bedford 5 solucionario

Problem 1.1 The value of is 3.14159265 If isthe circumference of a circle and is its radius, deter-mine the value of to four significant digits.

Solution: 2

12

0159154943.

To four significant digits we have

01592

Problem 1.2 The base of natural logarithms is 2718281828

(a) Express to five significant digits.(b) Determine the value of 2 to five significant digits.(c) Use the value of you obtained in part (a) to deter-

mine the value of 2 to five significant digits.

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

Solution: The value of e is: 2718281828

(a) To five significant figures 27183

(b) 2 to five significant figures is 2 73891

(c) Using the value from part (a) we find 2 73892 which isnot correct in the fifth digit.

Problem 1.3 A machinist drills a circular hole in apanel with a nominal radius 5 mm. The actual radiusof the hole is in the range 5 001 mm. (a) To whatnumber of significant digits can you express the radius?(b) To what number of significant digits can you expressthe area of the hole?

Solution:a) The radius is in the range 1 4.99 mm to 2 5.01 mm. These

numbers are not equal at the level of three significant digits, butthey are equal if they are rounded off to two significant digits.

Two: 50 mm

b) The area of the hole is in the range from 1 12 78.226 m2

to 2 22 78.854 m2. These numbers are equal only if rounded

to one significant digit:

One: 80 mm2

Problem 1.4 The opening in the soccer goal is 24 ftwide and 8 ft high, so its area is 24 ft0 8 ft 192 ft2.What is its area in m2 to three significant digits?

Solution:

192 ft21 m

3.281 ft

2 17.8 m2

17.8 m2

Problem 1.5 The Burj Dubai, scheduled for comple-tion in 2008, will be the world’s tallest building with aheight of 705 m. The area of its ground footprint will be8000 m2. Convert its height and footprint area to U.S.customary units to three significant digits.

Solution:

705 m3.281 ft

1 m 2310 103 ft

8000 m2 3.218 ft1 m

2 8610 104 ft2

2310 103 ft 861 0 104 ft2

c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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 widths 14 in, 1/2 in, 3/4 in, and 1 in, and the car has nutswith dimensions 5 mm, 10 mm, 15 mm, 20 mm,and 25 mm. Defining a wrench to fit if is no morethan 2% larger than , which of your wrenches can youuse?

Solution: Convert the metric size to inches, and compute thepercentage difference between the metric sized nut and the SAEwrench. The results are:

5 mm1 inch

254 mm 019685 in

019685 025019685

100

270%

10 mm1 inch

254 mm 03937 in

03937 0503937

100 270%

15 mm1 inch

254 mm 05905 in

05905 0505905

100 153%

20 mm1 inch

254 mm 07874 in

07874 07507874

100 47%

25 mm1 inch

254 mm 09843 in

09843 1009843

100 16%

A negative percentage implies that the metric nut is smaller than theSAE wrench; a positive percentage means that the nut is larger thenthe wrench. Thus within the definition of the 2% fit, the 1 in wrenchwill fit the 25 mm nut. The other wrenches cannot be used.

Problem 1.7 Suppose that the height of Mt. Everest isknown to be between 29,032 ft and 29,034 ft. Based onthis information, to how many significant digits can youexpress the height (a) in feet? (b) in meters?.

Solution:a) 1 29032 ft

2 29034 ft

The two heights are equal if rounded off to four significant digits.The fifth digit is not meaningful.

Four: 29,030 ft

b) In meters we have

1 29032 ft1 m

3.281 ft 8848.52 m

2 29034 ft1 m

3.281 ft 8849.13 m

These two heights are equal if rounded off to three significantdigits. The fourth digit is not meaningful.

Three: 8850 m

Problem 1.8 The maglev (magnetic levitation) trainfrom Shanghai to the airport at Pudong reaches a speedof 430 km/h. Determine its speed (a) in mi/h; (b) ft/s.

Solution:a) 430

kmh

0.6214 mi1 km

267 mi/h 267 mi/h

b) 430kmh

1000 m1 km

1 ft0.3048 m

1 h3600 s

392 ft/s

392 ft/s

Problem 1.9 In the 2006 Winter Olympics, the men’s15-km cross-country skiing race was won by AndrusVeerpalu of Estonia in a time of 38 minutes, 1.3 seconds.Determine his average speed (the distance traveleddivided by the time required) to three significant digits(a) in km/h; (b) in mi/h.

Solution:a)

15 km

38 1360

min

60 min1 h

23.7 km/h 23.7 km/h

b) 23.7 km/h1 mi

1.609 km 14.7 mi/h 14.7 mi/h


Problem 1.10 The Porsche’s engine exerts 229 ft-lb(foot-pounds) of torque at 4600 rpm. Determine thevalue of the torque in N-m (Newton-meters).

Solution:

229 ft-lb1 N

0.2248 lb1 m

3.281 ft 310 N-m 310 N-m

Problem 1.11 The of the man in ActiveExample 1.1 is defined by 1

2 2, where is his

mass and is his velocity. The man’s mass is 68 kgand he is moving at 6 m/s, so his kinetic energy is12 (68 kg)(6 m/s)2 1224 kg-m2/s2. What is his kineticenergy in U.S. Customary units?

Solution:

1224 kg-m2s2 1 slug14.59 kg

1 ft0.3048 m

2 903 slug-ft2s

903 slug-ft2s

Problem 1.12 The acceleration due to gravity at sealevel in SI units is 981 m/s2. By converting units,use this value to determine the acceleration due togravity at sea level in U.S. Customary units.

Solution: Use Table 1.2. The result is:

981ms2

1 ft03048 m

32185 fts2 322

fts2

Problem 1.13 A 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 nights). If you walk to classat 2 m/s, what is your speed in furlongs per fortnight tothree significant digits?

Solution:

2 m/s1 ft

0.3048 m1 furlong

660 ft3600 s

hr24 hr1 day

14 day1 fortnight

12000furlongsfortnight

Problem 1.14 Determine the cross-sectional area ofthe beam (a) in m2; (b) in in2.

Solution: 200 mm 2 2 80 mm 120 mm 20800 mm2

a) 20800 mm2 1 m1000 mm

2 0.0208 m2

0.0208 m2

b) 20800 mm2 1 in25.4 mm

2 32.2 in2

32.2 in2

Problem 1.15 The cross-sectional area of the C12030American Standard Channel steel beam is 881 in2.What is its cross-sectional area in mm2?

Solution:

881 in2 254 mm1 in

2 5680 mm2


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 ofthe Pascal unit. The result:

300lbin2

4448 N1 lb

12 in1 ft

2 1 ft03048 m

2

20683 106 Nm2 207 106 Pa

Problem 1.17 A horsepower is 550 ft-lb/s. A watt is1 N-m/s. Determine how many watts are generated bythe engines of the passenger jet if they are producing7000 horsepower.

Solution:

7000 hp550 ft-lb/s

1 hp1 m

3.281 ft1 N

0.2248 lb 5220 106 W

5220 106 W

Problem 1.18 Chapter 7 discusses distributed loadsthat are expressed in units of force per unit length. Ifthe value of a distributed load is 400 N/m, what is itsvalue in lb/ft?.

Solution:

400 N/m0.2248 lb

1 N1 m

3.281 ft 27.4 lb/ft 27.4 lb/ft

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

13 3

The dimensions of the area are 200 mm and 100 mm. Determine the value of to four significantdigits in terms of (a) mm4; (b) m4; (c) in4.

Solution:

(a) 13

200 mm 100 mm 3 6670 106 mm4

(b) 6670 106 mm4 1 m1000 mm

4 6670 10 6 m4

(c) 6670 106 mm4 1 in254 mm

4 160 in4

Problem 1.20 In Example 1.3, instead of Einstein’sequation consider the equation , where the mass is in kilograms and the velocity of light is in metersper second. (a) What are the SI units of ? (b) If thevalue of in SI units is 12, what is its value in U.S.Customary base units?

Solution:a) Units kg-m/s

b) 12 kg-m/s0.0685 slug

1 kg3.281 ft

1 m 2.70 slug-ft/s

2.70 slug-ft/s


Problem 1.21 The equation

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

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

(b) If 2000 N-m, 01 m, and 7010 5 m4, what is the value of in U.S. Customarybase units?

Solution:

(a)

(N-m)mm4

Nm2

(b)

2000 N-m 01 m70 10 5 m4

1 lb4448 N

03048 mft

2

59700lbft2

Problem 1.22 The acceleration due to gravity on thesurface of the moon is 1.62 m/s2. (a) What would themass of the C-clamp in Active Example 1.4 be on the surfaceof the moon? (b) What would the weight of the C-clampin newtons be on the surface of the moon?

Solution:a) The mass does not depend on location. The mass in kg is

0.0272 slug14.59 kg

1 slug 0.397 kg mass 0397 kg

b) The weight on the surface of the moon is 0.397 kg 1.62 m/s2 0.643 N 0643

Problem 1.23 The 1 ft 0 1 ft 0 1 ft cube of ironweighs 490 lb at sea level. Determine the weight innewtons of a 1 m 0 1 m 0 1 m cube of the samematerial at sea level.

Solution: The weight density is 490 lb1 ft3

The weight of the 1 m3 cube is:

490 lb1 ft3

1 m 3 1 ft0.3048 m

3 1 N0.2248 lb

77.0 kN

Problem 1.24 The area of the Pacific Ocean is64,186,000 square miles and its average depth is 12,925 ft.Assume that the weight per unit volume of ocean wateris 64 lb/ft3. Determine the mass of the Pacific Ocean(a) in slugs; (b) in kilograms

Solution: The volume of the ocean is

64186000 mi2 12925 ft5280 ft

1 mi

2 2312 0 1019 ft3

(a) 64 lb/ft3

322 ft/s2 23120 1019 ft3 460 0 1019 slugs

(b) 460 0 1019 slugs1459 kg

1 slug 6710 1020 kg

Problem 1.25 The acceleration due to gravity atsea level is 981 m/s2. The radius of the earthis 6370 km. The universal gravitational constant is 6670 10 11 N-m2kg2. Use this information todetermine the mass of the earth.

Solution: Use Eq (1.3)

2 . Solve for the mass,

2

981 m/s2 6370 km 2 103 mkm

2

667 10 11 N-m2

kg2

59679 1024 kg 597 1024 kg


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).

2

2

39603960 200

2

180 090616 163 lb

Problem 1.27 The acceleration due to gravity on thesurface of the moon is 1.62 m/s2. The moon’s radius is 1738 km.(a) What is the weight in newtons on the surface of

the moon of an object that has a mass of 10 kg?(b) Using the approach described in Example 1.5, deter-

mine the force exerted on the object by the gravityof the moon if the object is located 1738 km abovethe moon’s surface.

Solution:a) 10 kg 1.26 m/s2 12.6 N 12.6 N

b) Adapting equation 1.4 we have

2. The force is

then

10 kg 1.62 m/s2 1738 km1738 km 1738 km

2 4.05 N

4.05 N

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 981 m/s2. Theradius of the earth is 6370 km. The weight of an objectat sea level is , where is its mass. At what heightabove the earth does the weight of the object decreaseto 0.99 ?

Solution: Use a variation of Eq (1.5).

2 099 mg

Solve for the radial height,

1

0991 6370 10050378 10

3209 km 32100 m 321 km

Problem 1.29 The planet Neptune has an equatorialdiameter of 49,532 km and its mass is 102470 1026 kg.If the planet is modeled as a homogeneous sphere, whatis the acceleration due to gravity at its surface? (Theuniversal gravitational constant is 6670 10 11

N-m2kg2.)

Solution:We have:

2

2

2

Note that the radius of Neptune is 12 49,532 km 24,766 km

Thus 667 0 10 11 N-m2

kg2102470 1026 kg

24766 km 21 km

1000 m

2

11.1 m/s2 11.1 m/s2

Problem 1.30 At a point between the earth and themoon, the magnitude of the force exerted on an objectby the earth’s gravity equals the magnitude of the forceexerted on the object by the moon’s 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: Let be the distance from the Earth to the point wherethe gravitational accelerations are the same and let Mp be the distancefrom the Moon to that point. Then, Mp 383000 km.The fact that the gravitational attractions by the Earth and the Moonat this point are equal leads to the equation

2

Mp

2

where 383000 km. Substituting the correct numerical valuesleads to the equation

981ms2

6370 km

2 162

ms2

1738 km

2

where is the only unknown. Solving, we get 344770 km 345000 km.


Problem 2.1 In Active Example 2.1, suppose that thevectors U and V are reoriented as shown. The vectorV is vertical. The magnitudes are U 8 and V 3.Graphically determine the magnitude of the vectorU 2V.

Solution: Draw the vectors accurately and measure the resultant.

U 2V 57 57

Problem 2.2 Suppose that the pylon in Example 2.2 ismoved closer to the stadium so that the angle betweenthe forces F and F is 50 . Draw a sketch of thenew situation. The magnitudes of the forces are F 100 kN and F 60 kN. Graphically determine themagnitude and direction of the sum of the forces exertedon the pylon by the cables.

Solution: Accurately draw the vectors and measure the magnitudeand direction of the resultant

F F 146 kN

32


Problem 2.3 The magnitude F 80 lb and theangle 65 . The magnitude F F 120 lb.Graphically determine the magnitude of F.

Solution: Accurately draw the vectors and measure the magnitudeof F.

F 62 lb

Problem 2.4 The magnitudes F 40 N, F 50 N, and F 40 N. The angle 50 and 80 .Graphically determine the magnitude of F F F.

Solution: Accurately draw the vectors and measure the magnitudeof F F F.

F F F 83 N


Problem 2.5 The magnitudes F F F 100 lb, and the angles 30 . Graphically determinethe value of the angle for which the magnitudeF F F is a minimum and the minimum valueof F F F.

Solution: For a minimum, the vector F must point back to theorigin.

F F F 93.2 lb 165

Problem 2.6 The angle 50 . Graphically determinethe magnitude of the vector r.

Solution: Draw the vectors accurately and then measure r.

r 181 mm


Problem 2.7 The vectors F and F representthe forces exerted on the pulley by the belt.Their magnitudes are F 80 N and F 60 N.Graphically determine the magnitude of the total forcethe belt exerts on the pulley.

Solution: Draw the vectors accurately and then measure F F.

F F 134 N

Problem 2.8 The sum of the forces F F F 0. The magnitude F 100 N and the angle 60 . Graphically determine the magnitudes F and F.

Solution: Draw the vectors so that they add to zero.

F 86.6 N F 50.0 N

Problem 2.9 The sum of the forces F F F 0. The magnitudes F 100 N and F 80 N. Graphically determine the magnitude F and theangle .

Solution: Draw the vectors so that they add to zero.

