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Solução_Calculo_Stewart_6e

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F.<br />

TX.10 SECTION 8.3 APPLICATIONS TO PHYSICS AND ENGINEERING ¤ 363<br />

8.3 Applications to Physics and Engineering<br />

1. The weight density of water is δ =62.5lb/ft 3 .<br />

(a) P = δd ≈ (62.5lb/ft 3 )(3 ft) = 187.5lb/ft 2<br />

(b) F = PA ≈ (187.5lb/ft 2 )(5 ft)(2 ft) = 1875 lb.<br />

(A is the area of the bottom of the tank.)<br />

(c) As in Example 1, the area of the ith strip is 2(∆x) and the pressure is δd = δx i.Thus,<br />

F = 3<br />

0 δx · 2 dx ≈ (62.5)(2) 3<br />

0 xdx=125 1<br />

2 x2 3<br />

0 = 125 9<br />

2<br />

<br />

=562.5lb.<br />

In Exercises 3–9, n is the number of subintervals of length ∆x and x ∗ i is a sample point in the ith subinterval [x i−1 ,x i ].<br />

3. Set up a vertical x-axisasshown,withx =0at the water’s surface and x increasing in the<br />

downward direction. Then the area of the ith rectangular strip is 6 ∆x and the pressure on<br />

the strip is δx ∗ i (where δ ≈ 62.5lb/ft 3 ). Thus, the hydrostatic force on the strip is<br />

δx ∗ <br />

i · 6 ∆x and the total hydrostatic force ≈ n δx ∗ i · 6 ∆x. The total force<br />

F = lim<br />

n<br />

n→∞ i=1<br />

i=1<br />

δx ∗ i · 6 ∆x = 6<br />

δx · 6 dx =6δ 6<br />

xdx=6δ 1<br />

x2 6<br />

=6δ(18 − 2) = 96δ ≈ 6000 lb<br />

2 2 2 2<br />

5. Set up a vertical x-axis as shown. The base of the triangle shown in the figure<br />

has length 3 2 − (x ∗ i )2 ,sow i =2 9 − (x ∗ i )2 ,andtheareaoftheith<br />

rectangular strip is 2 9 − (x ∗ i )2 ∆x. Theith rectangular strip is (x ∗ i − 1) m<br />

below the surface level of the water, so the pressure on the strip is ρg(x ∗ i − 1).<br />

The hydrostatic force on the strip is ρg(x ∗ i − 1) · 2 9 − (x ∗ i )2 ∆x and the total<br />

<br />

force on the plate ≈ n ρg(x ∗ i − 1) · 2 9 − (x ∗ i )2 ∆x. The total force<br />

F =lim n <br />

i=1<br />

i=1<br />

ρg(x ∗ i − 1) · 2 9 − (x ∗ i )2 ∆x =2ρg 3<br />

1 (x − 1) √ 9 − x 2 dx<br />

=2ρg 3<br />

x √ 9 − x<br />

1 2 dx − 2ρg 3<br />

√ <br />

<br />

1 9 − x2 dx 30 3 x<br />

=2ρg − 1 (9 − √ x<br />

3<br />

3 x2 ) 3/2 − 2ρg 9 − x2 + 9<br />

1 2<br />

2 sin−1 3 1<br />

=2ρg √ <br />

0+ 1 3 8 8 − 2ρg 0+<br />

9 · π<br />

2 2 − 1<br />

√<br />

2 8+<br />

9<br />

sin−1 <br />

1<br />

2 3<br />

√<br />

= 32<br />

3 2 ρg −<br />

9π<br />

ρg +2√ 2<br />

2 ρg +9 sin −1 <br />

1<br />

3 ρg = 38<br />

3<br />

≈ 6.835 · 1000 · 9.8 ≈ 6.7 × 10 4 N<br />

√<br />

2 −<br />

9π<br />

2 +9sin−1 1<br />

3<br />

<br />

ρg<br />

Note: If you set up a typical coordinate system with the water level at y = −1,thenF = −1<br />

−3 ρg(−1 − y)2 9 − y 2 dy.

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