hw01 - Problem 1.11 The kinetic energy of a particle of...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 1.11 The kinetic energy of a particle of mass m is deFned to be 1 2 mv 2 , where v is the magnitude of the particle’s velocity. If the value of the kinetic energy of a particle at a given time is 200 when m is in kilograms and v is in meters per second, what is the value when m is in slugs and v is in feet per second? Solution: ± 200 kg-m 2 s 2 ²± 0 . 0685 slug 1 kg 1 ft 0 . 3048 m ² 2 = 147 . 46 = 147 slug-ft 2 s 2 Problem 1.12 The acceleration due to gravity at sea level in SI units is g =9 . 81 m/s 2 . By converting units, use this value to determine the acceleration due to gravity at sea level in U.S. Customary units. Solution: Use Table 1.2. The result is: g . 81 ± m s 2 1 ft 0 . 3048 m ² =32 . 185 ... ± ft s 2 ² . 2 ± ft s 2 ² Problem 1.13 A furlong per fortnight is a facetious unit of velocity, perhaps made up by a student as a satir- ical comment on the bewildering variety of units engi- neers must deal with. A furlong is 660 ft (1/8 mile). A fortnight is 2 weeks (14 days). If you walk to class at 2 m/s, what is your speed in furlongs per fortnight to three signiFcant digits? Solution: Convert the units using the given conversions. Record the Frst three digits on the left, and add zeros as required by the number of tens in the exponent. The result is: ± 5 ft s 1 furlong 660 ft 3600 s 1 hr 24 hr 1 day 14 day 1 fortnight ² = ± 9160 furlongs fortnight ² Problem 1.14 The cross-sectional area of a beam is 480 in 2 . What is its cross-section in m 2 ? Solution: Convert units using Table 1.2. The result: 480 in 2 ± 1 ft 12 in ² 2 ± 0 . 3048 m 1 ft ² 2 =0 . 30967 m 2 . 310 m 2
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Problem 1.15 At sea level, the weight density (weight per unit volume) of water is approximately 62 . 4 lb/ft 3 . 1 lb =4 . 448 N, 1 ft =0 . 3048 m , and g =9 . 81 m/s 2 . Using only this information, determine the mass density for water in kg/m 3 . Solution: Get wt. density in N/m 3 Frst. ± 62 . 4 lb ft 3 ²± 4 . 448 N 1 lb 1 ft 0 . 3048 m ² 3 = 9801 . 77 N m 3 (carry extra signiFcant Fgures till end—then round) weight = mass · g mass = weight g ± 9801 . 77 N m 3 s 2 9 . 81 m ² = 999 ± N-s 2 m 1 m 3 ² = 999 kg/m 3 Problem 1.16 A pressure transducer measures a value of 300 lb/in 2 . Determine the value of the pressure in pascals. A pascal (Pa) is one newton per meter squared. Solution: Convert the units using Table 1.2 and the deFnition of the Pascal unit. The result: 300 ± lb in 2 4 . 448 N 1 lb 12 in 1 ft ² 2 ± 1 ft 0 . 3048 m ² 2 =2 . 0683 ... (10 6 ) ± N m 2 ² . 07(10 6 ) Pa Problem 1.17 A horsepower is 550 ft-lb/s. A watt is 1 N-m/s. Determine the number of watts generated by (a) the Wright brothers’ 1903 airplane, which had a 12-horsepower engine; (b) a modern passenger jet with a power of 100,000 horsepower at cruising speed. Solution: Convert units using inside front cover of textbook derive the conversion between horsepower and watts. The result ( a )12 hp ± 746 watt 1 hp ² = 8950 watt ( b )1 0 5 hp ± 746 watt 1 hp ² =7 . 46(10 7 ) watt
Background image of page 2
Problem 2.5 The magnitudes | F A | = 100 lb and | F B | = 140 lb. If α can have any value, what are the minimum and maximum possible values of the magni- tude of the sum of the forces F = F A + F B , and what are the corresponding values of α ?
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 22

hw01 - Problem 1.11 The kinetic energy of a particle of...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online