F 50.4 N 525


Problem 2.10 The forces acting on the sailplane arerepresented by three vectors. The lift L and drag Dare perpendicular. The magnitude of the weight W is500 lb. The sum of the forces W L D 0. Graph-ically determine the magnitudes of the lift and drag.

Solution: Draw the vectors so that they add to zero. Then measurethe unknown magnitudes.

L 453 lbD 211 lb

Problem 2.11 A spherical storage tank is suspendedfrom cables. The tank is subjected to three forces, theforces F and F exerted by the cables and its weight W.The weight of the tank is W 600 lb. The vector sumof the forces acting on the tank equals zero. Graphicallydetermine the magnitudes of F and F.

40˚

F

W

F

20˚ 20˚

Solution: Draw the vectors so that they add to zero. Then measurethe unknown magnitudes.

F F 319 lb


Problem 2.12 The rope exerts forces F andF of equal magnitude on the block at . Themagnitude of the total force exerted on the block bythe two forces is 200 lb. Graphically determine F.

Solution: Draw the vectors accurately and then measure theunknown magnitudes.

F 174 lb

Problem 2.13 Two snowcats tow an emergency shelterto a new location near McMurdo Station, Antarctica.(The top view is shown. The cables are horizontal.)The total force F F exerted on the shelter is inthe direction parallel to the line and its magnitudeis 400 lb. Graphically determine the magnitudes of

and F.

Solution: Draw the vectors accurately and then measure theunknown magnitudes.

F 203 lbFB 311 lb

Problem 2.14 A surveyor determines that the horizon-tal distance from to is 400 m and the horizontaldistance from to is 600 m. Graphically determinethe magnitude of the vector r and the angle .

Solution: Draw the vectors accurately and then measure theunknown magnitude and angle.

r 390 m 212


Problem 2.15 The vector r extends from point tothe midpoint between points and . Prove that

r 12 r r

Solution: The proof is straightforward:

r r r and r r r

Add the two equations and note that r r 0, since the twovectors are equal and opposite in direction.

Thus 2r r r or r 12 r r

r

r

r

Problem 2.16 By drawing sketches of the vectors,explain why

U VW U V W

Solution: Additive associativity for vectors is usually given as anaxiom in the theory of vector algebra, and of course axioms are notsubject to proof. However we can by sketches show that associativityfor vector addition is intuitively reasonable: Given the three vectors tobe added, (a) shows the addition first of V W, and then the additionof U. The result is the vector U V W .

(b) shows the addition of U V, and then the addition of W, leadingto the result U V W.

The final vector in the two sketches is the same vector, illustrating thatassociativity of vector addition is intuitively reasonable.

(a)

U

W

V

U

W

V

V+W

U+V

U+[V+W]

[U+V]+W

(b)

Problem 2.17 A force F 40 i 20 j N . What isits magnitude F?

Strategy: The magnitude of a vector in terms of itscomponents is given by Eq. (2.8).

Solution: F

402 202 447 N

Problem 2.18 An engineer estimating the componentsof a force F i j acting on a bridge abutmenthas determined that 130 MN, F 165 MN, and is negative. What is ?

Solution:

F F 2 F 2

F F2 F 2 165 MN 2 130 MN 2 101.6 MN

102 MN


Problem 2.19 A support is subjected to a force F i 80j (N). If the support will safely support a forceof 100 N, what is the allowable range of values of thecomponent ?

Solution: Use the definition of magnitude in Eq. (2.8) and reducealgebraically.

100 2 80 2, from which 100 2 80 2

2.

Thus

3600, or 60 60 (N)

Problem 2.20 If F 600i 800j (kip) and F 200i 200j (kip), what is the magnitude of the forceF F 2F?

Solution: Take the scalar multiple of F, add the components ofthe two forces as in Eq. (2.9), and use the definition of the magnitude.F 600 2 200 i 800 2 200 j 200i 400j

F 200 2 400 2 4472 kip

Problem 2.21 The forces acting on the sailplane are itsweight W 500j lb , the drag D 200i 100j(lb)and the lift L. The sum of the forces W L D 0.Determine the components and the magnitude of L.

Solution:

L W D 500j 200i 100j 200i 400j lb

L 200 lb 2 400 lb 2 447 lb

L 200i 400j lb L 447 lb


Problem 2.22 Two perpendicular vectors U and V liein the - plane. The vector U 6i 8j and V 20.What are the components of V? (Notice that this problemhas two answers.)

Solution: The two possible values of V are shown in the sketch.The strategy is to (a) determine the unit vector associated with U,(b) express this vector in terms of an angle, (c) add 90 to thisangle, (d) determine the two unit vectors perpendicular to U, and(e) calculate the components of the two possible values of V. Theunit vector parallel to U is

e 6i

62 8 2

8j62 8 2

06i 08j

Expressed in terms of an angle,

e i cos j sin i cos 531 j sin 531

Add 90 to find the two unit vectors that are perpendicular to thisunit vector:

e1 i cos 1431 j sin 1431 08i 06j

e2 i cos 369 j sin 369 08i 06j

Take the scalar multiple of these unit vectors to find the two vectorsperpendicular to U.

V1 V 08i 06j 16i 12j

The components are 16 12

V2 V 08i 06j 16i 12j

The components are 16 12

6

8U

V2

V1

Problem 2.23 A fish exerts a 10-lb force on the linethat is represented by the vector F. Express F in termsof components using the coordinate system shown.

Solution: We can use similar triangles to determine thecomponents of F.

F 10 lb7

72 112

i11

72 112

j 537i 844j lb

F 537i 844j lb


Problem 2.24 A man exerts a 60-lb force F to push acrate onto a truck. (a) Express F in terms of componentsusing the coordinate system shown. (b) The weight ofthe crate is 100 lb. Determine the magnitude of the sumof the forces exerted by the man and the crate’s weight.

Solution:

(a) F 60 lb cos 20 i sin 20 j 564i 205j lb

F 564i 205j lb

(b) W 100 lb j

F W 564i [205 100]j lb 564i 795j lb

F W 56.4 lb 2 79.5 lb 2 97.4 lb

F W 97.4 lb

Problem 2.25 The missile’s engine exerts a 260-kNforce F. (a) Express F in terms of components using thecoordinate system shown. (b) The mass of the missileis 8800 kg. Determine the magnitude of the sum of theforces exerted by the engine and the missile’s weight.

Solution:

(a) We can use similar triangles to determine the components of F.

F 260 kN4

42 32

i 3

42 32

j 208i 156j kN

F 208i 156j kN

(b) The missile’s weight W can be expressed in component and thenadded to the force F.

W 8800 kg 9.81 m/s2 j 86.3 kN j

F W 208i [156 863]j kN 208i 697j kN

F W 208 kN 2 69.7 kN 2 219 kN

F W 219 kN

Problem 2.26 For the truss shown, express theposition vector r from point to point in terms ofcomponents. Use your result to determine the distancefrom point to point .

Solution: Coordinates (1.8, 0.7) m, (0, 0.4) m

r 0 18 m i 04 m 07 m j 18i 03j m

18 m 2 03 m 2 1825 m


Problem 2.27 The points , , are the joints ofthe hexagonal structural element. Let r be the positionvector from joint to joint , r the position vectorfrom joint to joint , and so forth. Determine thecomponents of the vectors r and r.

Solution: Use the coordinate system shown and find thelocations of and in those coordinates. The coordinates of thepoints in this system are the scalar components of the vectors r andr. For r, we have

r r r i j

i j

or r 2 0 i 0 0 j 2 cos 60 0 i

2 cos 60 0 j

giving

r 2 2 cos 60 i 2 sin 60 j For r we have

r i j

2 cos 60 0 i 2 sin 60 0 j

Problem 2.28 For the hexagonal structural element inProblem 2.27, determine the components of the vectorr r.

Solution: r r.

The angle between and the -axis is 60 .

r 2 cos 60 i 2 sin 60 j m

r 1i 173j m

r r 2i 1i 173j m

r r 1i 173j m

Problem 2.29 The coordinates of point are (1.8,3.0) ft. The coordinate of point is 0.6 ft. The vectorr has the same direction as the unit vector e 0616i 0788j. What are the components of r?

Solution: The vector r can be written two ways.

r r 0616i 0788j i j

Comparing the two expressions we have

06 30 ft 0788 r

r 2.4 ft0788

3.05 ft

Thus

r r 0616i 0788j 3.05 ft 0616i 0788j 188i 240j ft

r 188i 240j ft


Problem 2.30 (a) Express the position vector frompoint of the front-end loader to point in terms ofcomponents.

(b) Express the position vector from point to point in terms of components.

(c) Use the results of (a) and (b) to determine thedistance from point to point .

Solution: The coordinates are (50, 35); (98, 50); (45, 55).

(a) The vector from point to :

r 98 50 i 50 35 j 48i 15j (in)

(b) The vector from point to is

r 45 98 i 55 50 j 53i 5j (in)

(c) The distance from to is the magnitude of the sum of thevectors,

r r r 48 53 i 15 5 j 5i 20j

The distance from to is

r 5 2 20 2 2062 in


Problem 2.31 In Active Example 2.3, the cable exerts a 900-N force on the top of the tower. Supposethat the attachment point is moved in the horizontaldirection farther from the tower, and assume that themagnitude of the force F the cable exerts on the topof the tower is proportional to the length of the cable.(a) What is the distance from the tower to point Bif the magnitude of the force is 1000 N? (b) Expressthe 1000-N force F in terms of components using thecoordinate system shown.

Solution: In the new problem assume that point is located adistance away from the base. The lengths in the original problemand in the new problem are given by

original 40 m 2 80 m 2 8000 m2

new 2 80 m 2

(a) The force is proportional to the length. Therefore

1000 N 900 N2 80 m 2

8000 m2

8000 m2 1000 N900 N

280 m 2 59.0 m

59.0 m

(b) The force F is then

F 1000 N

2 80 m 2i

80 m2 80 m 2

j

593i 805j N

F 593i 805j N


Problem 2.32 Determine the position vector r interms of its components if (a) 30 , (b) 225 .

60 mm 150 mm

r

r

Solution:

(a) r 60 cos 30 i 60 sin 30 j or

r 5196i 30j mm. And

(b) r 60 cos 225 i 60 sin 225 j or

r 424i 424j mm

FF

60mm

150mm

Problem 2.33 In Example 2.4, the coordinates of thefixed point are (17, 1) ft. The driver lowers the bed ofthe truck into a new position in which the coordinatesof point are (9, 3) ft. The magnitude of the force exerted on the bed by the hydraulic cylinder when thebed is in the new position is 4800 lb. Draw a sketch ofthe new situation. Express in terms of components.

Solution:

tan 1 2 ft8 ft

1404

F 4800 lb cos i sin j

F 4660i 1160j lb


Problem 2.34 A surveyor measures the location ofpoint and determines that r 400i 800j (m). Hewants to determine the location of a point so thatr 400 m and r r 1200 m. What are thecartesian coordinates of point ?

Solution: Two possibilities are: The point lies west of point ,or point lies east of point , as shown. The strategy is to determinethe unknown angles , , and . The magnitude of is

r 400 2 800 2 8944

The angle is determined by

tan 800400

2 634

The angle is determined from the cosine law:

cos 8944 2 1200 2 400 2

2 8944 1200 09689

143 . The angle is 4912 , 7774 .

The two possible sets of coordinates of point are

r 1200 i cos 777 j sin 777 25467i 117266j (m)r 1200 i cos 491 j sin 491 78533i 90734j (m)

The two possibilities lead to (254.7 m, 1172.7 m) or (785.3 m,907.3 m)

0

Problem 2.35 The magnitude of the position vectorr from point to point is 6 m and the magnitude ofthe position vector r from point to point is 4 m.What are the components of r?

3 m

Solution: The coordinates are: , 0 0 , 3 m 0

Thus

r 0 i 0 j 6 m 2 2

2

r 3 m i 0 j 4 m 2 3 m 2 2

Solving these two equations, we find 4833 m, 3555 m.We choose the “-” sign and find

r 483i 356j m


Problem 2.36 In Problem 2.35, determine the compo-nents of a unit vector e that points from point towardpoint .

Strategy: Determine the components of r and thendivide the vector r by its magnitude.

Solution: From the previous problem we have

r 183i 356j m 1832 3562 m 356 m

Thus

e r

0458i 0889j

Problem 2.37 The and coordinates of points , ,and of the sailboat are shown.

(a) Determine the components of a unit vector thatis parallel to the forestay and points from toward .

(b) Determine the components of a unit vector thatis parallel to the backstay and points from toward .

Solution:

r i j

r i j

Points are: (0, 1.2), (4, 13) and (9, 1)

Substituting, we get

r 4i 118j m r 1246 m

r 5i 12j m r 13 m

The unit vectors are given by

e r

rand e

r


e 0321i 0947j

e 0385i 0923j


Problem 2.38 The length of the bar is 0.6 m.Determine the components of a unit vector e thatpoints from point toward point .

Solution: We need to find the coordinates of point

We have the two equations

03 m 2 2 06 m 2

2 2 04 m 2

Solving we find

0183 m 0356 m

Thus

e r

0183 m [ 03 m] i 0356 m j

0183 m 03 m 2 0356 m 2

0806i 0593j

A O

B

0.3 m

Problem 2.39 Determine the components of a unitvector that is parallel to the hydraulic actuator andpoints from toward .

1 m

0.6 m Scoop

0.15 m

0.6 m

1 m

Solution: Point is at (0.75, 0) and point is at (0, 0.6). Thevector

r i j

r 0 075 i 06 0 j m

r 075i 06j m

r 075 2 06 2 0960 m

e r

r

075096

i06

096j

e 0781i 0625j


Problem 2.40 The hydraulic actuator in Problem2.39 exerts a 1.2-kN force F on the joint at that isparallel to the actuator and points from toward .Determine the components of F.

Solution: From the solution to Problem 2.39,

e 0781i 0625j

The vector F is given by F Fe

F 12 0781i 0625j k N

F 937i 750j N

Problem 2.41 A surveyor finds that the length of theline is 1500 m and the length of line is 2000 m.

(a) Determine the components of the position vectorfrom point to point .

(b) Determine the components of a unit vector thatpoints from point toward point .

Solution: We need to find the coordinates of points and

r 1500 cos 60 i 1500 sin 60 j

r 750i 1299j m

Point is at (750, 1299) (m)

r 2000 cos 30 i 2000 sin 30 j m

r 1732i 1000j m

Point is at (1732, 1000) (m)

(a) The vector from to is

r i j

r 982i 299j m

(b) The unit vector e is

e

r

982i 299j10266

e 0957i 0291j


Problem 2.42 The magnitudes of the forces exertedby the cables are T1 2800 lb, T2 3200 lb, T3 4000 lb, and T4 5000 lb. What is the magnitude ofthe total force exerted by the four cables?

