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Chapter 1 Problem 1.1 In 1967, the International Committee of Weights and Measures defined one second to be the time required for 9,192,631,770 cycles of the transition between two quantum states of the cesium-133 atom. Express the number of cycles in two seconds to four significant digits.

Solution:

Problem 1.2 The base of natural logarithms is e = 2.71828183... . (a) Express e to three significant digits. (b) Determine the value of e 2 to three significant digits. (c) Use the value of e you obtained in part (a) to determine the value of e 2 to three significant digits.

Solution:

The number of cycles in two seconds is 2(9,192,631, 770) = 18,385, 263, 540. Expressed to four significant digits, this is 18,390, 000, 000 or 1.839E10 cycles. 18,390,000,000 or 1.839E10 cycles.

(a) The rounded-off value is e = 2.72. (b) e 2 = 7.38905610..., so to three significant digits it is e 2 = 7.39. (c) Squaring the three-digit number we obtained in part (a) and expressing it to three significant digits, we obtain e 2 = 7.40. (a) e = 2.72. (b) e 2 = 7.39. (c) e 2 = 7.40.

[Comparing the answers of parts (b) and (c) demonstrates the hazard of using rounded-off values in calculations.]

Problem 1.3 The base of natural logarithms (see Problem 1.2) is given by the infinite series 1 1 1 e = 2+ + + + . 2! 3! 4! Its value can be approximated by summing the first few terms of the series. How many terms are needed for the approximate value rounded off to five digits to be equal to the exact value rounded off to five digits?

Solution: The exact value rounded off to five significant digits is e = 2.7183. Let N be the number of terms summed. We obtain the results N

Sum

2.5

2.666...

2.708333...

2.71666...

2.7180555...

2.7182539...

We see that summing seven terms gives the rounded-off value 2.7183. Seven.

Problem 1.4 The opening in the soccer goal is 24 ft wide and 8 ft high, so its area is 24 ft × 8 ft = 192 ft 2 . What is its area in m 2 to three significant digits? Solution: A = 192 ft 2

1m ( 3.281 ) = 17.8 m ft

A = 17.8 m 2

Problem 1.4

Problem 1.5 In 2020, teams from China and Nepal, based on their independent measurements using GPS satellites, determined that the height of Mount Everest is 8848.86 meters. Determine the height of the mountain to three significant digits (a) in kilometers; (b) in miles.

Solution: (a) The height of the mountain in kilometers to three significant digits is 8848.86 m = 8848.86 m

1 km ( 1000 ) = 8.85 km. m

(b) Its height in miles to three significant digits is 8848.86 m = 8848.86 m

ft 1 mi ( 3.281 )( 5280 ) = 5.50 mi. 1m ft

(a) 8.85 km. (b) 5.50 mi.

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Problem 1.6 The distance D = 6 in (inches). The magnitude of the moment of the force F = 12 lb (pounds) about point P is defined to be the product M P = FD. What is the value of M P in N-m (newton-meters)?

Solution: The value of M P is M P = FD = (12 lb)(6 in) = 72 lb-in. Converting units, the value of M P in N-m is

72 lb-in = 72 lb-in

1N ( 0.2248 )( 1 m ) = 8.14 N-m. lb 39.37 in 8.14 N-m.

P D

Problem 1.6

Problem 1.7 The length of this Boeing 737 is 110 ft 4 in and its wingspan is 117 ft 5 in. Its maximum takeoff weight is 154,500 lb. Its maximum range is 3365 nautical miles. Express each of these quantities in SI units to three significant digits.

Solution: Because 1 ft = 12 in, the length in meters is m (110 + 124 ft )( 0.3048 ) = 33.6 m. 1 ft In the same way, the wingspan in meters is m (117 + 125 ft )( 0.3048 ) = 35.8 m. 1 ft The weight in newtons is N ( 4.448 ) = 687,000 N. 1 lb

( 154, 500 lb )

One nautical mile is 1852 meters. Therefore, the range in meters is (3365 nautical miles)

1852 m ( 1 nautical ) = 6.23E6 m. mile

Length = 33.6 m, wingspan = 35.8 m,

Problem 1.7

weight = 687 kN, range = 6230 km.

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

(

)

km 0.6214 mi = 267mi/h . h 1 km v = 267 mi/h

v = 430

(b)

(

km 1000 m h 1 km = 392 ft/s.

v = 430

1 ft )( 0.3048 )( 1 h ) m 3600 s

v = 392 ft/s Source: Courtesy of Qilai Shen/EPA/Shutterstock.

Problem 1.8

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Problem 1.9 In the 2006 Winter Olympics, the men’s 15-km cross-country skiing race was won by Andrus Veerpalu of Estonia in a time of 38 minutes, 1.3 seconds. Determine his average speed (the distance traveled divided by the time required) to three significant digits (a) in km/h; (b) in mi/h.

Solution: (a)

 60 min  15 km   1.3 min  1 h  38 + 60 = 23.7 km/h.

v =

(

)

v = 23.7 km/h

(b)

 1 mi  v = (23.7 km/h) = 14.7 mi/h.  1.609 km  v = 14.7 mi/h

Problem 1.10 The Porsche’s engine exerts 229 ft-lb (foot-pounds) of torque at 4600 rpm. Determine the value of the torque inN-m (newton-meters). Solution: T = 229 ft-lb

1N ( 0.2248 )( 1 m ) = 310 N-m. lb 3.281 ft

T = 310 N-m

Problem 1.10

Problem 1.11 The kinetic energy of the man in Practice Example 1.1 is defined by 12 mv 2, where m is his mass and υ is his velocity. The man’s mass is 68 kg and he is moving at 6 m/s, so his kinetic energy is 12 (68 kg)(6 m/s) 2 = 1224 kg-m 2 /s 2. What is his kinetic energy in US customary units?

Solution:

Problem 1.12 The acceleration due to gravity at sea level in SI units is g = 9.81 m/s 2 . By converting units, use this value to determine the acceleration due to gravity at sea level in US customary units.

Solution:

Problem 1.13 The value of the universal gravitational constant in SI units is G = 6.67E−11 m 3 /kg-s 2 . Use this value and convert units to determine the value of G in US customary units.

Solution:

2  1 slug  1 ft T = 1224 kg-m 2 /s 2    14.59 kg  0.3048 m

(

)

= 903 slug-ft 2 /s. T = 903 slug-ft 2 /s

Use Table 1.2. The result is: 1ft ( sm )( 0.3048m ) = 32.185...( sft ) = 32.2( sft ).

g = 9.81

Converting units, 6.67E − 11 m 3 /kg-s 2 = 6.67E − 11

m 3 3.281 ft 3 kg-s 2 1m

(

)

  1 kg  0.0685 slug  = 3.44E − 8 ft 3 /slug-s 2 . G = 3.44E − 8 ft 3 /slug-s 2

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Problem 1.14 The density (mass per unit volume) of aluminum is 2700 kg/m 3 . Determine its density in slug/ft 3 .

Solution: Converting units, the density is  0.0685 slug  1 m 3 2700 kg/m 3 = 2700 kg/m 3     3.281 ft 1 kg

(

)

5.24 slug/ft 3 . 5.24 slug/ft 3 .

Problem 1.15 The cross-sectional area of the C12×30 American Standard Channel steel beam is A = 8.81 in 2 . What is its cross-sectional area in mm 2 ?

Solution: A = 8.81 in 2

( 25.41 inmm ) = 5680 mm

Problem 1.15

Problem 1.16 A pressure transducer measures a value of 300 lb/in 2. Determine the value of the pressure in pascals. A pascal (Pa) is one newton per square meter.

Solution: Convert the units using Table 1.2 and the definition of the Pascal unit. The result: 300

N 12 in 1ft ( inlb )( 4.448 )( 1 ft ) ( 0.3048 ) 1 lb m

= 2.0683...(10 6 )

( mN ) = 2.07(10 ) Pa. 2

Problem 1.17 A horsepower is 550 ft-lb/s. A watt is 1 N-m/s. Determine how many watts are generated by the engines of the passenger jet if they are producing 7000 horsepower. Solution:  550 ft-lb/s  1 m P = 7000 hp    1 hp  3.28 1ft

(

1N )( 0.2248 ) lb

= 5.22 × 10 6 W. P = 5.22 × 10 6 W

Problem 1.17

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Problem 1.18 A typical speed of the head of a driver on the PGA Tour is 100 miles per hour. Determine the speed in ft/s, m/s, and km/h. Solution: The speed in ft/s is 100 mi/h = 100 mi/h The speed in m/s is

ft 1h ( 5280 )( 3600 ) = 147 ft/s. 1 mi s

147 ft/s = 147 ft/s The speed in km/h is

m ( 0.3048 ) = 44.7 m/s. 1 ft

100 mi/h = 100 mi/h

km ( 1.609 ) = 161 km/h. 1 mi

147 ft/s, 44.7 m/s, 161 km/h.

Source: Courtesy of Takanakai/123RF.

Problem 1.18

Problem 1.19 In 2000, a carbon nanotube was shown to support a tensile stress of 63 GPa (gigapascals). A pascal is 1 N/m 2 . Determine the value of this tensile stress in lb/in 2 . (This is the amount of force in pounds that could theoretically be supported in tension by a bar of this material with a cross-sectional area of one square inch.)

Solution:

Problem 1.20 In dynamics, the moments of inertia of an object are expressed in units of ( mass ) × ( length ) 2. If one of the moments of inertia of a particular object is 600 slug-ft 2, what is its value in kg-m 2?

Solution:

Converting units, the tensile stress is

(

N 1 lb m 2 4.448 N = 9.14E6 lb/in 2 .

63E9 N/m 2 = 63E9

1m )( 39.37 ) in

9.14E6 lb/in 2 .

Converting units, 2  14.59 kg  1 m 600 slug-ft 2 = 600 slug-ft 2    1 slug  3.281 ft

(

)

= 813 kg-m 2 . 813 kg-m 2

Problem 1.21 defined by

The airplane’s drag coefficient C D is

D , S 12 ρ v 2 where D is the drag force exerted on the airplane by the air, S is the wing area, ρ is the density (mass per unit volume) of the air, and v is the magnitude of the airplane’s velocity. At the instant shown, the drag force D = 5300 N, the wing area is S = 100 m 2 , the air density is ρ = 1.226 kg/m 3 , and the velocity is v = 60 m/s. (a) What is the value of the drag coefficient? (b) Determine the values of D, S, ρ, and v in terms of US customary units and use them to calculate the drag coefficient. CD =

Source: Courtesy of Fasttailwind/Shutterstock.

Solution: (a)

Problem 1.21

The drag coefficient is CD = =

(b)

D 1 S ρv 2 2

D = 5300 N 1 2

5300 N

the reference area is

( 100 m 2 ) ( 1.226 kg/m 3 )( 60 m/s ) 2

= 0.0240.

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Converting units, the drag force is

S = 100 m 2

lb ( 0.2248 ) = 1190 lb, 1N 2

ft ( 3.281 ) = 1080 ft , 1m 2

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1.21 (Continued) the density is  0.0685 slug  1 m 3 ρ = 1.226 kg/m 3     3.281 ft 1 kg

(

)

0.00238 slug/ft 3 ,

and the velocity is v = 60 m/s

ft ( 3.281 ) = 197 ft/s. 1m

Therefore, the drag coefficient is CD = =

D 1 S ρv 2 2 1 2

1190 lb

( 1080 ft 2 ) ( 0.00238 slug/ft 3 )( 197 ft/s ) 2

= 0.0240. (a) C D = 0.0240. (b) D = 1190 lb, S = 1080 ft 2 , ρ = 0.00238 slug/ft 3 , v = 197 ft/s, C D = 0.0240.

Problem 1.22 A person has a mass of 72 kg. (a) What is their weight at sea level in newtons? (b) What is their mass in slugs? (c) What is their weight at sea level in pounds?

(c)

Solution: (a)

W = mg = ( 4.93 slugs )( 32.2 ft/s 2 ) = 159 lb. Or we could use a conversion factor to convert their weight from newtons to pounds, obtaining

From Eq. (1.6), their weight is W = mg

lb ( 0.2248 ) 1N

W = ( 706 N )

= ( 72 kg )( 9.81 m/s 2 ) = 706 N. (b)

Using Eq. (1.6), their weight is

= 159 lb.

Using a unit conversion factor, their mass in slugs is

(a) 706 N. (b) 4.93 slugs. (c) 159 lb.

 0.0685 slug  m = ( 72 kg )    1 kg = 4.93 slugs.

Problem 1.23 The 1 ft × 1 ft × 1 ft cube of iron weighs 490 lb at sea level. Determine the weight in newtons of a 1 m × 1 m × 1 m cube of the same material at sea level.

1 ft

Solution:

The weight density is γ = 490 lb 1 ft 3

The weight of the 1 m 3 cube is: W = γV =

1 ft 3

lb 1 ft 1N ( 490 )(1 m) ( 0.3048 ) ( 0.2248 ) 1 ft m lb 3

1 ft

Problem 1.23

= 77.0 kN

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Problem 1.24 The area of the Pacific Ocean is 64,186,000 square miles and its average depth is 12,925 ft. Assume that the weight per unit volume of ocean water is 64 lb/ft 3. Determine the mass of the Pacific Ocean (a) in slugs; (b) in kilograms.

Solution: The volume of the ocean is V = (64,186, 000 mi 2 )(12, 925 ft)

( 5,1280mi ft )

= 2.312 × 10 19 ft 3 . (a) m = ρV =

64 lb/ft ( 32.2 )(2.312 × 10 ft ) ft/s 3

= 4.60 × 10 19 slugs (b)  14.59 kg  m = (4.60 × 10 19 slugs)   1slug  = 6.71 × 10 20 kg

Problem 1.25 Two oranges weighing 8 oz (ounces) each at sea level are 3 ft apart. (a) What is the magnitude of the gravitational force F they exert on each other? (b) If a stationary object of mass m is subjected to a constant force F and no other forces, the time required for it to reach a speed v is mv t = . F

Solution: (a)

m =

W 0.5 lb = = 0.0155 slug. g 32.2 ft/s 2

In U.S. customary units, the universal gravitational constant is G = 3.44E − 8 ft 3 /slug-s 2 . From Eq. (1.1), the gravitational force the oranges exert on each other is

If you subjected one of the oranges to a constant force equal to the force you determined in part (a) and no other forces, how long would it take to reach a speed of 1 ft/s?

3 ft

One pound equals 10 ounces, so the weight of each orange is W = 0.5 lb, and its mass is

Gm 2 r2 ( 3.44E − 8 ft 3 /slug-s 2 )( 0.0155 slug ) 2 = ( 3 ft ) 2 = 9.22E − 13 lb.

F =

(b)

From the given equation, the time required is mv F ( 0.0155 slug )( 1 ft/s ) = 9.22E − 13 lb = 1.68E10 s (534 years).

t =

Problem 1.25

(a) 9.22E − 13 lb. (b) 1.68E10 s (534 years).

Problem 1.26 A person weighs 180 lb at sea level. The radius of the Earth is 3960 mi. What force is exerted on the person by the gravitational attraction of the Earth if he is in a space station in orbit 200 mi above the surface of the Earth?

Solution: Use Eq. (1.5). 2

( Rr ) =  Wg  g R R+ H  3960 = W ( 3960 + 200 )

W = mg

= (180)(0.90616) = 163 lb.

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Problem 1.27 The acceleration due to gravity on the surface of the Moon is 1.62 m/s 2 . The Moon’s radius is R M = 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 Practice Example 1.6, determine the force exerted on the object by the gravity of the Moon if the object is located 1738 km above the Moon’s surface.

Solution: (a)

(b)

W = mg M = (10 kg)(1.26 m/s 2 ) = 12.6 N W = 12.6 N Adapting Eq. 1.4, we have a M = g M R M r

( ) . The force is then

F = ma M = (10 kg)(1.62 m/s 2 )

1738 km ( 1738 km + 1738 km )

= 4.05 N F = 4.05 N

Problem 1.28 If an object is near the surface of the Earth, the variation of its weight with distance from the center of the Earth can often be neglected. The acceleration due to gravity at sea level is g = 9.81 m/s 2. The radius of the Earth is 6370 km. The weight of an object at sea level is mg, where m is its mass. At what height above the surface of the Earth does the weight of the object decrease to 0.99 mg ?

Solution:

Problem 1.29 A person has a mass of 72 kg. The radius of the Earth is 6370 km. (a) What is his weight at sea level in newtons and in pounds? (b) If he is in a space station 420 km above the surface of the Earth, what force is exerted on him by the Earth’s gravitational attraction in newtons and in pounds?

(b)

Use a variation of Eq. (1.5).  R E  2 W = mg   = 0.99mg  R E + h  Solve for the radial height, h = RE

= 32.09 ... km = 32,100 m = 32.1 km.

To apply Eq. (1.5), we need his distance from the center of the earth: r = 6370 km + 420 km = 6790 km. Then from Eq. (1.5), the grativational force on him is W = mg

Solution: (a)

1 − 1 ) = (6370)(1.0050378 − 1.0) ( 0.99

From Eq. (1.6), his weight at sea level is

= (72 kg)(9.81 m/s 2 )

W = mg = (72 kg)(9.81 m/s 2 ) = 706 N. Using a conversion factor, his weight at sea level in pounds is

(

0.2248 lb W = 706 N 1N = 159 lb.

RE 2 r2

)

(6,370, 000 m) 2 (6, 790, 000 m) 2

= 622 N. The force on him in pounds is W = 622 N

lb ( 0.2248 ) 1N

= 140 lb. (a) 706 N, 159 lb. (b) 622 N, 140 lb.

