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

Problem 3.1 In ctive Eample 3.1, suppose that the angle between the ramp supporting the car is increased from 20 to 30. raw the free-bod diagram of the car showing the new geometr. Suppose that the cable from to must eert a 8500 N horizontal force on the car to hold it in place. etermine the car s weight in pounds. 20 he free-bod diagram is shown to the right. ppling the equilibrium equations F : N sin 30 0, F : N cos 30 mg 0 Setting 8500 N and solving ields N 17000 N, mg 14722 N Problem 3.2 he ring weighs 5 N and is in equilibrium. he force F 1 4.5 N. etermine the force F 2 and the angle. F 2 a F 1 30 he free-bod diagram is shown below the drawing. he equilibrium equations are F : F 1 cos 30 F 2 cos 0 F : F 1 sin 30 F 2 sin 5N 0 e can write these equations as F 2 sin 5N F 1 sin 30 F 2 cos F 1 cos 30 N ividing these equations and using the known value for F 1 we have. tan 5N 4.5 N sin 30 4.5 N cos 30 F 2 4.5 N cos 30 cos 4.77 N 0.706 ) 35.2 F 2 4.77 N, 35.2 84 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.3 In Eample 3.2, suppose that the attachment point is moved to the right and cable is etended so that the angle between cable and the ceiling decreases from 45 to 35. he angle between cable and the ceiling remains 60. hat are the tensions in cables and? 60 45 he free-bod diagram is shown below the picture. he equilibrium equations are: F : cos 35 cos 60 0 F : sin 35 sin 60 1962 N 0 Solving we find 1610 N, 985 N Problem 3.4 he 200-kg engine block is suspended b the cables and. he angle 40. he freebod diagram obtained b isolating the part of the sstem within the dashed line is shown. etermine the forces and. a a (200 kg) (9.81 m/s 2 ) 40 F : cos cos 0 α α F : sin sin 1962 N 0 Solving: 1.526 kn 1962 Ν c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 85

Problem 3.5 heav rope used as a mooring line for a cruise ship sags as shown. If the mass of the rope is 90 kg, what are the tensions in the rope at and? 55 40 he free-bod diagram is shown. he equilibrium equations are F : cos 40 cos 55 0 F : sin 40 sin 55 90 9.81 N 0 Solving: 679 N, 508 N Problem 3.6 phsiologist estimates that the masseter muscle of a predator, Martes, is capable of eerting a force M as large as 900 N. ssume that the jaw is in equilibrium and determine the necessar force that the temporalis muscle eerts and the force P eerted on the object being bitten. P 22 he equilibrium equations are M F : cos 22 M cos 36 0 36 F : sin 22 M sin 36 P 0 Setting M 900 N, and solving, we find 785 N,P 823 N 86 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.7 he two springs are identical, with unstretched lengths 250 mm and spring constants k 1200 N/m. (a) raw the free-bod diagram of block. (b) raw the free-bod diagram of block. (c) hat are the masses of the two blocks? 300 mm 280 mm he tension in the upper spring acts on block in the positive Y direction, Solve the spring force-deflection equation for the tension in the upper spring. ppl the equilibrium conditions to block. Repeat the steps for block. ( U 0i 1200 N ) 0.3 m 0.25 m j 0i 60j N m 300 mm Similarl, the tension in the lower spring acts on block in the negative Y direction ( L 0i 1200 N ) 0.28 m 0.25 m j 0i 36j N m he weight is 0i j jj 280 mm he equilibrium conditions are F F F 0, F U L 0 ension, upper spring ollect and combine like terms in i, j Solve F j j60 36 j 0 j j 60 36 24 N he mass of is m j Lj jgj 24 N 2.45 kg 2 9.81 m/s ension, lower spring eight, mass ension, lower spring he free bod diagram for block is shown. he tension in the lower spring L 0i 36j he weight: 0i j jj ppl the equilibrium conditions to block. eight, mass F L 0 ollect and combine like terms in i, j: F j j36 j 0 Solve: j j36 N he mass of is given b m j j jgj 36 N 3.67 kg 2 9.81 m/s c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 87