Solution: The -component of the total force is

T1 cos 9 T2 cos 29 T3 cos 40 T4 cos 51

2800 lb cos 9 3200 lb cos 29 4000 lb cos 40 5000 lb cos 51

11,800 lb

The -component of the total force is

T1 sin 9 T2 sin 29 T3 sin 40 T4 sin 51

2800 lb sin 9 3200 lb sin 29 4000 lb sin 40 5000 lb sin 51

8450 lb

The magnitude of the total force is

T 2

2 11,800 lb 2 8450 lb 2 14,500 lb T 14,500 lb


Problem 2.43 The tensions in the four cables are equal:T1 T2 T3 T4 . Determine the value of so that the four cables exert a total force of 12,500-lbmagnitude on the support.

Solution: The -component of the total force is

cos 9 cos 29 cos 40 cos 51

326

The -component of the total force is

sin 9 sin 29 sin 40 sin 51

206

The magnitude of the total force is

T 2

2 326 2 206 2 386 12,500 lb

Solving for we find 3240 lb


Problem 2.44 The rope exerts forces F andF on the block at . Their magnitudes are equal:F F . The magnitude of the total force exertedon the block at by the rope is F F 920 N.Determine F by expressing the forces F and Fin terms of components.

20

F

F

Solution:

F cos 20 i sin 20 j

F j

F F cos 20 i [sin 20 1]j

Therefore

920 N 2 2 cos2 20 [sin 20 1]2 802 N

20

Problem 2.45 The magnitude of the horizontal forceF1 is 5 kN and F1 F2 F3 0. What are the magni-tudes of F2 and F3?

%

%

Solution: Using components we have

: 5 kN 2 cos 45 3 cos 30 0

: 2 sin 45 3 sin 30 0

Solving simultaneously yields:

2 966 kN 3 1366 kN


Problem 2.46 Four groups engage in a tug-of-war. Themagnitudes of the forces exerted by groups , , and are F 800 lb, F 1000 lb, F 900 lb. If thevector sum of the four forces equals zero, what are themagnitude of F and the angle ?

F

7030

20

F

F

F

Solution: The strategy is to use the angles and magnitudes todetermine the force vector components, to solve for the unknown forceF and then take its magnitude. The force vectors are

F 800 i cos 110 j sin 110 2736i 75175j

F 1000 i cos 30 j sin 30 866i 500j

F 900 i cos 20 j sin 20 84572i 3078j

F F i cos 180 j sin 180

F i cos j sin

The sum vanishes:

F F F F i 14381 F cos

j 944 F sin 0

From which F 14381i 944j. The magnitude is

F 1438 2 944 2 1720 lb

The angle is: tan 944

1438 06565, or 333


Problem 2.47 In Example 2.5, suppose that the attach-ment point of cable is moved so that the angle betweenthe cable and the wall increases from 40 to 55 . Drawa sketch showing the forces exerted on the hook by thetwo cables. If you want the total force F F to havea magnitude of 200 lb and be in the direction perpen-dicular to the wall, what are the necessary magnitudesof F and F?

Solution: Let A and be the magnitudes of F and F.The component of the total force parallel to the wall must be zero. Andthe sum of the components perpendicular to the wall must be 200 lb.

cos 55 cos 20 0

sin 55 sin 20 200 lb

Solving we find 195 lb 119 lb


Problem 2.48 The bracket must support the two forcesshown, where F1 F2 2 kN. An engineer deter-mines that the bracket will safely support a total forceof magnitude 3.5 kN in any direction. Assume that 0

90 . What is the safe range of the angle ?

F2

F1

Solution:

: 2 kN 2 kN cos 2 kN 1 cos

: 2 kN sin

Thus the total force has a magnitude given by

2 kN 1 cos 2 sin 2 2 kN

2 2 cos 35 kN

Thus when we are at the limits we have

2 2 cos 35 kN2 kN

2

4916

cos 1732

579

In order to be safe we must have

579 90

F1

F1 + F2

F2


Problem 2.49 The figure shows three forces acting ona joint of a structure. The magnitude of F is 60 kN, andF F F 0. What are the magnitudes of F andF?

F

F

F

15

40

Solution: We need to write each force in terms of its components.

F F cos 40i F sin 40j kN

F F cos 195 i F sin 195j kN

F F cos 270 i F sin 270 j kN

Thus F 60j kN

Since F F F 0, their components in each direction must alsosum to zero.

0

0

Thus,

F cos 40 F cos 195 0 0

F sin 40 F sin 195 60 kN 0

Solving for F and F, we get

F 137 kN F 109 kN

F

F

F195

270

40

Problem 2.50 Four forces act on a beam. The vectorsum of the forces is zero. The magnitudes F 10 kN and F 5 kN. Determine the magnitudes ofF and F.

F30

F FF

Solution: Use the angles and magnitudes to determine the vectors,and then solve for the unknowns. The vectors are:

F F i cos 30 j sin 30 0866Fi 05Fj

F 0i 10j F 0i 5j F Fi 0j

Take the sum of each component in the - and -directions:

F 0866F F i 0

and F 05F 10 5 j 0

From the second equation we get F 10 kN . Using this value in

the first equation, we get F 87 kN


Problem 2.51 Six forces act on a beam that forms partof a building’s frame. The vector sum of the forcesis zero. The magnitudes F F 20 kN, F 16 kN, and F 9 kN. Determine the magnitudes ofF and F.

Solution: Write each force in terms of its magnitude and directionas

F F cos i F sin j

where is measured counterclockwise from the -axis.

Thus, (all forces in kN)

F F cos 110 i sin 110 j kN

F 20 cos 270 i 20 sin 270 j kN

F 16 cos 140 i 16 sin 140 j kN

F 9 cos 40 i 9 sin 40 j kN

F 20 cos 270 i 20 sin 270 j kN

F F cos 50 i F sin 50 j kN

We know that the components and components of the forces mustadd separately to zero.

Thus

0

0

F cos 110 0 1226 689 0 F cos 50 0

F sin 110 20 1028 579 20 F sin 50 0

Solving, we get

F 130 kN F 153 kN


Problem 2.52 The total weight of the man and parasailis W 230 lb. The drag force D is perpendicular tothe lift force L. If the vector sum of the three forces iszero, what are the magnitudes of L and D?

Solution: Let and be the magnitudes of the lift and dragforces. We can use similar triangles to express the vectors L and Din terms of components. Then the sum of the forces is zero. Breakinginto components we have

2

22 52

5

22 52 0

5

22 52

2

22 52 230 lb 0

Solving we find

D 85.4 lb L 214 lb

Problem 2.53 The three forces acting on the car areshown. The force T is parallel to the axis and themagnitude of the force W is 14 kN. If T W N 0,what are the magnitudes of the forces T and N?

Solution:

: sin 20 0

: cos 20 14 kN 0

Solving we find

1490 N 510 N


Problem 2.54 The cables , , and help support apillar that forms part of the supports of a structure. Themagnitudes of the forces exerted by the cables are equal:F F F. The magnitude of the vector sum ofthe three forces is 200 kN. What is F?

Solution: Use the angles and magnitudes to determine the vectorcomponents, take the sum, and solve for the unknown. The anglesbetween each cable and the pillar are:

tan 1 4 m6 m

337

tan 1 86

531

tan 1 126

634

Measure the angles counterclockwise form the -axis. The force vec-tors acting along the cables are:

F F i cos 3037 j sin 3037 05548Fi 08319Fj

F F i cos 3231 j sin 3231 07997F i 06004F j

F F i cos 3334 j sin 3334 08944F i 04472Fj

The sum of the forces are, noting that each is equal in magnitude, is

F 22489Fi 18795Fj

The magnitude of the sum is given by the problem:

200 F 22489 2 18795 2 2931F

from which F 6824 kN

Problem 2.55 The total force exerted on the top of themast by the sailboat’s forestay and backstay is180i 820j (N). What are the magnitudes of the forcesexerted at by the cables and ?

Solution: We first identify the forces:

F 40 mi 118 mj40 m 2 118 m 2

F 50 mi 120 mj

50 m 2 120 m 2

Then if we add the force we find

:4

15524

5

169

180 N

:118

15524

12

169

820 N

Solving simultaneously yields:

226 N 657 N


Problem 2.56 The structure shown forms part of atruss designed by an architectural engineer to supportthe roof of an orchestra shell. The members , ,and exert forces F, F, and F on the joint .The magnitude F 4 kN. If the vector sum of thethree forces equals zero, what are the magnitudes of Fand F?

F

(–4, 1) m

F

F

(–2, –3) m

(4, 2) m

Solution: Determine the unit vectors parallel to each force:

e 2

22 32

i3

22 32

j 05547i 08320j

e 4

42 12

i1

42 12

j 09701i 02425j

e 4

42 22

i2

42 22

j 089443i 04472j

The forces are F Fe F Fe,

F Fe 3578i 1789j. Since the vector sum of the forcesvanishes, the - and -components vanish separately:

F 05547F 09701F 3578 i 0 and

F 08320F 02425F 1789 j 0

These simultaneous equations in two unknowns can be solved by anystandard procedure. An HP-28S hand held calculator was used here:

The results: F 2108 kN , F 2764 kN

Problem 2.57 The distance 45 in.

(a) Determine the unit vector e that points from toward .

(b) Use the unit vector you obtained in (a) to determinethe coordinates of the collar .

Solution:

(a) The unit vector is the position vector from to divided by itsmagnitude

r [14 75]i [45 12]j in 61i 33j in

rBA 61 in 2 33 in 2 69.35 in

eBA 1

69.35 in61i 33j in 0880i 0476j

e 0880i 0476j

(b) To find the coordinates of point we will write a vector fromthe origin to point .

r r r r e 75i 12j in 45 in 0880i 0476j

r 354i 334j in

Thus the coordinates of are 354 334 in


Problem 2.58 In Problem 2.57, determine the and coordinates of the collar as functions of the distance .

Solution: The coordinates of the point are given by

0880 and 0476

Thus, the coordinates of point are 75 0880 in and 12 0476 in. Note from the solution of Problem 2.57 above, 0 694 in.

Problem 2.59 The position vector r goes from point to a point on the straight line between and . Itsmagnitude is r 6 ft. Express r in terms of scalarcomponents.

Solution: Determine the perpendicular vector to the line frompoint , and then use this perpendicular to determine the angular orien-tation of the vector r. The vectors are

r 7 3 i 9 5 j 4i 4j 56568

r 12 3 i 3 5 j 9i 2j 92195

r 12 7 i 3 9 j 5i 6j r 78102

The unit vector parallel to is

e r

r 06402i 07682j i cos 5019 j sin 5019

Add 90 to the angle to find the two possible perpendicular vectors:

e1 i cos 14019 j sin 14019 or

e2 i cos 398 j sin 398

Choose the latter, since it points from to the line.

Given the triangle defined by vertices , , , then the magnitude ofthe perpendicular corresponds to the altitude when the base is the line

. The altitude is given by 2 area

base. From geometry, the area of

a triangle with known sides is given by

area

r r r

where is the semiperimeter, 12 r r r . Substi-

tuting values, 11343, and area 220 and the magnitude of the

perpendicular is r 2 22

78102 56333. The angle between the

vector r and the perpendicular r is cos 1 563336

201 . Thusthe angle between the vector r and the -axis is 398 201 591 or 197 . The first angle is ruled out because it causes the vectorr to lie above the vector r , which is at a 45 angle relative to the-axis. Thus:

r 6 i cos 197 j sin 197 565i 202j

B[7,9]

[3,5][12,3]

r


Problem 2.60 Let r be the position vector from point to the point that is a distance meters along thestraight line between and . Express r in terms ofcomponents. (Your answer will be in terms of ).

Solution: First define the unit vector that points from to .

r [10 3]i [9 4]j 7i 5j m

r 7 m 2 5 m 2

74 m

e 1

74

7i 5j

Let be the point that is a distance along the line from to . Thecoordinates of point are

3 m 7

74

3 0814 m

4 m 5

74

4 0581 m

The vector r that points from to is then

r [3 0814 9]i [4 0581 3]j m

r [0814 6]i [0581 1]j m

Problem 2.61 A vector U 3i 4j 12k. What is itsmagnitude?

Strategy: The magnitude of a vector is given in termsof its components by Eq. (2.14).

Solution: Use definition given in Eq. (14). The vector magni-tude is

U 32 4 2 12 2 13

Problem 2.62 The vector e 13 i

23 j k is a unit

vector. Determine the component . (Notice that thereare two answers.)

Solution:

e 13i

23j k

13

2

23

2

2 1 2 49

Thus

23

or 23

Problem 2.63 An engineer determines that an attach-ment point will be subjected to a force F 20i j45k kN . If the attachment point will safely support aforce of 80-kN magnitude in any direction, what is theacceptable range of values for ?

Solution:

802 2 2

2

802 202 2 45 2

To find limits, use equality.

2LIMIT

802 202 45 2

2LIMIT

3975

LIMIT 630 630 kN

LIMIT 630 kN 630 kN 630 kN


Problem 2.64 A vector U ijk. Itsmagnitude is U 30. Its components are related bythe equations 2 and 4 . Determine thecomponents. (Notice that there are two answers.)

Solution: Substitute the relations between the components, deter-mine the magnitude, and solve for the unknowns. Thus

U i 2 j 4 2 k 1i 2j 8k

where can be factored out since it is a scalar. Take the magnitude,noting that the absolute value of must be taken:

30

12 22 82 831

Solving, we get 3612, or 361. The two possiblevectors are

U 361i 2 361 j 4 2 361 k

361i 722j 289k

U 361i 2 361 j

4 2 361 k 361i 722j 289k

Problem 2.65 An object is acted upon by twoforces F1 20i 30j 24k (kN) and F2 60i20j 40k (kN). What is the magnitude of the total forceacting on the object?

Solution:

F1 20i 30j 24k kN

F2 60i 20j 40k kN

F F1 F2 40i 50j 16k kN

Thus

40 kN 2 50 kN 2 16 kN 2 66 kN

Problem 2.66 Two vectors U 3i 2j 6k andV 4i 12j 3k.

(a) Determine the magnitudes of U and V.(b) Determine the magnitude of the vector 3U 2V.

Solution: The magnitudes:

(a) U

32 22 62 7 and V

42 122 32 13

The resultant vector

3U 2V 9 8 i 6 24 j 18 6 k

17i 18j 12k

(b) The magnitude 3U 2V

172 182 122 2751


Problem 2.67 In Active Example 2.6, suppose thatyou want to redesign the truss, changing the positionof point so that the magnitude of the vector r frompoint to point is 3 m. To accomplish this, let thecoordinates of point be 2 1 m, and determinethe value of so that r 3 m. Draw a sketch ofthe truss with point in its new position. What are thenew directions cosines of r?