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Problem 1.30 If you model the Earth as a homogeneous sphere, at what height above sea level does your weight decrease to one-half of its value at sea level? Determine the answer (a) in kilometers; (b) in miles. (The radius of the Earth is 6370 km. )

Gm E . ( RE + H ) 2 Dividing this equation by Eq. (1) yields

Solution:

Solving for H, we obtain

(a)

0.5g =

0.5 =

Your weight will decrease to 0.5 times its value at sea level at the height at which the earth’s acceleration due to gravity decreases to 0.5 times its value at sea level. From Eq. (1.3), the acceleration due to gravity at a distance r from the center of the earth is Gm E , r2 where m E is the earth’s mass. The acceleration due to gravity at sea level is a =

Gm E , (1) R E2 where R E is the earth’s radius. Let H be the height above sea level at which the acceleration due to gravity equals 0.5g. That is, g =

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R E2 . ( RE + H ) 2

1 − 1) R ( 0.5 1 = ( − 1 ) (6370 km) 0.5

H =

= 2640 km. (b)

In terms of miles, the value of H is H = 2640 km

1 mi ( 1.609 ) km

= 1640 mi. (a) 2640 km. (b) 1640 mi.

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Chapter 2 Problem 2.1 Consider vectors U and V oriented as shown. Their magnitudes are U = 8 and V = 3. Graphically determine the magnitude of the vector 2U − 3V.

Solution: The vector 2U has the same direction as U and magnitude 16. The vector −3V has the direction opposite to V and magnitude 9. We choose a scale and add these vectors to obtain the vector 2U − 3 V: 45o

208 458

2U U V

2U – 3V 20o 3V

By measuring its magnitude, we estimate that 2U − 3V = 21.5. 2U − 3V = 21.5.

Problem 2.2 Suppose that the pylon in Example 2.2 is moved closer to the stadium so that the angle between the forces F AB and F AC is 45 °. Draw a sketch of the structure with the cables in their new orientation. The magnitudes of the forces are F AB = 100 kN and F AC = 60 kN. Graphically determine the magnitude and direction of the sum of the forces exerted on the pylon at A by the two cables.

Solution: FAB

45o

FAB + FAC

100 kN

28o 60 kN

FAC

From the diagram we estimate that F AB + F AC = 148 kN at 48 ° relative to the horizontal. 148 kN at 28 ° from horizontal.

Problem 2.3 Two unit vectors e 1 and e 2 are oriented as shown. Graphically determine the magnitude of the vector U = 2e 1 + 3e 2 .

Solution: The vector 2e 1 has the same direction as e 1 and magnitude 2. The vector 3e 2 has the same direction as e 2 and magnitude 3. We choose a scale and add these vectors to obtain the vector 2e 1 + 3 e 2 :

2e1 1 3e2 3e2

608

2e1 e1

By measuring its magnitude, we estimate that 2e 1 + 3 e 2 = 4.4. U = 4.4.

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Problem 2.4 Two unit vectors e 1 and e 2 are oriented as shown. Graphically determine the magnitude of the vector U = 4 e 1 − 3e 2 .

Solution: The vector 4e 1 has the same direction as e 1 and magnitude 4. The vector −3e 2 has the direction opposite to e 2 and magnitude 3. We choose a scale and add these vectors to obtain the vector 4 e 1 − 3 e 2 : 4e1

608

]3e2

4e1 ] 3e2

By measuring its magnitude, we estimate that 4 e 1 − 3 e 2 = 3.6. U = 3.6.

Problem 2.5 Three forces act on the ring. Their sum is zero: F A + FB + FC = 0. The magnitudes FB = 100 N and FC = 160 N. Graphically determine the magnitude of F A and the angle α.

Solution: If the sum of the three forces equals zero, the three force vectors form a closed triangle when placed head-to-tail. We choose a scale and draw the force triangle: o

FB o

408

FA a 808 By measuring, we estimate that F A = 140 lb, α = 42 °. F A = 140 lb, α = 42 °.

Problem 2.6 Three forces act on the ring. Their sum is zero: F A + FB + FC = 0. The magnitude FC = 200 lb. Graphically determine the value of the angle α for which the magnitude of F A is a minimum. If α has that value, what are the magnitudes of F A and FB ?

Solution: The magnitude and direction of FC are known. The direction of FB is known. If the sum of the three forces equals zero, the three force vectors form a closed triangle when placed head-to-tail. Choosing a scale and drawing the force triangle, we can see that the magnitude of F A is a minimum when it is perpendicular to the line of action of FB . Constructing the triangle in this way,

408

40 FA

808 a

and measuring, we estimate that α = 50 °, F A = 173 lb, FB = 100 lb. α = 50 °, F A = 173 lb, FB = 100 lb.

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Problem 2.7 A T-shaped structural member is suspended from cables. The member is subjected to three forces: the forces F A and FB exerted by the cables and its weight W. The weight of the member is W = 1000 lb. The sum of the forces is zero. Graphically determine the magnitudes of F A and FB .

Solution:

The directions of all three vectors and the magnitude of W are known. If the sum of the three forces equals zero, the three force vectors form a closed triangle when placed head-to-tail. We choose a scale and draw the force triangle:

FB 308 W

408 408

308

408 F A

308

By measuring, we estimate that F A = 530 lb, FB = 690 lb. F A = 530 lb, FB = 690 lb.

Problem 2.8 A T-shaped structural member is suspended from cables. The member is subjected to three forces: the forces F A and FB exerted by the cables and its weight W. The sum of the forces is zero. Suppose that the magnitude of the force exerted by either cable must not exceed 1200 lb. Graphically determine the largest magnitude of the weight W that can be suspended in this way.

Solution:

The directions of all three vectors are known. If the sum of the three forces equals zero, the three force vectors form a closed triangle when placed head-to-tail. We draw the force triangle:

FB 308 W

408

308

The force FB has the larger magnitude of the two cable forces. It must not exceed 1200 lb. Assuming it has that value and measuring, we estimate that the magnitude of W is 1740 lb.

408 F A

308

1740 lb.

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Problem 2.9 The 2700-kg tractor is at rest on the inclined surface. The angle α = 30 ° relative to the horizontal. The forces exerted on the tractor’s tires by the inclined surface are represented by the normal force N, which is perpendicular to the surface, and the friction force f, which is tangential to the surface. The sum of the normal force, the friction force, and the tractor’s weight is zero: N + f + W = 0. Graphically determine the magnitudes of N and f in newtons.

Solution: The tractor’s weight is (2700 kg)(9.81 m/s 2 ) = 26.5 kN. The directions of all three vectors and the magnitude of W are known. If the sum of the three forces equals zero, the three force vectors form a closed triangle when placed head-to-tail. We choose a scale and draw the force triangle: f 308

W N 308

By measuring, we estimate that N = 23.0 kN, f = 13.5 kN. W

N = 23.0 kN, f = 13.5 kN.

Source: Courtesy of Tka4ko/Shutterstock.

Problem 2.10 The 2700-kg tractor is at rest on the inclined surface. The forces exerted on the tractor’s tires by the inclined surface are represented by the normal force N, which is perpendicular to the surface, and the friction force f, which is tangential to the surface. The sum of the normal force, the friction force, and the tractor’s weight is zero: N + f + W = 0. Graphically determine the value of the angle α relative to the horizontal for which the magnitudes of N and f are equal. What is their magnitude?

Solution:

The tractor’s weight is (2700 kg)(9.81 m/s 2 ) = 26.5 kN. If the sum of the three forces equals zero, the three force vectors form a closed triangle when placed head-to-tail. The vector f is normal (perpendicular) to the vector N, so these two vectors can be of equal magnitude only if they have the following configuration:

f 458 W

N 458 We see immediately that α = 45 °. Measuring, we estimate that N = f = 18.8 kN. f a

α = 45 °, N = f = 18.8 kN. W N

Source: Courtesy of Tka4ko/Shutterstock.

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Problem 2.11 A spherical storage tank is suspended from cables. The tank is subjected to three forces: the forces F A and FB exerted by the cables and its weight W. The weight of the tank is W = 800 lb. The vector sum of the forces acting on the tank equals zero. Graphically determine the magnitudes of F A and FB .

Solution:

Measuring a vertical distance of 8 units and drawing two lines at 20 ° to where they intersect, we obtain the diagram

FB 208 208

408

208

Measuring the magnitudes, we estimate that F A = FB = 424 lb. F A = FB = 424 lb. W

Solution:

Problem 2.12 The rope ABC exerts forces FBA and FBC of equal magnitude on the block at B. The magnitude of the total force exerted on the block by the two forces is 200 lb. Graphically determine FBA .

Draw the vectors accurately and then measure the unknown

magnitudes.

|FBA| 5 174 lb |FBC|

FBC

208 208

B |FBA| |R| 5 200 lb FBA

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Problem 2.13 Two snowcats tow an emergency shelter to a new location near McMurdo Station, Antarctica. (The top view is shown. The cables are horizontal.) The total force F A + FB exerted on the shelter is in the direction of the line L and has a magnitude of 300 lb. Graphically determine the magnitudes of F A and FB .

Solution: Measuring a vertical distance of 3 units and drawing two lines at 30 ° and 50 ° as shown to identify where they intersect, we obtain the diagram FA FA 1 FB

508

300 lb

308

508

Measuring the magnitudes, we estimate that F A = 150 lb, FB = 234 lb. F A = 150 lb, FB = 234 lb.

Top View

Problem 2.14 A surveyor determines that the horizontal distance from A to B is 400 m and the horizontal distance from A to C is 600 m. Graphically determine the magnitude of the vector rBC and the angle α.

Solution: Draw the vectors accurately and then measure the unknown magnitude and angle. |rBC| 5 390 m a 5 21.28

North B

A a

|rBC|

rBC C 608 208

East

Problem 2.15 The figure shows three forces acting on a joint of a structure. The magnitude of FC is 60 kN, and F A + FB + FC = 0. Graphically determine the magnitudes of F A and FB .

Solution: Measuring a vertical distance of 6 units and drawing two lines at 40 ° and 15 ° as shown to identify where they intersect, we obtain the diagram

y FC

158

FB 158

408 FA

60 kN FC 408 Measuring the magnitudes, FB = 110 kN.

estimate

that

F A = 138 kN,

F A = 138 kN, FB = 110 kN.

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Problem 2.16 explain why

By drawing sketches of the vectors,

First we add U and V, then add W:

U + ( V + W) = (U + V) + W. Solution:

Consider three arbitrary vectors:

U U1V

U1V V

(U 1 V) 1 W

From this diagram we can also see that U + ( V + W) = (U + V) + W:

U V 1 W

W U 1 (V 1 W)

Problem 2.17 Two forces are given in terms of their components by F A = 60 i − 20 j (lb), FB = 30 i + 40 j (lb). (a) What are the magnitudes of the forces? (b) What is the magnitude of their sum F A + FB?

Solution: (a) FA =

(60 lb) 2 + (−20 lb) 2

= 63.2 lb. FB =

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

(30 lb) 2 + (40 lb) 2

= 50.0 lb. (b) F A + FB = (60 + 30) i + (−20 + 40) j (lb) = 90 i + 20 j (lb). F A + FB =

(90 lb) 2 + (20 lb) 2

= 92.2 lb. (a) F A = 63.2 lb, FB = 50.0 lb. (b) F A + FB = 92.2 lb.

Problem 2.18 Consider three vectors A = 12 i + 8 j, B = B x i + B y j, and C = C x i + C y j. Their sum is zero, A + B + C = 0, and the components of B and C satisfy the relations B x = −2 B y and C x = 3C y . What are the magnitudes of B and C?

Solving these equations together with the two given equations, we obtain

Solution:

If the sum of the three vectors is zero, that implies that each component of their sum is zero. That yields the two equations

B =

12 + B x + C x = 0,

C = (−16.8) 2 + (−5.6) 2 = 17.7.

B = 4.8i − 2.4 j, C = −16.8i − 5.6 j. Therefore

8 + B y + C y = 0.

(4.8) 2 + (−2.4) 2

= 5.37,

B = 5.37, C = 17.7.

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Problem 2.19 The vector r = xi + yj extends to a point on the curve described by the equation shown. Determine the values of x and y for which the magnitude of the vector is a minimum. What is the minimum magnitude? y

We equate the derivative of this expression to zero, −2 x +

1 3 x = 0, 4

obtaining the roots x = 0 and x = ± 8. Substituting these values into the expression for the magnitude in terms of x, we find that r = 4 at x = 0 and r = 3.46 at x = ± 8. So the minimum magnitude occurs at x = ± 8. At both of these points, y = 2. If we examine a graph of the magnitude as a function of x, 8 7.5

(x, y)

Solution: r =

1 4 + x . 16

We want to find the value of x for which the magnitude of r is a minimum. We can simplify things a bit by seeking the value of x for which the square of the magnitude of r is a minimum: r 2 = 16 − x 2 +

5.5 5

x2 + y2.

Substituting the relation y = 4 − (1/4) x 2 into this relation gives r =

4.5

The magnitude of r is

16 − x 2

Magnitude

6.5 y 5 4 2 1 x2 4

1 4 x . 16

3 –6

–4

–2

0 x

we see that this is an interesting case, x = 0 is a stationary point but is not the minimum magnitude. x = ± 8, y = 2, magnitude = 3.46.

Problem 2.20 The forces F A = 90 i (kN) and FB = 60 j (kN). The force FC = c ( −2 i − j ), where c is a parameter. Determine the value of c so that the magnitude of the total force exerted on the beam by the three forces is a minimum. What is the minimum magnitude? y FB

3.5

FC FA

Solution:

The total force is

F A + FB + FC = (90 − 2c) i + (60 − c) j (kN). Its magnitude is F A + FB + FC =

(90 − 2c) 2 + (60 − c) 2

8100 − 360 c + 4 c 2 + 3600 − 120 c + c 2

11,700 − 480 c + 5c 2 .

We can seek the value of c that makes the square of the magnitude a minimum. We set d (11,700 − 480 c + 5c 2 ) = −480 + 10 c = 0, dc obtaining c = 48 kN. Substituting this value into the expression for the magnitude gives F A + FB + FC = 13.4 kN. c = 48 kN, magnitude = 13.4 kN.

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Problem 2.21 The forces acting on the sailplane are its weight W = −500 j (lb), the lift L, and the drag D. The angle between the lift vector L and the vertical y-axis is 20°, and the drag vector is perpendicular to the lift vector. At the instant shown, the sum of the forces on the sailplane is zero: L + D + W = 0. Determine the magnitudes of L and D.

Solution: y L

208

208 W

x L Let L and D denote the magnitudes of the lift and drag. Then D

L = L sin 20 °i + L cos 20 ° j, D = −D cos 20 °i + D sin 20 ° j. We have

L + D + W = ( L sin 20 ° − D cos 20 °) i + ( L cos 20 ° + D sin 20 ° − 500 lb) j

= 0. Each component must equal zero, resulting in two equations: L sin 20 ° − D cos 20 ° = 0, L cos 20 ° + D sin 20 ° = 500 lb. Solving, we obtain L = 470 lb, D = 171 lb. L = 470 lb, D = 171 lb.

Problem 2.22 Two unit vectors e 1 and e 2 are oriented as shown. Determine the magnitude of the vector U = 2e 1 + 3e 2 . Strategy: Introduce a cartesian coordinate system and express e 1 and e 2 in terms of their components.

Solution:

We introduce the coordinate system shown: y

e2 608 e1 e2

608

The unit vectors have magnitude one, so we can express them as e 1 = i, e 2 = cos60 °i + sin 60 ° j. e1

With these expressions, the vector U = 2e 1 + 3e 2 = (2 + 3cos60 °) i + 3sin 60 ° j. Its magnitude is U = (2 + 3cos60 °) 2 + (3sin 60°) 2 = 4.36. U = 4.36.

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Problem 2.23 Two forces act on the support. Their magnitudes are F A = 300 lb, FB = 200 lb. Determine the magnitude F A + FB of the total force exerted on the support by the two forces. y

Solution:

Using similar triangles to determine the components, the force F A in terms of its components is 11 (300 lb) i + (11) 2 + (5) 2 = 273i + 124 j (lb).

FA =

5 (300 lb) j (11) 2 + (5) 2

The force FB in terms of its components is

FB FA 5

7 (200 lb) i + (7) 2 + (5) 2 = −163i + 116 j (lb).

FB = −

5 (200 lb) j (7) 2 + (5) 2

The total force is x

F A + FB = 110 i + 240 j (lb), and the magnitude of the total force is F A + FB =

(110 lb) 2 + (240 lb) 2

= 265 lb. F A + FB = 265 lb.

Problem 2.24 The person exerts a 30-lb force F to push the crate onto a truck. (a) Express F in terms of components using the coordinate system shown. (b) The weight of the crate is 120 lb. What is the magnitude of the sum of the two forces exerted by the person and the crate’s weight?

Solution: y

F Fy

208 Fx

W F

208

(a) The force F in terms of its components is

F = Fx i + Fy j x

= (30 lb) cos 20 ° i + (30 lb)sin 20 ° j = 28.2 i + 10.3 j (lb). (b) The sum of the forces is F + W = 28.2 i + 10.3 j − 120 j (lb) = 28.2 i − 110 j (lb). Its magnitude is F + W = (28.2 lb) 2 + (−110 lb) 2 = 113 lb. (a) F = 28.2 i − 110 j (lb). (b) 113 lb.

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Problem 2.25 The missile’s engine exerts a 130-kN thrust F. (a) Express F in terms of components using the coordinate system shown. (b) The mass of the missile is 1800 kg. What is the magnitude of the sum of the thrust F and the missile’s weight?

