was surprised at breakfast, without arms and
at a considerable distance from their quarters, and fifty of them were
massacred. About thirty fought their way bare handed through the
mob, each man of which had a bolo or a gun, and lived to tell the tale.
It was a disaster so ghastly in its details, so undreamed of under the
conditions of almost universal peace which had been established,
that it created absolute panic. Men began to go about their everyday
occupations in Manila carrying pistols conspicuously displayed, and
half the people one met could talk of nothing else but their
conviction that the whole archipelago was a smouldering volcano
and that we were all liable to be murdered in our beds any night. Of
course this made the Army officers more certain than ever that the
Islands should have remained under military control indefinitely,
and I cannot deny that, at the time, their arguments seemed to have
some foundation. It was a frightful nervous strain and it took several
months of tranquillity to restore confidence. If it had been a regular
engagement in which the Americans had sustained a reverse it could
have been accepted with some philosophy, but it was a plain
massacre of a company of defenceless men by many times their
number who had gotten into the town with the consent of the
American authorities, and in conspiracy with the local headman and
the native parish priest, on the pretext of bringing in for surrender a
band of insurrectos.
The man, Lucban, who was in command of the Samar ladrones
who committed this atrocity, is now a prominent politico in Manila,
and it is interesting to know that only last year, in a campaign
speech, he referred with dramatic intensity to “our glorious victory of
Balangiga.” He was appealing to an ignorant electorate, many of
whom, as he knew, wore the scar of the awful Katipunan “blood
pact,” but it is just to record that the average Filipino is not proud of
the Balangiga “victory.”
Shortly before these unhappy events my sister Maria was called
back to America by the illness of our mother, and I was left to face
the tragic excitements of the month of September without her
comforting companionship. By October I began to feel that I would
have to get out of the Philippine Islands or suffer a nervous
breakdown, so my husband and I agreed that it would be well for me
to “run up to China,” as they express it out there. Running up to
China at that time of year meant getting out of tropic heat into
bracing autumn weather with a nip of real winter in it, and there was
nothing that I needed more.
Mrs. Wright and Mrs. Moses were both anxious to see something
of China before leaving the Orient, and as this seemed an excellent
opportunity to make the trip, they decided to go with me. The Boxer
Insurrection had just been suppressed and the Dowager Empress
had not yet returned from the West, whither she had fled during the
siege of Peking. We were used to the alarums of war and we thought
we were likely to see more of China “from the inside” than if we
visited the country during a period of complete calm. Then there
were wonderful tales of valuable “loot” which interested us. Not
necessarily illegitimate loot, but curios and art treasures in the hands
of Chinese themselves who were selling things at ridiculously low
figures and, sometimes, with a fascinating air of great mystery. There
is some allurement in the idea of bargaining for priceless porcelains,
ivories, silks and Russian sables behind closed and double-locked
doors, in the dark depths of some wretched Chinese hovel. Our Army
officers who had helped to relieve Peking brought us stories of this
kind of adventure, and I secretly hoped that we should be able to
have just some such experience. But being the wives of American
officials I thought likely we should be “taken care of” every hour of
every twenty-four. And so we were.
We sailed to Shanghai and went from there straight to Peking,
where we became the guests of Colonel and Mrs. Robertson, who had
gone in with the American troops in the Allied Armies and were
quartered in no less a place than the Temple of Heaven. The casual
tourist looking now upon that glorious collection of ancestral shrines
would find it difficult to believe that they once served as barracks for
American soldiers. Most people who visit the Temple of Heaven find
in it an atmosphere of peace and serenity such as is achieved by few
structures in the world, and to have this deep calm invaded by
business-like “foreign-devil” troops must have ruffled the spirits of
the high gods. But the soldiers had to be quartered somewhere and
this great, clean, tree-sheltered enclosure in the heart of the Chinese
city offered ample space.
Mr. Conger was then our Minister to China, and after spending a
few very busy days sightseeing we went to the Legation to visit him.
The Legation quarter, which had been laid in ruins during the Boxer
troubles, had not yet begun to assume an aspect of orderliness, and
many were the evidences of the weeks of horror through which the
besieged foreign representatives had lived.