Solution: The vector r and the magnitude r are

r [2 m 4 m]i [ 0]j [1 m 0]k 2 m i j

1 m k

r 2 m 2 2 1 m 2 3 m

Solving we find 3 m 2 2 m 2 1 m 2 2 m 2 m

The new direction cosines of r .

cos 23 0667cos 23 0667cos 13 0333

Problem 2.68 A force vector is given in terms of itscomponents by F 10i 20j 20k (N).

(a) What are the direction cosines of F?(b) Determine the components of a unit vector e that

has the same direction as F.

Solution:

F 10i 20j 20k N

10 N 2 20 N 2 20 N 2 30 N

(a)cos

10 N30 N

0333 cos 20 N

30 N0667

cos 20 N

30 N0667

(b) e 0333i 0667j 0667k


Problem 2.69 The cable exerts a force F on the hookat whose magnitude is 200 N. The angle between thevector F and the axis is 40 , and the angle betweenthe vector F and the axis is 70 .

(a) What is the angle between the vector F and the axis?

(b) Express F in terms of components.

Strategy: (a) Because you know the angles betweenthe vector F and the and axes, you can use Eq. (2.16)to determine the angle between F and the axis.(Observe from the figure that the angle between F andthe axis is clearly within the range 0 180 .) (b)The components of F can be obtained with Eqs. (2.15).

70

40

F

Solution:

(a) cos 40 2 cos 70 2 cos 2 1 570

(b)F 200 N cos 40 i cos 70 j cos 570 k

F 1532i 684j 1088k N

Problem 2.70 A unit vector has direction cosinescos 05 and cos 02. Its component is posi-tive. Express it in terms of components.

Solution: Use Eq. (2.15) and (2.16). The third direction cosine is

cos 1 05 2 02 2 08426

The unit vector is

u 05i 02j 08426k

Problem 2.71 The airplane’s engines exert a total thrustforce T of 200-kN magnitude. The angle between T andthe axis is 120 , and the angle between T and the axisis 130 . The component of T is positive.

(a) What is the angle between T and the axis?(b) Express T in terms of components.

Solution: The - and -direction cosines are

cos 120 05 cos 130 06428

from which the -direction cosine is

1 05 2 06428 2 05804

Thus the angle between T and the -axis is

(a) cos 1 05804 545 , and the thrust is

T 200 05i 06428j 05804k or:

(b) T 100i 1286j 1161k (kN)


Problem 2.72 Determine the components of the posi-tion vector r from point to point . Use your resultto determine the distance from to .

(5, 0, 3) m

(6, 0, 0) m

(4, 3, 1) m

Solution: We have the following coordinates: 0 0 0 ,5 0 3 m, 6 0 0 m, 4 3 1 m

r 4 m 5 m i 3 m 0 j 1 m 3 m k

i 3j 2k m

1 m 2 3 m 2 2 m 2 374 m

Problem 2.73 What are the direction cosines of theposition vector r from point to point ?

Solution:

cos 1 m

374 m0267 cos

3 m374 m

0802

cos 2 m

374 m 0535

Problem 2.74 Determine the components of the unitvector e that points from point toward point .

Solution: We have the following coordinates: 0 0 0 ,5 0 3 m, 6 0 0 m, 4 3 1 m

r 4 m 6 m i 3 m 0 j 1 m 0 k 2i 3j 1k

2 m 2 3 m 2 1 m 2 374 m

Thus

e 1

374 m2i 3j k m 0535i 0802j 0267k

Problem 2.75 What are the direction cosines of theunit vector e that points from point toward point ?

Solution: Using Problem 2.74

cos 0535 cos 0802 cos 0267


Problem 2.76 In Example 2.7, suppose that thecaisson shifts on the ground to a new position. Themagnitude of the force F remains 600 lb. In the newposition, the angle between the force F and the axisis 60 and the angle between F and the axis is 70 .Express F in terms of components.

Solution: We need to find the angle between the force F andthe axis. We know that

cos2 cos2

cos2 1

cos 1 cos2 cos2

1 cos2 60 cos2 70 07956

cos 1 07956 373 or 1427

We will choose 373 because the picture shows the force pointingup. Now

600 lb cos 60 300 lb

600 lb cos 373 477 lb

600 lb cos 70 205 lb

Thus F 300i 477j 205k lb

Problem 2.77 Astronauts on the space shuttle use radarto determine the magnitudes and direction cosines of theposition vectors of two satellites and . The vector rfrom the shuttle to satellite has magnitude 2 km, anddirection cosines cos 0768, cos 0384, cos 0512. The vector r from the shuttle to satellite hasmagnitude 4 km and direction cosines cos 0743,cos 0557, cos 0371. What is the distancebetween the satellites?

r

r

Solution: The two position vectors are:

r 2 0768i 0384j 0512k 1536i 0768j 1024k (km)

r 4 0743i 0557j 0371k 2972i 2228j 1484k (km)

The distance is the magnitude of the difference:

r r

1536 2927 2 0768 2228 2 1024 1484 2

324 (km)


Problem 2.78 Archaeologists measure a pre-Colum-bian ceremonial structure and obtain the dimensionsshown. Determine (a) the magnitude and (b) thedirection cosines of the position vector from point topoint .

Solution:(a) The coordinates are (0, 16, 14) m and (10, 8, 4) m.

r [10 0]i [8 16]j [4 14]k m 10i 8j 10k m

rAB 102 82 102 m

264 m 16.2 m

r 16.2 m

(b)cos

10

264 0615

cos 8

264

0492

cos 10

264

0615

Problem 2.79 Consider the structure described inProblem 2.78. After returning to the United States,an archaeologist discovers that a graduate student haserased the only data file containing the dimension .But from recorded GPS data he is able to calculate thatthe distance from point to point is 16.61 m.

(a) What is the distance ?(b) Determine the direction cosines of the position

vector from to .

Solution: We have the coordinates (10 m, 8 m, 4 m), (10 m b, 0 18 m).

r 10 m 10 m i 0 8 m j 18 m 4 m k

r i 8 m j 14 m k

(a) We have 16.61 m 2 2 8 m 2 14 m 2 3.99 m(b) The direction cosines of r are

cos 3.99 m

16.61 m 0240

cos 8 m

16.61 m0482

cos 14 m

16.61 m 0843


Problem 2.80 Observers at and use theodolites tomeasure the direction from their positions to a rocketin flight. If the coordinates of the rocket’s position at agiven instant are (4, 4, 2) km, determine the directioncosines of the vectors r and r that the observerswould measure at that instant.

Solution: The vector r is given by

r 4i 4j 2k km

and the magnitude of r is given by

r 4 2 4 2 2 2 km 6 km

The unit vector along is given by

u rr

Thus, u 0667i 0667j 0333k

and the direction cosines are

cos 0667 cos 0667 and cos 0333

The vector r is given by

r i j k km

4 5 i 4 0 j 2 2 k km

and the magnitude of r is given by

r 1 2 4 2 0 2 km 412 km

The unit vector along is given by

e

Thus, u 0242i 0970j 0k

and the direction cosines are

cos 0242 cos 0970 and cos 00


Problem 2.81 In Problem 2.80, suppose that the coor-dinates of the rocket’s position are unknown. At a giveninstant, the person at determines that the directioncosines of r are cos 0535, cos 0802, andcos 0267, and the person at determines that thedirection cosines of r are cos 0576, cos 0798, and cos 0177. What are the coordinates ofthe rocket’s position at that instant.

Solution: The vector from to is given by

r i j k or

r 5 0 i 0 0 j 2 0 k 5i 2k km.

The magnitude of r is given by r 5 2 2 2 539 km.The unit vector along , u , is given by

u rr 0928i 0j 0371k km.

The unit vector along the line ,

u cos i cos j cos k 0535i 0802j 0267k

Similarly, the vector along , u 0576i 0798 0177k.From the diagram in the problem statement, we see that r r r. Using the unit vectors, the vectors r and r can bewritten as

r 0535ri 0802rj 0267rk and

r 0576ri 0798rj 0177rk

Substituting into the vector addition r r r and equatingcomponents, we get, in the direction, 0535r 0576r , andin the direction, 0802r 0798r . Solving, we get that r 4489 km. Calculating the components, we get

r re 0535 4489 i 0802 4489 j 0267 4489 k

Hence, the coordinates of the rocket, , are (2.40, 3.60, 1.20) km.

Problem 2.82* The height of Mount Everest was orig-inally measured by a surveyor in the following way.He first measured the altitudes of two points and thehorizontal distance between them. For example, supposethat the points and are 3000 m above sea leveland are 10,000 m apart. He then used a theodolite tomeasure the direction cosines of the vector r frompoint to the top of the mountain and the vector rfrom point to . Suppose that the direction cosines ofr are cos 05179 cos 06906, and cos 05048, and the direction cosines of r are cos

03743 cos 07486, and cos 05472. Usingthis data, determine the height of Mount Everest abovesea level.

Solution: We have the following coordinates 0 0 3000 m,10 000 0 3000 m,

Then

r i j 3000 m k 05179i 06906j 05048k

r 10000 m i j 3000 m k

03743i 07486j 05472k

Equating components gives us five equations (one redundant) whichwe can solve for the five unknowns.

05179

06906

3000 m 05048 8848 m

10000 m 07486

05472


Problem 2.83 The distance from point to point is20 ft. The straight line is parallel to the axis, andpoint is in the - plane. Express the vector r interms of scalar components.

Strategy: You can resolve r into a vector from to and a vector from to . You can then resolve thevector form to into vector components parallel tothe and axes. See Example 2.8.

30

r

60

Solution: See Example 2.8. The length is, from the right triangle,

r r sin 30 20 05 10 ft

Similarly, the length is

r r cos 30 20 0866 1732 ft

The vector r can be resolved into components along the axes by theright triangles and and the condition that it lies in the -plane.Hence,

r r i cos 30 j cos 90 k cos 60 or

r 15i 0j 866k

The vector r can be resolved into components from the conditionthat it is parallel to the -axis. This vector is

r r i cos 90 j cos 0 k cos 90 0i 10j 0k

The vector r is given by r r r , from which

r 15i 10j 866k (ft)

r

60

30

Problem 2.84 The magnitudes of the two force vectorsare F 140 lb and F 100 lb. Determine the mag-

nitude of the sum of the forces F F.

Solution: We have the vectors

F 140 lb [cos 40 sin 50 ]i [sin 40 ]j [cos 40 cos 50 ]k

F 822i 900j 689k lb

F 100 lb [ cos 60 sin 30 ]i [sin 60 ]j [cos 60 cos 30 ]k

F 250i 866j 433k lb

Adding and taking the magnitude we have

F F 572i 1766j 1122k lb

FA F 57.2 lb 2 176.6 lb 2 112.2 lb 2 217 lb

F F 217 lb


Problem 2.85 Determine the direction cosines of thevectors F and F.

Solution: We have the vectors

F 140 lb [cos 40 sin 50 ]i [sin 40 ]j [cos 40 cos 50 ]k

F 822i 900j 689k lb

F 100 lb [ cos 60 sin 30 ]i [sin 60 ]j [cos 60 cos 30 ]k

F 250i 866j 433k lb

The direction cosines for FA are

cos 82.2 lb140 lb

0587 cos 90.0 lb140 lb

0643

cos 68.9 lb140 lb

0492

The direction cosines for F are

cos 25.0 lb

100 lb0250 cos

86.6 lb100 lb

0866

cos 43.3 lb100 lb

0433

F : cos 0587 cos 0643 cos 0492F : cos 0250 cos 0866 cos 0433

Problem 2.86 In Example 2.8, suppose that a changein the wind causes a change in the position of the balloonand increases the magnitude of the force F exerted onthe hook at to 900 N. In the new position, the anglebetween the vector component F and F is 35 , andthe angle between the vector components F and F is40 . Draw a sketch showing the relationship of theseangles to the components of F. Express F in terms of itscomponents.

Solution: We have

F 900 N sin 35 516 N

F 900 N cos 35 737 N

F F sin 40 474 N

F F cos 40 565 N

Thus

F 474i 516j 565k N


Problem 2.87 An engineer calculates that the magni-tude of the axial force in one of the beams of a geodesicdome is P 765 kN. The cartesian coordinates ofthe endpoints and of the straight beam are ( 12.4,22.0, 18.4) m and ( 9.2, 24.4, 15.6) m, respectively.Express the force P in terms of scalar components.

Solution: The components of the position vector from to are

r i j k

124 92 i 220 244 j

184 156 k

32i 24j 28k m

Dividing this vector by its magnitude, we obtain a unit vector thatpoints from toward :

e 0655i 0492j 0573k

Therefore

P Pe

765 e

501i 376j 439k kN

Problem 2.88 The cable exerts an 8-kN force Fon the bar at .

(a) Determine the components of a unit vector thatpoints from toward point .

(b) Express F in terms of components.

(5, 6, 1) m

(3, 0, 4) m

F

Solution:

(a) e r

r i j k

2

2 2

e 2i 6j 3k

22 62 32

27i

67j

37k

e 0286i 0857j 0429k

(b) F Fe 8e 229i 686j 343k kN


Problem 2.89 A cable extends from point topoint . It exerts a 50-lb force T on plate that isdirected along the line from to . Express T in termsof components.

Solution: Find the unit vector e and multiply it times the magni-tude of the force to get the vector in component form,

e r

r i j k

2

2 2

The coordinates of point are 4 4 sin 20 4 cos 20 or4 137 376 (ft) The coordinates of point are (0, 2, 6) (ft)

e 0 4 i 2 137 j 6 376 k

42 3372 2242

e 0703i 0592j 0394k

T 50e lb

T 352i 296j 197k lb

20 4 ft

4 ft

6 ft

2 ft

T

Problem 2.90 In Example 2.9, suppose that the metalloop at is moved upward so that the vertical distance to increases from 7 ft to 8 ft. As a result, the magnitudesof the forces F and F increase to F F 240 lb. What is the magnitude of the total force F F exerted on the loop by the rope?

Solution: The new coordinates of point are (6, 8, 0) ft. Theposition vectors are

r 4i 8j 4k ft

r 4i 8j 6k ft

The forces are

F 240 lbr

r 980i 196j 980k lb

F 240 lbr

r 891i 178j 1340k lb

The sum of the forces is

F F F 885i 374j 232k lb

The magnitude is F 440 lb


Problem 2.91 The cable exerts a 200-lb force Fat point that is directed along the line from to .Express F in terms of components.