(a) The thrust F in terms of its components is F = Fx i + Fy j 4 3 (130 kN) i + (130 kN) j 5 5 = 104 i + 78 j (kN). =

(b) The missile’s weight is

W = mg = (1800 kg)(9.81 m/s 2 ) = 17.7 kg.

The sum of the forces is

F + W = 104 i + 78 j − 17.7 j (kN) = 104 i + 60.3 j (kN). x

Its magnitude is

Solution:

F + W = (104 kN) 2 + (60.3 kN) 2 = 120 kN.

(a) F = 104 i + 78 j (kN). (b) 120 kN.

F Fy

Fx W

Problem 2.26 The magnitude of the force F A is 8 kN. The magnitude of the vertical force FB is 2 kN. For what value of the angle α in the range 0 ≤ α ≤ 90 ° is the magnitude of the sum of the two forces equal to 9 kN ?

Solution: In terms of the coordinate system shown, the two forces in terms of their components are F A = (8 kN)(cos αi + sin α j), FB = (2 kN) j. The magnitude of their sum is

F A + FB = = FA

68 + 32sin α (kN).

We want to determine the value of α for which F A + FB =

(8cos α) 2 + (8sin α + 2) 2 (kN)

68 + 32sin α (kN) = 9 kN.

Squaring both sides gives the equation 68 + 32sin α = 81. Solving, we obtain α = 24.0 °. α = 24.0 °. y

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Problem 2.27 The x-axis is parallel to the bar AB. The x ′-axis is parallel to the bar BC. The magnitude of the vector F is 200 lb. (a) Express F in terms of its components in terms of the x – y coordinate system. Use these components to determine the m agnitude of F. (b) Express F in terms of its components in terms of the x ′ – y ′ coordinate system. Use these components to determine the magnitude of F. y9

5 ft

x9 x

(a) The angle between bar AB and bar BC is arctan

( 105 ) = 26.6°.

Therefore, the angle between the vector F and the x-axis is 26.6 ° + 40 ° = 66.6 °. The vector F expressed in terms of the x -y coordinate system is F = (200 lb) cos66.6 °i + (200 lb)sin 66.6 ° j = 79.5i + 183.5 j (lb). Using these components to determine the magnitude of F yields

F 408

Solution:

F = (79.5 lb) 2 + (183.5 lb) 2 = 200 lb. (b) The vector F expressed in terms of the x ′-y ′ coordinate system is F = (200 lb) cos 40 °i′ + (200 lb)sin 40 ° j′ = 153.2 i′ + 128.6 j′ (lb).

10 ft

Using these components to determine the magnitude of F yields F = (153.2 lb) 2 + (128.6 lb) 2 = 200 lb. (a) F = 79.5i + 183.5 j (lb), F = 200 lb. (b) F = 153.2 i′ + 128.6 j′ (lb), F = 200 lb.

Problem 2.28 Suppose that the magnitude of the component of the vector F parallel to the x ′-axis is 150 lb. (a) What is the magnitude of F? (b) Express F in terms of its components in terms of the x – y coordinate system. y9

5 ft

F cos 40 ° = 150 lb for the magnitude of F yields F = 196 lb. (b) The angle between bar AB and bar BC is

F 408

x9 x

Solution: (a) The component of F parallel to the x ′-axis is F cos 40 °. Solving

10 ft

arctan

( 105 ) = 26.6°.

Therefore the angle between the vector F and the x-axis is 26.6 ° + 40 ° = 66.6 °. The vector F expressed in terms of the x -y coordinate system is F = (196 lb) cos66.6 °i + (196 lb)sin 66.6 ° j = 77.9 i + 180 j (lb). (a) F = 196 lb. (b) F = 77.9 i + 180 j (lb).

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Problem 2.29 The magnitudes A = 12 and B = 10. (a) Express A in terms of its components in terms of the x ′ – y ′ coordinate system. (b) Express B in terms of its components in terms of the x – y coordinate system. y9

Solution: (a) The angle between the vector A and the x ′-axis is 20 ° + 40 ° = 60 °. The vector A expressed in terms of the x ′-y ′ coordinate system is A = 12 cos60 °i′ + 12sin 60 ° j′ = 6 i′ + 10.4 j′. (b) The angle between the vector B and the x-axis is 15 ° − 40 ° = −25 °. The vector B expressed in terms of the x -y coordinate system is B = 10 cos 25 °i − 10 sin 25 ° j = 9.06 i − 4.23 j.

158

408

(a) A = 6 i′ + 10.4 j′. (b) B = 9.06 i − 4.23 j.

A 208

A = 12

Problem 2.30 The magnitudes B = 10. Determine A + B . y9

and

Solution:

To determine the sum of A and B, we must express them in terms of their components in terms of the same coordinate system. The angle between the vector A and the x ′-axis is 20 ° + 40 ° = 60 °. The vector A expressed in terms of the x ′-y ′ coordinate system is A = 12 cos60 °i′ + 12sin 60 ° j′ = 6 i′ + 10.4 j′.

The vector B expressed in terms of the x ′-y ′ coordinate system is 158

408

B x9

B = 10 cos15 °i′ + 10 sin15 ° j′ = 9.66 i′ + 2.59 j′. The sum of A and B is A + B = (6 i′ + 10.4 j′) + (9.66 i′ + 2.59 j′) = 15.7 i′ + 13.0 j′.

Therefore A 208

A + B = (15.7) 2 + (13.0) 2 = 20.3. A + B = 20.3.

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Problem 2.31 In Practice Example 2.3, the cable AB exerts a 900-N force on the top of the tower. Suppose that the attachment point B is moved in the horizontal direction farther from the tower, and assume that the magnitude of the force F the cable exerts on the top of the tower is proportional to the length of the cable. (a) What is the distance from the tower to point B if the magnitude of the force is 1000 N? (b) Express the 1000-N force F in terms of components using the coordinate system shown.

Solution:

In the new problem assume that point B is located a distance d away from the base. The lengths in the original problem and in the new problem are given by L original = L new =

(40 m) 2 + (80 m) 2 =

8000 m 2

d 2 + (80 m) 2

(a) The force is proportional to the length. Therefore 1000 N = (900 N) d =

d 2 + (80 m) 2 8000 m 2

(8000 m 2 )

N ( 1000 ) − (80 m) = 59.0 m. 900 N 2

d = 59.0 m (b) The force F is then  80 m j  d 2 + (80 m) 2 

 d F = (1000 N)  i −  d 2 + (80 m) 2 = (593i − 805 j) N. F = (593i − 805 j) N

Solution:

Problem 2.32 Determine the position vector r AB in terms of its components if (a) θ = 30 °; (b) θ = 225 °.

(a) r AB = (60) cos(30 °) i + (60)sin(30 °) j, or r AB = 51.96 i + 30 j mm. And (b) r AB = (60) cos(225 °)i + (60)sin (225 °)j or r AB = −42.4 i − 42.4 j mm.

150 mm

60 mm

y rAB B u A

150 mm

60 mm

rBC C

FAB A

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FBC C

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

Solution:

Problem 2.34 Using laser instruments, a surveyor determines that the coordinates of point B are x = 120 m, y = 122 m. He knows that rOA = 125 m and r AB = 62 m. Determine the vector rOA .

Solution:

θ = tan −1

( 82 ftft ) = 14.04°

F = 4800 lb(− cos θ i + sin θ j). F = (−4660 i + 1160 j) lb

a B A

N b x

rAB

We can apply geometry. The angle β = arctan (122 m/120 m) = 45.5 °. The side b = (122 m) 2 + (120 m) 2 = 171 m, and we are told that a = 125 m, c = 62 m. Applying the law of cosines (Appendix A), rOA

c 2 = a 2 + b 2 − 2ab cos α,

Proposed roadway

we obtain cos α = 0.956, so α = 16.3 °. We can now determine the position vector of A: x

rOA = (125 m) cos(α + β ) i + (125 m)sin(α + β ) j = 59.1i + 110 j (m). rOA = 59.1i + 110 j (m).

Problem 2.35 The magnitude of the position vector rBA from point B to point A is 6 m and the magnitude of the position vector rCA from point C to point A is 4 m. What are the components of rBA? Solution:

y 3m B

The coordinates are: A( x A , y A ), B(0, 0), C (3 m, 0) Thus

rBA = ( x A − 0) i + ( y A − 0) j ⇒ (6 m) 2 = x 2A + y 2A rCA = ( x A − 3 m) i + ( y A − 0) j ⇒ (4 m) 2 = ( x A − 3 m) 2 + y 2A Solving these two equations, we find x A = 4.833 m, y A = ±3.555 m. We choose the “-” sign and find

rBA = (4.83i − 3.56 j) m

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Problem 2.36 In Problem 2.35, determine the components of a unit vector e CA that points from point C toward point A. Strategy: Determine the components of rCA and then divide the vector rCA by its magnitude. Solution:

y 3m B

From the previous problem, we have

rCA = (1.83i − 3.56 j) m, rCA =

1.83 2 + 3.56 2 m = 3.56 m.

Thus e CA =

rCA = (0.458i − 0.889 j) rCA

Problem 2.37 The hydraulic cylinder BC exerts a force F on the boom of the crane at C. The force points from B toward C, and its magnitude is F = 300 kN. Determine the components of F.

Solution: y

2.4 m

a C 2.4 m 1m

1.2 m x

1.8 m

The angle α = arctan(2.4/1.2) = 63.4 °, so F = (300 kN)(cos αi + sin α j) = 134 i + 268 j (kN).

1.2 m

Alternatively, the distance from B to C is

7m x

(1.2 m) 2 + (2.4 m) 2 = 2.68 m, so 1.2 m 2.4 m (300 kN) i + (300 kN) j 2.68 m 2.68 m = 134 i + 268 j (kN).

F =

F = 134 i + 268 j (kN).

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Problem 2.38 The length of the bar AB is 0.6 m. Determine the components of a unit vector e AB that points from point A toward point B.

Solution:

We need to find the coordinates of point B( x, y ) We have the two equations (0.3 m + x ) 2 + y 2 = (0.6 m) 2 x 2 + y 2 = (0.4 m) 2 B

Solving we find x = 0.183 m, y = 0.356 m.

e AB =

0.6

0.4

Thus r AB (0.183 m − [−0.3 m]) i + (0.356 m) j = r AB (0.183 m + 0.3 m) 2 + (0.356 m) 2

= (0.806 i + 0.593 j)

Problem 2.39 The link AC exerts a force F AC on the hydraulic cylinder at C. The force points from C toward A, and its magnitude is F AC = 650 N. Determine the components of F AC .

0.3 m

Solution: y

C FAC

1m D

0.6 m a

Hydraulic cylinder

1m 0.6 m

0.15 m x

0.6 m 0.15 m

Scoop

The angle α = arctan(0.15/0.6) = 14.0 °, so x

F AC = (650 N)(sin αi − cos α j) = 158i − 631 j (N). Alternatively, the distance from A to C is (0.15 m) 2 + (0.6 m) 2 = 0.618 m, so 0.15 m 0.6 m (650 N) i − (650 N) j 0.618 m 0.618 m = 158i − 631 j (N).

F AC =

F AC = 158i − 631 j (N).

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Problem 2.40 The link CD exerts a force FCD on the hydraulic cylinder at C. The force points from C toward D, and its magnitude is FCD = 600 N. Determine the components of FCD .

1m D C

Hydraulic cylinder

Solution:

The angle α = arctan(0.4/1.0) = 21.8 °, so

0.6 m

FCD = (600 N)(cos αi + sin α j) = 557 i + 223 j (N).

0.6 m

Alternatively, the distance from C to D is

Scoop

0.15 m

(1 m) 2 + (0.4 m) 2 = 1.077 m, so

1.0 m 0.4 m FCD = (600 N) i + (600 N) j 1.077 m 1.077 m = 557 i + 223 j (N).

FAC

FCD = 557 i + 223 j (N).

0.6 m a

A 0.15 m x

Problem 2.41 A surveyor finds that the length of the line OA is 1500 m and the length of the line OB is 2000 m. (a) Determine the components of the position vector from point A to point B. (b) Determine the components of a unit vector that points from point A toward point B.

Solution:

We need to find the coordinates of points A and B

rOA = 1500 cos60 °i + 1500 sin 60 ° j rOA = 750 i + 1299 j (m) Point A is at (750, 1299) (m) rOB = 2000 cos30 °i + 2000 sin 30 ° j (m) rOB = 1732 i + 1000 j (m) Point B is at (1732, 1000) (m)

(a) The vector from A to B is r AB = ( x B − x A ) i + ( y B − y A ) j

Proposed A bridge

r AB = 982 i − 299 j (m). (b) The unit vector e AB is

e AB =

r AB 982 i − 299 j = 1026.6 r AB

e AB = 0.957 i − 0.291 j. 608 308 O

River x

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

518

408

T2 298

Solution:

The x-component of the total force is

T x = T1 cos 9 ° + T2 cos 29 ° T3 cos 40 ° + T4 cos 51 ° T x = (2800 lb) cos 9 ° + (3200 lb) cos 29 ° + (4000 lb) cos 40 ° + (5000 lb) cos 51 ° T x = 11,800 lb The y-component of the total force is T y = T1 sin 9 ° + T2 sin 29 ° + T3 sin 40 ° + T4 sin 51 ° T y = (2800 lb)sin 9 ° + (3200 lb)sin 29 ° + (4000 lb)sin 40 ° + (5000 lb)sin 51 ° T y = 8450 lb The magnitude of the total force is T =

T x2 + T y2 =

(11,800 lb) 2 + (8450 lb) 2 = 14,500 lb

T = 14,500 lb

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Problem 2.43 The tensions in the four cables are T1 = T , T2 = 1.5T , T3 = 2.5T , and T4 = 3 T . Determine the value of T so that the four cables exert a total force of 24 kN magnitude on the support. y T4

518

408

T2 298

Solution:

The sum of the x components of the forces is

T cos 9 ° + 1.5T cos 29 ° + 2.5T cos 40 ° + 3T cos 51 ° = 6.10 T . The sum of the y components of the forces is T sin 9 ° + 1.5T sin 29 ° + 2.5T sin 40 ° + 3T sin 51 ° = 4.82T . The magnitude of the total force is (6.10T ) 2 + (4.82T ) 2 = 7.78T . Equating this expression to 24 kN, we obtain T = 3.09 kN.

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Problem 2.44 The rope ABC exerts forces FBA and FBC on the block at B. Their magnitudes are equal: FBA = FBC . The magnitude of the total force exerted on the block at B by the rope is FBA + FBC = 920 N. Determine FBA by expressing the forces FBA and FBC in terms of components.

Solution: FBC = F (cos 20 °i + sin 20 ° j) FBA = F (− j) FBC + FBA = F (cos 20 °i + [sin 20 ° − 1] j) Therefore (920 N) 2 = F 2 (cos 2 20 ° + [sin 20 ° − 1] 2 ) ⇒ F = 802 N

FBC C

208 FBC

B B 208 FBA A

FBA

Problem 2.45 The magnitude of the horizontal force F1 is 5 kN and F1 + F2 + F3 = 0. What are the magnitudes of F2 and F3? Solution:

F3 308 F1

Using components we have

∑ Fx : 5 kN + F2 cos 45 ° − F3 cos30 ° = 0 458

∑ Fy : −F2 sin 45 ° + F3 sin 30 ° = 0 Solving simultaneously yields: ⇒ F2 = 9.66 kN, F3 = 13.66 kN

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Problem 2.46 Four groups engage in a tug-of-war. The magnitudes of the forces exerted by groups B, C, and D are FB = 800 lb, FC = 1000 lb, and FD = 900 lb. If the vector sum of the four forces equals zero, what is the magnitude of F A and the angle α ?

y FB

708

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

FB = 800( i cos110 ° + j sin110 °) = −273.6 i + 751.75 j

308 208 FD

FC = 1000( i cos30 ° + j sin 30 °) = 866 i + 500 j x

FD = 900( i cos( −20°) + j sin(−20°)) = 845.72 i − 307.8 j F A = F A ( i cos(180 + α) + j sin(180 + α)) = F A (−i cos α − j sin α) The sum vanishes: F A + FB + FC + FD = i(1438.1 − F A cosα) + j(944 − F A sin α) = 0 From which F A = 1438.1i + 944 j. The magnitude is FA =

(1438) 2 + (944) 2 = 1720 lb

The angle is: tan α =

944 = 0.6565, or α = 33.3 ° 1438

Problem 2.47 In Example 2.5, suppose that the attachment point of cable A is moved so that the angle between the cable and the wall increases from 40° to 55 °. Draw a sketch showing the forces exerted on the hook by the two cables. If you want the total force F A + FB to have a magnitude of 200 lb and be in the direction perpendicular to the wall, what are the necessary magnitudes of F A and FB?

Solution: Let FA and FB be the magnitudes of FA and FB . The component of the total force parallel to the wall must be zero. And the sum of the components perpendicular to the wall must be 200 lb. FA cos 55 ° − FB cos 20 ° = 0 FA sin 55 ° + FB sin 20 ° = 200 lb Solving we find F A 5 195 lb F B 5 119 lb

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Problem 2.48 The bracket must support the two forces shown, where F1 = F2 = 2 kN. An engineer determines that the bracket will safely support a total force of magnitude 3.5 kN in any direction. Assume that 0 ≤ α ≤ 90 °. What is the safe range of the angle α ?

