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Solution Manual For Engineering Mechanics: Statics, 5th Edition

Solution Manual For Engineering Mechanics: Statics, 5th Edition offers textbook solutions that are easy to follow, helping you ace your assignments.

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    Problem 1.1 The value of  is 3.14159265 . . . If C is the circumference of a circle and r is its radius, deter- mine the value of r/C to four signi fi cant digits. Solution: C D 2 r ) r C D 1 2  D 0 . 159154943. To four signi fi cant digits we have r C D 0 . 1592 Problem 1.2 The base of natural logarithms is e D 2 . 718281828 . . . (a) Express e to fi ve signi fi cant digits. (b) Determine the value of e 2 to fi ve signi fi cant digits. (c) Use the value of e you obtained in part (a) to deter- mine the value of e 2 to fi ve signi fi cant digits. [Part (c) demonstrates the hazard of using rounded-off values in calculations.] Solution: The value of e is: e D 2 . 718281828 (a) To fi ve signi fi cant fi gures e D 2 . 7183 (b) e 2 to fi ve signi fi cant fi gures is e 2 D 7 . 3891 (c) Using the value from part (a) we fi nd e 2 D 7 . 3892 which is not correct in the fi fth digit. Problem 1.3 A machinist drills a circular hole in a panel with a nominal radius r D 5 mm. The actual radius of the hole is in the range r D 5 š 0 . 01 mm. (a) To what number of signi fi cant digits can you express the radius? (b) To what number of signi fi cant digits can you express the area of the hole? Solution: a) The radius is in the range r 1 D 4.99 mm to r 2 D 5.01 mm. These numbers are not equal at the level of three signi fi cant digits, but they are equal if they are rounded off to two signi fi cant digits. Two: r D 5 . 0 mm b) The area of the hole is in the range from A 1 D r 1 2 D 78.226 m 2 to A 2 D r 2 2 D 78.854 m 2 . These numbers are equal only if rounded to one signi fi cant digit: One: A D 80 mm 2 Problem 1.4 The opening in the soccer goal is 24 ft wide and 8 ft high, so its area is 24 ft ð 8 ft D 192 ft 2 . What is its area in m 2 to three signi fi cant digits? Solution: A D 192 ft 2 ( 1 m 3.281 ft ) 2 D 17.8 m 2 A D 17.8 m 2 Problem 1.5 The Burj Dubai, scheduled for comple- tion in 2008, will be the world’s tallest building with a height of 705 m. The area of its ground footprint will be 8000 m 2 . Convert its height and footprint area to U.S. customary units to three signi fi cant digits. Solution: h D 705 m ( 3.281 ft 1 m ) D 2 . 31 ð 10 3 ft A D 8000 m 2 ( 3.218 ft 1 m ) 2 D 8 . 61 ð 10 4 ft 2 h D 2 . 31 ð 10 3 ft , A D 8 . 61 ð 10 4 ft 2