Solution: The coordinates of are (0,6,8). The position vectorfrom to is

r 0 6 i 6 0 j 8 10 k 6i 6j 2k

The magnitude is r

62 62 22 8718 ft.

The unit vector is

u 6

8718i

68718

j2

8718k

or

u 06882i 06882j 02294k

F Fu 200 06882i 06882j 02294k

The components of the force are

F Fu 200 06882i 06882j 02294k or

F 1376i 1376j 459k

Problem 2.92 Consider the cables and wall describedin Problem 2.91. Cable exerts a 200-lb force Fat point that is directed along the line from to .The cable exerts a 100-lb force F at point thatis directed along the line from to . Determine themagnitude of the total force exerted at point by thetwo cables.

Solution: Refer to the figure in Problem 2.91. From Problem 2.91the force F is

F 1376i 1376j 459k

The coordinates of are (8,6,0). The position vector from to is

r 8 6 i 6 0 j 0 10 k 2i 6j 10k

The magnitude is r

22 62 102 1183 ft.The unit vector is

u 2

1183i

61183

j10

1183k 01691i 05072j 08453k

The force is

F Fu 100u 169i 507j 845k

The resultant of the two forces is

F F F 1376 169 i 1376 507 j

845 459 k

F 1207i 1883j 1304k

The magnitude is

F

12072 18832 13042 2590 lb


Problem 2.93 The 70-m-tall tower is supported bythree cables that exert forces F, F, and F on it.The magnitude of each force is 2 kN. Express the totalforce exerted on the tower by the three cables in termsof components.

40 m

60 m

40 m 40 m

60 m

FF

F

Solution: The coordinates of the points are (0, 70, 0), (40, 0, 0), ( 40, 0, 40) ( 60, 0, 60).

The position vectors corresponding to the cables are:

r 60 0 i 0 70 j 60 0 k

r 60i 70k 60k

r 40 0 i 0 70 j 40 0 k

r 40i 70j 40k

r 40 0 i 0 70 j 0 0 k

r 40i 70j 0k

The unit vectors corresponding to these position vectors are:

u r

r

60110

i70

110j

60110

k

05455i 06364j 05455k

u r

r4090i

7090j

4090k

04444i 07778j 04444k

u r

r

40806

i70

806j 0k 04963i 08685j 0k

The forces are:

F Fu 09926i 1737j 0k

F Fu 08888i 15556j 08888

F Fu 10910i 12728j 10910k

The resultant force exerted on the tower by the cables is:

F F F F 09875i 45648j 02020k kN


Problem 2.94 Consider the tower described in Pro-blem 2.93. The magnitude of the force F is 2 kN. The and components of the vector sum of the forcesexerted on the tower by the three cables are zero. Whatare the magnitudes of F and F?

Solution: From the solution of Problem 2.93, the unit vectors are:

u r

r4090i

7090j

4090k

04444i 07778j 04444k

u r

r

60110

i70

110j

60110

05455i 06364j 05455k

From the solution of Problem 2.93 the force F is

F Fu 09926i 1737j 0k

The forces F and F are:

F Fu F 04444i 07778j 04444k

F Fu F 05455i 06364j 05455k

Taking the sum of the forces:

F F F F 09926 04444F 05455F i

1737 07778F 06364F j

04444F 05455F k

The sum of the - and -components vanishes, hence the set of simul-taneous equations:

04444F 05455F 09926 and

04444F 05455F 0

These can be solved by means of standard algorithms, or by the use ofcommercial packages such as TK Solver Plus or Mathcad . Herea hand held calculator was used to obtain the solution:

F 11163 kN F 09096 kN

Problem 2.95 In Example 2.10, suppose that thedistance from point to the collar is increased from0.2 m to 0.3 m, and the magnitude of the force Tincreases to 60 N. Express T in terms of its components.

Solution: The position vector from to is nowr 0.3 m e 0137i 0205j 0171kThe position vector form the origin to is

r r r 04i 03j m 0137i 0205j 0171k m

r 0263i 00949j 0171k m

The coordinates of are (0.263, 0.0949, 0.171) m.The position vector from to is

r [0 0263]i [05 00949]j [015 0171]k m

r 0263i 0405j 0209k m

The force T is

T 60 Nr

r 327i 503j 260k N

T 327i 503j 260k N


Problem 2.96 The cable exerts a 32-lb force T onthe collar at . Express T in terms of components.

6 ft

4 ft

4 ft

7 ft

4 ft

TSolution: The coordinates of point are (0, 7, 4). The vectorposition of is r 0i 7j 4k.

The vector from point to point is given by

r r r

From Problem 2.95, r 267i 233j 267k. Thus

r 0 267 i 7 233 j 4 267 j

r 267i 467j 133k

The magnitude is

r

2672 4672 1332 554 ft

The unit vector pointing from to is

u r

r04819i 08429j 02401k

The force T is given by

T Tu 32u 154i 270j 77k (lb)

Problem 2.97 The circular bar has a 4-m radius andlies in the - plane. Express the position vector frompoint to the collar at in terms of components.

4 m

4 m

3 m

20

Solution: From the figure, the point is at (0, 4, 3) m. The coor-dinates of point are determined by the radius of the circular barand the angle shown in the figure. The vector from the origin to

is r 4 cos 20 i 4 sin 20 j m. Thus, the coordinates of point are (3.76, 1.37, 0) m. The vector from to is given by r

i j k 376i 263j 3k m. Finally, thescalar components of the vector from to are (3.76, 2.63, 3) m.


Problem 2.98 The cable in Problem 2.97 exerts a60-N force T on the collar at that is directed along theline from toward . Express T in terms of components.

Solution: We know r 376i 263j 3k m from Problem2.97. The unit vector u . The unit vector is u

0686i 0480j 0547k. Hence, the force vector T is given by

TT 0686i 0480j 0547k N 411i 288j 328k N

Problem 2.99 In Active Example 2.11, suppose thatthe vector V is changed to V 4i 6j 10k.

(a) What is the value of UV?(b) What is the angle between U and V when they are

placed tail to tail?

Solution: From Active Example 2.4 we have the expression for U.Thus

U 6i 5j 3kV 4i 6k 10k

U V 6 4 5 6 3 10 84

cos U VVV

84

62 5 2 3 2 42 6 2 10 2 0814

cos 1 0814 355

U V 84 355

Problem 2.100 In Example 2.12, suppose that the coor-dinates of point are changed to (6, 4, 4) m. What isthe angle between the lines and ?

Solution: Using the new coordinates we have

r 2i j 2k m r 3 m

r 4i 5j 2k m r 6.71 m

cos r rrr

2 4 1 5 2 2 m2

3 m 6.71 m 0845

cos 1 0845 324

324

Problem 2.101 What is the dot product of the positionvector r 10i 25j (m) and the force vector

F 300i 250j 300k N ?

Solution: Use Eq. (2.23).

F r 300 10 250 25 300 0 3250 N-m

Problem 2.102 Suppose that the dot product of twovectors U and V is U V 0. If U 0, what do youknow about the vector V?

Solution:Either V 0 or V U


Problem 2.103 Two vectors are givenin terms of their components by

U i 4j 6k

and V 3i 2j 3k

Use the dot product to determine the component .

Solution: When the vectors are perpendicular, U V 0.

Thus

U V 0

3 4 2 6 3 0

3 26

867

Problem 2.104 Three vectors

U i 3j 2k

V 3i j 3k

W 2i 4jk

are mutually perpendicular. Use the dot product to deter-mine the components , and .

Solution: For mutually perpendicular vectors, we have threeequations, i.e.,

U V 0

U W 0

V W 0

Thus

3 3 6 0

2 12 2 0

6 4 3 0

3 Eqns3 Unknowns

Solving, we get

2857 0857 3143

Problem 2.105 The magnitudes U 10 and V 20.

(a) Use the definition of the dot product to determineU V.

(b) Use Eq. (2.23) to obtain U V.

Solution:

(a) The definition of the dot product (Eq. (2.18)) is

U V UV cos Thus

U V 10 20 cos 45 30 1932

(b) The components of U and V are

U 10 i cos 45 j sin 45 707i 707j

V 20 i cos 30 j sin 30 1732i 10j

From Eq. (2.23) U V 707 1732 707 10 1932


Problem 2.106 By evaluating the dot product U V,prove the identity cos 1 2 cos 1 cos 2 sin 1sin 2.

Strategy: Evaluate the dot product both by usingEq. (2.18) and by using Eq. (2.23).

Solution: The strategy is to use the definition Eq. (2.18) and theEq. (2.23). From Eq. (2.18) and the figure,

U V UV cos 1 2 From Eq. (2.23) and the figure

U U i cos 1 j sin 2 V V i cos 2 j sin 2

and the dot product is U V UV cos 1 cos 2 sin 1 sin 2 .

Equating the two results:

U V UV cos 1 2 UV cos 1 cos 2 sin 1 sin 2

from which if U 0 and V 0, it follows that

cos 1 2 cos 1 cos 2 sin 1 sin 2 Q.E.D.

Problem 2.107 Use the dot product to determine theangle between the forestay (cable ) and the backstay(cable ).

Solution: The unit vector from to is

e r

r0321i 0947j

The unit vector from to is

e r

r 0385i 0923j

From the definition of the dot product, e e 1 1 cos , whereis the angle between and . Thus

cos 0321 0385 0947 0923

cos 0750

413


Problem 2.108 Determine the angle between thelines and (a) by using the law of cosines (seeAppendix A); (b) by using the dot product.

Solution:(a) We have the distances:

42 32 12 m

26 m

52 12 32 m

35 m

5 4 2 1 3 2 3 1 2 m

33 m

The law of cosines gives

2 2 2 2 cos

cos 2 2 2

2 0464 623

(b) Using the dot product

r 4i 3j k m r 5i j 3k m

r r 4 m 5 m 3 m 1 m 1 m 3 m 14 m2

r r cos

Therefore

cos 14 m2

26 m

35 m

0464 623

Problem 2.109 The ship measures the positions ofthe ship and the airplane and obtains the coordinatesshown. What is the angle between the lines of sight and ?

Solution: From the coordinates, the position vectors are:

r 6i 0j 3k and r 4i 4j 4k

The dot product: r r 6 4 0 4 3 4 12

The magnitudes: r

62 02 32 671 km and

r

42 42 42 693 km

From Eq. (2.24) cos r rrr

02581, from which 75 .

From the problem and the construction, only the positive angle makessense, hence 75


Problem 2.110 Astronauts on the space shuttle useradar to determine the magnitudes and direction cosinesof the position vectors of two satellites and . Thevector r from the shuttle to satellite has magnitude2 km and direction cosines cos 0768, cos 0384, cos 0512. The vector r from the shuttleto satellite has magnitude 4 km and direction cosinescos 0743, cos 0557, cos 0371. Whatis the angle between the vectors r and r?

r

r

Solution: The direction cosines of the vectors along r and rare the components of the unit vectors in these directions (i.e.,u cos i cos j cos k, where the direction cosines are thosefor r). Thus, through the definition of the dot product, we can findan expression for the cosine of the angle between r and r .

cos cos cos cos cos cos cos

Evaluation of the relation yields

cos 0594 535

Problem 2.111 In Example 2.13, if you shift yourposition and the coordinates of point where you applythe 50-N force become (8, 3, 3) m, what is the vectorcomponent of F parallel to the cable ?

Solution: We use the following vectors to define the force F.

r 8i 3j 3k m

e r

r 0833i 0331j 0331k

F 50 N e 442i 166j 166k N

Now we need the unit vector e.

r 10i 2j 3k m

e r

r 0941i 0188j 0282k

To find the vector component parallel to we use the dot productin the following manner

F e 44.2 N 0941 16.6 N 0188 16.6 N 0282 33.8 N

F F e e 33.8 N 0941i 0188j 0282k

F 318i 635j 953k N


Problem 2.112 The person exerts a force F 60i40j (N) on the handle of the exercise machine. UseEq. (2.26) to determine the vector component of F thatis parallel to the line from the origin to where theperson grips the handle.

Solution: The vector r from the to where the person grips thehandle is

r 250i 200j 150k mm

r 354 mm

To produce the unit vector that is parallel to this line we divide by themagnitude

e r

r

250i 200j 150k mm354 mm

0707i 0566j 0424k

Using Eq. (2.26), we find that the vector component parallel to theline is

F e F e [ 0707 60 N 0566 40 N ] 0707i

0566j 0424k

F 140i 112j 84k N

Problem 2.113 At the instant shown, the Harrier’sthrust vector is T 17000i 68000j 8000k (N)and its velocity vector is v 73i 18j 06k (m/s).The quantity Tpv, where Tp is the vectorcomponent of T parallel to v, is the power currentlybeing transferred to the airplane by its engine. Determinethe value of .

T

v

Solution:

T 17000i 68000j 8000k N

v 73i 18j 06k m/s

Power T v 17000 N 73 m/s 68000 N 18 m/s

8000 N 06 m/s

Power 251000 Nm/s 251 kW


Problem 2.114 Cables extend from to and from to . The cable exerts a 1000-lb force F at .

(a) What is the angle between the cables and ?(b) Determine the vector component of F parallel to

the cable . F

(0, 7, 0) ft

(14, 0, 14) ft(0, 0, 10) ft

Solution: Use Eq. (2.24) to solve.

(a) From the coordinates of the points, the position vectors are:

r 0 0 i 0 7 j 10 0 k

r 0i 7j 10k

r 14 0 i 0 7 j 14 0 k

r 14i 7j 14k

The magnitudes are:

r

72 102 122 (ft) and

r

142 72 142 21

The dot product is given by

r r 14 0 7 7 10 14 189

The angle is given by

cos 189

122 21 07377

from which 425 . From the construction: 425

(b) The unit vector associated with is

e r

r 0i 05738j 08197k

The unit vector associated with is

e rr 06667i 03333j 06667k

Thus the force vector along is

F Fe 6667i 3333j 6667k

The component of this force parallel to is

F e e 7375 e 0i 4228j 6045k (lb)


Problem 2.115 Consider the cables and shownin Problem 2.114. Let r be the position vector frompoint to point . Determine the vector component ofr parallel to the cable .

Solution: From Problem 2.114, r 0i 7j 10k, and e 06667i 03333j 06667k. Thus r e 9, and r e e 6i 3j 6k ft.

Problem 2.116 The force F 10i 12j 6k N .Determine the vector components of F parallel and nor-mal to line .

(0, 6, 4) m

F

Solution: Find e r

r

Then

F F e e

and F F F

e 0i 6j 4k

62 42

6j 4k

52

e 6

721j

4721

k 0832j 0555k

F [ 10i 12j 6k 0832j 0555k ]e

F [6656]e 0i 554j 369k N

F F F

F 10i 12 554 j 6 369k

F 10i 646j 969k N


Problem 2.117 The rope exerts a 50-N force T oncollar . Determine the vector component of T parallelto the bar .