F2 a

Solution:

∑ Fx : (2 kN) + (2 kN) cos α = (2 kN)(1 + cos α) ∑ Fy : (2 kN)sin α F2

Thus the total force has a magnitude given by F1

F = 2 kN (1 + cos α) 2 + (sin α) 2 = 2 kN 2 + 2 cos α = 3.5 kN

Thus when we are at the limits we have 2 + 2 cos α =

F1 + F2

17 49 ⇒ cos α = ⇒ α = 57.9 ° ( 3.52 kNkN ) = 16 32

In order to be safe we must have 57.9 ° ≤ α ≤ 90 °

Problem 2.49 The figure shows three forces acting on a joint of a structure. The magnitude of FC is 80 kN, and F A + FB + FC = 0. What are the magnitudes of F A and FB?

y FC FB 158

Solution:

Denote the magnitudes of F A and FB by A and B. Writing the vectors in terms of their magnitudes,

408 FA

F A = A cos 40 °i + A sin 40 ° j, FB = −B cos15 °i − B sin15 ° j, FC = −(80 kN) j. We know that F A + FB + FC = ( A cos 40 ° − B cos15 °) i + [ A sin 40 ° − B sin15 ° − (80 kN) ] j = 0 . If a vector equals zero, each of its components equals zero, so we obtain the two equations

A cos 40 ° − B cos15 ° = 0, A sin 40 ° − B sin15 ° − (80 kN) = 0.

Solving them we obtain A = F A = 183 kN, B = FB = 145 kN. 408 F A = 183 kN, FB = 145 kN.

158 FB

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Problem 2.50 Four coplanar forces act on a beam. The forces FB and FC are vertical. The vector sum of the forces is zero. The magnitudes FB = 10 kN and FC = 5 kN. Determine the magnitudes of F A and FD .

308 FA FB

Solution:

Use the angles and magnitudes to determine the vectors, and then solve for the unknowns. The vectors are: F A = F A ( i cos30 ° + j sin 30°) = 0.866 F A i + 0.5 F A j FB = 0 i − 10 j, FC = 0 i + 5 j, FD = − FD i + 0 j.

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

∑ Fx = (0.866 FA − FD )i = 0 and

∑ Fy = (0.5 FA − (10 − 5)) j = 0. From the second equation we get F A = 10 kN . Using this value in the first equation, we get FD = 8.7 kN

Problem 2.51 Six forces act on a beam that forms part of a building’s frame. The vector sum of the forces is zero. The magnitudes FB = FE = 20 kN, FC = 16 kN, and FD = 9 kN. Determine the magnitudes of F A and FG .

Solution:

Write each force in terms of its magnitude and direction as

F = F cos θ i + F sin θ j where θ is measured counterclockwise from the +x-axis. Thus, (all forces in kN) F A = F A cos110 °i + FA sin110 ° j (kN) FB = 20 cos 270 °i + 20 sin 270 ° j (kN) FC = 16 cos140 °i + 16sin140° j (kN) FD = 9 cos 40°i + 9sin 40° j (kN) FE = 20 cos 270°i + 20 sin 270° j (kN) FG = FG cos 50°i + FG sin 50° j (kN) We know that the x components and y components of the forces must add separately to zero. Thus

FA 708

FD 408

 FAx + FBx + FCx + FDx + FEx + FGx = 0   FAy + FBy + FCy + FDy + FEy + FGy = 0 

FG 508

408

 F A cos110 ° + 0 − 12.26 + 6.89 + 0 + FG cos 50 ° = 0   F A sin110 ° − 20 + 10.28 + 5.79 − 20 + FG sin 50 ° = 0  Solving, we get F A = 13.0 kN

FG = 15.3 kN y

U x

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

Solution: Let L and D be the magnitudes of the lift and drag forces. We can use similar triangles to express the vectors L and D in terms of components. Then the sum of the forces is zero. Breaking into components we have 2 L− 22 + 52

5 D = 0 22 + 52

5 L− 22 + 52

2 D − 230 lb = 0 22 + 52

y L 5

Solving we find

D = 85.4 lb, L = 214 lb

Problem 2.53 The three forces acting on the car are shown. The force T is parallel to the x-axis and the magnitude of the force W is 14 kN. If T + W + N = 0, what are the magnitudes of the forces T and N ?

Solution: ∑ Fx : T − N sin 20 ° = 0 ∑ Fy : N cos 20 ° − 14 kN = 0 Solving we find N = 14.90 N, T = 5.10 N

208

208 N

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Problem 2.54 The sum of the three forces acting on the beam is zero: F A + FB + FC = 0. The magnitude F A = 800 lb. Determine the magnitudes of FC and FB .

Solution: y FA

458

508

458

808 FB

808

Let FA , FB , FC denote the magnitudes of the forces. The sum of the x components of the forces is

ΣFx = −FA cos 45 ° − FB cos80 ° + FC cos 50 ° = 0. The sum of the x components of the forces is ΣFy = FA sin 45 ° − FB sin80 ° + FC sin 50 ° = 0. Solving these two equations, we obtain FB = 1590 lb, FC = 1310 lb. FB = 1590 lb, FC = 1310 lb.

Problem 2.55 The total force exerted on the top of the mast B by the sailboat’s forestay AB and backstay BC is 180 i − 820 j (N). What are the magnitudes of the forces exerted at B by the cables AB and BC ?

Solution:

We first identify the forces:

F AB = T AB

(−4.0 mi − 11.8 mj) (−4.0 m) 2 + (11.8 m) 2

FBC = TBC

(5.0 mi − 12.0 mj) (5.0 m) 2 + (−12.0 m) 2

y B (4, 13) m

Then if we add the force we find ∑ Fx : −

4 T AB + 155.24

5 TBC = 180 N 169

∑ Fy : −

11.8 T AB − 155.24

12 TBC = −820 N 169

Solving simultaneously yields: ⇒ T AB = 226 N, T AC = 657 N

A (0, 1.2) m

C (9, 1) m

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Problem 2.56 The structure shown forms part of a truss designed by an architectural engineer to support the roof of an orchestra shell. The members AB, AC, and AD exert forces F AB , F AC , and F AD on the joint A. The magnitude F AB = 4 kN. If the vector sum of the three forces equals zero, what are the magnitudes of F AC and F AD? Solution:

−2 i+ 22 + 32

−3 j = −0.5547 i − 0.8320 j 22 + 32

e AC =

−4 i+ 4 2 + 12

1 j = −0.9701i + 0.2425 j 4 2 + 12

e AB =

4 i+ 42 + 22

2 j = 0.89443i + 0.4472 j 42 + 22

FAB

FAC

FAD

(4, 2) m x

Determine the unit vectors parallel to each force:

e AD =

(24, 1) m

(22, 23) m

The forces are F AD = F AD e AD , F AC = F AC e AC , C

F AB = F AB e AB = 3.578i + 1.789 j. Since the vector sum of the forces vanishes, the x- and y-components vanish separately:

A D

∑ Fx = (−0.5547 FAD − 0.9701 FAC + 3.578)i = 0, and

∑ Fy = (−0.8320 FAD + 0.2425 FAC + 1.789) j = 0 These simultaneous equations in two unknowns can be solved by any standard procedure. An HP-28S hand held calculator was used here: The results: F AC = 2.108 kN , F AD = 2.764 kN

Problem 2.57 The distance s = 45 in. (a) Determine the unit vector e BA that points from B toward A. (b) Use the unit vector you obtained in (a) to determine the coordinates of the collar C.

A (14, 45) in

Solution: (a) The unit vector is the position vector from B to A divided by its magnitude rBA = ([14 − 75]i + [45 − 12] j)in = (−61i + 33 j)in rBA =

(−61 in) 2 + (33 in) 2 = 69.35 in 1 e BA = (−61i + 33 j) in = (−0.880 i + 0.476 j) 69.35 in

s B (75, 12) in x

e BA = (−0.880 i + 0.476 j) (b) To find the coordinates of point C we will write a vector from the origin to point C. rC = r A + r AC = r A + se BA = (75i + 12 j) in + (45 in)(−0.880 i + 0.476 j) rC = (35.4 i) + 33.4 j) in Thus the coordinates of C are C (35.4, 33.4) in

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Problem 2.58 Determine the x and y coordinates of the collar C as functions of the distance s. y

Solution:

The coordinates of the point C are given by

x C = x B + s(−0.880) and y C = y B + s(0.476). Thus, the coordinates of point C are x C = 75 − 0.880 s in and y C = 12 + 0.476s in. Note from the solution of Problem 2.57 above, 0 ≤ s ≤ 69.4 in.

A (14, 45) in s

B (75, 12) in x

Problem 2.59 The vector U can be expressed in terms of its components in terms of either the x–y coordinate system or the x′–y′ coordinate system:

Solution: y y9

U = U xi + U y j = U x ′ i ′ + U y ′ j′. It can be shown that the component U x′ is given in terms of the components U x and U y and the angle θ by U x ′ = U x cos θ + U y sin θ. Determine the component U y′ in terms of the components U x and U y and the angle θ . Strategy: Draw a diagram showing the components U x , U y , U x′ , and U y′ and apply trigonometry to determine U y′ . y9

Ux u Ux sinu

u Uy sinu

Uy9

From the diagram, U y ' = U y cos θ − U x sin θ. U

U y ' = U y cos θ − U x sin θ.

x9 u

Problem 2.60 Let r be the position vector from point C to the point that is a distance s meters from point A along the straight line between A and B. Express r in terms of components. (Your answer will be in terms of s.)

Solution:

rB /A = ([10 − 3]i + [9 − 4] j) m = (7 i + 5 j) m r B /A = e B /A =

y B

First define the unit vector that points from A to B.

(7 m) 2 + (5 m) 2 = 1 (7 i + 5 j) 74

74 m

Let P be the point that is a distance s along the line from A to B. The coordinates of point P are

(10, 9) m

xp = 3 m + s

s r

( 774 ) = (3 + 0.814s) m

yp = 4 m + s

A (3, 4) m

( 574 ) = (4 + 0.581s) m.

The vector r that points from C to P is then

C (9, 3) m

r = ([3 + 0.814 s − 9]i + [4 + 0.581s − 3] j) m x r = ([0.814 s − 6]i + [0.581s + 1] j) m

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Problem 2.61 A vector U = 3i − 4 j − 12k. What is its magnitude? Strategy: The magnitude of a vector is given in terms of its components by Eq. (2.14).

Solution:

Problem 2.62 Two forces are given in terms of their components by F A = 60 i − 20 j + 30 k (lb), FB = 30 i + 40 j − 10 k (lb). (a) What are the magnitudes of the forces? (b) What is the magnitude of their sum F A + FB? Strategy: The magnitude of a vector in terms of its components is given by Eq. (2.14).

Solution:

U =

Use definition given in Eq. (14). The vector magnitude is

3 2 + (−4) 2 + (−12) 2 = 13

(a) The magnitudes of the vectors are FA =

(60 lb) 2 + (−20 lb) 2 + (30 lb) 2 = 70.0 lb,

FB =

(30 lb) 2 + (40 lb) 2 + (−10 lb) 2 = 51.0 lb.

(b) The sum of the vectors is F A + FB = 90 i + 20 j + 20 k (lb). The magnitude of the sum is F A + FB =

(90 lb) 2 + (20 lb) 2 + (20 lb) 2 = 94.3 lb.

(a) F A = 70.0 lb, FB = 51.0 lb. (b) F A + FB = 94.3 lb.

Problem 2.63 Consider three vectors A = 12 i + 8 j − 16k, B = B x i + B y j + B z k, and C = C x i + C y j + C z k. Their sum is zero, A + B + C = 0, and the components of B and C satisfy the relations B x = −2 B y , B y = −B z , and C x = 3C y . What are the magnitudes of B and C?

Solution:

The sum of the vectors is zero, which yields the three

equations ΣFx = 12 + B x + C x = 0, ΣFy = 8 + B y + C y = 0, ΣFz = −16 + B z + C z = 0. We also have the three given equations B x = −2 B y , B y = −B z , C x = 3C y . Solving these six equations, we obtain B x = 4.8, B y = −2.4, B z = 2.4, C x = −16.8, C y = −5.6, C z = 13.6. The magnitudes of the vectors are B =

Bx 2 + By 2 + Bz 2

= (4.8) 2 + (−2.4 lb) 2 + (2.4) 2 = 5.88, C =

Cx 2 + Cy 2 + Cz 2

= (−16.8) 2 + (−5.6) 2 + (13.6) 2 = 22.3. B = 5.88, C = 22.3.

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Problem 2.64 A vector U = U x i + U y j + U z k. Its magnitude U = 30. Its components are related by the equations U y = −2U x and U z = 4U y . Determine the components.

Solution:

Substitute the relations between the components, determine the magnitude, and solve for the unknowns. Thus U = U x i + (−2U x ) j + (4(−2U x ))k = U x (1i − 2 j − 8k) where U x can be factored out since it is a scalar. Take the magnitude, noting that the absolute value of U x must be taken: 30 = U x

1 2 + 2 2 + 8 2 = U x (8.31).

Solving, we get U x = 3.612, or U x = ±3.61. The two possible vectors are U = +3.61i + (−2(3.61)) j + (4(−2)(3.61))k = 3.61i − 7.22 j − 28.9k U = −3.61i + (−2(−3.61) j + 4(−2)(−3.61)k = −3.61i + 7.22 j + 28.9k

Problem 2.65 An object is acted on by two forces F1 = 20 i + 30 j − 24 k (kN) and F2 = −60 i + 20 j + 40 k (kN). What is the magnitude of the total force acting on the object?

Solution: F1 = (20 i + 30 j − 24 k) kN F2 = (−60 i + 20 j + 40 k) kN F = F1 + F2 = (−40 i + 50 j + 16k) kN Thus F =

Problem 2.66 The position vector r AB extends from point A to point B. (a) Express r AB in terms of components. (b) Determine its magnitude r AB .

(−40 kN) 2 + (50 kN) 2 + (16 kN) 2 = 66 kN

y (22, 6, 3) ft A rAB

Solution:

(a) The components of r AB are r AB = (12 ft − (−2 ft)) i + (−2 ft − 6 ft) j + (8 ft − 3 ft)k = 14 i − 8 j + 5k (ft). (b)

(12, 22, 8) ft z

Its magnitude is r AB =

(14 ft) 2 + (−8 ft) 2 + (5 ft) 2

= 16.9 ft. (a) r AB = 14 i − 8 j + 5k (ft). (b) r AB = 16.9 ft.

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Problem 2.67 The position vector r AB extends from point A to point B. (a) Determine the direction cosines of r AB . (b) Determine the components of a unit vector that has the same direction as r AB .

(b) The direction cosines are the components of the desired unit vector. (Notice that the direction cosines are the components of r AB divided by its magnitude.) Therefore, e = 0.829 i − 0.474 j + 0.296k. (a) cos θ x = 0.829, cos θ y = −0.474, cos θ z = 0.296.

(b) e = 0.829 i − 0.474 j + 0.296k.

(22, 6, 3) ft A rAB x (12, 22, 8) ft B

Solution: (a) The components of r AB are r AB = (12 ft − (−2 ft)) i + (−2 ft − 6 ft) j + (8 ft − 3 ft)k = 14 i − 8 j + 5k (ft), and its magnitude is r AB =

(14 ft) 2 + (−8 ft) 2 + (5 ft) 2

= 16.9 ft. Its direction cosines are 14 ft = 0.829, 16.9 ft −8 ft cos θ y = = −0.474, 16.9 ft 5 ft cos θ z = = 0.296. 16.9 ft cos θ x =

Problem 2.68 A force vector is given in terms of its components by F = 10 i − 20 j − 20 k (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 = (10 i − 20 j − 20 k) N F =

(10 N) 2 + (−20 N) 2 + (−20 N) 2 = 30 N

(a)

cos θ x =

(b)

e = (0.333i − 0.667 j − 0.667k)

10 N −20 N = 0.333, cos θ y = = −0.667, 30 N 30 N −20 N cos θ z = = −0.667 30 N

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Problem 2.69 Express the position vector rBD from point B to point D in terms of components. What are its direction cosines?

Solution: y

(4, 3, 1) m

(0, 5, 5) m

(0, 4, 23) m C

rBD

B (4, 3, 1) m

D (0, 5, 5) m

x x A E

z The vector rBD from point B to point D is rBD = (0 − 4 m) i + (5 m − 3 m) j + (5 m − 1 m) = −4 i + 2 j + 4 k (m). Its magnitude is rBD =

(−4 m) 2 + (2 m) 2 + (4 m) 2

= 6 m. Its direction cosines are −4 m = −0.667, 6m 2m cos θ y = = 0.333, 6m 4m cos θ z = = 0.667. 6m cos θ x =

rBD = −4 i + 2 j + 4 k (m), cos θ x = −0.667, cos θ y = 0.333, cos θ z = 0.667.

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Problem 2.70 The cable from B to C exerts a force on the bar AB at B. The force points from B toward C and its magnitude is 80 N. Express the force in terms of components.

Solution: y (0, 4, 23) m C

rBC

(0, 4, 23) m C

(4, 3, 1) m B

B (4, 3, 1) m

D (0, 5, 5) m

x A

z E

The vector rBC from point B to point C is rBC = (0 − 4 m) i + (4 m − 3 m) j + (−3 m − 1 m) = −4 i + j − 4 k (m). Its magnitude is rBC =

(−4 m) 2 + (1 m) 2 + (−4 m) 2

= 5.74 m. By dividing the vector by its magnitude, we obtain a unit vector that points from point B toward point C: −4 i + j − 4 k (m) 5.74 m = −0.696 i + 0.174 j − 0.696k.

e BC =

y (0, 4, 23) m C eBC

(4, 3, 1) m B

z We obtain the force in terms of its components by multiplying the unit vector by 80 N: (80 N)e BC = −0.55.7 i + 13.9 j − 55.7k (N). −0.55.7 i + 13.9 j − 55.7k (N).