Problem 3.8 he two springs in Problem 3.7 are identical, with unstretched lengths of 250 mm. Suppose that their spring constant k is unknown and the sum of the masses of blocks and is 10 kg. etermine the value of k and the masses of the two blocks. ll of the forces are in the vertical direction so we will use scalar equations. First, consider the upper spring supporting both masses (10 kg total mass). he equation of equilibrium for block the entire assembl supported b the upper spring is is U m m g 0, where U k l U 0.25 N. he equation of equilibrium for block is U m g 0, where U k l L 0.25 N. he equation of equilibrium for block alone is U L m g 0 where L U. Using g 9.81 m/s 2, and solving simultaneousl, we get k 1962 N/m, m 4 kg, and m 6kg. Problem 3.9 he inclined surface is smooth (Remember that smooth means that friction is negligble). he two springs are identical, with unstretched lengths of 250 mm and spring constants k 1200 N/m. hat are the masses of blocks and? 300 mm 280 mm 30 m g F 1 1200 N/m 0.3 0.25 m 60 N F 1 F 2 1200 N/m 0.28 0.25 m 36 N F &: F 2 m g sin 30 0 F 2 F 2 m g F &: F 1 F 2 m g sin 30 0 N Solving: m 4.89 kg, m 7.34 kg N 88 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.10 he mass of the crane is 20,000 kg. he crane s cable is attached to a caisson whose mass is 400 kg. he tension in the cable is 1 kn. (a) (b) etermine the magnitudes of the normal and friction forces eerted on the crane b the level ground. etermine the magnitudes of the normal and friction forces eerted on the caisson b the level ground. 45 Strateg: o do part (a), draw the free-bod diagram of the crane and the part of its cable within the dashed line. 45 (a) F : N crane 196.2 kn 1 kn sin 45 0 F : F crane 1 kn cos 45 0 196.2 kn 1 kn N crane 196.9 kn,f crane 0.707 kn (b) F : N caisson 3.924 kn 1 kn sin 45 0 Fcrane F : 1 kn cos 45 F caisson 0 N crane N caisson 3.22 kn, F caisson 0.707 kn 1 kn 3.924 kn 45 Fcaisson Ncaisson Problem 3.11 he inclined surface is smooth. he 100-kg crate is held stationar b a force applied to the cable. (a) raw the free-bod diagram of the crate. (b) etermine the force. (a) he F 981 Ν 60 Ν (b) 60 F -: 981 N sin 60 0 850 N c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 89

Problem 3.12 sloping road. he 1200-kg car is stationar on the (a) (b) If 20, what are the magnitudes of the total normal and friction forces eerted on the car s tires b the road? he car can remain stationar onl if the total friction force necessar for equilibrium is not greater than 0.6 times the total normal force. hat is the largest angle for which the car can remain stationar? a (a) 20 11.772 kn F% : N 11.772 kn cos 0 F- : F 11.772 kn sin 0 N 11.06 kn, F 4.03 kn (b) F 0.6 N F% : N 11.772 kn cos 0 ) 31.0 α F F- : F 11.772 kn sin 0 N Problem 3.13 he 450 N crate is in equilibrium on the smooth surface. he spring constant is k 6000 N/m. Let S be the stretch of the spring. Obtain an equation for S (in metre) as a function of the angle. a he free-bod diagram is shown. he equilibrium equation in the direction parallel to the inclined surface is ks 450 N sin 0 450 N Solving for S and using the given value for k we find S 0.075 m sin 90 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.14 600 N bo is held in place on the smooth bed of the dump truck b the rope. (a) If 25, what is the tension in the rope? (b) If the rope will safel support a tension of 400 N, what is the maimum allowable value of? α Isolate the bo. Resolve the forces into scalar components, and solve the equilibrium equations. he eternal forces are the weight, the tension in the rope, and the normal force eerted b the surface. he angle between the ais and the weight vector is 90 (or 270 ). he weight vector is α jj i sin j cos 600 i sin j cos he projections of the rope tension and the normal force are j ji 0j N 0i jn jj he equilibrium conditions are F N 0 N Substitute, and collect like terms α F 600 sin j j i 0 F 600 cos jn j j 0 Solve for the unknown tension when For 25 j j600 sin 253.6 N. For a tension of 400 N, (600 sin 400 0. Solve for the unknown angle sin 400 600 0.667 or 41.84 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 91

Problem 3.15 he 80-N bo is held in place on the smooth inclined surface b the rope. etermine the tension in the rope and the normal force eerted on the bo b the inclined surface. 30 50 he equilibrium equations (in terms of a coordinate sstem with the ais parallel to the inclined surface) are N F : 80 N sin 50 cos 50 0 F : N 80 N cos 50 sin 50 0 Solving: 95.34 N, N 124 N Problem 3.16 he 1360-kg car and the 2100-kg tow truck are stationar. he mudd surface on which the car s tires rest eerts negligible friction forces on them. hat is the tension in the tow cable? 10 18 26 F of the car being towed F- : cos 8 13.34 kn sin 26 0 18 13.34 kn 5.91 kn 26 N 92 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.17 Each bo weighs 40 N. he angles are measured relative to the horizontal. he surfaces are smooth. etermine the tension in the rope and the normal force eerted on bo b the inclined surface. 70 45 20 he free-bod diagrams are shown. he equilibrium equations for bo are N F : 40 N sin 20 cos 25 0 F : N 40 N cos 20 sin 25 0 he equilibrium equations for bo are F : 40 N sin 70 cos 25 0 F : N 40 N cos 70 sin 25 0 N Solving these four equations ields: 51.2 N, 15.1 N, N 7.30 N, N 31.2 N hus 51.2 N, N 7.30 N Problem 3.18 10-kg painting is hung with a wire supported b a nail. he length of the wire is 1.3 m. (a) (b) hat is the tension in the wire? hat is the magnitude of the force eerted on the nail b the wire? 1.2 m (a) F :98.1 N 2 5 13 0 98.1 N 128 N (b) Force 98.1 N 5 12 12 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 93