Solution: We have the following vectors

r 02i 03j 025k m

e r

r 0456i 0684j 0570k

r 05j 015k m

r 04i 03j m

r r 0.2 m e 0309i 0163j 0114k m

r r r 0309i 0337j 0036k m

e r

r 0674i 0735j 0079k

We can now write the force T and determine the vector componentparallel to .

T 50 N e 337i 367j 393k N

T e T e 343i 514j 429k N

T 343i 514j 429k N

Problem 2.118 In Problem 2.117, determine the vectorcomponent of T normal to the bar .

Solution: From Problem 2.117 we have

T 337i 367j 393k N

Tp 343i 514j 429k N

The normal component is then

T T T

T 371i 316j 822k N


Problem 2.119 The disk is at the midpoint of thesloped surface. The string from to exerts a 0.2-lbforce F on the disk. If you express F in terms of vectorcomponents parallel and normal to the sloped surface,what is the component normal to the surface?

F

(0, 6, 0) ft

2 ft

8 ft

10 ft

Solution: Consider a line on the sloped surface from perpendic-ular to the surface. (see the diagram above) By SIMILAR triangles wesee that one such vector is r 8j 2k. Let us find the componentof F parallel to this line.The unit vector in the direction normal to the surface is

e r

r

8j 2k

82 22 0970j 0243k

The unit vector e can be found by

e i j h

2

2 2

Point is at (0, 6, 0) (ft) and is at (5, 1, 4) (ft).


e 0615i 0615j 0492k

Now F Fe 02 e

F 0123i 0123j 00984k lb

The component of F normal to the surface is the component parallelto the unit vector e.

FNORMAL F e e 0955 e

FNORMAL 0i 00927j 00232k lb

8

8

2

2

Problem 2.120 In Problem 2.119, what is the vectorcomponent of F parallel to the surface?

Solution: From the solution to Problem 2.119,

F 0123i 0123j 00984k lb and

FNORMAL 0i 00927j 00232k lb

The component parallel to the surface and the component normal tothe surface add to give F F FNORMAL Fparallel .

Thus

Fparallel F FNORMAL


Fparallel 01231i 00304j 01216k lb


Problem 2.121 An astronaut in a maneuvering unitapproaches a space station. At the present instant, thestation informs him that his position relative to the originof the station’s coordinate system is r 50i 80j180k (m) and his velocity is v 22j 36k (m/s).The position of the airlock is r 12i 20k (m).Determine the angle between his velocity vector and theline from his position to the airlock’s position.

Solution: Points and are located at : (50, 80, 180) mand : ( 12, 0, 20) m. The vector r is r i

j k 12 50 i 0 80 j 20 180 k m. Thedot product between v and r is v r cos , where is the angle between v and r. Substitutingin the numerical values, we get 197

Problem 2.122 In Problem 2.121, determine the vec-tor component of the astronaut’s velocity parallel to theline from his position to the airlock’s position.

Solution: The coordinates are ( 12, 0, 20) m, (50, 80, 180) m.Therefore

r 62i 80j 160k m

e r

r 0327i 0423j 0845k

The velocity is given as

v 22j 36k m/s

The vector component parallel to the line is now

v e v e [ 0423 22 0845 36 ]e

130i 168j 336k m/s


Problem 2.123 Point is at longitude 30 W and lati-tude 45 N on the Atlantic Ocean between NovaScotia and France. Point is at longitude 60 E andlatitude 20 N in the Arabian Sea. Use the dot product todetermine the shortest distance along the surface of theearth from to in terms of the radius of the earth .

Strategy: Use the dot product to detrmine the anglebetween the lines and ; then use the definition ofan angle in radians to determine the distance along thesurface of the earth from to .

Solution: The distance is the product of the angle and the radius ofthe sphere, , where is in radian measure. From Eqs. (2.18)and (2.24), the angular separation of and is given by

cos P QPQ

The strategy is to determine the angle in terms of the latitude andlongitude of the two points. Drop a vertical line from each point and to and on the equatorial plane. The vector position of is the sumof the two vectors: P r r. The vector r r i cos 0j k sin . From geometry, the magnitude is r cos .The vector r r 0i 1j 0k . From geometry, the magnitudeis r sin . Substitute and reduce to obtain:

P r r i cos cos j sin k sin cos

A similar argument for the point yields

Q r r i cos cos j sin k sin cos

Using the identity cos2 sin2 1, the magnitudes are

P Q

The dot product is

P Q 2 cos cos cos sin sin

Substitute:

cos P QPQ

cos cos cos sin sin

Substitute 30 , 60 , 45 , 20 , to obtaincos 02418, or 1326 radians. Thus the distance is 1326

30

45

6020

QP

Problem 2.124 In Active Example 2.14, suppose thatthe vector V is changed to V 4i 6j 10k.(a) Determine the cross product U 0 V. (b) Use the dotproduct to prove that U 0 V is perpendicular to V.

Solution: We have U 6i 5j kV 4k 6j 10k

(a) U 0 V i j k6 5 14 6 10

44i 56j 16k

U 0 V 44i 56j 16k

(b) U 0 V V 44 4 56 6 16 10 0 U 0 V V


Problem 2.125 Two vectors U 3i 2j and V 2i 4j.

(a) What is the cross product U 0 V?(b) What is the cross product V 0 U?

Solution: Use Eq. (2.34) and expand into 2 by 2 determinants.

U 0 V i j k3 2 02 4 0

i 2 0 4 0 j 3 0 2 0

k 3 4 2 2 8k

V 0 U i j k2 4 03 2 0

i 4 0 2 0 j 2 0 3 0

k 2 2 3 4 8k

Problem 2.126 The two segments of the L-shaped barare parallel to the and axes. The rope exertsa force of magnitude F 500 lb on the bar at .Determine the cross product r 0 F, where r is theposition vector form point to point .

Solution: We need to determine the force F in terms of itscomponents. The vector from to is used to define F.

r 2i 4j k ft

F 500 lbr

r 500 lb

2i 4j k2 2 4 2 1 2

F 218i 436j 109k lb

Also we have r 4i 5k ftTherefore

r 0 F i j k4 0 5

218 436 109 2180i 1530j 1750k ft-lb

r 0 F 2180i 1530j 1750k ft-lb


Problem 2.127 The two segments of the L-shaped barare parallel to the and axes. The rope exertsa force of magnitude F 500 lb on the bar at .Determine the cross product r 0 F, where r is theposition vector form point to point . Compare youranswers to the answer to Problem 2.126.

Solution: We need to determine the force F in terms of its compo-nents. The vector from to is used to define F.

r 2i 4j k ft

F 500 lbr

r 500 lb

2i 4j k2 2 4 2 1 2

F 218i 436j 109k lb

Also we have r 6i 4j 4k ftTherefore

r 0 F i j k6 4 4

218 436 109 2180i 1530j 1750k ft-lb

r 0 F 2180i 1530j 1750k ft-lb

The answer is the same for 2.126 and 2.127 because the positionvectors just point to different points along the line of action of theforce.

Problem 2.128 Suppose that the cross product of twovectors U and V is U 0 V 0. If U 0, what do youknow about the vector V?

Solution:Either V 0 or VU

Problem 2.129 The cross product of two vectors Uand V is U0 V 30i 40k. The vector V 4i2j 3k. The vector U 4ij k . Determine and .

Solution: From the given information we have

U 0 V i j k

4

4 2 3

3 2 i 4 12 j 8 4 k

U 0 V 30 40k

Equating the components we have

3 2 30 4 12 0 8 4 40

Solving any two of these three redundant equations gives

12 3


Problem 2.130 The magnitudes U 10 and V 20.

(a) Use the definition of the cross product to determineU0 V.

(b) Use the definition of the cross product to determineV0 U.

(c) Use Eq. (2.34) to determine U0 V.

(d) Use Eq. (2.34) to determine V0 U.

U

V

4530

Solution: From Eq. (228) U 0 V UV sin e. From the sketch,the positive -axis is out of the paper. For U 0 V, e 1k (points intothe paper); for V 0 U, e 1k (points out of the paper). The angle

15 , hence (a) U 0 V 10 20 02588 e 518e 518k.Similarly, (b) V 0 U 518e 518k (c) The two vectors are:

U 10 i cos 45 j sin 45 707i 0707j

V 20 i cos 30 j sin 30 1732i 10j

U 0 V i j k

707 707 01732 10 0

i 0 j 0 k 707 12245

518k

(d) V 0 U i j k

1732 10 0707 707 0

i 0 j 0 k 12245 707

518k

Problem 2.131 The force F 10i 4j (N). Deter-mine the cross product r 0 F.

r

(6, 3, 0) m

(6, 0, 4) m

F

Solution: The position vector is

r 6 6 i 0 3 j 4 0 k 0i 3j 4k

The cross product:

r 0 F i j k0 3 4

10 4 0 i 16 j 40 k 30

16i 40j 30k (N-m)

r

(6, 3, 0)

(6, 0, 4) F


Problem 2.132 By evaluating the cross productU0 V, prove the identity sin 1 2 sin 1 cos 2cos 1 sin 2.

V

1

2

U

Solution: Assume that both U and V lie in the - plane. Thestrategy is to use the definition of the cross product (Eq. 2.28) and theEq. (2.34), and equate the two. From Eq. (2.28) U 0 V UV sin 1

2 e. Since the positive -axis is out of the paper, and e points intothe paper, then e k. Take the dot product of both sides with e, andnote that k k 1. Thus

sin 1 2U 0 V k

UV

The vectors are:

U U i cos 1 j sin 2 and V V i cos 2 j sin 2

The cross product is

U 0 V i j k

U cos 1 U sin 1 0V cos 2 V sin 2 0

i 0 j 0 k UV cos 1 sin 2 cos 2 sin 1

Substitute into the definition to obtain: sin 1 2 sin 1 cos 2cos 1 sin 2. Q.E.D.

U

V

1

2

Problem 2.133 In Example 2.15, what is the minimumdistance from point to the line ?

Solution: Let be the angle between r and r. Then theminimum distance is

r sin

Using the cross product, we have

r 0 r rr sin r r 0 r

r

We have

r 10i 2j 3k m

r 6i 6j 3k m

r 0 r i j k

10 2 36 6 3

12i 48j 72k m2

Thus

12 m2 48 m2 2 72 m2 2

10 m 2 2 m 2 3 m 2 8.22 m

8.22 m


Problem 2.134 (a) What is the cross product r 0r? (b) Determine a unit vector e that is perpendicularto r and r.

(6, –2, 3) m

(4, 4, –4) m

r

r

Solution: The two radius vectors are

r 4i 4j 4k r 6i 2j 3k

(a) The cross product is

r 0 r i j k6 2 34 4 4

i 8 12 j 24 12

k 24 8

4i 36j 32k m2

The magnitude is

r 0 r

42 362 322 4833 m2

(b) The unit vector is

e r 0 r

r 0 r 00828i 07448j 06621k

(Two vectors.)

Problem 2.135 For the points , , and in Pro-blem 2.134, use the cross product to determine the lengthof the shortest straight line from point to the straightline that passes through points and .

Solution:

r 6i 2j 3k (m)

r 4i 4j 4k m

r 0 r C

(C is to both r and r)

C i j k6 2 34 4 4

8 12 i

12 24 j 24 8 k

C 4i 36j 32k

C is to both r and r. Any line to the plane formed by C andr will be parallel to the line on the diagram. C 0 r is such aline. We then need to find the component of r in this direction andcompute its magnitude.

C 0 r i j k4 36 326 2 3

C 172i 204j 208k

The unit vector in the direction of C is

e C

C 0508i 0603j 0614k

(The magnitude of C is 338.3)

We now want to find the length of the projection, , of line indirection e.

r e

4i 4j 4k e

690 m

(6, –2, 3) m

(4, 4, –4) m

r

rP


Problem 2.136 The cable exerts a 1000-lb force Fon the hook at . Determine r 0 F.

r

r

F6 ft

8 ft

4 ft

4 ft 12 ft

Solution: The coordinates of points , , and are (16, 0, 12), (4, 6, 0), (4, 0, 8). The position vectors are

r 16i 0j 12k r 4i 6j 0k r 4i 0j 8k

The force F acts along the unit vector

e r

r

r rr r

rr

Noting r r 4 4 i 0 6 j 8 0 k 0i 6j 8k

r r

62 82 10. Thus

e 0i 06j 08k and F Fe 0i 600j 800k (lb)

The vector

r 4 16 i 6 0 j 0 12 k 12i 6j 12k

Thus the cross product is

r 0 F i j k12 6 120 600 800

2400i 9600j 7200k (ft-lb)

r

6 ft

8 ft

4 ft

4 ft 12 ft

Problem 2.137 The force vector F points along thestraight line from point to point . Its magnitudeis F 20 N. The coordinates of points and are 6 m 8 m 4 m and 8 m 1 m 2 m.

(a) Express the vector F in terms of its components.(b) Use Eq. (2.34) to determine the cross products

r 0 F and r 0 F.

Solution: We have r 6i 8j 4k m r 8i j 2k m,

(a)

F 20 N8 6 mi 1 8 mj 2 4 mk

2 m 2 7 m 2 6 m 2

20 N

892i 7j 6k

(b)

r 0 F 20 N

89

i j k6 m 8 m 4 m

2 7 6

424i 933j 1230k Nm

r 0 F 20 N

89

i j k8 m 1 m 2 m

2 7 6

424i 933j 1230k Nm

Note that both cross products give the same result (as they must).


Problem 2.138 The rope exerts a 50-N force Ton the collar at . Let r be the position vector frompoint to point . Determine the cross product r 0 T.

Solution: We define the appropriate vectors.

r 02i 03j 025k m

r 0.2 mr

r 0091i 0137j 0114k m

r 05j 015k m

r 04i 03j m

r r r r 061i 122j 0305k m

T 50 Nr

r 337i 367j 393k N

Now take the cross product

r 0 T i j l

0091 0137 0114337 367 393

472i 348j 796k N-m

r 0 T 472i 348j 796k N-m

Problem 2.139 In Example 2.16, suppose that theattachment point is moved to the location (0.3, 0.3,0) m and the magnitude of T increases to 600 N. Whatis the magnitude of the component of T perpendicularto the door?

Solution: We first develop the force T.

r 03i 01j m

T 600 Nr

r 569i 190j N

From Example 2.16 we know that the unit vector perpendicular to thedoor is

e 0358i 0894j 0268k

The magnitude of the force perpendicular to the door (parallel to e) isthen

T T e 569 N 0358 190 N 0894 373 N

T 373 N


Problem 2.140 The bar is 6 m long and is perpen-dicular to the bars and . Use the cross product todetermine the coordinates , , of point .