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    Problem 1.6 Suppose that you have just purchased a Ferrari F355 coupe and you want to know whether you can use your set of SAE (U.S. Customary Units) wrenches to work on it. You have wrenches with widths w D 1 / 4 in, 1/2 in, 3/4 in, and 1 in, and the car has nuts with dimensions n D 5 mm, 10 mm, 15 mm, 20 mm, and 25 mm. De fi ning a wrench to fi t if w is no more than 2% larger than n , which of your wrenches can you use? n Solution: Convert the metric size n to inches, and compute the percentage difference between the metric sized nut and the SAE wrench. The results are: 5 mm ( 1 inch 25 . 4 mm ) D 0 . 19685 .. in , ( 0 . 19685  0 . 25 0 . 19685 ) 100 D  27 . 0% 10 mm ( 1 inch 25 . 4 mm ) D 0 . 3937 .. in , ( 0 . 3937  0 . 5 0 . 3937 ) 100 D  27 . 0% 15 mm ( 1 inch 25 . 4 mm ) D 0 . 5905 .. in , ( 0 . 5905  0 . 5 0 . 5905 ) 100 D C 15 . 3% 20 mm ( 1 inch 25 . 4 mm ) D 0 . 7874 .. in , ( 0 . 7874  0 . 75 0 . 7874 ) 100 D C 4 . 7% 25 mm ( 1 inch 25 . 4 mm ) D 0 . 9843 .. in , ( 0 . 9843  1 . 0 0 . 9843 ) 100 D  1 . 6% A negative percentage implies that the metric nut is smaller than the SAE wrench; a positive percentage means that the nut is larger then the wrench. Thus within the de fi nition of the 2% fi t, the 1 in wrench will fi t the 25 mm nut. The other wrenches cannot be used. Problem 1.7 Suppose that the height of Mt. Everest is known to be between 29,032 ft and 29,034 ft. Based on this information, to how many signi fi cant digits can you express the height (a) in feet? (b) in meters?. Solution: a) h 1 D 29032 ft h 2 D 29034 ft The two heights are equal if rounded off to four signi fi cant digits. The fi fth digit is not meaningful. Four: h D 29,030 ft b) In meters we have h 1 D 29032 ft ( 1 m 3.281 ft ) D 8848.52 m h 2 D 29034 ft ( 1 m 3.281 ft ) D 8849.13 m These two heights are equal if rounded off to three signi fi cant digits. The fourth digit is not meaningful. Three: h D 8850 m 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) ft/s. Solution: a) v D 430 km h ( 0.6214 mi 1 km ) D 267 mi/h v D 267 mi/h b) v D 430 km h ( 1000 m 1 km ) ( 1 ft 0.3048 m ) ( 1 h 3600 s ) D 392 ft/s v D 392 ft/s 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 signi fi cant digits (a) in km/h; (b) in mi/h. Solution: a) v D 15 km ( 38 C 1 . 3 60 ) min ( 60 min 1 h ) D 23.7 km/h v D 23.7 km/h b) v D  23.7 km/h  ( 1 mi 1.609 km ) D 14.7 mi/h v D 14.7 mi/h

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    Problem 1.10 The Porsche’s engine exerts 229 ft-lb (foot-pounds) of torque at 4600 rpm. Determine the value of the torque in N-m (Newton-meters). Solution: T D 229 ft-lb ( 1 N 0.2248 lb ) ( 1 m 3.281 ft ) D 310 N-m T D 310 N-m Problem 1.11 The kinetic energy of the man in Active Example 1.1 is de fi ned by 1 2 m v 2 , where m is his mass and v is his velocity. The man’s mass is 68 kg and he is moving at 6 m/s, so his kinetic energy is 1 2 (68 kg)(6 m/s) 2 D 1224 kg-m 2 /s 2 . What is his kinetic energy in U.S. Customary units? Solution: T D 1224 kg-m 2 / s 2 ( 1 slug 14.59 kg ) ( 1 ft 0.3048 m ) 2 D 903 slug-ft 2 / s T D 903 slug-ft 2 / s Problem 1.12 The acceleration due to gravity at sea level in SI units is g D 9 . 81 m/s 2 . By converting units, use this value to determine the acceleration due to gravity at sea level in U.S. Customary units. Solution: Use Table 1.2. The result is: g D 9 . 81 ( m s 2 ) ( 1 ft 0 . 3048 m ) D 32 . 185 . . . ( ft s 2 ) D 32 . 2 ( ft s 2 ) Problem 1.13 A furlong per fortnight is a facetious unit of velocity, perhaps made up by a student as a satirical comment on the bewildering variety of units engineers must deal with. A furlong is 660 ft (1/8 mile). A fortnight is 2 weeks (14 nights). If you walk to class at 2 m/s, what is your speed in furlongs per fortnight to three signi fi cant digits? Solution: v D 2 m/s ( 1 ft 0.3048 m ) ( 1 furlong 660 ft ) ( 3600 s hr ) ( 24 hr 1 day ) ( 14 day 1 fortnight ) v D 12 , 000 furlongs fortnight Problem 1.14 Determine the cross-sectional area of the beam (a) in m 2 ; (b) in in 2 . 120 mm x y 40 mm 40 mm 40 mm 200 mm Solution: A D  200 mm  2  2  80 mm  120 mm  D 20800 mm 2 a) A D 20800 mm 2 ( 1 m 1000 mm ) 2 D 0.0208 m 2 A D 0.0208 m 2 b) A D 20800 mm 2 ( 1 in 25.4 mm ) 2 D 32.2 in 2 A D 32.2 in 2 Problem 1.15 The cross-sectional area of the C12 ð 30 American Standard Channel steel beam is A D 8 . 81 in 2 . What is its cross-sectional area in mm 2 ? x y A Solution: A D 8 . 81 in 2 ( 25 . 4 mm 1 in ) 2 D 5680 mm 2

