statics, chap 01 - bedford

5/24/2018 Statics, Chap 01 - BedFord

1/30

Problem 1.1 Express the fractions 13

and 23

to threesignificant digits.

Solution:

1/3 D 0.3333. .D 0.333

2/3 D 0.6666. .D 0.667

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2/30

Problem 1.2 The base of natural logarithms is e D2.718281828 . . .

(a) Express e to five significant digits.(b) Determine the value of e2 to five significant digits.(c) Use the value of eyou obtained in part (a) to deter-

mine the value of e2 to five significant digits.

[Part (c) demonstrates the hazard of using rounded-offvalues in calculations.]

Solution: The value of e is: e D 2.718281828

(a) To five significant figures e D 2.7183

(b) e2 to five significant figures is e2 D 7.3891

(c) Using the value from part (a) we find e2 D 7.3892 which is

not correct in the fifth digit.




3/30

Problem 1.3 A machinist drills a circular hole in apanel with radius rD 5 mm. Determine the circumfer-ence C and area A of the hole to four significant digits.

Solution:

C D 2rD 10 D 31.42 mm

A D r2 D 25 D 78.54 mm2




4/30

Problem 1.4 The opening in a soccer goal is 24 ftwide and 8 ft high. Use these values to determine itsdimensions in meters to three significant digits.

Solution: The conversion between feet and meters, found inside

the front cover of the textbook, is 1 m D 3.281 ft. The goal width,

w D 24 ft

1 m

3.281 ft

D 7.3148 m D 7.31 m.

The goal height is given by

h D 8 ft

1 m

3.281 ft

D 2.438 m D 2.44 m.




5/30

Problem 1.5 The coordinates (in meters) of point Aare xAD 3, yAD 7, and the coordinates of point B arexBD 10, yBD 2. Determine the length of the straightline from A to B to three significant digits.

x

y

A

B

Solution: The length is

L D

10 32 C 2 72 m Dp

74 m D 8.602325 m

to three significant figures this is

L D 8.60 m

c

2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they



6/30

Problem 1.6 Suppose that you have just purchaseda Ferrari F355 coupe and you want to know whetheryou can use your set of SAE (U.S. Customary Units)wrenches to work on it. You have wrenches with widthsw D 1/4 in, 1/2 in, 3/4 in, and 1 in, and the car has nutswith dimensions n D 5 mm, 10 mm, 15 mm, 20 mm,and 25 mm. Defining a wrench to fit if w is no morethan 2% larger than n, which of your wrenches can youuse?

n

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

D 0.19685.. in,

0.19685 0.25

0.19685

100

D 27.0%

10 mm

1 inch

25.4 mm

D 0.3937.. in,

0.3937 0.5

0.3937

100 D 27.0%

15 mm

1 inch

25.4 mm

D 0.5905.. in,

0.5905 0.5

0.5905

100 D C15.3%

20 mm

1 inch

25.4 mm

D 0.7874.. in,

0.7874 0.75

0.7874

100 D C4.7%

25 mm

1 inch

25.4 mm

D 0.9843.. in,

0.9843 1.0

0.9843

100 D 1.6%

A negative percentage implies that the metric nut is smaller than the

SAE wrench; a positive 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.




7/30

Problem 1.7 On August 20, 1974, Nolan Ryan threwthe first baseball pitch measured at over 100 mi/h. Themeasured speed was 100.9 mi/h. Determine the speed ofthe pitch to four significant digits (a) in ft/s; (b) in km/h.

Solution:

(a) v D 100.9mi

h

5280 ft

1 mi

1 h

3600 s

D 148.0

ft

s

(b)v D 100.9

mi

h

5280 ft

mi

0.3048 m

ft

1 km

1000 m

D 162.4 km/h




8/30

Problem 1.8 On March 18, 1999, an experimentalMaglev (magnetic levitation) train in Japan reached amaximum speed of 552 km/h. What was its velocity inmi/h to three significant digits?

Solution:

v D 552km

h

1000 m

km

1 ft

0.3048 m

1 mi

5280 ft

D 343 mi/h




9/30

Problem 1.9 In May, 1963, in the last flight ofProject Mercury, Astronaut L. Gordon Cooper traveleda distance of 546,167 miles in 1 day, 10 hours, 19minutes, and 49 seconds. Determine his average speed(the distance traveled divided by the time required) tothree significant digits (a) in mi/h; (b) in km/h.

Solution:

(a)v D

546167 mi34C

19

60C

49

3600

h

D 15,900 mi/h

(b)v D 15,900

mi

h

5280 ft

1 mi

0.3048 m

ft

1 km

1000 m

D 25,600 km/h




10/30

Problem 1.10 Engineers who study shock wavessometimes express velocity in millimeters per micro-second (mm/s). Suppose the velocity of a wavefrontis measured and determined to be 5 mm/s. Determineits velocity: (a) in m/s; (b) in mi/s.

Solution: Convert units using Tables 1.1 and 1.2. The results:

(a) 5

mm

s

1 m

1000 mm

106 s

1 s

D 5000

ms

.

