SECTION 7. 6
7.6
6. The masses m 1 4 and m 2 8 are located at the points
5x 2 1 pounds at a point x feet from the origin. Find the
work done in moving the particle from the origin to a distance
of 10 ft.
P11, 2 and P22, 4. Find the moments Mx and My and the
center of mass of the system.
7–8 ■ Sketch the region bounded by the curves, and visually estimate the location of the centroid. Then find the exact coordinates
of the centroid.
2. A uniform cable hanging over the edge of a tall building is
40 ft long and weighs 60 lb. How much work is required to
pull 10 ft of the cable to the top?
7. y x 2 ,
The end of a tank containing water is vertical and has the
indicated shape. Explain how to approximate the hydrostatic force
against the end of the tank by a Riemann sum. Then express the
force as an integral and evaluate it.
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4.
10 m
5m
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4 ft
h
b
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x2
x9
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Find the centroid of the region bounded by the curves.
10. y ln x,
5.
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y 0,
9. y sin 2x,
4 ft
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9–10
water
level
water
level
y 0,
8. y sx ,
6 ft
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1
S Click here for solutions.
1. A particle is moved along the x-axis by a force that measures
3.
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APPLICATIONS TO PHYSICS AND ENGINEERING
A Click here for answers.
3–5
APPLICATIONS TO PHYSICS AND ENGINEERING
y 0,
y 0,
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x 2
x 0,
xe
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2
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SECTION 7. 6
APPLICATIONS TO PHYSICS AND ENGINEERING
7. 6
ANSWERS
E Click here for exercises.
1. 5030
ft-lb
3
2. 525 ft-lb
6
3. 1.23 × 10 N
3
4. 1.56 × 10 lb
2
5. 13 ρgbh
6. 40, 12, 1, 10
3
7. (1.5, 1.2)
8.
9.
Copyright © 2013, Cengage Learning. All rights reserved.
10.
27
5
π
4
,
, 98
π
2
8
e +1 e−2
,
4
2
S Click here for solutions.
SECTION 7. 6
7. 6
APPLICATIONS TO PHYSICS AND ENGINEERING
■
3
SOLUTIONS
E Click here for exercises.
4. Place an x-axis as in Problem 3. Using similar
1. By Equation 4,
b
W =
triangles,
f (x) dx
a
10 2
= 0 5x + 1 dx
10
= 53 x3 + x 0 = 5000
+ 10
3
5030
3
=
of the ith rectangular strip is 46 x∗i ∆x. The pressure on the
strip is δ (x∗i − 2) [δ = ρg = 62.5 lb/ft3 ] and the hydrostatic
force is δ (x∗i − 2) 46 x∗i ∆x.
n
6
δ (x∗i − 2) 46 x∗i ∆x = 2 δ (x − 2) 23 x dx
F = lim
ft-lb
n→∞ i=1
∗
2. Each part of the top 10 ft of cable is lifted a distance xi equal
to its distance from the top. The cable weighs
60
40
= 1.5 lb/ft,
so the work done on the ith subinterval is 32 x∗i ∆x. The
remaining 30 ft of cable is lifted 10 ft. Thus,
W =
=
=
=
lim
n 3
n→∞ i=1
10
0
3
3
x dx
2
2
x∗i ∆x +
+
40
3
10 2
3
2
· 10 ∆x
· 10 dx
10
x2 0 + [15x]40
10
4
3
4
4 ft wide
w ft wide
= ∗
, so w = 46 x∗i and the area
6 ft high
xi ft high
(100) + 15 (30)
= 75 + 450
6
6
= 23 δ 2 x2 − 2x dx = 23 δ 13 x3 − x2 2
= 23 δ 36 − − 43 = 224
δ ≈ 1.56 × 103 lb
9
5. This is like Problem 4, except that the area of the ith
rectangular strip is
δx∗i .