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    Problem 1.16 A pressure transducer measures a value of 300 lb/in 2 . Determine the value of the pressure in pascals. A pascal (Pa) is one newton per meter squared. Solution: Convert the units using Table 1.2 and the de fi nition of the Pascal unit. The result: 300 ( lb in 2 ) ( 4 . 448 N 1 lb ) ( 12 in 1 ft ) 2 ( 1 ft 0 . 3048 m ) 2 D 2 . 0683 . . .  10 6  ( N m 2 ) D 2 . 07  10 6  Pa 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: P D 7000 hp ( 550 ft-lb/s 1 hp ) ( 1 m 3.281 ft ) ( 1 N 0.2248 lb ) D 5 . 22 ð 10 6 W P D 5 . 22 ð 10 6 W Problem 1.18 Chapter 7 discusses distributed loads that are expressed in units of force per unit length. If the value of a distributed load is 400 N/m, what is its value in lb/ft?. Solution: w D 400 N/m ( 0.2248 lb 1 N ) ( 1 m 3.281 ft ) D 27.4 lb/ft w D 27.4 lb/ft Problem 1.19 The moment of inertia of the rectan- gular area about the x axis is given by the equation I D 1 3 bh 3 . The dimensions of the area are b D 200 mm and h D 100 mm. Determine the value of I to four signi fi cant digits in terms of (a) mm 4 ; (b) m 4 ; (c) in 4 . h b x y Solution: (a) I D 1 3  200 mm  100 mm  3 D 66 . 7 ð 10 6 mm 4 (b) I D 66 . 7 ð 10 6 mm 4 ( 1 m 1000 mm ) 4 D 66 . 7 ð 10  6 m 4 (c) I D 66 . 7 ð 10 6 mm 4 ( 1 in 25 . 4 mm ) 4 D 160 in 4 Problem 1.20 In Example 1.3, instead of Einstein’s equation consider the equation L D mc , where the mass m is in kilograms and the velocity of light c is in meters per second. (a) What are the SI units of L ? (b) If the value of L in SI units is 12, what is its value in U.S. Customary base units? Solution: a) L D mc ) Units L D kg-m/s b) L D 12 kg-m/s ( 0.0685 slug 1 kg ) ( 3.281 ft 1 m ) D 2.70 slug-ft/s L D 2.70 slug-ft/s