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Problem 2.71* The suspended weight E exerts an 800-N downward force on bar AB at B. The cable from B to C exerts a force on the bar at B that points from B toward C, but its magnitude is unknown. The cable from B to D exerts a force on the bar that points from B toward D, and its magnitude is also unknown. If the weight of bar AB is negligible, the sum of the three forces exerted on it at B is a vector that is parallel to the bar. Use this condition to determine the magnitudes of the forces exerted by cables BC and BD.

Let FBC be the force exerted at B by the cable BC. It can be expressed as FBC = FBC e BC = FBC (−0.696 i + 0.174 j − 0.696k). y

(4, 3, 1) m

(0, 5, 5) m rBC

y (0, 4, 23) m C

x B (4, 3, 1) m

D (0, 5, 5) m

x A

z The vector rBD from point B to point D is rBD = (0 − 4 m) i + (5 m − 3 m) j + (5 m − 1 m)k

= −4 i + 2 j + 4 k (m).

Its magnitude is rBD =

Solution:

(−4 m) 2 + (2 m) 2 + (4 m) 2

= 6 m. y

Dividing rBD by its magnitude, we obtain a unit vector that points from point B toward point D:

(0, 4, 23) m

−4 i + 2 j + 4 k (m) 6m = −0.667 i + 0.333 j + 0.667k.

e BD = rBC

(4, 3, 1) m B

Let FBD be the force exerted at B by the cable BD. It can be expressed as x

FBD = FBD e BD = FBD (−0.667 i + 0.333 j + 0.667k). y

z The vector rBC from point B to point C is

(4, 3, 1) m

rBC = (0 − 4 m) i + (4 m − 3 m) j + (−3 m − 1 m) = −4 i + j − 4 k (m). Its magnitude is rBC =

rBA (0, 0, 0)

(−4 m) 2 + (1 m) 2 + (−4 m) 2

= 5.74 m.

By dividing rBC by its magnitude, we obtain a unit vector that points from point B toward point C: −4 i + j − 4 k (m) 5.74 m = −0.696 i + 0.174 j − 0.696k.

e BC =

z The vector rBA from point B to point A is rBA = (0 − 4 m) i + (0 − 3 m) j + (0 − 1 m)k = −4 i − 3 j − k (m).

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2.71 (Continued) Its magnitude is

where F is an unknown constant. Notice that this condition yields three equations. Its i component is

rBA =

−0.696 FBC − 0.667 FBD = −0.784 F ,

(−4 m) 2 + (−3 m) 2 + (−1 m) 2

= 5.10 m.

its j component is

Dividing rBA by its magnitude, we obtain a unit vector that points from point B toward point A:

0.174 FBC + 0.333 FBD − 800 N = −0.588F , and its k component is −0.696 FBC + 0.667 FBD = −0.196 F .

−4 i − 3 j − k (m) 5.10 m = −0.784 i − 0.588 j − 0.196k.

e BA =

Solving these three equations yields FBC = 657 N, FBD = 411 N, and F = 932 N.

We can write the given condition on the forces at B as

BC : 657 N, BD : 411 N.

FBC + FBD − (800 N) j = Fe BA ,

Problem 2.72 Determine the components of the position vector rBD from point B to point D. Use your result to determine the distance from B to D.

Solution:

We have the following B(5, 0, 3) m, C (6, 0, 0) m, D(4, 3, 1) m

coordinates:

A(0, 0, 0),

rBD = (4 m − 5 m) i + (3 m − 0) j + (1 m − 3 m)k

= (−i + 3 j − 2k) m

D (4, 3, 1) m

rBD =

(−1 m) 2 + (3 m) 2 + (−2 m) 2 = 3.74 m

A C (6, 0, 0) m x z

B (5, 0, 3) m

Problem 2.73 What are the direction cosines of the position vector rBD from point B to point D ?

Solution: cos θ x =

y D (4, 3, 1) m

−1 m 3m = −0.267, cos θ y = = 0.802, 3.74 m 3.74 m −2 m cos θ z = = 0.535 3.74 m

A C (6, 0, 0) m x z

B (5, 0, 3) m

Problem 2.74 Determine the components of the unit vector e CD that points from point C toward point D.

Solution:

We have the following B(5, 0, 3) m, C (6, 0, 0) m, D(4, 3, 1) m

= (−2 i + 3 j + 1k)

D (4, 3, 1) m

rCD = A

(−2 m) 2 + (3 m) 2 + (1 m) 2 = 3.74 m

Thus C (6, 0, 0) m x

A(0, 0, 0),

rCD = (4 m − 6 m) i + (3 m − 0) j + (1 m − 0)k

coordinates:

B (5, 0, 3) m

1 (−2 i + 3 j + k) m 3.74 m = (−0.535i + 0.802 j + 0.267k)

e CD =

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Problem 2.75 What are the direction cosines of the unit vector e CD that points from point C toward point D?

Solution:

Using Problem 2.74

cos θ x = −0.535, cos θ y = 0.802, cos θ z = 0.267

y D (4, 3, 1) m

A C (6, 0, 0) m x z

B (5, 0, 3) m

Problem 2.76 In Example 2.7, suppose that the caisson shifts on the ground to a new position. The magnitude of the force F remains 600 lb. In the new position, the angle between the force F and the x-axis is 60° and the angle between F and the z-axis is 70 °. Express F in terms of components. Solution:

We need to find the angle θ y between the force F and the y-axis. We know that cos 2 θ x + cos 2 θ y + cos 2 θ z = 1 cos θ y = ± 1 − cos 2 θ x − cos 2 θ z = ± 1 − cos 2 60 ° − cos 2 70 ° = ±0.7956 θ y = ± cos −1 (0.7956) = 37.3 ° or 142.7°

We will choose θ y = 37.3 ° because the picture shows the force pointing up. Now Fx = (600 lb) cos60 ° = 300 lb Fy = (600 lb) cos 37.3 ° = 477 lb Fz = (600 lb) cos 70 ° = 205 lb Thus F = (300 i + 477 j + 205k) lb

Problem 2.77 Astronauts on the space shuttle use radar to determine the magnitudes and direction cosines of the position vectors of two satellites A and B. The vector r A from the shuttle to satellite A has magnitude 2 km and direction cosines cos θ x = 0.768, cos θ y = 0.384, cos θ z = 0.512. The vector rB from the shuttle to satellite B has magnitude 4 km and direction cosines cos θ x = 0.743, cos θ y = 0.557, cos θ z = −0.371. What is the distance between the satellites? Solution:

rB x

The two position vectors are:

r A = 2(0.768i + 0.384 j + 0.512k) = 1.536 i + 0.768 j + 1.024 k (km)

rB = 4(0.743i + 0.557 j − 0.371k) = 2.972 i + 2.228 j − 1.484 k (km) The distance is the magnitude of the difference: r A − rB =

(1.536 − 2.927) 2 + (0.768 − 2.228) 2 + (1.024 − (−1.484)) 2

= 3.24 (km)

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Problem 2.78 Archaeologists measure a pre-Columbian ceremonial structure and obtain the dimensions shown. Determine (a) the magnitude and (b) the direction cosines of the position vector from point A to point B.

y 4m

10 m

10 m 8m

Solution:

(a) The coordinates are A (0, 16, 14) m and B (10, 8, 4) m. r AB = ([10 − 0]i + [8 − 16] j + [4 − 14]k) m = (10 i − 8 j − 10 k) m r AB =

10 2 + 8 2 + 10 2 m =

z 264 m = 16.2 m

r AB = 16.2 m (b)

10 = 0.615 264 −8 = −0.492 cos θ y = 264 −10 = −0.615 cos θ z = 264 cos θ x =

Problem 2.79 Consider the structure described in Problem 2.78. After returning to the United States, an archaeologist discovers that a graduate student has erased the only data file containing the dimension b. But from recorded GPS data he is able to calculate that the distance from point B to point C is 16.61 m. (a) What is the distance b? (b) Determine the direction cosines of the position vector from B to C. Solution:

We have the coordinates B (10 m, 8 m, 4 m), C (10 m + b, 0 18 m). rBC = (10 m + b − 10 m) i + (0 − 8 m) j + (18 m − 4 m)k rBC = (b) i + (−8 m) j + (14 m)k

(a) We have (16.61 m) 2 = b 2 + (−8 m) 2 + (14 m) 2 ⇒

b = 3.99 m

(b) The direction cosines of rBC are 3.99 m = 0.240 16.61 m −8 m cos θ y = = −0.482 16.61 m 14 m cos θ z = = 0.843 16.61 m cos θ x =

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Problem 2.80 Observers at A and B use theodolites to measure the direction from their positions to a rocket in flight. If the coordinates of the rocket’s position at a given instant are (4, 4, 2) km, determine the direction cosines of the vectors r AR and rBR that the observers would measure at that instant. Solution:

rAR

The vector r AR is given by

rBR A

r AR = 4 i + 4 j + 2k km and the magnitude of r AR is given by r AR =

x z

B (5, 0, 2) km

(4) 2 + (4) 2 + (2) 2 km = 6 km.

The unit vector along AR is given by u AR = r AR r AR . Thus, u AR = 0.667 i + 0.667 j + 0.333k and the direction cosines are cos θ x = 0.667, cos θ y = 0.667, and cos θ z = 0.333. The vector rBR is given by rBR = ( x R − x B ) i + ( y R − y B ) j + ( z R − z B )k km = (4 − 5) i + (4 − 0) j + (2 − 2)k km and the magnitude of rBR is given by rBR =

(1) 2 + (4) 2 + (0) 2 km = 4.12 km.

The unit vector along BR is given by e BR = rBR rBR . Thus, u BR = −0.242 i + 0.970 j + 0 k and the direction cosines are cos θ x = −0.242, cos θ y = 0.970, and cos θ z = 0.0.

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Problem 2.81 Point p lies in the x–z plane. The line from point p to the tip of the vector r is parallel to the y-axis. If r = 12 i + 6 j + 10 k (m), what are the angles φ and ψ in degrees?

Solution:

The coordinates of the tip of the vector r are (12, 6, 10) m, so the coordinates of point p are (12, 0, 10) m. The length of the line from the origin to point p is

c x f

(12 m) 2 + (10 m) 2 = 15.6 m:

tan ψ =

The tangent of ψ is 6m = 0.384, 15.6 m

which yields θ = 21.0 °. The tangent of φ is tan φ =

(12,6,10) m

10 m = 0.833, 12 m

which yields φ = 39.8 °. φ = 39.8 °, ψ = 21.0 °.

x f

10 m

15.6 m 12 m

Problem 2.82 The height of Mount Everest was originally measured by a surveyor in the following way. He first measured the altitudes of two points and the horizontal distance between them. For example, suppose that the points A and B are 3000 m above sea level and are 10,000 m apart. He then used a theodolite to measure the direction cosines of the vector r AP from point A to the top of the mountain P and the vector rBP from point B to P. Suppose that the direction cosines of r AP are cos θ x = 0.5179, cos θ y = 0.6906, and cos θ z = 0.5048, and the direction cosines of rBP are cos θ x = −0.3743, cos θ y = 0.7486, and cos θ z = 0.5472. Using this data, determine the height of Mount Everest above sea level. z

Solution:

We have the following coordinates A(0, 0, 3000) m, B(10,000, 0, 3000) m, P( x, y, z )

Then r AP = xi + yj + ( z − 3000 m)k = r AP (0.5179 i + 0.6906 j + 0.5048k) rBP = ( x − 10,000 m) i + yj + ( z − 3000 m)k = rBP (−0.3743i + 0.7486 j + 0.5472k) Equating components gives us five equations (one redundant) which we can solve for the five unknowns. x = r AP 0.5179 y = r AP 0.6906 z − 3000 m = r AP 0.5048 ⇒ z = 8848 m x − 10000 m = −rBP − 0.7486 y = rBP 0.5472

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Problem 2.83 The distance from point O to point A is 20 ft. The straight line AB is parallel to the y-axis, and point B is in the x–z plane. Express the vector rOA in terms of components. Strategy: You can express rOA as the sum of a vector from O to B and a vector from B to A. You can then express the vector from O to B as the sum of vector components parallel to the x- and z-axes. See Example 2.8. Solution:

A rOA

308 608

See Example 2.8. The length BA is, from the right

triangle OAB, r AB = rOA sin 30 ° = 20(0.5) = 10 ft.

Similarly, the length OB is rOB = rOA cos30 ° = 20(0.866) = 17.32 ft The vector rOB can be resolved into components along the axes by the right triangles OBP and OBQ and the condition that it lies in the x -z plane. Hence,

rOB = rOB ( i cos30 ° + j cos 90 ° + k cos60 °) or

rOB = 15i + 0 j + 8.66k.

A rOA 308

x Q

z P

308

The vector rBA can be resolved into components from the condition that it is parallel to the y-axis. This vector is rBA = rBA ( i cos 90 ° + j cos 0 ° + k cos 90 °) = 0 i + 10 j + 0 k. The vector rOA is given by rOA = rOB + rBA , from which rOA = 15i + 10 j + 8.66k (ft)

Problem 2.84 The magnitudes of the two force vectors are F A = 140 lb and FB = 100 lb. Determine the magnitude of the sum of the forces F A + FB . Solution:

y FB

We have the vectors

F A = 140 lb([ cos 40 ° sin 50 °]i + [sin 40 °] j + [cos 40 ° cos 50 °]k) F A = (82.2 i + 90.0 j + 68.9k) lb

608

FB = 100 lb([− cos60 ° sin 30 °]i + [sin 60 °] j + [cos60 ° cos30 °]k)

308

FB = (−25.0 i + 86.6 j + 43.3k) lb Adding and taking the magnitude we have

408

508

F A + FB = (57.2 i + 176.6 j + 112.2k) lb FA + FB =

(57.2 lb) 2 + (176.6 lb) 2 + (112.2 lb) 2 = 217 lb

F A + FB = 217 lb

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Problem 2.85 Determine the direction cosines of the vectors F A and FB . Solution:

y FB

We have the vectors

F A = 140 lb([ cos 40 ° sin 50 °]i + [sin 40 °] j + [cos 40 ° cos 50 °]k) F A = (82.2 i + 90.0 j + 68.9k) lb

608

FB = 100 lb([ −cos60 ° sin 30 °]i + [sin 60 °] j + [cos60 ° cos30 °]k)

408

308

FB = (−25.0 i + 86.6 j + 43.3k) lb The direction cosines for FA are

508

82.2 lb 90.0 lb cos θ x = = 0.643, = 0.587, cos θ y = 140 lb 140 lb 68.9 lb cos θ z = = 0.492 140 lb The direction cosines for FB are 86.6 lb −25.0 lb = 0.866, = −0.250, cos θ y = 100 lb 100 lb 43.3 lb cos θ z = = 0.433 100 lb cos θ x =

F A : cos θ x = 0.587, cos θ y = 0.643, cos θ z = 0.492 FB : cos θ x = −0.250, cos θ y = 0.866, cos θ z = 0.433

Problem 2.86 The positions of two airplanes A and B relative to a control tower are determined by radar. Their distances from the control tower are r A = 16 km and rB = 12 km. Determine the components of the position vector r AB from plane A to plane B (that is, the vector that specifies the position of plane B relative to plane A). What is the straight-line distance between the two planes?

Solution:

The components of r A are

r A = (16 km)(cos 20 ° cos30 °i + sin 20 ° j − cos 20 ° sin 30 °k) = 13.0 i + 5.47 j − 7.52k (km). The components of are rB = (12 km)(− cos60 ° sin 40 °i + sin 60 ° j + cos60 ° cos 40 °k) = −3.86 i + 10.4 j + 4.60 k (km). y

y B

B 608

rA 208

408 z

rAB

A rA

308

The vector from plane A to plane B is r AB = rB − r A = −16.9 i + 4.92 j + 12.1k (km). The distance between the planes is r AB =

(−16.9 km) 2 + (4.92 km) 2 + (12.1 km) 2

= 21.3 km. r AB = −16.9 i + 4.92 j + 12.1k (km), distance = 21.3 km.

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Problem 2.87 An engineer calculates that the magnitude of the axial force in one of the beams of a geodesic dome is P = 7.65 kN. The cartesian coordinates of the endpoints A and B 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 components.

Solution:

The components of the position vector from B to A are

rBA = ( x A − x B ) i + ( y A − y B ) j + ( z A − z B )k = (−12.4 + 9.2) i + (22.0 − 24.4) j + (−18.4 + 15.6)k = −3.2 i − 2.4 j − 2.8k (m). Dividing this vector by its magnitude, we obtain a unit vector that points from B toward A: e BA = −0.655i − 0.492 j − 0.573k.

Therefore P = P e BA

P A

= 7.65 e BA = −5.01i − 3.76 j − 4.39k (kN).

Problem 2.88 The cable BC exerts an 8-kN force F on the bar AB at B. (a) Determine the components of a unit vector that points from point B toward point C. (b) Express F in terms of components.

Solution: (a) e BC = rBC = rBC e BC =

( x C − x B ) i + ( y C − y B ) j + ( z C − z B )k ( x C − x B ) 2 + ( y C − y B ) 2 + (z C − z B ) 2

−2 i − 6 j + 3k 2 6 3 = − i− j+ k 7 7 7 22 + 62 + 32

e BC = −0.286 i − 0.857 j + 0.429k

B (5, 6, 1) m

C (3, 0, 4) m z

(b) F = F e BC = 8e BC = −2.29 i − 6.86 j + 3.43k (kN)

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Problem 2.89 A cable extends from point C to point E. It exerts a 50-lb force T on the plate at C that is directed along the line from C to E. Express T in terms of components.