Problem 3.19 10-kg painting is hung with a wire supported b two nails. he length of the wire is 1.3 m. (a) (b) hat is the tension in the wire? hat is the magnitude of the force eerted on each nail b the wire? (ssume that the tension is the same in each part of the wire.) 0.4 m 0.4 m 0.4 m ompare our answers to the answers to Problem 3.18. (a) Eamine the point on the left where the wire is attached to the picture. his point supports half of the weight R 27.3 F : sin 27.3 49.05 N 0 (b) 107 N Eamine one of the nails F : R cos 27.3 0 49.05 N R F : R sin 27.3 0 R R 2 R 2 R 27.3 R 50.5 N Problem 3.20 ssume that the 750 N climber is in equilibrium. hat are the tensions in the rope on the left and right sides? 14 15 F R cos 15 L cos 14 0 F R sin 15 L sin 14 750 0 Solving, we get L 1490 N, R 1500 N 14 L R 15 750 N 94 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.21 If the mass of the climber shown in Problem 3.20 is 80 kg, what are the tensions in the rope on the left and right sides? F R cos 15 L cos 14 0 F R sin 15 L sin 14 mg 0 Solving, we get L 1.56 kn, R 1.57 kn R L 14 15 mg = (80) (9.81) N Problem 3.22 he construction worker eerts a 90 N force on the rope to hold the crate in equilibrium in the position shown. hat is the weight of the crate? 5 30 he free-bod diagram is shown. he equilibrium equations for the part of the rope sstem where the three ropes are joined are F : 90 N cos 30 sin 5 0 F : 90 N sin 30 cos 5 0 Solving ields 935.9 N 90 N c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 95

Problem 3.23 construction worker on the moon, where the acceleration due to gravit is 1.62 m/s 2, holds the same crate described in Problem 3.22 in the position shown. hat force must she eert on the cable to hold the crate ub equilibrium (a) in newtons; (b) in pounds? 5 30 he free-bod diagram is shown. From Problem 3.22 we know that the weight is 188 lb. herefore its mass is 188 lb m 5.84 slug 2 32.2 ft/s ( ) 14.59 kg m 5.84 slug 85.2 kg slug he equilibrium equations for the part of the rope sstem where the three ropes are joined are F : F cos 30 sin 5 0 F : F sin 30 cos 5 mg m 0 where g m 1.62 m/s 2. Solving ields F 3.30 lb 14.7 N a F 14.7 N, b F 3.30 lb 96 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.24 he person wants to cause the 200-N crate to start sliding toward the right. o achieve this, the horizontal component of the force eerted on the crate b the rope must equal 0.35 times the normal force eerted on the crate b the floor. In Fig.a, the person pulls on the rope in the direction shown. In Fig.b, the person attaches the rope to a support as shown and pulls upward on the rope. hat is the magnitude of the force he must eert on the rope in each case? (a) 20 10 (b) he friction force F fr is given b F fr 0.35N N (a) For equilibrium we have F : cos 20 0.35N 0 F : sin 20 200 N N 0 Solving: 66.1 N (b) he person eerts the force F. Using the free-bod diagram of the crate and of the point on the rope where the person grabs the rope, we find N F : L 0.35N 0 F : N 200 N 0 F : L R cos 10 0 F : F R sin 10 0 Solving we find F 12.3 N c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 97

Problem 3.25 traffic engineer wants to suspend a 500 N traffic light above the center of the two right lanes of a four-lane thoroughfare as shown. etermine the tensions in the cables and. 5 m 40 m 10 m 15 m F : p 6 p 2 0 37 5 1 F : p p 1 500 N 0 37 5 Solving: 760 N, 838 N 6 1 2 1 500 N Problem 3.26 able is 3 m long and cable is 4 m long. he mass of the suspended object is 350 kg. etermine the tensions in cables and. 5m F : 3 5 4 5 0 F : 4 5 3 5 3.43 kn 0 4 3 4 3 2.75 kn, 2.06 kn 3.43 kn 98 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.27 In Problem 3.26, the length of cable is adjustable. If ou don t want the tension in either cable or cable to eceed 3 kn, what is the minimum acceptable length of cable? onsider the geometr: 5 e have the constraints hese constraint impl L 2 2 2, 4m 2 5 m 2 2 L 4 m 10 m 2 9m 2 L 10 m 9m 2 Now draw the F and write the equations in terms of F : p 5 10 9 4 0 4 5 F : p p 10 2 9 10 p 2 9 3.43 kn 0 10 9 4 If we set 3 kn and solve for we find 1.535, 2.11 kn < 3kN 3.43 kn Using this value for we find that L 2.52 m c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 99