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    Problem 1.21 The equation  D My I is used in the mechanics of materials to determine normal stresses in beams. (a) When this equation is expressed in terms of SI base units, M is in newton-meters (N-m), y is in meters (m), and I is in meters to the fourth power (m 4 ). What are the SI units of  ? (b) If M D 2000 N-m, y D 0 . 1 m, and I D 7 ð 10  5 m 4 , what is the value of  in U.S. Customary base units? Solution: (a)  D My I D (N-m)m m 4 D N m 2 (b)  D My I D  2000 N-m  0 . 1 m  7 ð 10  5 m 4 ( 1 lb 4 . 448 N ) ( 0 . 3048 m ft ) 2 D 59 , 700 lb ft 2 Problem 1.22 The acceleration due to gravity on the surface of the moon is 1.62 m/s 2 . (a) What would the mass of the C-clamp in Active Example 1.4 be on the surface of the moon? (b) What would the weight of the C-clamp in newtons be on the surface of the moon? Solution: a) The mass does not depend on location. The mass in kg is 0.0272 slug ( 14.59 kg 1 slug ) D 0.397 kg mass D 0 . 397 kg b) The weight on the surface of the moon is W D mg D  0.397 kg  1.62 m/s 2  D 0.643 N W D 0 . 643 N Problem 1.23 The 1 ft ð 1 ft ð 1 ft cube of iron weighs 490 lb at sea level. Determine the weight in newtons of a 1 m ð 1 m ð 1 m cube of the same material at sea level. 1 ft 1 ft 1 ft Solution: The weight density is  D 490 lb 1 ft 3 The weight of the 1 m 3 cube is: W D V D ( 490 lb 1 ft 3 )  1 m  3 ( 1 ft 0.3048 m ) 3 ( 1 N 0.2248 lb ) D 77.0 kN Problem 1.24 The area of the Paci fi c 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 Paci fi c Ocean (a) in slugs; (b) in kilograms Solution: The volume of the ocean is V D  64 , 186 , 000 mi 2  12 , 925 ft  ( 5 , 280 ft 1 mi ) 2 D 2 . 312 ð 10 19 ft 3 (a) m D V D ( 64 lb/ft 3 32 . 2 ft/s 2 )  2 . 312 ð 10 19 ft 3  D 4 . 60 ð 10 19 slugs (b) m D  4 . 60 ð 10 19 slugs  ( 14 . 59 kg 1 slug ) D 6 . 71 ð 10 20 kg Problem 1.25 The acceleration due to gravity at sea level is g D 9 . 81 m/s 2 . The radius of the earth is 6370 km. The universal gravitational constant is G D 6 . 67 ð 10  11 N-m 2 / kg 2 . Use this information to determine the mass of the earth. Solution: Use Eq (1.3) a D Gm E R 2 . Solve for the mass, m E D gR 2 G D  9 . 81 m/s 2  6370 km  2 ( 10 3 m km ) 2 6 . 67  10  11  ( N-m 2 kg 2 ) D 5 . 9679 . . .  10 24  kg D 5 . 97  10 24  kg