6 ft A

Solution:

Find the unit vector e CE and multiply it times the magnitude of the force to get the vector in component form, r e CE = CE = rCE

208

C 4 ft

The coordinates of point C are (4, −4 sin 20 °, 4 cos 20 °) or (4, −1.37, 3.76) (ft) The coordinates of point E are (0, 2, 6) (ft) e CE =

(0 − 4) i + (2 − (−1.37)) j + (6 − 3.76)k 4 2 + 3.37 2 + 2.24 2

6 ft

e CE = −0.703i + 0.592 j + 0.394 k

T = 50 e CE (lb)

T = −35.2 i + 29.6 j + 19.7k (lb)

2 ft z

4 ft C

4 ft

The vector rBD from point B to point D is

rBD = (18 in − 8 in) i + (−8 in − 0) j + (7 in − 0)

= 10 i − 8 j + 7k (in).

8 in

14 in

Its magnitude is rBD =

208

Problem 2.90 A cable extends from point B to point D. It exerts a 30-lb force on the bar at B that is directed along the line from B to D. Express the force in terms of components. Solution:

4 ft

2 ft

( x E − x C ) i + ( y E − y C ) j + ( z E − z C )k ( x E − x C ) 2 + ( y E − y C ) 2 + (z E − z C ) 2

(10 in) 2 + (−8 in) 2 + (7 in) 2

= 14.6 in.

(18, 28, 7) in

By dividing this vector by its magnitude, we obtain a unit vector that points from point B toward point D: 10 i − 8 j + 7k (in) 14.6 in = 0.685i − 0.548 j + 0.480 k.

e BD =

(8, 0, 0) in

We obtain the force in terms of its components by multiplying this unit vector by 30 lb:

eBD z

(30 lb)e BD = 20.6 i − 16.4 j + 14.4 k (lb).

(18, 28, 7) in D

20.6 i − 16.4 j + 14.4 k (lb). y

(8, 0, 0) in

B rBD z D

(18, 28, 7) in

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Problem 2.91 The cable AB exerts a 200-lb force F AB at point A that is directed along the line from A to B. Express F AB in terms of components. Solution:

8 ft

The coordinates of B are B(0, 6, 8). The position vector

from A to B is r AB = (0 − 6) i + (6 − 0) j + (8 − 10)k = −6 i + 6 j − 2k The magnitude is r AB =

6 ft B

6 2 + 6 2 + 2 2 = 8.718 ft.

The unit vector is u AB =

−6 6 2 i+ j− k 8.718 8.718 8.718

FAC

FAB

u AB = −0.6882 i + 0.6882 j − 0.2294 k.

A (6, 0, 10) ft

F AB = F AB u AB = 200(−0.6882 i + 0.6882 j − 0.2294 k) The components of the force are F AB = F AB u AB = 200(−0.6882 i + 0.6882 j − 0.2294 k) or F AB = −137.6 i + 137.6 j − 45.9k

Problem 2.92 The cable CD exerts a 6-kN force on the bar at C that is directed from C toward D. Express the force in terms of components.

y C

(0, 2, –1) m

y 1m

rCD

D (1.5, 0, 0) m x

z 2m y A z

C 1.5 m

(0, 2, –1) m eCD

D x

Solution: The vector rCD from point B to point D is rCD = (1.5 m − 0) i + (0 − 2 m) j + (0 − (−1 m))k = 1.5i − 2 j + k (m). Its magnitude is rCD =

D (1.5, 0, 0) m x

(1.5 m) 2 + (−2 m) 2 + (1 m) 2

= 2.69 m. By dividing this vector by its magnitude, we obtain a unit vector that points from point C toward point D: 1.5i − 2 j + k (m) 2.69 m = 0.557 i − 0.743 j + 0.371k.

e CD =

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We obtain the force in terms of its components by multiplying this unit vector by 6 kN: (6 kN)e CD = 3.34 i − 4.46 j + 2.23k (kN). 3.34 i − 4.46 j + 2.23k (kN).

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

Solution:

The coordinates of the points B (40, 0, 0), C (−40, 0, 40) D (−60, 0, −60).

are

A (0, 70, 0),

The position vectors corresponding to the cables are: r AD = (−60 − 0) i + (0 − 70) j + (−60 − 0)k r AD = −60 i − 70 k − 60 k r AC = (−40 − 0) i + (0 − 70) j + (40 − 0)k r AC = −40 i − 70 j + 40 k

FAD

r AB = (40 − 0) i + (0 − 70) j + (0 − 0)k A

r AB = 40 i − 70 j + 0 k

FAB

FAC

The unit vectors corresponding to these position vectors are: D

u AD = 60 m

60 m

= −0.5455i − 0.6364 j − 0.5455k B

40 m

C 40 m

r AD −60 70 60 = i− j− k r AD 110 110 110

u AC =

r AC 40 70 40 = − i− j+ k r AC 90 90 90

= −0.4444 i − 0.7778 j + 0.4444 k

40 m z

u AB =

r AB 40 70 = i− j + 0 k = 0.4963i − 0.8685 j + 0 k 80.6 80.6 r AB

The forces are: F AB = F AB u AB = 0.9926 i − 1.737 j + 0 k F AC = F AC u AC = −0.8888i − 1.5556 j + 0.8888 F AD = F AD u AD = −1.0910 i − 1.2728 j − 1.0910 k The resultant force exerted on the tower by the cables is: FR = F AB + F AC + F AD = −0.9875i − 4.5648 j − 0.2020 k kN

Problem 2.94 The magnitudes of the two force vectors are F A = 4.6 kN and FB = 5.2 kN. Determine the magnitude of the sum of the forces F A + FB . (Notice that the angle between the vector FB and the positive x-axis is 180 ° − 57 ° = 123 °. ) Solution:

y FB 408 578 708

448

758

The vectors in terms of their direction cosines are 508

F A = (4.6 kN)(cos θ Ax i + cos θ Ay j + cos θ Az k) = (4.6 kN)(cos 50 °i + cos 44 ° j + cos 75 °k) = 2.96 i + 3.31 j + 1.19k (kN), FB = (5.2 kN)(cos θ Bx i + cos θ By j + cos θ Bz k)

z x

= (5.2 kN)(cos123 °i + cos 40 ° j + cos 70 °k) = −2.83i + 3.98 j + 1.78k (kN). The magnitude of their sum is F A + FB =

(2.96 − 2.83) 2 + (3.31 + 3.98) 2 + (1.19 + 1.78) 2 (kN)

= 7.87 kN. F A + FB = 7.87 kN.

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Problem 2.95 The rope BD is attached to the bar ABC at B. The distance from A to B is 3 m. Determine the components of the position vector rBD from point B to point D. y

Solution:

The vector from A to C is

r AC = 3 i + 4 j − 6 k (m). Dividing this vector by its magnitude gives a unit vector that points from A toward C. We can obtain the vector from A to B by multiplying that unit vector by 3 m: r AB = (3 m)

  3 i + 4 j − 6 k (m) = (3 m)    (3 m) 2 + (4 m) 2 + (−6 m) 2  = 1.15i + 1.54 j − 2.30 k (m).

r AC r AC

B A

The vector from A to D is 4m

3m z

2m x

r AD = 7 i − 4 k (m). Because r AD = r AB + rBD , we obtain rBD = r AD − r AB = 7 i − 4 k − (1.15i + 1.54 j − 2.30 k) (m) = 5.85i − 1.54 j − 1.70 k (m). rBD = 5.85i − 1.54 j − 1.70 k (m).

Problem 2.96 The rope BD is attached to the bar ABC at B. The distance from A to B is 3 m. The rope exerts a 2-kN force on the bar at B that is directed from B toward D. Determine the components of the force. y

Dividing this vector by its magnitude gives a unit vector that points from A toward C. We can obtain the vector from A to B by multiplying that unit vector by 3 m: r AC r AC

  3 i + 4 j − 6 k (m) = (3 m)    (3 m) 2 + (4 m) 2 + (−6 m) 2  = 1.15i + 1.54 j − 2.30 k (m).

4m B

A 3m z

The vector from A to C is

r AC = 3 i + 4 j − 6 k (m).

r AB = (3 m)

C 6m

Solution:

The vector from A to D is 4m

2m x

r AD = 7 i − 4 k (m). Because r AD = r AB + rBD , we obtain rBD = r AD − r AB = 7 i − 4 k − (1.15i + 1.54 j − 2.30 k) (m) = 5.85i − 1.54 j − 1.70k (m). Using this result, the force exerted on the bar at B is F = (2 kN)

rBD rBD

  5.85i − 1.54 j − 1.70k (m) (2 kN)    (5.85 m) 2 + (−1.54 m) 2 + (−1.70 m) 2  = 1.86 i − 0.489j − 0.540k (kN). 1.86 i − 0.489 j − 0.540 k (kN).

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Problem 2.97 The circular bar has a 4-m radius and lies in the x–y plane. Express the position vector from point B to the collar at A in terms of components.

Solution:

From the figure, the point B is at (0, 4, 3) m. The coordinates of point A are determined by the radius of the circular bar and the angle shown in the figure. The vector from the origin to A is rOA = 4 cos(20 °)i + 4 sin(20 °) j m. Thus, the coordinates of point A are (3.76, 1.37, 0) m. The vector from B to A is given by rBA = ( x A − x B ) i + ( y A − y B ) j + ( z A − z B )k = 3.76 i − 2.63 j −3k m. Finally, the scalar components of the vector from B to A are (3.76, −2.63, −3) m.

4m 208

4m z

Problem 2.98 The cable AB exerts a 60-N force T on the collar at A that is directed along the line from A toward B. Express T in terms of components.

Solution:

We know rBA = 3.76 i − 2.63 j − 3k m from Problem 2.97. The unit vector u AB = −rBA rBA . The unit vector is u AB = −0.686 i + 0.480 j + 0.547k. Hence, the force vector T is given by

T = T (−0.686 i + 0.480 j + 0.547k) N = −41.1i + 28.8 j + 32.8k N

4m 208 4m

Problem 2.99 In Practice Example 2.11, suppose that the vector V is changed to V = 4 i − 6 j − 10 k. (a) What is the value of U ⋅ V ? (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 = 6 i − 5 j − 3k, V = 4 i − 6k − 10 k U ⋅ V = (6)(4) + (−5)(−6) + (−3)(−10) = 84 cos θ =

U⋅V = V V

84 = 0.814 6 2 + (−5) 2 + (−3) 2 4 2 + (−6) 2 + (−10) 2

θ = cos −1 (0.814) = 35.5 ° (a) U ⋅ V = 84, (b) θ = 35.5 °

Problem 2.100 In Example 2.12, suppose that the coordinates of point B are changed to (6, 4, 4) m. What is the angle θ between the lines AB and AC?

Solution:

Using the new coordinates we have

r AB = (2 i + j + 2k) m, r AB = 3 m r AC = (4 i + 5 j + 2k) m, r AC = 6.71 m cos θ =

r AB ⋅ r AC ((2)(4) + (1)(5) + (2)(2)) m 2 = = 0.845 r AB r AC (3 m)(6.71 m)

θ = cos −1 (0.845) = 32.4 ° θ = 32.4 °

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Problem 2.101

Solution:

Consider the vectors

Applying Eq. (2.23), the dot products are

U ⋅ V = (12)(−6) + (4)(8) + (−8)(14)

U = 12 i + 4 j − 8k, V = −6 i + 8 j + 14 k, W = 8i − 8 j + 8k.

= −152, U ⋅ W = (12)(8) + (4)(−8) + (−8)(8)

Determine the values of the dot products U ⋅ V, U ⋅ W, and V ⋅ W. What can you deduce about the three vectors from the results?

= 0, V ⋅ W = (−6)(8) + (8)(−8) + (14)(8) = 0. U ⋅ V = −152, U ⋅ W = 0, V ⋅ W = 0. U and V are not perpendicular. W is perpendicular to both U and V.

Problem 2.102 Consider the position vectors P and Q. (a) Determine the dot product P ⋅ Q by using the definition, Eq. (2.18). (b) Determine the dot product P ⋅ Q by using Eq. (2.23).

Solution: (a) The two vectors in terms of their components are P = 12 i + 10 k (m), Q = 8i (m). Their magnitudes are

P = (12 m) 2 + (10 m) 2 = 15.6 m, Q = 8 m. The angle between the two vectors is

θ = arctan

(8, 0, 0) m

m ( 10 ) = 39.8°. 12 m

From Eq. (2.18),

P · Q = (15.6 m)(8 m) cos39.8 °

= 96 m 2 .

z (12, 0, 10) m

(b)

From Eq. (2.23), P · Q = (12 m)(8 m) + (0)(0) + (10 m)(0) = 96 m 2 . (a), (b) P · Q = 96 m 2 .

Problem 2.103 Two perpendicular vectors are given in terms of their components by U = U x i − 4 j + 6k and V = 3i + 2 j − 3k. Use the dot product to determine the component U x .

Solution:

When the vectors are perpendicular, U ⋅ V ≡ 0.

Thus U ⋅ V = U xV x + U yV y + U zVz = 0 = 3U x + (−4)(2) + (6)(−3) = 0 3U x = 26 U x = 8.67

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Problem 2.104 The vectors P, Q, and F = Fx i + Fy j lie in the x–y plane. The magnitudes of P and Q are 4 m and 5 m, respectively. The dot products P ⋅ F = 50 N-m (newton-meters) and Q ⋅ F = 60 N-m. Determine the components Fx and Fy .

Solution: The components of P and Q are P = (4 m)(cos15 °i + sin15 ° j), Q = (5 m)(sin 30 °i + cos30 ° j). The two given conditions are P ⋅ F = [ (4 m) cos15 ° ] Fx + [ (4 m)sin15 ° ] Fy = 50 N-m,

Q ⋅ F = [ (5 m)sin 30 ° ] Fx + [ (5 m) cos30 ° ] Fy = 60 N-m.

Solving these two equations, we obtain Fx = 10.9 N, Fy = 7.55 N.

F 308 P 158

Problem 2.105 The magnitudes U = 10 and V = 20. (a) Use Eq. (2.18) to determine U ⋅ V. (b) Use Eq. (2.23) to determine U ⋅ V.

Solution: (a) The definition of the dot product (Eq. (2.18)) is U ⋅ V = U V cos θ. Thus U ⋅ V = (10)(20) cos(45 ° − 30 °) = 193.2

(b) The components of U and V are

U = 10( i cos 45 ° + j sin 45 °) = 7.07 i + 7.07 j U

V = 20( i cos30 ° + j sin 30 °) = 17.32 i + 10 j

458

Problem 2.106 OA and OB?

From Eq. (2.23) U ⋅ V = (7.07)(17.32) + (7.07)(10) = 193.2

308 x

What is the angle between the ropes B

Solution: Let e OA be a unit vector that points from the origin O toward A. In terms of the angles shown, its components are

A 608

e OA = cos35 ° cos15 °i + sin 35 ° j − cos35 ° sin15 °k. 408

The components of a unit vector e OB that points from O toward B are e OB = − cos60 ° sin 40 °i + sin 60 ° j + cos60 ° cos 40 °k.

358 158 x

We can use definition of the dot product to determine the angle between OA and OB: cos θ =

e OA ⋅ e OB e OA e OB

(cos35 ° cos15 °)(− cos60 ° sin 40 °) + (sin 35 °)(sin 60 °) + (− cos35 ° sin15 °)(cos60 ° cos 40 °) (1)(1) = 0.161.

We obtain θ = 80.7 °. 80.7 °.

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Problem 2.107 Use the dot product to determine the angle between the forestay (cable AB) and the backstay (cable BC ) of the sailboat.

Solution:

The unit vector from B to A is rBA e BA = = −0.321i − 0.947 j rBA

The unit vector from B to C is y e BC =

B (4, 13) m

rBC = 0.385i − 0.923 j rBC

From the definition of the dot product, e BA ⋅ e BC = 1 ⋅ 1 ⋅ cosθ, where θ is the angle between BA and BC. Thus cos θ = (−0.321)(0.385) + (−0.947)(−0.923) cos θ = 0.750 θ = 41.3 °

C (9, 1) m

A (0, 1.2) m

Problem 2.108 Determine the angle θ between the lines AB and AC (a) by using the law of cosines (see Appendix A); (b) by using the dot product. y

Solution: (a) We have the distances: AB =

4 2 + 3 2 + 12 m =

AC =

BC =

(5 − 4) 2

m =

+ (−1 − 3) 2

26 m 35 m + (3 + 1) 2 m =

33 m

BC 2 = AB 2 + AC 2 − 2( AB)( AC ) cos θ cos θ = x

u z

+ 32

The law of cosines gives

B (4, 3, 21) m

+ 12

(5, 21, 3) m C

AB 2 + AC 2 − BC 2 = 0.464 ⇒ θ = 62.3 ° 2( AB)( AC )

(b) Using the dot product r AB = (4 i + 3 j − k) m,

r AC = (5i − j + 3k) m

r AB ⋅ r AC = (4 m)(5 m) + (3 m)(−1 m) + (−1 m)(3 m) = 14 m 2 r AB ⋅ r AC = ( AB)( AC ) cos θ Therefore cos θ =

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14 m 2 = 0.464 ⇒ θ = 62.3 ° 26 m 35 m

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Problem 2.109 The ship O measures the positions of the ship A and the airplane B and obtains the coordinates shown. What is the angle θ between the lines of sight OA and OB? y

Solution:

From the coordinates, the position vectors are: rOA = 6 i + 0 j + 3k and rOB = 4 i + 4 j − 4 k The dot product:

rOA ⋅ rOB = (6)(4) + (0)(4) + (3)(−4) = 12 The magnitudes: rOA = rOA =

B (4, 4, 24) km

6 2 + 0 2 + 3 2 = 6.71 km and 4 2 + 4 2 + 4 2 = 6.93 km.

rOA ⋅ rOB = 0.2581, from which θ = ±75 °. rOA rOB From the problem and the construction, only the positive angle makes sense, hence θ = 75 ° From Eq. (2.24) cos θ =

u x

O A (6, 0, 3) km

Problem 2.110 Astronauts on the space shuttle use radar to determine the magnitudes and direction cosines of the position vectors of two satellites A and B. The vector r A from the shuttle to satellite A has magnitude 2 km and direction cosines cos θ x = 0.768, cos θ y = 0.384, cos θ z = 0.512. The vector rB from the shuttle to satellite B has magnitude 4 km and direction cosines cos θ x = 0.743, cos θ y = 0.557, cos θ z = −0.371. What is the angle θ between the vectors r A and rB?