Next, use this result to get (b):

(b) 5000m

s

1 ft0.3048 m

1 mi

5280 ft

D 3.10685 . . .

mi

s

D 3.11mi

s




11/30

Problem 1.11 The kinetic energy of a particle of massmis defined to be 1

2 mv

2, where v is the magnitude of theparticles velocity. If the value of the kinetic energy ofa particle at a given time is 200 when m is in kilogramsand vis in meters per second, what is the value when mis in slugs and v is in feet per second?

Solution:

200

kg-m2

s2

0.0685 slug

1 kg

1 ft

0.3048 m

2

D 147.46 D 147slug-ft2

s2




12/30

Problem 1.12 The acceleration due to gravity at sea

level in SI units is g D 9.81 m/s2. By converting units,use this value to determine the acceleration due to gravityat sea level in U.S. Customary units.

Solution: Use Table 1.2. The result is:

g D 9.81m

s2

1 ft0.3048 m

D 32.185 . . .

ft

s2

D 32.2

ft

s2




13/30

Problem 1.13 A furlong per fortnight is a facetiousunit of velocity, perhaps made up by a student as asatirical comment on the bewildering variety of unitsengineers must deal with. A furlong is 660 ft (1/8 mile).A fortnight is 2 weeks (14 days). If you walk to classat 2 m/s, what is your speed in furlongs per fortnight tothree significant digits?

Solution: Convert the units using the given conversions. Record

the first 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

D

9160

furlongs

fortnight




14/30

Problem 1.14 The cross-sectional area of a beam is480 in2. What is its cross-section in m2?

Solution: Convert units using Table 1.2. The result:

480 in2

1 ft

12 in

2 0.3048 m

1 ft

2D 0.30967 . . . m2 D 0.310 m2




15/30

Problem 1.15 The cross-sectional area of the C1230American Standard Channel steel beam is A D 8.81 in2.What is its cross-sectional area in mm2?

x

y

A

Solution:

A D 8.81 in2

25.4 mm

1 in

2D 5680 mm2




16/30

Problem 1.16 A pressure transducer measures a valueof 300 lb/in2. Determine the value of the pressure inpascals. A pascal (Pa) is one newton per meter squared.

Solution: Convert the units using Table 1.2 and the definition of

the Pascal unit. The result:

300

lb

in2

4.448 N

1 lb

12 in

1 ft

2 1 ft

0.3048 m

2

D 2.0683 . . . 106

N

m2D 2.0710

6

Pa




17/30

Problem 1.17 A horsepower is 550 ft-lb/s. A watt is1 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 apower of 100,000 horsepower at cruising speed.

Wright

Brothers' Flier

(shown to scale)

Boeing 747

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

D 8950 watt

(b) 105 hp

746 watt

1 hp

D 7.46107 watt




18/30

Problem 1.18 In SI units, the universal gravitationalconstant G D 6.67 1011 N-m2/kg2. Determine thevalue ofG in U.S. Customary units.

Solution: Convert units using Table 1.2. The result:

6.671011

N-m2

kg2

1 lb

4.448 N

1 ft

0.3048 m

2 14.59 kg

1 slug

2

D 3.43590 . . . 108

lb-ft2

slug2

D 3.44108

lb-ft2

slug2




19/30

Problem 1.19 The moment of inertia of the rectan-gular area about the xaxis is given by the equation

I D 13bh3.

The dimensions of the area are b D 200 mm and h D100 mm. Determine the value of I to four significant

digits in terms of (a) mm4; (b) m4; (c) in4.

h

bx

y

Solution:

(a) I D1

3200 mm100 mm3 D 66.7 106 mm4

(b) I D 66.7 106 mm4

1 m

1000 mm

4D 66.7 106 m4

(c) I D 66.7 106 mm4 1 in

25.4 mm

4

D 160 in4




20/30

Problem 1.20 In the equation

T D 12I2,

the term I is in kg-m2 and is in s1.

(a) What are the SI units of T?(b) If the value ofT is 100 when I is in kg-m2 and is

in s1, what is the value ofT when it is expressedin U.S. Customary base units?

Solution: For (a), substitute the units into the expression forT:

(a) T D

1

2

I kg-m2s12 D

kg-m2

s2

For (b), convert units using Table 1.2. The result:

(b) 100

kg-m2

s2

1 slug

14.59 kg

1 ft

0.3048 m

2

D 73.7759 . . .

slug-ft2

s2D 73.8

slug-ft

2

s2




21/30

Problem 1.21 The equation

DMy

I

is used in the mechanics of materials to determinenormal stresses in beams.

(a) When this equation is expressed in terms of SI baseunits,M is in newton-meters (N-m), y is in meters(m), and I is in meters to the fourth power (m4).

What are the SI units of?(b) If M D 2000 N-m, yD 0.1 m, and I D 7

105 m4, what is the value of in U.S. Customarybase units?

Solution:

(a) DMy

ID

(N-m)m

m4 D

N

m2

(b)

DMy

ID

2000 N-m0.1 m

7 105 m4

1 lb

4.448 N

0.3048 m

ft

2

D 59,700lb

ft2




22/30

Problem 1.22 Let W be your weight at sea level inpounds. (a) What is your weight at sea level in newtons?(b) What is your mass in kilograms?