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    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). W D mg ( R E r ) 2 D ( W E g ) g ( R E R E C H ) 2 D W E ( 3960 3960 C 200 ) 2 D  180  0 . 90616  D 163 lb 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 D 1738 km. (a) What is the weight in newtons on the surface of the moon of an object that has a mass of 10 kg? (b) Using the approach described in Example 1.5, deter- mine the force exerted on the object by the gravity of the moon if the object is located 1738 km above the moon’s surface. Solution: a) W D mg M D  10 kg  1.26 m/s 2  D 12.6 N W D 12.6 N b) Adapting equation 1.4 we have a M D g M ( R M r ) 2 . The force is then F D ma M D  10 kg  1.62 m/s 2  ( 1738 km 1738 km C 1738 km ) 2 D 4.05 N F D 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 acceler- ation due to gravity at sea level is g D 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 earth does the weight of the object decrease to 0.99 mg ? Solution: Use a variation of Eq (1.5). W D mg ( R E R E C h ) 2 D 0 . 99 mg Solve for the radial height, h D R E ( 1 p 0 . 99  1 ) D  6370  1 . 0050378  1 . 0  D 32 . 09 . . . km D 32 , 100 m D 32 . 1 km Problem 1.29 The planet Neptune has an equatorial diameter of 49,532 km and its mass is 1 . 0247 ð 10 26 kg. If the planet is modeled as a homogeneous sphere, what is the acceleration due to gravity at its surface? (The universal gravitational constant is G D 6 . 67 ð 10  11 N-m 2 / kg 2 .) Solution: We have: W D G m N m r N 2 D ( G m N r 2 ) m ) g N D G m N r N 2 Note that the radius of Neptune is r N D 1 2  49,532 km  D 24,766 km Thus g N D ( 6 . 67 ð 10  11 N-m 2 kg 2 ) ( 1 . 0247 ð 10 26 kg  24766 km  2 ) ( 1 km 1000 m ) 2 D 11.1 m/s 2 g N D 11.1 m/s 2 Problem 1.30 At a point between the earth and the moon, the magnitude of the force exerted on an object by the earth’s gravity equals the magnitude of the force exerted on the object by the moon’s gravity. What is the distance from the center of the earth to that point to three signi fi cant digits? The distance from the center of the earth to the center of the moon is 383,000 km, and the radius of the earth is 6370 km. The radius of the moon is 1738 km, and the acceleration due to gravity at its surface is 1.62 m/s 2 . Solution: Let r Ep be the distance from the Earth to the point where the gravitational accelerations are the same and let r Mp be the distance from the Moon to that point. Then, r Ep C r Mp D r EM D 383 , 000 km. The fact that the gravitational attractions by the Earth and the Moon at this point are equal leads to the equation g E ( R E r Ep ) 2 D g M ( R M r Mp ) 2 , where r EM D 383 , 000 km. Substituting the correct numerical values leads to the equation 9 . 81 ( m s 2 ) ( 6370 km r Ep ) 2 D 1 . 62 ( m s 2 ) ( 1738 km r EM  r Ep ) 2 , where r Ep is the only unknown. Solving, we get r Ep D 344 , 770 km D 345 , 000 km.

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    Problem 2.1 In Active Example 2.1, suppose that the vectors U and V are reoriented as shown. The vector V is vertical. The magnitudes are j U j D 8 and j V j D 3. Graphically determine the magnitude of the vector U C 2 V . V U 45  Solution: Draw the vectors accurately and measure the resultant. R D j U C 2 V j D 5 . 7 R D 5 . 7 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 50 ° . Draw a sketch of the new situation. The magnitudes of the forces are j F AB j D 100 kN and j F AC j D 60 kN. Graphically determine the magnitude and direction of the sum of the forces exerted on the pylon by the cables. Solution: Accurately draw the vectors and measure the magnitude and direction of the resultant j F AB C F AC j D 146 kN ˛ D 32 °

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    Problem 2.3 The magnitude j F A j D 80 lb and the angle ˛ D 65 ° . The magnitude j F A C F B j D 120 lb. Graphically determine the magnitude of F B . F B F A a F C  Solution: Accurately draw the vectors and measure the magnitude of F B . j F B j D 62 lb Problem 2.4 The magnitudes j F A j D 40 N, j F B j D 50 N, and j F C j D 40 N. The angle ˛ D 50 ° and ˇ D 80 ° . Graphically determine the magnitude of F A C F B C F C . F B F A a F C  Solution: Accurately draw the vectors and measure the magnitude of F A C F B C F C . R D j F A C F B C F C j D 83 N

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    Problem 2.5 The magnitudes j F A j D j F B j D j F C j D 100 lb, and the angles ˛ D 30 ° . Graphically determine the value of the angle ˇ for which the magnitude j F A C F B C F C j is a minimum and the minimum value of j F A C F B C F C j . F B F A a F C  Solution: For a minimum, the vector F C must point back to the origin. R D j F A C F B C F C j D 93.2 lb ˇ D 165 ° Problem 2.6 The angle  D 50 ° . Graphically determine the magnitude of the vector r AC . 60 mm 150 mm A C B r AB r BC r AC  Solution: Draw the vectors accurately and then measure j r AC j . j r AC j D 181 mm