As the Empress Dowager and her court had not yet returned, we
hoped to be able to see all the mysteries of the Forbidden City, but
order had been restored to a point where it was possible to make the
palaces once more “forbidden,” so we were shown only enough to
whet our curiosity. But the wonderful walls and the temples, the
long, unbelievable streets and the curious life of the people were
sufficient to save us from any feeling of disappointment in our visit.
At a dinner given for us by our Minister we met a number of men and
women who had been through the siege, and I sat next to Sir Robert
Hart, of the Imperial Chinese Customs, the most interesting man,
perhaps, that the great occidental-oriental co-operation has ever
produced.
When we returned to Shanghai on our way down from Peking I
was greeted by two cablegrams. It just happened that I opened them
in the order of their coming and the first one contained the
information that my husband was very ill and said that I had better
return at once to Manila, while the second read that he was much
better and that there was no cause for alarm. There was no way of
getting to Manila for several days, because there were no boats going.
So I decided to take a trip up the Yangtse River on the house-boat
belonging to the wife of the American Consul. If I had been doing
this for pleasure instead of for the purpose of “getting away from
myself” I should have enjoyed it exceedingly, but as it was I have but
a vague recollection of a very wide and very muddy river; great
stretches of clay flats, broken here and there by little clumps of round
mounds which I knew were Chinese graves, and bordered by distant,
low hills; an occasional quaint grey town with uptilted tile roofs; and
a few graceful but dreary-looking pagodas crowning lonesome hilltops. And in addition to all of this there was a seething mass of very
dirty and very noisy humanity which kept out of our way and
regarded us with anything but friendly looks.
I had left my husband apparently perfectly well, but I subsequently
learned that the night after I left Manila he developed the first
symptoms of his illness. It was diagnosed at first as dengue fever, a
disease quite common in the Philippines which, though exceedingly
disagreeable, is not regarded as dangerous. It was about two weeks
before a correct diagnosis was made, and it was then discovered that
he was suffering from an abscess which called for a serious
emergency operation. He was taken to the First Reserve Army
hospital and the operation was performed by Dr. Rhoads, the Army
surgeon who afterward became his aide when he was President.
The children must have been much frightened. They had never
seen their father ill before, and he told me afterward that he should
never forget the way they looked as he was being carried out of
Malacañan on a stretcher borne by six stalwart American policemen.
They were all huddled together in the great hall as he passed
through, and while Bob and Charlie were gazing at the proceedings
in open-eyed astonishment, Helen was weeping.
For twenty-four hours after the operation the doctors were not at
all certain that their patient would live, nor did their anxiety end at
that time. The abscess was of long growth, the wound had to be made
a terrible one, and there was great danger of blood poisoning. Mr.
Taft rallied but a second operation was necessary. By the time I
reached Manila he was well on the way to recovery, though even then
there was no prospect of his being able to move for many weeks to
come.
He used to lie on his cot in the hospital and recite to his visitors a
verse of Kipling’s which he thought fitted his case exactly:
“Now it is not well for the white man
To hurry the Aryan brown,
For the white man riles and the Aryan smiles,
And it weareth the white man down.
And the end of the fight is a tombstone white
With the name of the late deceased,
And the epitaph drear: ‘A fool lies here
Who tried to hurry the East.’”
It was decided at once by everybody, including the doctors, Mr.
Root and President Roosevelt, that Mr. Taft must leave the Islands as
soon as he was able to travel, and there were several reasons, besides
those connected with health, why it seemed best for us to return to
the United States. The principal one was that Congress was
becoming very active with regard to Philippine matters, and as Mr.
Taft was anxious that the right kind of legislation should be passed,
he wished to go to Washington and present the facts about the
situation as he had found them during his long hand-to-hand
struggle with the problem. Mr. Root cabled him that his presence in
Washington was necessary and granted him a three months’ leave of
absence from his duties as Governor, while General Wright was
appointed vice-Governor to fill his place for the time being.