Solution:

(a) W (lb)

4.448 N

lb

D 4.448W N

(b) m D

4.448W N

9.81 m/s2 D 0.453W kg




23/30

Problem 1.23 The acceleration due to gravity is1.62 m/s2 on the surface of the moon and 9.81 m/s2

on the surface of the earth. A female astronauts massis 57 kg. What is the maximum allowable mass of herspacesuit and equipment if the engineers dont want thetotal weight on the moon of the woman, her spacesuitand equipment to exceed 180 N?

Solution: Find the mass which weighs 180 N on the moon.

m Dw

gD

180 N-s2

1.62 mD 111.1 kg

This is the total allowable mass. Thus, the suit & equipment can have

mass of

mS/E D 111.1 kg 57 kg D 54.1 kg




24/30

Problem 1.24 A person has a mass of 50 kg.

(a) The acceleration due to gravity at sea level isg D

9.81 m/s2. What is the persons weight at sea level?(b) The acceleration due to gravity on the moon isg D

1.62 m/s2. What would the person weigh on themoon?

Solution: Use Eq (1.6).

(a) We D 50 kg

9.81m

s2

D 490.5 N D 491 N, and

(b) Wmoon D 50 kg

1.62m

s2

D 81 N.




25/30

Problem 1.25 The acceleration due to gravity at

sea level is g D 9.81 m/s2. The radius of the earthis 6370 km. The universal gravitational constant is

G D 6.67 1011 N-m2/kg2. Use this information todetermine the mass of the earth.

Solution: Use Eq (1.3) a DGmE

R2 . Solve for the mass,

mE DgR2

GD

9.81 m/s26370 km2

103 m

km

2

6.671011

N-m2

kg2

D 5.9679 . . . 1024 kg D 5.971024 kg




26/30

Problem 1.26 A person weighs 180 lb at sea level. Theradius of the earth is 3960 mi. What force is exerted onthe person by the gravitational attraction of the earth ifhe is in a space station in orbit 200 mi above the surfaceof the earth?

Solution: Use Eq (1.5).

WDmg

RE

r

2D

WE

g

g

RE

RE CH

2DWE

3960

3960C 200

2

D 1800.90616 D 163 lb




27/30

Problem 1.27 The acceleration due to gravity on thesurface of the moon is 1.62 m/s2. The radius of the moonis RM D 1738 km. Determine the acceleration due togravity of the moon at a point 1738 km above its surface.

Strategy: Write an equation equivalent to Eq. (1.4) forthe acceleration due to gravity of the moon.

Solution: Use Eq (1.4), rewritten to apply to the Moon. . . a D

gM

RM

r

2

a D 1.62 m/s2

RM

RMCRM

2D 1.62 m/s2

1

2

2D 0.405 m/s2




28/30

Problem 1.28 If an object is near the surface of theearth, the variation of its weight with distance from thecenter of the earth can often be neglected. The acceler-ation due to gravity at sea level is g D 9.81 m/s2. Theradius of the earth is 6370 km. The weight of an objectat sea level is mg, where m is its mass. At what heightabove the earth does the weight of the object decreaseto 0.99 mg?

Solution: Use a variation of Eq (1.5).

W D mg

RE

RE C h

2D 0.99 mg

Solve for the radial height,

h D RE

1p0.99

1D 63701.0050378 1.0

D32.09 . . . km

D32,100 m

D32.1 km

c

2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they



29/30

Problem 1.29 The centers of two oranges are 1 mapart. The mass of each orange is 0.2 kg. Whatgravitational force do they exert on each other? (Theuniversal gravitational constant is G D 6.67 1011 N-m2/kg2.)

Solution: Use Eq (1.1) F DGm1m2

r2 . Substitute:

F D6.6710110.20.2

12 D 2.6681012 N




30/30

Problem 1.30 At a point between the earth and themoon, the magnitude of the force exerted on an objectby the earths gravity equals the magnitude of the forceexerted on the object by the moons gravity. What isthe distance from the center of the earth to that pointto three significant digits? The distance from the centerof the earth to the center of the moon is 383,000 km,and the radius of the earth is 6370 km. The radius of themoon is 1738 km, and the acceleration due to gravity atits surface is 1.62 m/s2.

Solution: LetrEpbe the distance from the Earth to the point where

the gravitational accelerations are the same and let rMp be the distance

from the Moon to that point. Then, rEp C rMp D rEM D 383,000 km.

The fact that the gravitational attractions by the Earth and the Moon

at this point are equal leads to the equation

gE

RE

rEp

2D gM

RM

rMp

2,

where rEM D 383,000 km. Substituting the correct numerical values

leads to the equation

9.81m

s2

6370 kmrEp

2D 1.62

ms2

1738 kmrEM rEp

2,

whererEp is the only unknown. Solving, we get rEp D 344,770 km D

345,000 km.



statics, chap 01 - bedford

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