b ∗
x ∆x and the pressure on the strip is
h i
n
b
ρgx∗i x∗i ∆x =
n→∞ i=1
h
h
ρgb 3
= 13 ρgbh2
=
x
3h
0
F =
lim
h
0
ρgx ·
b
x dx
h
= 525 ft-lb
3. In the middle of the figure, draw a vertical x-axis that
increases in the downward direction. The area of the ith
rectangular strip is 2 100 − (x∗i )2 ∆x and the pressure on
the strip is ρg (x∗i − 5) [ρ = 1000 kg/m3 and
g = 9.8 m/s2 ]. Thus, the hydrostatic force on the ith strip is
the product ρg (x∗i − 5) 2 100 − (x∗i )2 ∆x.
n
F = lim
ρg (x∗i − 5) 2 100 − (x∗i )2 ∆x
6. m1 = 4, m2 = 8; P1 (−1, 2), P2 (2, 4).
m=
2
i=1
n→∞ i=1
√
ρg (x − 5) · 2 100 − x2 dx
5
10 √
10 √
100 − x2 dx
= ρg 5 2x 100 − x2 dx − 10ρg 5
3/2 10
30
= −ρg 23 100 − x2
5
1 √
10
− 10ρg 2 x 100 − x2 + 50 sin−1 (x/10) 5
√
= 23 ρg (75)3/2 − 10ρg 50 π2 − 52 75 − 50 π6
√
≈ 1.23 × 106 N
= 250ρg 3 2 3 − 2π
3
Copyright © 2013, Cengage Learning. All rights reserved.
=
10
Mx =
My =
mi = m1 + m2 = 12.
2
i=1
2
i=1
mi yi = 4 · 2 + 8 · 4 = 40;
mi xi = 4 · (−1) + 8 · 2 = 12;
x = My /m = 1 and y = Mx /m = 10
, so the center of
3
10 mass of the system is (x, y) = 1, 3 .
4
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SECTION 7. 6
2
2
APPLICATIONS TO PHYSICS AND ENGINEERING
1
x3
2
= 83 ,
2
2
x = A−1 0 x · x2 dx = 38 14 x4 0 = 38 · 4 = 32 ,
2
2 2
3
y = A−1 0 21 x2 dx = 38 · 12 15 x5 0 = 16
· 32
=
5
3 6
Centroid (x, y) = 2 , 5 = (1.5, 1.2)
7. A = 0 x dx =
3
9. From the figure we see that x = π
4
0
6
5
(halfway from x = 0 to π2 ). Now
π/2
= − 12 (−1 − 1) = 1
A = 0 sin 2x dx = − 12 [cos 2x]π/2
0
so
π/2 1
π/2
1
y= A
sin2 2x dx = 11 0 12 · 12 (1 − cos 4x) dx
2
0
π/2
= 14 x − 14 sin 4x 0 = π8 .
Centroid (x, y) = π4 , π8 .
e
e
10. A = 1 ln x dx = [x ln x − x]1 = 0 − (−1) = 1,
1
A
1
x=
9
9√
8. A = 0
x dx = 23 x3/2 =
x=
1
A
9 √
x x dx =
0
1
18
0
2 5/2
x
5
2
3
9
0
=
1
18
·
9
√ 2
1
( x) dx = 18
· 12 12 x2 0 =
, 9 = (5.4, 1.125)
Centroid (x, y) = 27
5 8
y=
1
A
9
1
0 2
2
5
· 243 =
81
72
= 98 .
27
,
5
e
x ln x dx =
1
1
2
x2 ln x − 14 x2
e
1
e2 + 1
e2 − 14 e2 − − 14 =
4
1 e (ln x)2
1 e
y=
(ln x)2 dx. To evaluate
dx =
A 1
2
2 1
(ln x)2 dx, take u = ln x and dv = ln x dx, so that
du = 1/x dx and v = x ln x − x. Then
(ln x)2 dx = x (ln x)2 − x (ln x) − (x ln x − x) x1 dx
= x (ln x)2 − x (ln x) − (ln x − 1) dx
=
· 27 = 18,
2
= x (ln x)2 − x ln x − x ln x + x + x + C
= x (ln x)2 − 2x ln x + 2x + C
Thus
y =
1
2
=
1
2
e
x (ln x)2 − 2x ln x + 2x 1
[(e − 2e + 2e) − (0 − 0 + 2)] =
Copyright © 2013, Cengage Learning. All rights reserved.
So (x, y) =
e2 + 1 e − 2
,
.
4
2
e−2
2