Mr. Worcester was the ranking member of the Commission, but
my husband felt that he had not quite the same talent for genially
dealing with every kind of person, whether evasive Filipino or
dictatorial Army officer, which General Wright so conspicuously
displayed, and, moreover, Mr. Worcester was entirely engrossed with
the problems of his department, which included health and
sanitation and the satisfactory adjustment of the difficulties
connected with the government of the non-Christian tribes. These
were matters which appealed to Mr. Worcester’s scientific mind and
which he vastly preferred to the uncongenial task of administering
the routine of government, so he was only too willing not to be
encumbered with the duties of Governor. This, I understand, was Mr.
Worcester’s attitude throughout his thirteen years as Secretary of the
Interior, during which time he was always the ranking Commissioner
with the first right, under a promotion system, to the Governorship
whenever a vacancy occurred in that office.
The transport Grant was assigned for our use by General Chaffee,
and we made our preparations for an extended absence.
One incident of my husband’s convalescence in the hospital I think
I must relate. In an adjoining room General Frederick Funston was
recovering from an operation for appendicitis and he was sufficiently
far advanced to be able to walk around, so he used to call on Mr. Taft
quite often. Now General Funston, for the benefit of those who have
no mental picture of him, is by no means gigantic. He has the bearing
of a seven-foot soldier, but the truth is he is not more than five feet
three or four inches in height.
One day there was an earthquake of long duration and extended
vibration which would have been sufficient to destroy Manila had it
not lacked a certain upward jerk calculated to unbalance swaying
walls. One gets used to earthquakes in the Orient in a way, but no
amount of familiarity can make the sensation a pleasant one. My
husband was alone at the time and he had decided to hold hard to his
bed and let the roof come down on him if it had to. The hospital was
a one story wooden building and he really thought he was as safe in it
as he would be anywhere. Moreover, he was quite unable to walk, so
his fortitude could hardly be called voluntary, but he had scarcely
had time to steel himself for the worst when his door was thrown
open and in rushed General Funston.
“We must carry out the Governor!” he shouted; “we must carry out
the Governor!”
“But how are you going to do that, General?” asked Mr. Taft.
He knew quite well that General Funston, in his weakened
condition, would be incapable of carrying an infant very far.
“Oh, I have my orderly with me,” responded the doughty General,
and by this time he had begun to get a firm grasp on the mattress
while behind him hurried a soldier, shorter even than his chief, but
with the same look of dauntless determination in his eye.
In spite of the straining on the rafters, Mr. Taft burst out laughing
and flatly refused to let them try to move him. Fortunately for them
all the upward jerk necessary to bring down the roof didn’t occur, so
there is no way of telling whether or not, for once in his life, General
Funston started something that he couldn’t finish.
We sailed from Manila on Christmas Eve, 1901, and, much as I had
enjoyed my life and experiences in our new world of the Philippines,
I was glad to see the tropic shores fade away and to feel that we were
to have a few months in our own land and climate, and among our
own old friends, before I sighted them again.
CHAPTER XI
A TRIP TO ROME
The winter of 1902, the greater part of which we spent in
Cincinnati, is memorable only as a period of bereavement and
protracted illnesses. Perhaps such a record has no place in a
narrative wherein it is my wish to dwell on pleasant memories only,
or, at least, to touch as lightly as possible upon those incidents
which, for one’s peace, may better be forgotten, but a whole winter
filled with grief and worry is not so easily torn from the leaves of the
calendar rolled back.
In the first place, when I left Manila in December, 1901, I was very
near to a nervous breakdown. This was due to the long strain of a
peculiarly exacting official life in a trying climate, and an added
weight of uneasiness about my husband’s illness.
Then, too, my mother was very ill. She had suffered a stroke of
paralysis the year before from which she had never rallied and I was
extremely anxious to be with her in Cincinnati.
When we arrived in San Francisco a terrible mid-winter storm was
sweeping the country from one end to the other and we were strongly
advised to delay our trip across the continent, but we were both eager
to go on so we started East at once over the Union Pacific.
When we passed Ogden we found ourselves in the midst of the
worst blizzard I ever saw. The snow piled up ahead of us, delaying us
hour by hour; the bitter wind fairly shook the heavy train; and to
turn mere discomfort into misery the water pipes in the cars froze
solid and we were left without heat of any kind. There was nothing to
do but to go to bed; but even so, with all the blankets available piled
on top of us, we shivered through interminable hours while the train
creaked and puffed and struggled over the icy tracks.