(0, 0, 3) m (4, 0, 0) m

(0, 3, 0) m

(, , )

Solution: The strategy is to determine the unit vector perpendic-ular to both and , and then determine the coordinates that willagree with the magnitude of . The position vectors are:

r 0i 3j 0k r 0i 0j 3k and

r 4i 0j 0k. The vectors collinear with the bars are:

r 0 0 i 0 3 j 3 0 k 0i 3j 3k

r 4 0 i 0 3 j 0 0 k 4i 3j 0k

The vector collinear with r is

R r 0 r i j k0 3 34 3 0

9i 12j 12k

The magnitude R 1921 (m). The unit vector is

e R

R 04685i 06247j 06247k

Thus the vector collinear with is

r 6e 2811i 375j 375k

Using the coordinates of point :

281 0 281 (m)

375 3 675 (m)

375 0 375 (m)

Problem 2.141* Determine the minimum distancefrom point to the plane defined by the three points, , and .

(3, 0, 0) m

(0, 5, 0) m

(0, 0, 4) m

(9, 6, 5) mSolution: The strategy is to find the unit vector perpendicular to

the plane. The projection of this unit vector on the vector : r e isthe distance from the origin to along the perpendicular to the plane.The projection on e of any vector into the plane (r e, r e, orr e) is the distance from the origin to the plane along this sameperpendicular. Thus the distance of from the plane is

r e r e

The position vectors are: r 3i, r 5j, r 4k and r 9i 6j 5k. The unit vector perpendicular to the plane is foundfrom the cross product of any two vectors lying in the plane. Noting:r r r 5j 4k, and r r r 3i 5j. Thecross product:

r 0 r i j k0 5 43 5 0

20i 12j 15k

The magnitude is r 0 r 2773, thus the unit vector is e 07212i 04327j 05409k. The distance of point from the planeis r e r e 11792 2164 963 m. The second termis the distance of the plane from the origin; the vectors r, or rcould have been used instead of r.

[0,5,0]

[9,6,5]

[3,0,0]

[0,0,4]


Problem 2.142* The force vector F points alongthe straight line from point to point . UseEqs. (2.28)–(2.31) to prove that

r 0 F r 0 F

Strategy: Let r be the position vector from point to point . Express r in terms of of r and r . Noticethat the vectors r and F are parallel.

Solution: We have

r r r

Therefore

r 0 F r r 0 F r 0 F r 0 F

The last term is zero since rF.

Thereforer 0 F r 0 F

Problem 2.143 For the vectors U 6i 2j 4k,V 2i 7j, and W 3i 2k, evaluate the followingmixed triple products: (a) U V0W ; (b) W V0U ; (c) V W0 U .

Solution: Use Eq. (2.36).

(a) U V 0 W 6 2 42 7 03 0 2

6 14 2 4 4 21 160

(b) W V 0 U 3 0 22 7 06 2 4

3 28 0 2 4 42 160

(c) V W0 U 2 7 03 0 26 2 4

2 4 7 12 12 0 160


Problem 2.144 Use the mixed triple product to calcu-late the volume of the parallelepiped.

(140, 90, 30) mm

(200, 0, 0) mm

(160, 0, 100) mm

Solution: We are given the coordinates of point . From the geom-etry, we need to locate points and . The key to doing this is to notethat the length of side is 200 mm and that side is the axis.Sides , , and are parallel to the axis and the coordinatesof the point pairs ( and ), ( and ), and ( and ) differ only by200 mm in the coordinate. Thus, the coordinates of point are ( 60,90, 30) mm and the coordinates of point are ( 40, 0, 100) mm.Thus, the vectors r, r, and r are r 200i mm, r

60i 90j 30k mm, and r 40i 0j 100k mm. The mixedtriple product of the three vectors is the volume of the parallelepiped.The volume is

r r 0 r 60 90 3040 0 100

200 0 0

60 0 90 200 100 30 0 mm3

1800000 mm3

(140, 90, 30)mm

(200, 0, 0)mm

(160, 0, 100)mm

Problem 2.145 By using Eqs. (2.23) and (2.34), showthat

U V0W

.

Solution: One strategy is to expand the determinant in terms ofits components, take the dot product, and then collapse the expansion.Eq. (2.23) is an expansion of the dot product: Eq. (2.23): U V . Eq. (2.34) is the determinant representationof the cross product:

Eq. (2.34) U 0 i j k

For notational convenience, write P U 0 V . Expand the determi-nant about its first row:

P i

j

k

Since the two-by-two determinants are scalars, this can be written inthe form: P i j k where the scalars , , and arethe two-by-two determinants. Apply Eq. (2.23) to the dot product ofa vector Q with P. Thus Q P . Substitute, , and into this dot product

Q P

But this expression can be collapsed into a three-by-three determinantdirectly, thus:

Q U 0 V

. This completes the demonstration.


Problem 2.146 The vectors U ij 4k, V 2i j 2k, and W 3i j 2k are coplanar (theylie in the same plane). What is the component ?

Solution: Since the non-zero vectors are coplanar, the cross pro-duct of any two will produce a vector perpendicular to the plane, andthe dot product with the third will vanish, by definition of the dotproduct. Thus U V 0 W 0, for example.

U V 0 W 1 42 1 23 1 2

1 2 2 4 6 4 2 3

10 20 0

Thus 2

Problem 2.147 The magnitude of F is 8 kN. ExpressF in terms of scalar components.

F

(7, 2) m

(3, 7) m

Solution: The unit vector collinear with the force F is developedas follows: The collinear vector is r 7 3 i 2 7 j 4i 5j

The magnitude: r

42 52 6403 m. The unit vector is

e r

r 06247i 07809j. The force vector is

F Fe 4998i 6247j 5i 625j (kN)

Problem 2.148 The magnitude of the vertical force Wis 600 lb, and the magnitude of the force B is 1500 lb.Given that A BW 0, determine the magnitude ofthe force A and the angle .

50

B W

A

Solution: The strategy is to use the condition of force balance todetermine the unknowns. The weight vector is W 600j. The vectorB is

B 1500 i cos 50 j sin 50 9642i 11491j

The vector A is A A i cos 180 j sin 180

A A i cos j sin . The forces balance, hence A B W 0, or 9642 A cos i 0, and 11491 600 A sin j 0. Thus A cos 9642, and A sin 5491. Take the ratio of thetwo equations to obtain tan 05695, or 297 . Substitute thisangle to solve: A 1110 lb


Problem 2.149 The magnitude of the vertical forcevector A is 200 lb. If A B C 0, what are the mag-nitudes of the force vectors B and C?

C

70 in. 100 in.

50 in.E

F

DB

A

Solution: The strategy is to express the forces in terms of scalarcomponents, and then solve the force balance equations for the un-knowns. C C i cos j sin , where

tan 5070

07143 or 355

Thus C C 08137i 05812j . Similarly, B Bi, and A 200j. The force balance equation is A B C 0. Substituting,

08137C B i 0, and 05812C 200 j 0. Solving,C 3441 lb, B 280 lb

Problem 2.150 The magnitude of the horizontal forcevector D in Problem 2.149 is 280 lb. If D E F 0,what are the magnitudes of the force vectors E and F?

Solution: The strategy is to express the force vectors in terms ofscalar components, and then solve the force balance equation for theunknowns. The force vectors are:

E E i cos j sin where tan 50

10005 or 266

Thus

E E 08944i 04472j

D 280i and F Fj

The force balance equation is D E F 0. Substitute and resolveinto two equations:

08944E 280 i 0 and 04472E F j 0

Solve: E 3131 lb, F 140 lb

Problem 2.151 What are the direction cosines of F?

Refer to this diagram when solving Problems 2.151–2.157.

Solution: Use the definition of the direction cosines and theensuing discussion.

The magnitude of F: F

202 102 102 245.

The direction cosines are cos

F

20245

08165,

cos

F

10245

04082

cos

F

10245

04082


Problem 2.152 Determine the scalar components ofa unit vector parallel to line that points from toward .

Solution: Use the definition of the unit vector, we get

The position vectors are: r 4i 4j 2k, r 8i 1j 2k. Thevector from to is r 8 4 i 1 4 j 2 2 k 4i 3j 4k. The magnitude: r

42 32 42 64. The unit

vector is

e r

r

464

i3

64j

464

k 06247i 04685j 06247k

Problem 2.153 What is the angle between the line and the force F?

Solution: Use the definition of the dot product Eq. (2.18), andEq. (2.24):

cos r FrF

From the solution to Problem 2.130, the vector parallel to is r 4i 3j 4k, with a magnitude r 64. From Problem 2.151, theforce is F 20i 10j 10k, with a magnitude of F 245. The dotproduct is r F 4 20 3 10 4 10 90. Substi-

tuting, cos 90

64 245 0574, 55

Problem 2.154 Determine the vector component of Fthat is parallel to the line .

Solution: Use the definition in Eq. (2.26): U e U e, where eis parallel to a line . From Problem 2.152 the unit vector parallel toline is e 06247i 04688j 06247k. The dot product is

e F 06247 20 04688 10 06247 10 14053

The parallel vector is

e F e 14053 e 878i 659j 878k (lb)

Problem 2.155 Determine the vector component of Fthat is normal to the line .

Solution: Use the Eq. (2.27) and the solution to Problem 2.154.

F F F 20 878 i 10 659 j 10 878 k

1122i 1659j 122k (lb)

Problem 2.156 Determine the vector r 0 F, wherer is the position vector from to .

Solution: Use the definition in Eq. (2.34). Noting r r,from Problem 2.155 r 4i 3j 4k. The cross product is

r 0 F i j k4 3 4

20 10 10 30 40 i 40 80 j

40 60

70i 40j 100k (ft-lb)


Problem 2.157 (a) Write the position vector r frompoint to point in terms of components.

(b) A vector R has magnitude R 200 lb and isparallel to the line from to . Write R in terms ofcomponents.

Solution:

(a) r [8 4]i [1 4]j [ 2 2]k ft

r 4i 3j 4k ft

(b) R 200 Nr

r 125i 937j 125k N

R 125i 963j 125k N

Problem 2.158 The rope exerts a force of magnitudeF 200 lb on the top of the pole at .

(a) Determine the vector r 0 F, where r is theposition vector from to .

(b) Determine the vector r 0 F, where r is theposition vector from to .

(5, 6, 1) ft

(3, 0, 4) ft

F

Solution: The strategy is to define the unit vector pointing from

to , express the force in terms of this unit vector, and take the crossproduct of the position vectors with this force. The position vectors

r 5i 6j 1k r 3i 0j 4k

r 3 5 i 0 6 j 4 1 k 2i 6j 3k

The magnitude r

22 62 32 7. The unit vector is

e r

r02857i 08571j 04286k

The force vector is

F Fe 200e 5714i 17142j 8572k

The cross products:

r 0 F i j k5 6 1

5714 17142 8572

68574i 48574j 51426k

6857i 4857j 5143k (ft-lb)

r 0 F i j k3 0 4

5714 17142 8572

68568i 48572j 51426k

6857i 4857j 5143k (ft-lb)


Problem 2.159 The pole supporting the sign is parallelto the axis and is 6 ft long. Point is contained in the– plane. (a) Express the vector r in terms of compo-nents. (b) What are the direction cosines of r?

Solution: The vector r is

r r sin 45 i cos 45 sin 60 j cos 45 cos 60 k

The length of the pole is the component of r. Therefore

r sin 45 6 ft r 6 ft

sin 45 8.49 ft

(a) r 600i 520j 300k ft

(b) The direction cosines are

cos

r 0707 cos

r 0612 cos

r 0354

cos 0707 cos 0612 cos 0354

Problem 2.160 The component of the force F is80 lb. (a) Express F in terms of components. (b) whatare the angles , and between F and the positivecoordinate axes?

Solution: We can write the force as

F F cos 20 sin 60 i sin 20 j cos 20 cos 60 k

We know that the component is 80 lb. Therefore

F cos 20 cos 60 80 lb F 170 lb

(a) F 139i 582j 80k lb

(b) The direction cosines can be found:

cos 1 139170

355

cos 1 582170

700

cos 1 80170

620

355 700 620


Problem 2.161 The magnitude of the force vector Fis 2 kN. Express it in terms of scalar components.

(4, 3, 1) m

(6, 0, 0) m

(5, 0, 3) m

FF

F

F

Solution: The strategy is to determine the unit vector collinearwith F and then express the force in terms of this unit vector.

The radius vector collinear with F is

r 4 5 i 3 0 j 1 3 k or r 1i 3j 2k

The magnitude is

r

12 32 22 374

The unit vector is

e r

r02673i 08018j 05345k

The force is

F Fe 2e (kN) F 05345i 16036j 10693k

053i 160j 107k (kN)

F

(4,3,1)

(6,0,0)

(5,0,3)

A

F

F

F

Problem 2.162 The magnitude of the vertical forcevector F in Problem 2.161 is 6 kN. Determine the vectorcomponents of F parallel and normal to the line from to .

Solution: The projection of the force F onto the line from

to is F F e e. The vertical force has the componentF 6j (kN). From Problem 2.139, the unit vector pointing from to is e 02673i 08018j 05345k. The dot product isF e 4813. Thus the component parallel to the line is F

4813e 129i 386j 257k (kN). The component perpen-dicular to the line is: F F F . Thus F 129i 214j257k (kN)


Problem 2.163 The magnitude of the vertical forcevector F in Problem 2.161 is 6 kN. Given that F F F F 0, what are the magnitudes of F, F, andF?

Solution: The strategy is to expand the forces into scalar compo-nents, and then use the force balance equation to solve for the un-knowns. The unit vectors are used to expand the forces into scalarcomponents. The position vectors, magnitudes, and unit vectors are:

r 4i 3j 1k r

26 51

e 07845i 05883j 01961k

r 1i 3j 2k r

14 374

e 02673i 08018j 05345k

r 2i 3j 1k r

14 374

e 05345i 08018j 02673k

The forces are:

F FeF FeF FeF 6j (kN)

Substituting into the force balance equation

F F F F 0

07843F 02674F 05348F i 0

05882F 08021F 08021F 6 j

0 01961F 05348F 02674F k 0

These simple simultaneous equations can be solved a standard method(e.g., Gauss elimination) or, conveniently, by using a commercialpackage, such as TK Solver , Mathcad , or other. An HP-28S handheld calculator was used here: F 283 (kN), F 249 (kN),F 291 (kN)

Problem 2.164 The magnitude of the vertical force Wis 160 N. The direction cosines of the position vector from to are cos 0500, cos 0866, and cos 0, and the direction cosines of the position vector from to are cos 0707, cos 0619, and cos

0342. Point is the midpoint of the line from to .Determine the vector r 0W, where r is the positionvector from to .

W

Solution: Express the position vectors in terms of scalar compo-nents, calculate r, and take the cross product. The position vectorsare: r 06 5i 0866j 0k r 03i 05196j 0k,

r 03 0707i 0619j 0342k

r 02121i 01857j 01026k

r r r 05121i 07053j 01026k

W 160j

r 0 W i j k

05121 07053 010260 160 0

1644i 0j 8195k 164i 0j 82k (N m)

W

600 mm

600 mm


Problem 2.165 The rope exerts a 500-N force Ton the hinged door.