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    Problem 2.7 The vectors F A and F B represent the forces exerted on the pulley by the belt. Their magnitudes are j F A j D 80 N and j F B j D 60 N. Graphically determine the magnitude of the total force the belt exerts on the pulley. 45  F A F B 10  Solution: Draw the vectors accurately and then measure j F A C F B j . j F A C F B j D 134 N Problem 2.8 The sum of the forces F A C F B C F C D 0 . The magnitude j F A j D 100 N and the angle ˛ D 60 ° . Graphically determine the magnitudes j F B j and j F C j . 30 F B F A F C a Solution: Draw the vectors so that they add to zero. j F B j D 86.6 N , j F C j D 50.0 N Problem 2.9 The sum of the forces F A C F B C F C D 0 . The magnitudes j F A j D 100 N and j F B j D 80 N. Graphically determine the magnitude j F C j and the angle ˛ . 30 F B F A F C a Solution: Draw the vectors so that they add to zero. j F C j D 50.4 N , ˛ D 52 . 5 °

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    Problem 2.10 The forces acting on the sailplane are represented by three vectors. The lift L and drag D are perpendicular. The magnitude of the weight W is 500 lb. The sum of the forces W C L C D D 0 . Graph- ically determine the magnitudes of the lift and drag. W D L 25  Solution: Draw the vectors so that they add to zero. Then measure the unknown magnitudes. j L j D 453 lb j D j D 211 lb Problem 2.11 A spherical storage tank is suspended from cables. The tank is subjected to three forces, the forces F A and F B exerted by the cables and its weight W . The weight of the tank is j W j D 600 lb. The vector sum of the forces acting on the tank equals zero. Graphically determine the magnitudes of F A and F B . 40˚ F A W F B 20˚ 20˚ Solution: Draw the vectors so that they add to zero. Then measure the unknown magnitudes. j F A j D j F B j D 319 lb

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    Problem 2.12 The rope ABC exerts forces F BA and F BC 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 j F BA j . 20  F BC F BA B C A B Solution: Draw the vectors accurately and then measure the unknown magnitudes. j F BA j D 174 lb 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 C F B exerted on the shelter is in the direction parallel to the line L and its magnitude is 400 lb. Graphically determine the magnitudes of F A and F B . L Top View F A F B 50  30  Solution: Draw the vectors accurately and then measure the unknown magnitudes. j F A j D 203 lb j F B j D 311 lb Problem 2.14 A surveyor determines that the horizon- tal 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 r BC and the angle ˛ . East North 60  20  C B A r BC a Solution: Draw the vectors accurately and then measure the unknown magnitude and angle. j r BC j D 390 m ˛ D 21 . 2 °

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    Problem 2.15 The vector r extends from point A to the midpoint between points B and C . Prove that r D 1 2  r AB C r AC . A C B r AC r r AB r AB Solution: The proof is straightforward: r D r AB C r BM , and r D r AC C r CM . Add the two equations and note that r BM C r CM D 0, since the two vectors are equal and opposite in direction. Thus 2 r D r AC C r AB , or r D ( 1 2 )  r AC C r AB  r AC r AB r A B M C Problem 2.16 By drawing sketches of the vectors, explain why U C  V C W  D  U C V  C W . Solution: Additive associativity for vectors is usually given as an axiom in the theory of vector algebra, and of course axioms are not subject to proof. However we can by sketches show that associativity for vector addition is intuitively reasonable: Given the three vectors to be added, (a) shows the addition fi rst of V C W , and then the addition of U . The result is the vector U C  V C W  . (b) shows the addition of U C V , and then the addition of W , leading to the result  U C V  C W . The fi nal vector in the two sketches is the same vector, illustrating that associativity of vector addition is intuitively reasonable. (a) U W V U W V V+W U+V U+[V+W] [U+V]+W (b) Problem 2.17 A force F D 40 i  20 j  N  . What is its magnitude j F j ? Strategy: The magnitude of a vector in terms of its components is given by Eq. (2.8). Solution: j F j D p 40 2 C 20 2 D 44 . 7 N Problem 2.18 An engineer estimating the components of a force F D F x i C F y j acting on a bridge abutment has determined that F x D 130 MN, j F j D 165 MN, and F y is negative. What is F y ? Solution: j F j D √ j F x j 2 C j F y j 2 j F y j D √ j F j 2  j F x j 2 D √  165 MN  2   130 MN  2 D 101.6 MN F y D  102 MN

