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p0112 - F =(1 kslug X 10 3 slug kslug ~(1 AU w 93 10 6 mi...

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1.12. CHAPTER 1, PROBLEM 12 13 1.12 Chapter 1, Problem 12 Problem: What force is required to impart an acceleration of 1 AU/yr 2 to a 1 kiloslug mass? AU denotes Astronomical Unit ( 93 · 10 6 miles) and yr denotes year ( 365 1 4 days). Solution: Since force ( F ) has units of mass ( M ) times length ( L ) divided by time ( T ) squared, we can say F = ML T 2 So, noting that there are 10 3 slugs in a kiloslug, 93 · 10 6 miles in 1 AU, 5280 feet in a mile, 365 . 25 days in a year, 24 hours in a day, and 3600 seconds in an hour, there follows,
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Unformatted text preview: F = (1 kslug) X 10 3 slug kslug ~ (1 AU) w 93 · 10 6 mi AU W X 5280 ft mi ~ ^ (1 yr) X 365 . 25 day yr ~ X 24 hr day ~ w 3600 sec hr W ± 2 = 4 . 9104 · 10 14 slug · ft [3 . 15576 · 10 7 sec] 2 = 0 . 493 slug · ft sec 2 Thus, since 1 pound equals 1 slug · ft/sec 2 , we conclude that F = 0 . 493 lb...
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