Solution:

The direction cosines of the vectors along r A and rB are the components of the unit vectors in these directions (i.e., u A = cos θ x i + cos θ y j + cos θ z k, where the direction cosines are those for r A ). Thus, through the definition of the dot product, we can find an expression for the cosine of the angle between r A and rB . cos θ = cos θ x A cos θ x B + cos θ y A cos θ y B + cos θ z A cos θ z B . Evaluation of the relation yields cos θ = 0.594 ⇒ θ = 53.5 °.

rB x u

rA z

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Problem 2.111 Segment AB of the bar is parallel to the x-axis. Determine the angle θ between the straight segments AB and BC.

A (0, 3, 1) m

Solution:

The coordinates of point B are (5, 3, 1) m. The vector from B to A is rBA = −5i (m). The vector from B to C is

rBC = (2 m − 5 m) i + (0 − 3 m) j + (4 m − 1 m)k = −3i − 3 j + 3k (m).

The magnitudes of these vectors are

(2, 0, 4) m

rBA = 5 m, rBC =

(−3 m) 2 + (−3 m) 2 + (3 m) 2

= 5.20 m. The value of their dot product is rBA · rBC = (−5 m)(−3 m) + (0)(−3 m) + (0)(3 m) = 15 m 2 . Using the definition of the dot product, cos θ =

rBA · rBC rBA rBC

15 m 2 (5 m)(5.20 m) = 0.577. =

This yields θ = 54.7 °.

Problem 2.112 The person exerts a force F = 60 i − 40 j (N) on the handle of the exercise machine. Use Eq. (2.26) to determine the vector component of F that is parallel to the line from the origin O to where the person grips the handle. Solution:

150 mm

The vector r from the O to where the person grips the

handle is

r = (250 i + 200 j − 150 k) mm, r = 354 mm To produce the unit vector that is parallel to this line we divide by the magnitude e =

200 mm

(250 i + 200 j − 150 k) mm r = = (0.707 i + 0.566 j − 0.424 k) r 354 mm

250 mm

Using Eq. (2.26), we find that the vector component parallel to the line is F p = (e ⋅ F)e = [(0.707)(60 N) + (0.566)(−40 N)](0.707 i + 0.566 j − 0.424 k) F p = (14.0 i + 11.2 j + 8.4 k) N

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Problem 2.113 At the instant shown, the Harrier’s thrust vector is T = 4.22 i + 21.6 j − 3.30 k (kip) and its velocity vector is v = 24.4 i + 8.86 j + 2.44 k (ft/s). (a) Determine the components of the thrust vector parallel and normal (perpendicular) to the velocity vector. (b) The term T ⋅ v is the power being transferred to the plane by its engine. What is the power in horsepower?

Solution: (a) We divide the velocity vector by its magnitude to obtain a unit vector with the same direction: e =

v v

T x

24.4 i + 8.86 j + 2.44 k (ft/s) (24.4 ft/s) 2 + (8.86 ft/s) 2 + (2.44 ft/s) 2 = 0.936 i + 0.340 j + 0.0936k. =

The component of T parallel to the velocity vector is Tp = (e ⋅ T)e. The dot product of e and T is e ⋅ T = (0.936)(4.22 kip) + (0.340)(21.6 kip) + (0.0936)(−3.30 kip) = 11.0 kip. The parallel component is Tp = (e ⋅ T)e = (11.0 kip)(0.936 i + 0.340 j + 0.0936k) = 10.3i + 3.73 j + 1.03k (kip). The component of T normal to the velocity vector is Tn = T − Tp = −6.06 i + 17.9 j − 4.33k (kip). (b) The dot product of T and v is T ⋅ v = (4.22 kip)(24.4 ft/s) + (21.6 kip)(8.86 ft/s) + (−3.30 kip)(2.44 ft/s) = 286 kip-ft/s = 286,000 ft-lb/s. Converting this to horsepower, 286,000 ft-lb/s = 286,000 ft-lb/s

horsepower ( 1 550 ft-lb/s )

= 521 horsepower. (a) Tp = 10.3i + 3.73 j + 1.03k (kip), Tn = −6.06 i + 17.9 j − 4.33k (kip). (b) 521 horsepower.

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Problem 2.114 For the three-dimensional truss shown, determine the angle between members AB and AD. Solution:

F A (4, 3, 4) ft

The vector from A to B is

D (6, 0, 0) ft x

r AB = −4 i − 3 j − 4 k (ft). The vector from A to D is r AD = (6 ft − 4 ft) i + (0 − 3 ft) j + (0 − 4 ft)k

C (5, 0, 6) ft

= 2 i − 3 j − 4 k (ft). The magnitudes of these vectors are r AB =

(−4 ft) 2 + (−3 ft) 2 + (−4 ft) 2

= 6.40 ft, r AD =

(2 ft) 2 + (−3 ft) 2 + (−4 ft) 2

= 5.39 ft. The value of their dot product is r AB ⋅ r AD = (−4 ft)(2 ft) + (−3 ft)(−3 ft) + (−4)(−4 ft) = 17 ft 2 . Using the definition of the dot product, cos θ =

rBA ⋅ rBC rBA rBC

17 ft 2 (6.40 ft)(5.39 ft) = 0.493. =

This yields 60.5 °.

Problem 2.115 For the three-dimensional truss shown, determine the angle between members AC and CD. Solution:

F A (4, 3, 4) ft

The vector from C to A is

D (6, 0, 0) ft x

rCA = (4 ft − 5 ft) i + (3 ft − 0) j + (4 ft − 6 ft)k = −i + 3 j − 2k (ft). z

The vector from C to D is

C (5, 0, 6) ft

rCD = (6 ft − 5 ft) i + (0 − 0) j + (0 − 6 ft)k = i − 6k (ft). The magnitudes of these vectors are rCA =

(−1 ft) 2

+ (3 ft) 2

+ (−2 ft) 2

= 3.74 ft, rCD =

Using the definition of the dot product, cos θ =

(1 ft) 2 + (0) 2 + (−6 ft) 2

= 6.08 ft.

= 11 ft 2 .

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11 ft 2 (3.74 ft)(6.08 ft)

= 0.483.

The value of their dot product is rCA ⋅ rCD = (−1 ft)(1 ft) + (3 ft)(0) + (−2)(−6 ft)

rCA ⋅ rCD rCA rCD

This yields 61.1 °.

12/04/23 4:03 PM

Problem 2.116 The three-dimensional truss is subjected to a force F = 25i − 100 j − 30 k (lb) at A. Determine the components of F parallel and normal (perpendicular) to member AD. Solution:

F A (4, 3, 4) ft

D (6, 0, 0) ft x

The vector from A to D is

r AD = (6 ft − 4 ft) i + (0 − 3 ft) j + (0 − 4 ft)k

C (5, 0, 6) ft

= 2 i − 3 j − 4 k (ft). Dividing this vector by its magnitude yields a unit vector that points from A toward D: e AD =

r AD r AD 2 i − 3 j − 4 k (ft) (2 ft) 2 + (−3 ft) 2 + (−4 ft) 2

= 0.371i − 0.557 j − 0.743k. The component of F parallel to member AD is Fp = (e AD ⋅ F)e AD . The dot product of e AD and F is e AD ⋅ F = (0.371)(25 lb) + (−0.557)(−100 lb) + (−0.743)(−30 lb) = 87.3 lb. The parallel component is Fp = (e AD ⋅ F)e AD = (87.3 lb)(0.371i − 0.557 j − 0.743k) = 32.4 i − 48.6 j − 64.8k (lb). The component of F normal to member AD is Fn = F − Fp = −7.41i − 51.4 j + 34.8k (lb). Fp = 32.4 i − 48.6 j − 64.8k (lb), Fn = −7.41i − 51.4 j + 34.8k (lb).

Problem 2.117 The rope AB exerts a 50-N force T on collar A. Determine the vector component of T parallel to the bar CD. Solution:

0.15 m

We have the following vectors

0.4 m

rCD = (−0.2 i − 0.3 j + 0.25k) m e CD =

rCD = (−0.456 i − 0.684 j + 0.570 k) rCD

A 0.5 m

rOB = (0.5 j + 0.15k) m

rOC = (0.4 i + 0.3 j) m

r e AB = AB = (0.674 i + 0.735 j + 0.079k) r AB

x D

rOA = rOC + (0.2 m)e CD = (0.309 i + 0.163 j + 0.114 k) m r AB = rOB − rOA = (−0.309 i + 0.337 j + 0.036k) m

0.2 m 0.3 m

0.25 m

0.2 m

We can now write the force T and determine the vector component parallel to CD. T = (50 N)e AB = (−33.7 i + 36.7 j + 3.93k) N T p = (e CD ⋅ T)e CD = (3.43i + 5.14 j − 4.29k) N T p = (3.43i + 5.14 j − 4.29k) N

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Problem 2.118 In Problem 2.117, determine the vector component of T normal to the bar CD. Solution:

0.15 m

From Problem 2.117 we have

0.4 m

T = (−33.7 i + 36.7 j + 3.93k) N

Tp = (3.43i + 5.14 j − 4.29k) N

The normal component is then

0.2 m 0.3 m

0.5 m

Tn = T − T p

Tn = (−37.1i + 31.6 j + 8.22k) N.

x D

0.2 m

Problem 2.119 The magnitudes of the two force vectors are F A = 200 lb and FB = 160 lb. What is the angle between the two vectors?

Solution:

0.25 m

The components of F A are

F A = (200 lb)(cos 40 ° sin 50 °i + sin 40 ° j + cos 40 ° cos 50 °k) = 117 i + 129 j + 98.5k (lb). The components of FB are

FB = (160 lb)(− cos60 ° sin 30 °i + sin 60 ° j + cos60° cos30°k) = −40.0 i + 139 j + 69.3k (lb).

FB FA 608 308

408

Their dot product is F A ⋅ FB = (117 lb)(−40.0 lb) + (129 lb)(139 lb) + (98.5 lb)(69.3 lb) = 19,900 lb 2 .

508

Using the definition of the dot product, cos θ = =

F A · FB F A FB 19,900 lb 2 (200 lb)(160 lb)

= 0.623. This yields 51.5 °.

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Problem 2.120 The magnitudes of the two force vectors are F A = 200 lb and FB = 160 lb. Determine the components of F A parallel and normal to FB .

Solution:

The components of F A are

F A = (200 lb)(cos 40 ° sin 50 °i + sin 40 ° j + cos 40 ° cos 50 °k) = 117 i + 129 j + 98.5k (lb). The components of FB are

FB = (160 lb)(− cos60 ° sin 30 °i + sin 60 ° j + cos60° cos30°k) = −40.0 i + 139 j + 69.3k (lb).

FB FA 608 308

408

We divide FB by its magnitude to determine the components of a unit vector with the same direction: x

eB =

508

FB FB

−40.0 i + 139 j + 69.3k (lb) (−40.0 lb) 2 + (139 lb) 2 + (69.3 lb) 2 = −0.250 i + 0.866 j + 0.433k. =

The dot product of e B with F A is e B ⋅ F A = (−0.250)(117 lb) + (0.866)(129 lb) + (0.433)(98.5 lb) = 125 lb. The component of F A that is parallel to FB is F Ap = (e B ⋅ F A )e B = (125 lb)(−0.250 i + 0.866 j + 0.433k) = −31.2 i + 108 j + 54.0 k (lb). The component of F A that is normal to FB is F An = F − F Ap = 149 i + 20.6 j + 44.5k (lb). F Ap = −31.2 i + 108 j + 54.0 k (lb), F An = 149 i + 20.6 j + 44.5k (lb).

Problem 2.121 An astronaut in a maneuvering unit approaches the ISS. At the present instant, the station informs him that his position relative to the origin of the station’s coordinate system is rG = 50 i + 80 j + 180 k (m) and his velocity is v = −2.2 j − 3.6k (m/s). The position of an airlock is r A = −12 i + 20 k (m). Determine the angle between his velocity vector and the line from his position to the airlock’s position.

Solution:

Points G and A are located at G: (50, 80, 180) m and A: (−12, 0, 20) m. The vector rGA is rGA = ( x A − x G ) i + ( y A − y G ) j + ( z A − z G )k = (−12 − 50) i + (0 − 80) j + (20 − 180)k m. The dot product between v and rGA is v ⋅ rGA = v rGA cos θ = v x x GA + v y y GA + v z z GA , where θ is the angle between v and rGA . Substituting in the numerical values, we get θ = 19.7 °.

G A z

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Problem 2.122 In Problem 2.121, determine the vector component of the astronaut’s velocity parallel to the line from his position to the airlock’s position.

Solution:

The coordinates are A (−12, 0, 20) m, G (50, 80, 180) m.

Therefore rGA = (−62 i − 80 j − 160 k) m e GA =

rGA = (−0.327 i − 0.423 j − 0.845k) rGA

The velocity is given as v = (−2.2 j − 3.6k) m/s The vector component parallel to the line is now v p = (e GA ⋅ v)e GA = [(−0.423)(−2.2) + (−0.845)(−3.6)]e GA v p = (−1.30 i − 1.68 j − 3.36k) m/s

Problem 2.123 Point P is at longitude 30°W and latitude 45°N on the Atlantic Ocean between Nova Scotia and France. Point Q is at longitude 60°E and latitude 20°N in the Arabian Sea. Use the dot product to determine the shortest distance along the surface of the earth from P to Q in terms of the radius of the earth R E . Strategy: Use the dot product to determine the angle between the lines OP and OQ; then use the definition of an angle in radians to determine the distance along the surface of the earth from P to Q.

P = rOB + rBP = R E ( i cos λ P cos θ P + j sin θ P + k sin λ P cos θ P ). A similar argument for the point Q yields Q = rOC + rCQ = R E ( i cos λ Q cos θQ + j sin θQ + k sin λ Q cos θQ ) Using the identity cos 2 β + sin 2 β = 1, the magnitudes are P = Q = RE The dot product is P ⋅ Q = R E 2 (cos(λ P − λ Q ) cos θ P cos θQ + sin θ P sin θQ )

Substitute:

cos θ =

P⋅Q = cos(λ P − λ Q ) cos θ P cos θQ + sin θ P sin θQ P Q

P Q 458

308

208

Substitute λ P = +30 °, λ Q = −60 °, θ p = +45 °, θQ = +20 °, to obtain cos θ = 0.2418, or θ = 1.326 radians. Thus the distance is d = 1.326 R E

608

G P

Equator

458 x

Solution:

The distance is the product of the angle and the radius of the sphere, d = R E θ, where θ is in radian measure. From Eqs. (2.18) and (2.24), the angular separation of P and Q is given by

RE b 308 x

608

Q 208 c

 P ⋅ Q  . cos θ =   P Q  The strategy is to determine the angle θ in terms of the latitude and longitude of the two points. Drop a vertical line from each point P and Q to b and c on the equatorial plane. The vector position of P is the sum of the two vectors: P = rOB + rBP . The vector rOB = rOB ( i cos λ P + 0 j + k sin λ P ). From geometry, the magnitude is rOB = R E cos θ P . The vector rBP = rBP (0 i + 1 j + 0 k). From geometry, the magnitude is rBP = R E sin θ P . Substitute and reduce to obtain:

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Problem 2.124

Consider the vectors U = 12 i + 4 j − 8k, V = −6 i + 8 j + 14 k.

(a) Determine the components of the vector W = U × V. (b) Evaluate the dot products U ⋅ W and V ⋅ W. What is the explanation of these two results? Solution: (a) The vector W is W = U×V i = 12

j 4

k −8

−6

= [ (4)(14) − (−8)(8) ] i − [ (12)(14) − (−8)(−6) ] j + [ (12)(8) − (4)(−6) ] k = 120 i − 120 j + 120 k. (b) The dot products are U ⋅ W = (12)(120) + (4)(−120) + (−8)(120) = 0, V ⋅ W = (−6)(120) + (8)(−120) + (14)(120) = 0. (a) W = 120 i − 120 j + 120 k. (b) U ⋅ W = 0, V ⋅ W = 0. From the definition of the cross product, W is perpendicular to both U and V.

Problem 2.125 Two vectors U = 3i + 2 j and V = 2 i + 4 j. (a) What is the cross product U × V? (b) What is the cross product V × U?

Solution:

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

U×V =

i j k 3 2 0 2 4 0

= i((2)(0) − (4)(0)) − j((3)(0) − (2)(0)) + k((3)(4) − (2)(2)) = 8k

V×U =

i j k 2 4 0 3 2 0

= i((4)(0) − (2)(0)) − j((2)(0) − (3)(0)) + k((2)(2) − (3)(4)) = −8k

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Problem 2.126 Consider the position vectors P and Q. (a) Determine the cross product P × Q by using the definition, Eq. (2.28). (b) Determine the dot product P ⋅ Q by using Eq. (2.34).

The angle between the two vectors is θ = arctan

m ( 10 ) = 39.8°. 12 m

From Eq. (2.28), P × Q = (15.6 m)(8 m)sin 39.8 °e = (80 m 2 )e.