When we reached Omaha I received a telegram telling me that my
mother had died the day before, and I found it no longer possible to
brace myself against the inevitable collapse. We hurried on to
Cincinnati and arrived in time for my mother’s funeral, but I was too
ill to be present. It was two months before I began to recover.
In the meantime Mr. Taft left us and went on to Washington for
consultation with the President and Mr. Root and to appear before
the Philippine Committees of the House and Senate which were then
conducting minute inquiries into conditions in the Islands
preparatory to passing a much needed governmental bill. For a
whole month he was subjected to a hostile cross-examination, but he
was able to place before the Committees more first-hand and
accurate information on the subject of their deliberations than they
had theretofore received. This was exactly what he wanted to come to
the United States for, and he would greatly have enjoyed it had he
been in his usual form, but he was not. During his stay in
Washington he was the guest of Secretary and Mrs. Root and only
their friendly care and solicitude enabled him to continue so long. In
March he was compelled to return to Cincinnati for another
operation, the third in five months. Everything considered, it seemed
to me the Taft family had fallen upon evil days.
However, the weeks passed, I began to improve, and as soon as my
husband had fairly set his feet on earth again we began to make plans
for our return to the Philippines. There could be no thought of
abandoning the work in the Islands just when it was beginning to
assume an ordered and encouraging aspect, nor was it possible just
then to shift the responsibility to other shoulders. This would have
been too much like “changing horses in the middle of a stream.”
My husband was able while he was in Washington to present to
President Roosevelt and Secretary Root a very clear outline of
Philippine affairs, together with such details as could never be
conveyed by cable, and the inevitable conclusion reached was that no
solution of the problem was possible which did not include the
settlement of the Friar controversy. The four monastic orders, the
Franciscan, the Dominican, the Augustinian and the Recoleto, which
held four hundred thousand acres of the best agricultural land in the
Islands, had won the lasting enmity of the Filipino people and it was
absolutely impossible to establish permanent peace while the Friars
remained and persisted in an attempt to return to their parishes.
Hundreds of them were living in practical imprisonment in the
monasteries of Manila, and that they should not be allowed to return
to their churches throughout the Islands, from which they had been
driven, was the one stand taken by the Filipinos from which they
could not by any form of persuasion be moved.
The solution of the difficulty proposed by Mr. Taft and his
colleagues in the Philippine government was that the United States
purchase the Friars’ lands and turn them into a public domain on the
condition that the orders objected to by the people be withdrawn
from the Islands.
As soon as President Roosevelt recognised the importance of
accomplishing these things he decided, with characteristic
directness, that somebody should go at once to Rome and open
negotiations with the Vatican, and after considering various men for
this delicate mission he concluded that Mr. Taft was the man best
fitted to undertake it.
The prospect of another novel experience was exceedingly
gratifying to me and I began at once to look forward with interest to a
renewal of my acquaintance with Rome and to the trip back to the
East by the Suez Canal, the Red Sea and the Indian Ocean which,
according to Kipling, “sits an’ smiles, so sof’, so bright, so bloomin’
blue.” So my feet no longer lagged in my preparations for a long trip
with my three children and another extended residence in the
tropics.
To assist Mr. Taft in his negotiations with the Vatican, and to make
up a dignified and formidable looking Commission, the President
appointed Bishop O’Gorman of the Catholic diocese of South Dakota,
and General James F. Smith, at that time a member of the Philippine
judiciary and in later years Philippine Commissioner and GovernorGeneral of the Islands. His rank of General he attained as an officer
of volunteers in the Army of Pacification in the Philippines, but, a
lawyer in the beginning, after he was appointed to the Bench he
became known as Judge Smith, and Judge we always called him. He
is an Irish Catholic Democrat and a man of very sane views and
exceptional ability. Major John Biddle Porter was made SecretaryInterpreter to the Commission, and Bishop Brent, Episcopal Bishop
of the Philippines, on his way to Manila, decided to go with Mr. Taft,