(a) Express T in terms of components.(b) Determine the vector component of T parallel to

the line from point to point .

Solution: We have

r 02i 02j 01k m

T 500 Nr

r 333i 333j 167k N

(a) T 333i 333j 167k N

(b) We define the unit vector in the direction of and then use thisvector to find the component parallel to .

r 015i 02k m

e r

r 06i 08k

T e T e [ 06][333 N] [08][ 167 N] 06i 08k

T 200i 267k N

Problem 2.166 In Problem 2.165, let r be the posi-tion vector from point to point . Determine the crossproduct r 0 T.

Solution: From Problem 2.165 we know that

T 333i 333j 167k N

The vector r is

r 035i 02j 02k m

The cross product is

r 0 T i j k

035 02 02333 333 137

333i 125j 183k Nm

r 0 T 333i 125j 183k Nm


Problem 3.1 In Active Example 3.1, suppose that theangle between the ramp supporting the car is increasedfrom 20 to 30 . Draw the free-body diagram of the carshowing the new geometry. Suppose that the cable from to must exert a 1900-lb horizontal force on the carto hold it in place. Determine the car’s weight in pounds.

Solution: The free-body diagram is shown to the right.Applying the equilibrium equations

: sin 30 0

: cos 30 0

Setting 1900 lb and solving yields

3800 lb 3290 lb

Problem 3.2 The ring weighs 5 lb and is in equilib-rium. The force 1 4.5 lb. Determine the force 2 andthe angle .

Solution: The free-body diagram is shown below the drawing. Theequilibrium equations are

: 1 cos 30 2 cos 0

: 1 sin 30 2 sin 5 lb 0

We can write these equations as

2 sin 5 lb 1 sin 30

2 cos 1 cos 30

Dividing these equations and using the known value for 1 we have.

tan 5 lb 4.5 lb sin 30

4.5 lb cos 30 0706 352

2 4.5 lb cos 30

cos 4.77 lb

2 4.77 lb 352


Problem 3.3 In Example 3.2, suppose that the attach-ment point is moved to the right and cable isextended so that the angle between cable and theceiling decreases from 45 to 35 . The angle betweencable and the ceiling remains 60 . What are thetensions in cables and ?

Solution: The free-body diagram is shown below the picture.The equilibrium equations are:

: cos 35 cos 60 0

: sin 35 sin 60 1962 N 0

Solving we find

1610 N 985 N

Problem 3.4 The 200-kg engine block is suspendedby the cables and . The angle 40 . The free-body diagram obtained by isolating the part of the systemwithin the dashed line is shown. Determine the forces and .

Solution:

40

: cos cos 0

: sin sin 1962 N 0

Solving: 1526 kN


Problem 3.5 A heavy rope used as a mooring line fora cruise ship sags as shown. If the mass of the rope is90 kg, what are the tensions in the rope at and ?

Solution: The free-body diagram is shown.The equilibrium equations are

: cos 40 cos 55 0

: sin 40 sin 55 90 981 N 0

Solving: 679 N 508 N

Problem 3.6 A physiologist estimates that themasseter muscle of a predator, , is capable ofexerting a force as large as 900 N. Assume thatthe jaw is in equilibrium and determine the necessaryforce that the temporalis muscle exerts and the force exerted on the object being bitten.

Solution: The equilibrium equations are

: cos 22 cos 36 0

: sin 22 sin 36 0

Setting 900 N, and solving, we find

785 N 823 N


Problem 3.7 The two springs are identical, with un-stretched lengths 250 mm and spring constants 1200 N/m.

(a) Draw the free-body diagram of block .(b) Draw the free-body diagram of block .(c) What are the masses of the two blocks?

300 mm

280 mm

Solution: The tension in the upper spring acts on block A in thepositive Y direction, Solve the spring force-deflection equation forthe tension in the upper spring. Apply the equilibrium conditions toblock . Repeat the steps for block .

T 0i 1200Nm

03 m 025 m j 0i 60j N

Similarly, the tension in the lower spring acts on block A in the nega-tive Y direction

T 0i 1200Nm

028 m 025 m j 0i 36j N

The weight is W 0i Wj

The equilibrium conditions are

F F F 0 F W T T 0

Collect and combine like terms in i, j

F W 60 36 j 0

Solve W 60 36 24 N

The mass of is

Wg

24 N

981 m/s2 245 kg

The free body diagram for block B is shown.

The tension in the lower spring T 0i 36j

The weight: W 0i WjApply the equilibrium conditions to block .

F W T 0

Collect and combine like terms in i, j:

F W 36 j 0

Solve: W 36 N

The mass of is given by Wg

36 N

981 m/s2 367 kg

300 mm

280 mm

Tension,upper spring

Tension,lowerspring

Weight,mass

Tension,lower spring

Weight,mass


Problem 3.8 The two springs in Problem 3.7 are iden-tical, with unstretched lengths of 250 mm. Suppose thattheir spring constant is unknown and the sum of themasses of blocks and is 10 kg. Determine the valueof and the masses of the two blocks.

Solution: All of the forces are in the vertical direction so we willuse scalar equations. First, consider the upper spring supporting bothmasses (10 kg total mass). The equation of equilibrium for block theentire assembly supported by the upper spring is is 0, where 025 N. The equation of equilibriumfor block is 0, where 025 N. Theequation of equilibrium for block alone is 0where . Using 981 m/s2, and solving simultane-ously, we get 1962 N/m 4 kg, and 6 kg .

Problem 3.9 The inclined surface is smooth (Remem-ber that “smooth” means that friction is negligble). Thetwo springs are identical, with unstretched lengths of250 mm and spring constants 1200 N/m. What arethe masses of blocks and ?

Solution:

1 1200 N/m 03 025 60 N

2 1200 N/m 028 025 36 N

: 2 sin 30 0

: 1 2 sin 30 0

Solving: 489 kg 734 kg

1

2

2

N

N


Problem 3.10 The mass of the crane is 20,000 kg. Thecrane’s cable is attached to a caisson whose mass is400 kg. The tension in the cable is 1 kN.

(a) Determine the magnitudes of the normal andfriction forces exerted on the crane by thelevel ground.

(b) Determine the magnitudes of the normal andfriction forces exerted on the caisson by thelevel ground.

Strategy: To do part (a), draw the free-body diagramof the crane and the part of its cable within thedashed line.

45

Solution:

(a) : crane 1962 kN 1 kN sin 45 0

: crane 1 kN cos 45 0

crane 1969 kN crane 0707 kN

(b) : caisson 3924 kN 1 kN sin 45 0

: 1 kN cos 45 caisson 0

caisson 322 kN caisson 0707 kN

crane

crane

196.2 kN1 kN

45

45

1 kN kN

caisson

caisson

Problem 3.11 The inclined surface is smooth. The100-kg crate is held stationary by a force applied tothe cable.

(a) Draw the free-body diagram of the crate.(b) Determine the force .

Solution:

(a) The FBD

60

(b) : 981 N sin 60 0

850 N


Problem 3.12 The 1200-kg car is stationary on thesloping road.

(a) If 20 , what are the magnitudes of the totalnormal and friction forces exerted on the car’s tiresby the road?

(b) The car can remain stationary only if the totalfriction force necessary for equilibrium is notgreater than 0.6 times the total normal force.What is the largest angle for which the car canremain stationary?

Solution:

(a) 20

: 11772 kN cos 0

: 11772 kN sin 0

1106 kN 403 kN

(b) 06 N

: 11772 kN cos 0 310

: 11772 kN sin 0

11.772 kN

Problem 3.13 The 100-lb crate is in equilibrium on thesmooth surface. The spring constant is 400 lb/ft. Let be the stretch of the spring. Obtain an equation for (in feet) as a function of the angle .

Solution: The free-body diagram is shown.The equilibrium equation in the direction parallel to the inclinedsurface is

100 lb sin 0

Solving for and using the given value for we find

0.25 ft sin


Problem 3.14 A 600-lb box is held in place on thesmooth bed of the dump truck by the rope .

(a) If 25 , what is the tension in the rope?(b) If the rope will safely support a tension of 400 lb,

what is the maximum allowable value of ?

Solution: Isolate the box. Resolve the forces into scalar compo-nents, and solve the equilibrium equations.

The external forces are the weight, the tension in the rope, and thenormal force exerted by the surface. The angle between the axis andthe weight vector is 90 (or 270 ). The weight vector is

W W i sin j cos 600 i sin j cos

The projections of the rope tension and the normal force are

T T i 0j N 0i N j

The equilibrium conditions are

F W N T 0

Substitute, and collect like terms

F 600 sin T i 0

F 600 cos N j 0

Solve for the unknown tension when

For 25

T 600 sin 2536 lb.

For a tension of 400 lb, (600 sin 400 0. Solve for the unknownangle

sin 400600

0667 or 4184

T

N

W


Problem 3.15 The 80-lb box is held in place onthe smooth inclined surface by the rope . Determinethe tension in the rope and the normal force exertedon the box by the inclined surface.

Solution: The equilibrium equations (in terms of a coordinatesystem with the axis parallel to the inclined surface) are

: 80 lb sin 50 cos 50 0

: 80 lb cos 50 sin 50 0

Solving: 95.34 lb 124 lb

Problem 3.16 The 1360-kg car and the 2100-kg towtruck are stationary. The muddy surface on which thecar’s tires rest exerts negligible friction forces on them.What is the tension in the tow cable?

Solution: FBD of the car being towed

: cos 8 1334 kN sin 26 0

591 kN

kN

18

26


Problem 3.17 Each box weighs 40 lb. The angles aremeasured relative to the horizontal. The surfaces aresmooth. Determine the tension in the rope and thenormal force exerted on box by the inclined surface.

Solution: The free-body diagrams are shown.The equilibrium equations for box are

: 40 lb sin 20 cos 25 0

: 40 lb cos 20 sin 25 0

The equilibrium equations for box are

: 40 lb sin 70 cos 25 0

: 40 lb cos 70 sin 25 0

Solving these four equations yields:

51.2 lb 15.1 lb 7.30 lb 31.2 lb

Thus 51.2 lb 7.30 lb

Problem 3.18 A 10-kg painting is hung with a wiresupported by a nail. The length of the wire is 1.3 m.

(a) What is the tension in the wire?(b) What is the magnitude of the force exerted on the

nail by the wire?

Solution:

(a) : 981 N 25

13 0

128 N

(b) Force 981 N

98.1 N

5

12 12


Problem 3.19 A 10-kg painting is hung with a wiresupported by two nails. The length of the wire is 1.3 m.

(a) What is the tension in the wire?(b) What is the magnitude of the force exerted on each

nail by the wire? (Assume that the tension is thesame in each part of the wire.)

Compare your answers to the answers to Problem 3.18.

Solution:

(a) Examine the point on the left where the wire is attached to thepicture. This point supports half of the weight

: sin 273 4905 N 0

107 N

(b) Examine one of the nails

: cos 273 0

: sin 273 0

2

2

505 N

27.3

49.05 N

27.3

Problem 3.20 Assume that the 150-lb climber is inequilibrium. What are the tensions in the rope on theleft and right sides?

Solution:

cos 15 cos 14 0

sin 15 sin 14 150 0

Solving, we get 299 lb 300 lb

14 15

150 lb


Problem 3.21 If the mass of the climber shown inProblem 3.20 is 80 kg, what are the tensions in the ropeon the left and right sides?

Solution:

cos 15 cos 14 0

sin 15 sin 14 0

Solving, we get

156 kN 157 kN

14 15

(80) (9.81) N

Problem 3.22 The construction worker exerts a 20-lbforce on the rope to hold the crate in equilibrium in theposition shown. What is the weight of the crate?

5

30

Solution: The free-body diagram is shown.The equilibrium equations for the part of the rope system where thethree ropes are joined are

: 20 lb cos 30 sin 5 0

: 20 lb sin 30 cos 5 0

Solving yields 188 lb


Problem 3.23 A construction worker on the moon,where the acceleration due to gravity is 162 m/s2, holdsthe same crate described in Problem 3.22 in the positionshown. What force must she exert on the cable to holdthe crate ub equilibrium (a) in newtons; (b) in pounds?

Solution: The free-body diagram is shown.From Problem 3.22 we know that the weight is 188 lb. Thereforeits mass is

188 lb

32.2 ft/s2 5.84 slug

5.84 slug14.59 kg

slug 85.2 kg

The equilibrium equations for the part of the rope system where thethree ropes are joined are

: cos 30 sin 5 0

: sin 30 cos 5 0

where 1.62 m/s2.Solving yields 3.30 lb 14.7 N

14.7 N 3.30 lb


Problem 3.24 The person wants to cause the 200-lbcrate to start sliding toward the right. To achieve this,the of the force exerted on thecrate by the rope must equal 0.35 times the normal forceexerted on the crate by the floor. In Fig.a, the personpulls on the rope in the direction shown. In Fig.b, theperson attaches the rope to a support as shown and pullsupward on the rope. What is the magnitude of the forcehe must exert on the rope in each case?

Solution: The friction force is given by

035

(a) For equilibrium we have

: cos 20 035 0

: sin 20 200 lb 0

Solving: 66.1 lb

(b) The person exerts the force . Using the free-body diagram ofthe crate and of the point on the rope where the person grabs therope, we find

: 035 0

: 200 lb 0

: cos 10 0

: sin 10 0

Solving we find 12.3 lb


Problem 3.25 A traffic engineer wants to suspend a200-lb traffic light above the center of the two rightlanes of a four-lane thoroughfare as shown. Determinethe tensions in the cables and .

Solution:

:6

37

2

5 0

:1

37

1

5 200 lb 0

Solving: 304 lb 335 lb

61

2

1

200 lb

Problem 3.26 Cable is 3 m long and cable is4 m long. The mass of the suspended object is 350 kg.Determine the tensions in cables and .

Solution:

:35

45 0

:45

35 343 kN 0

275 kN 206 kN

4

4

33

3.43 kN


bedford 5 solucionario

Engineering

ve signicant digits

number of signicant

ve signicant gures

ft wide

ft high

footprint area

value of e

sae wrench