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    Problem 2.19 A support is subjected to a force F D F x i C 80 j (N). If the support will safely support a force of 100 N, what is the allowable range of values of the component F x ? Solution: Use the de fi nition of magnitude in Eq. (2.8) and reduce algebraically. 100 ½ √ F x  2 C  80  2 , from which  100  2   80  2 ½ F x  2 . Thus j F x j  p 3600, or  60  F x   C 60 (N) Problem 2.20 If F A D 600 i  800 j (kip) and F B D 200 i  200 j (kip), what is the magnitude of the force F D F A  2 F B ? Solution: Take the scalar multiple of F B , add the components of the two forces as in Eq. (2.9), and use the de fi nition of the magnitude. F D  600  2  200  i C   800  2   200  j D 200 i  400 j j F j D √  200  2 C   400  2 D 447 . 2 kip Problem 2.21 The forces acting on the sailplane are its weight W D  500 j  lb  , the drag D D  200 i C 100 j (lb) and the lift L . The sum of the forces W C L C D D 0 . Determine the components and the magnitude of L . y x W D L Solution: L D  W  D D    500 j     200 i C 100 j  D 200 i C 400 j  lb  j L j D √  200 lb  2 C  400 lb  2 D 447 lb L D 200 i C 400 j  lb , j L j D 447 lb

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    Problem 2.22 Two perpendicular vectors U and V lie in the x - y plane. The vector U D 6 i  8 j and j V j D 20. What are the components of V ? (Notice that this problem has two answers.) Solution: The two possible values of V are shown in the sketch. The strategy is to (a) determine the unit vector associated with U , (b) express this vector in terms of an angle, (c) add š 90 ° to this angle, (d) determine the two unit vectors perpendicular to U , and (e) calculate the components of the two possible values of V . The unit vector parallel to U is e U D 6 i √ 6 2 C   8  2  8 j √ 6 2 C   8  2 D 0 . 6 i  0 . 8 j Expressed in terms of an angle, e U D i cos ˛  j sin ˛ D i cos  53 . 1 °   j sin  53 . 1 °  Add š 90 ° to fi nd the two unit vectors that are perpendicular to this unit vector: e p 1 D i cos  143 . 1 °   j sin  143 . 1 °  D  0 . 8 i  0 . 6 j e p 2 D i cos   36 . 9 °   j sin   36 . 9 °  D 0 . 8 i C 0 . 6 j Take the scalar multiple of these unit vectors to fi nd the two vectors perpendicular to U . V 1 D j V j   0 . 8 i  0 . 6 j  D  16 i  12 j . The components are V x D  16 , V y D  12 V 2 D j V j  0 . 8 i C 0 . 6 j  D 16 i C 12 j . The components are V x D 16 , V y D 12 y x 6 8 U V 2 V 1 Problem 2.23 A fi sh exerts a 10-lb force on the line that is represented by the vector F . Express F in terms of components using the coordinate system shown. y x F 7 11 Solution: We can use similar triangles to determine the components of F . F D  10 lb  ( 7 p 7 2 C 11 2 i  11 p 7 2 C 11 2 j ) D  5 . 37 i  8 . 44 j  lb F D  5 . 37 i  8 . 44 j  lb

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