Solucionario Dinamica 5 Ed. Bedford

Problem 12.1 The value of π is 3.1415962654. . . . . If C is the circumference of a circle and r is its radius, determine the value of r/C to four significant digits.

Solution:
C = 2π r ⇒ 1 r = = 0.159154943. C 2π r = 0.1592 C

To four significant digits we have

Problem 12.2 The base of natural logarithms is e = 2.718281828 . . . (a) (b) (c) Express e to five significant digits. Determinethe value of e2 to five significant digits. Use the value of e you obtained in part (a) to determine the value of e2 to five significant digits.

Solution: The value of e is: e = 2.718281828
(a) (b) (c) To five significant figures e = 2.7183 e2 to five significant figures is e2 = 7.3891 Using the value from part (a) we find e2 = 7.3892 which is not correct in the fifth digit.

[Part (c) demonstrates thehazard of using rounded-off values in calculations.] Problem 12.3 A machinist drills a circular hole in a panel with a nominal radius r = 5 mm. The actual radius of the hole is in the range r = 5 ± 0.01 mm. (a) To what number of significant digits can you express the radius? (b) To what number of significant digits can you express the area of the hole? Solution:
(a) The radius is in the range r1 =4.99 mm to r2 = 5.01 mm. These numbers are not equal at the level of three significant digits, but they are equal if they are rounded off to two significant digits. Two: r = 5.0 mm (b)
2 The area of the hole is in the range from A1 = π r1 = 78.226 m2 2 = 78.854 m2 . These numbers are equal only if rounded to A2 = π r2 to one significant digit:

5 mm

One: A = 80 mm2

Problem 12.4 The opening inthe soccer goal is 25 ft wide and 8 ft high, so its area is 24 ft × 8 ft = 192 ft2 . What is its area in m2 to three significant digits?

Solution:
A = 192 ft2 1m 3.281 ft
2

= 17.8 m2

A = 17.8 m2

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Problem 12.5 The Burj Dubai, scheduled for completion in 2008, will be the world’s tallest building with a height of 705 m. The area of its ground footprint will be 8000 m2 . Convert its height and footprint area to U.S. customary units to threesignificant digits.

Solution:
h = 705 m 3.281 ft 1m = 2.31 × 103 ft
2

A = 8000 m2

3.218 ft 1m

= 8.61 × 104 ft2

h = 2.31 × 103 ft,

A = 8.61 × 104 ft2

Problem 12.6 Suppose that you have just purchased a Ferrari F355 coupe and you want to know whether you can use your set of SAE (U.S. Customary Units) wrenches to work on it. You have wrenches with widths w = 1/4 in, 1/2 in, 3/4in, and 1 in, and the car has nuts with dimensions n = 5 mm, 10 mm, 15 mm, 20 mm, and 25 mm. Defining a wrench to fit if w is no more than 2% larger than n, which of your wrenches can you use?

Solution: Convert the metric size n to inches, and compute the
percentage difference between the metric sized nut and the SAE wrench. The results are: 5 mm 1 inch 25.4 mm = 0.19685.. in, 0.19685 − 0.25 1000.19685 = −27.0% 10 mm 1 inch 25.4 mm 1 inch 25.4 mm 1 inch 25.4 mm 1 inch 25.4 mm = 0.3937.. in, 0.3937 − 0.5 100 = −27.0% 0.3937 0.5905 − 0.5 100 = +15.3% 0.5905 0.7874 − 0.75 100 = +4.7% 0.7874 0.9843 − 1.0 100 = −1.6% 0.9843

15 mm

= 0.5905.. in,

n
20 mm

= 0.7874.. in,

25 mm

= 0.9843.. in,

A negative percentage implies that the metric nut is smaller than the SAE wrench; apositive percentage means that the nut is larger then the wrench. Thus within the definition of the 2% fit, the 1 in wrench will fit the 25 mm nut. The other wrenches cannot be used.

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