The unit vector e is defined to be perpendicular to P and perpendicular to Q. Either j or −j satisfies those criteria. The remaining criterion states that P, Q, e form a right-handed system, which requires that e = j. Therefore,

(8, 0, 0) m

P × Q = 80 j (m 2 ).

(b) From Eq. (2.34),

z P×Q =

(12, 0, 10) m

Solution:

12 m 0 10 m 8m 0 0

= [(0)(0) − (10 m)(0)]i − [(12 m)(0) − (10 m)(8 m)] j

(a) The two vectors in terms of their components are

+ [(12 m)(0) − (0)(8 m)]k

P = 12 i + 10 k (m), Q = 8i (m).

= 80 j (m 2 ).

Their magnitudes are

(a), (b) P × Q = 80 j (m 2 ).

P = (12 m) 2 + (10 m) 2 = 15.6 m, Q = 8 m.

Problem 2.127 Consider the position vectors P and Q. (a) Determine the cross product Q × P by using the definition, Eq. (2.28). (b) Determine the dot product Q ⋅ P by using Eq. (2.34).

Solution: (a) The two vectors in terms of their components are P = 12 i + 10 k (m), Q = 8i (m). Their magnitudes are P = (12 m) 2 + (10 m) 2 = 15.6 m, Q = 8 m.

The angle between the two vectors is θ = arctan Q

m ( 10 ) = 39.8°. 12 m

From Eq. (2.28),

(8, 0, 0) m x

Q × P = (8 m)(15.6 m)sin 39.8 °e = (80 m 2 )e.

The unit vector e is defined to be perpendicular to Q and p erpendicular to P. Either j or −j satisfies those criteria. The remaining criterion states that Q, P, e form a right-handed system, which requires that e = − j. Therefore,

z (12, 0, 10) m

Q × P = −80 j (m 2 ). (b) From Eq. (2.34),

Q×P =

i j k 8m 0 0 12 m 0 10 m

= [ (0)(10 m) − (0)(0) ] i − [ (8 m)(10 m) − (0)(12 m) ] j + [ (8 m)(0) − (0)(12 m) ] k = −80 j (m 2 ). (a), (b) Q × P = −80 j (m 2 ).

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Problem 2.128 Suppose that the cross product of two vectors U and V is U × V = 0. If U ≠ 0, what do you know about the vector V ?

Solution:

Problem 2.129 The cross product of two vectors U and V is U × V = −30 i + 40 k. The vector V = 4 i − 2 j + 3k. The vector U = 4 i + U y j + U z k. Determine U y and U z .

Solution:

From the given information we have

U×V =

i j 4 Uy

k Uz

4 −2

Either V = 0 or V U

= (3U y + 2U z ) i + (4U z − 12) j + (−8 − 4U y )k U × V = (−30i + 40 k) Equating the components we have 3U y + 2U z = −30, 4U z − 12 = 0,

−8 − 4U y = 40.

Solving any two of these three redundant equations gives U y = −12, U z = 3.

Problem 2.130 (a) Determine the components of a unit vector e p that is parallel to the line AB and points from A toward B. (b) Determine the components of a unit vector e n that is normal (perpendicular) to the line AB. (Make sure your result is a unit vector.) Confirm that it is perpendicular to the line by showing that e p ⋅ e n = 0.

Dividing this vector by its magnitude yields the unit vector e p : ep =

r AB r AB

4 i − 18 j − 6k (ft) (4 ft) 2 + (−18 ft) 2 + (−6 ft) 2 = 0.206 i − 0.928 j − 0.309k. =

(b) The cross product of e p with any non-zero vector that is not parallel to the line will yield a vector that is perpendicular to the line, from the definition of the cross product. For example, the vector

y A

e p × i = (0.206 i − 0.928 j − 0.309k) × i = 0.206( i × i) − 0.928( j × i) − 0.309(k × i) = 0.928k − 0.309 j

(22, 12, 2) ft

is perpendicular to e p . However, although it is the cross product of two unit vectors, it is not a unit vector: ep × i =

= 0.978.

To obtain the desired unit vector, we must divide the vector e p × i by its magnitude:

(2, 26, 24) ft

Solution:

en =

y A

B (2, –6, –4) ft

x z

B (2, –6, –4) ft

(a) The position vector from point A to point B is r AB = (2 ft − (−2 ft)) i + (−6 ft − 12 ft) j + (−4 ft − 2 ft)k = 4 i − 18 j − 6k (ft).

ep × i ep × i −0.309 j + 0.928k 0.978

= −0.316 j + 0.949k.

(–2, 12, 2) ft rAB

(–2, 12, 2) ft

(0.928) 2 + (−0.309) 2

The dot product of this vector with e p is zero, confirming that the two vectors are perpendicular: e p ⋅ e n = (0.206)(0) + (−0.928)(−0.316) + (−0.309)(0.949) = 0. (a) e p = 0.206 i − 0.928 j − 0.309k. (b) For example, e n = −0.316 j + 0.949k.

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Problem 2.131 The force F = 10 i − 4 j (N). Determine the cross product r AB × F.

(6, 3, 0) m A rA B x

(6, 0, 4) m B F

Solution:

The position vector is

A(6, 3, 0)

r AB = (6 − 6) i + (0 − 3) j + (4 − 0)k = 0 i − 3 j + 4 k The cross product: r AB × F =

i j k 0 −3 4 10 −4 0

rA B = i(16) − j(−40) + k(30)

x z

= 16 i + 40 j + 30 k (N-m)

Problem 2.132 By evaluating the cross product U × V, prove the identity sin(θ1 − θ 2 ) = sin θ1 cos θ 2 − cos θ1 sin θ 2 .

B(6, 0, 4)

Solution: Assume that both U and V lie in the x -y plane. The strategy is to use the definition of the cross product (Eq. 2.28) and the Eq. (2.34), and equate the two. From Eq. (2.28) U × V = U V sin (θ1 − θ 2 )e. Since the positive z -axis is out of the paper, and e points into the paper, then e = −k. Take the dot product of both sides with e, and note that k ⋅ k = 1. Thus

( (U ×U VV) ⋅ k )

sin (θ1 − θ 2 ) = − The vectors are: U

U = U (i cos θ1 + j sin θ 2 ), and V = V (i cos θ 2 + j sin θ 2 ).

The cross product is

u1 u2

U×V =

i U cos θ1

j U sin θ1

k 0

V cos θ 2

V sin θ 2

= i(0) − j(0) + k( U V )(cos θ1 sin θ 2 − cos θ 2 sin θ1 ) Substitute into the definition to obtain: sin (θ1 − θ 2 ) = sin θ1 cos θ 2 − cos θ1 sin θ 2 . Q.E.D.

y U V

U1 U2

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Problem 2.133 In Example 2.15, what is the minimum distance from point B to the line OA?

Solution: Let θ be the angle between rOA and rOB . Then the minimum distance is d = rOB sin θ Using the cross product, we have rOA × rOB = rOA rOB sin θ = rOA d ⇒ d =

rOA × rOB rOA

We have rOA = (10 i − 2 j + 3k) m rOB = (6 i + 6 j − 3k) m

rOA × rOB =

i j k 10 −2 3 6 6 −3

= (−12 i + 48 j + 72k) m 2

Thus d =

(−12 m 2 ) + (48 m 2 ) 2 + (72 m 2 ) 2 = 8.22 m (10 m) 2 + (−2 m) 2 + (3 m) 2

d = 8.22 m

Problem 2.134 (a) What is the cross product rOA × rOB? (b) Determine a unit vector e that is perpendicular to rOA and rOB . Solution:

y B (4, 4, 24) m

The two radius vectors are rOB

rOB = 4 i + 4 j − 4 k, rOA = 6 i − 2 j + 3k (a) The cross product is rOA × rOB =

i j k 6 −2 3 4 4 −4

O = i(8 − 12) − j(−24 − 12) + k(24 + 8)

= −4 i + 36 j + 32k (m 2 )

x rOA

A (6, 22, 3) m

The magnitude is rOA × rOB =

4 2 + 36 2 + 32 2 = 48.33 m 2

(b) The unit vector is  r × rOB  e = ± OA  = ± (−0.0828i + 0.7448 j + 0.6621k)  rOA × rOB  (Two vectors.)

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Problem 2.135 Use the cross product to determine the length of the shortest straight line from point B to the straight line that passes through points O and A. y

i j k −4 +36 32 −2 3 6

C × rOA =

C = 172 i + 204 j − 208k

B (4, 4, 24) m

The unit vector in the direction of C is eC =

rOB

C = 0.508i + 0.603 j − 0.614 k C

(The magnitude of C is 338.3)

We now want to find the length of the projection, P, of line OB in direction e c . P = rOB ⋅ e C

rOA

= (4 i + 4 j − 4 k) ⋅ e C

A (6, 22, 3) m

P = 6.90 m y

Solution:

B( 4, 4, –4) m

rOA = 6 i − 2 j + 3k (m) rOB = 4 i + 4 j − 4 k (m) rOA × rOB = C

r OB

(C is ⊥ to both rOA and rOB ) C =

i j k 6 −2 3 4 4 −4

= (+8 − 12) i + (12 + 24) j + (24 + 8)k

C = −4 i + 36 j + 32k

x r OA

P A(6, –2, 3) m

C is ⊥ to both rOA and rOB . Any line ⊥ to the plane formed by C and rOA will be parallel to the line BP on the diagram. C × rOA is such a line. We then need to find the component of rOB in this direction and compute its magnitude.

Problem 2.136 The cable BC exerts a 1000-lb force F on the hook at B. Determine r AB × F. Solution:

The coordinates of points A, B, and C are A (16, 0, 12), B (4, 6, 0), C (4, 0, 8). The position vectors are

F 6 ft

rOA = 16 i + 0 j + 12k, rOB = 4 i + 6 j + 0 k, rOC = 4 i + 0 j + 8k.

8 ft

rBC r − rOB r = OC = AB rBC rOC − rOB r AB

Noting rOC − rOB = (4 − 4) i + (0 − 6) j + (8 − 0)k = 0 i − 6 j + 8k rOC − rOB =

rAB x

The force F acts along the unit vector e BC =

C rAC

4 ft 4 ft

e BC = 0 i − 0.6 j + 0.8k, and F = F e BC = 0 i − 600 j + 800 k (lb).

The vector

6 ft

r AB = (4 − 16) i + (6 − 0) j + (0 − 12)k = −12 i + 6 j − 12k Thus the cross product is r AB × F =

8 ft = −2400 i + 9600 j + 7200 k (ft-lb)

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12 ft

6 2 + 8 2 = 10. Thus

i j k −12 −12 6 −600 800 0

r x

4 ft 4 ft

12 ft

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Problem 2.137* The force F = 200 i + 180 j + 100 k ( lb ). Determine the vector component of F that is normal (perpendicular) to the flat plate ACD. y

The vector from A to C is r AC = −6 i + 10 j (ft). The vector from A to D is r AD = −2 i + 12k (ft). The cross product r AC × r AD is perpendicular to the plate.

(0, 10, 0) ft

i j k −6 10 0 −2 0 12

r AC × r AD =

= 120 i + 72 j + 20 k(ft 2 ). A

(6, 0, 0) ft z

Dividing this vector by its magnitude yields a unit vector that is perpendicular to the plate: e =

(4, 0, 12) ft

r AC × r AD r AC × r AD

= 0.849 i + 0.509 j + 0.141k. From Eq. (2.26), the vector component of F that is parallel to e, and so is perpendicular to the plate, is

Solution: y

Fn = (e ⋅ F)e = [(0.849)(200 lb) + (0.509)(180 lb)

+ (0.141)(100 lb)](0.849 i + 0.509 j + 0.141k)

(0, 10, 0) ft

= 234 i + 140 j + 39k(lb).

rAC

Fn = 234 i + 140 j + 39k(lb). A

rAD z

(6, 0, 0) ft

D (4, 0, 12) ft y

Problem 2.138 The rope AB exerts a 50-N force T on the collar at A. Let rCA be the position vector from point C to point A. Determine the cross product rCA × T.

0.15 m

Solution:

We define the appropriate vectors.

0.4 m C

rCD = (−0.2 i − 0.3 j + 0.25k) m

r rCA = (0.2 m) CD = (−0.091i − 0.137 j + 0.114 k) m rCD

A 0.5 m

rOB = (0.5 j + 0.15k) m

rOC = (0.4 i + 0.3 j) m

x D

r AB = rOB − (rOC + rCA ) = (0.61i − 1.22 j − 0.305k) m T = (50 N)

0.2 m 0.3 m

0.25 m

0.2 m

r AB = (−33.7 i + 36.7 j + 3.93k) N r AB

Now take the cross product rCA × T =

i j l −0.091 −0.137 0.114 36.7 3.93 −33.7

= (−4.72 i − 3.48 j + −7.96k) N-m

rCA × T = (−47.2 i − 3.48 j + −7.96k) N-m

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

Solution:

We first develop the force T.

rCE = (0.3i + 0.1 j) m r T = (600 N) CE = (569 i + 190 j) N rCE From Example 2.16 we know that the unit vector perpendicular to the door is e = (0.358i + 0.894 j + 0.268k) The magnitude of the force perpendicular to the door (parallel to e ) is then Tn = T ⋅ e = (569 N)(0.358) + (190 N)(0.894) = 373 N Tn = 373 N

Problem 2.140 The slender bar AB is 14 ft in length and is perpendicular to the flat plate ACD. (a) Determine the components of a unit vector e that is perpendicular to the plate ACD and points in the direction from A toward B. (b) Determine the coordinates ( x B , y B , z B ) of point B.

The cross product r AC × r AD is perpendicular to the plate and points in the direction from A toward B. i r AC × r AD =

−6 10

−2

= 120 i + 72 j + 20 k (ft 2 ). Dividing this vector by its magnitude yields the desired unit vector: (0, 10, 0) ft

B (xB, yB, zB)

x (6, 0, 0) ft D

r AC × r AD r AC × r AD

= 0.849 i + 0.509 j + 0.141k. (b) The bar is 14 ft long, so we obtain the vector from A to B by multiplying e by 14 ft:

e =

r AB = (14 ft)e = 11.9 i + 7.13 j + 1.98k (ft). Adding the coordinates of point A to the components of this vector, we obtain the coordinates of point B:

(4, 0, 12) ft

( x B , y B , z B ) = (17.9, 7.13, 1.98) ft.

Solution:

(a) e = 0.849 i + 0.509 j + 0.141k.

(b) ( x B , y B , z B ) = (17.9, 7.13, 1.98) ft. (0, 10, 0) ft

rAC A

rAB x

rAD z

(6, 0, 0) ft

D (4, 0, 12) ft

(a) The vector from A to C is r AC = −6 i + 10 j (ft). The vector from A to D is r AD = −2 i + 12k (ft).

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Problem 2.141* Determine the minimum distance from point P to the plane defined by the three points A, B, and C.

y B (0, 5, 0) m P (9, 6, 5) m

Solution:

The strategy is to find the unit vector perpendicular to the plane. The projection of this unit vector on the vector OP : rOP ⋅ e is the distance from the origin to P along the perpendicular to the plane. The projection on e of any vector into the plane (rOA ⋅ e, rOB ⋅ e, or rOC ⋅ e) is the distance from the origin to the plane along this same perpendicular. Thus the distance of P from the plane is

A (3, 0, 0) m C

d = rOP ⋅ e − rOA ⋅ e.

rBC × rBA =

i j k 0 −5 4 3 −5 0

P[9, 6, 5]

B[0, 5, 0] = 20 i + 12 j + 15k.

Problem 2.142 Two cartesian coordinate systems are shown. The unit vectors i′ and j′ of the x ′y ′z ′ coordinate system are given in terms of their components in the xyz coordinate system by

A[3, 0, 0] z

C[0, 0, 4]

Solution: (a) The vector k ′ is k′ = i′ × j′

i′ = 0.743i + 0.557 j + 0.371k, j′ = −0.408i + 0.816 j − 0.408k.

(a) Determine the unit vector k′ of the x ′y ′z ′ coordinate system in terms of its components in the xyz coordinate system. (b) The force F = 40 i + 80 j + 20 k (N). Use the dot product to determine the components of F in terms of the x ′y ′z ′ coordinate system.

The magnitude is rBC × rBA = 27.73, thus the unit vector is e = 0.7212 i + 0.4327 j + 0.5409k. The distance of point P from the plane is d = rOP ⋅ e − rOA ⋅ e = 11.792 − 2.164 = 9.63 m. The second term is the distance of the plane from the origin; the vectors rOB , or rOC could have been used instead of rOA .

(0, 0, 4) m

The position vectors are: rOA = 3i, rOB = 5 j, rOC = 4 k and rOP = 9 i + 6 j + 5k. The unit vector perpendicular to the plane is found from the cross product of any two vectors lying in the plane. Noting: rBC = rOC − rOB = −5 j + 4 k, and rBA = rOA − rOB = 3i − 5 j. The cross product:

i j k 0.743 0.557 0.371 −0.408 0.816 −0.408

= −0.530 i + 0.152 j + 0.834 k. (b) The x ′ component of F is Fx ′ = F ⋅ i′ = (40 N)(0.743) + (80 N)(0.557) + (20 N)(0.371) = 81.7 N, the y ′ component is

Fy ′ = F ⋅ j′

= (40 N)(−0.408) + (80 N)(0.816) + (20 N)(−0.408) = 40.8 N,

and the z ′ component is Fz ′ = F ⋅ k ′

Therefore F = 81.7 i′ + 40.8 j′ + 7.61k ′ (N).

z9 z

= (40 N)(−0.530) + (80 N)(0.152) + (20 N)(0.834) = 7.61 N.

(a) k ′ = −0.530 i + 0.152 j + 0.834 k. (b) F = 81.7 i′ + 40.8 j′ + 7.61k ′ (N).

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