Problem 3.28 hat are the tensions in the upper and lower cables? (Your answers will be in terms of. Neglect the weight of the pulle.) 45 30 Isolate the weight. he frictionless pulle changes the direction but not the magnitude of the tension. he angle between the right hand upper cable and the ais is, hence UR j U j i cos j sin. he angle between the positive and the left hand upper pulle is 180 ˇ, hence L β U U α UL j U j i cos 180 ˇ j sin 180 ˇ j U j i cos ˇ j sin ˇ. he lower cable eerts a force: L j L ji 0j he weight: 0i jjj he equilibrium conditions are F UL UR L 0 Substitute and collect like terms, F j U j cos ˇ j U j cos j L j i 0 F j U j sin j U j sin ˇ jj j 0. Solve: ( j U j jj sin sin ˇ ), j L jj U j cos cos ˇ. From which j L jjj ( ) cos cos ˇ. sin sin ˇ For 30 and ˇ 45 j U j0.828jj, j L j0.132jj 100 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.29 wo tow trucks lift a 100 kg motorccle out of a ravine following an accident. If the motorccle is in equilibrium in the position shown, what are the tensions in cables and? (3, 8) m (10, 9) m (6, 4.5) m he angles are ( ) tan 1 8 4.5 49. 4 6 3 ( ) ˇ tan 1 10 4.5 54 10 6 Now from equilibrium we have F : cos cos ˇ 0 F : sin sin ˇ 981 N 0 981 N Solving ields 658 N, 645 N c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 101

Problem 3.30 n astronaut candidate conducts eperiments on an airbearing platform. hile she carries out calibrations, the platform is held in place b the horizontal tethers,, and. he forces eerted b the tethers are the onl horizontal forces acting on the platform. If the tension in tether is 2 N, what are the tensions in the other two tethers? OP VIE 4.0 m 3.5 m 3.0 m 1.5 m Isolate the platform. he angles and ˇ are lso, tan tan ˇ ( ) 1.5 0.429, 23.2. 3.5 ( ) 3.0 0.857, ˇ 40.6. 3.5 3.0 m 1.5 m 3.5 m 4.0 m he angle between the tether and the positive ais is 180 ˇ, hence j j i cos 180 ˇ j sin 180 ˇ j j i cos ˇ j sin ˇ. β α he angle between the tether and the positive ais is 180. he tension is j j i cos 180 j sin 180 j j i cos j sin. he tether is aligned with the positive ais, j ji 0j. he equilibrium condition: Solve: j j ( ) sin j j, sin ˇ ( ) j j sin ˇ j j. sin ˇ F 0. Substitute and collect like terms, For j j2n, 23.2 and ˇ 40.6, j j1.21 N, j j2.76 N F j j cos ˇ j j cos j j i 0, F j j sin ˇ j j sin j 0. 102 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.31 he bucket contains concrete and weighs 5800 N. hat are the tensions in the cables and? (5, 34) m (20, 34) m (12, 16) m he angles are ( ) 34 16 tan 1 68.7 12 5 ( ) 34 16 ˇ tan 1 66.0 20 12 Now from equilibrium we have F : cos cos ˇ 0 F : sin sin ˇ 5800 N 0 5800N Solving ields 3397 N, 2983 N Problem 3.32 he slider is in equilibrium and the bar is smooth. hat is the mass of the slider? 20 200 N 45 he pulle does not change the tension in the rope that passes over it. here is no friction between the slider and the bar. 20 = 200 N Eqns. of Equilibrium: F sin 20 N cos 45 0 200 N F N sin 45 cos 20 mg 0 g 9.81 m/s 2 Substituting for and g, we have two eqns in two unknowns (N and m). Solving, we get N 96.7 N, m 12.2 kg. N 45 mg = (9.81) g c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 103

Problem 3.33 he 20-kg mass is suspended from three cables. able is equipped with a turnbuckle so that its tension can be adjusted and a strain gauge that allows its tension to be measured. If the tension in cable is 40 N, what are the tensions in cables and? 0.64 m 0.4 m 0.4 m 0.48 m 40 N F : 5 p 89 5 p 89 11 p 185 0 8 8 5 8 F : 8 p p 8 p 8 196.2 N 0 89 89 185 5 11 Solving: 144.1 N, 68.2 N 196.2 N Problem 3.34 he structural joint is in equilibrium. If F 1000 Nand F 5000 N, what are F and F? F 80 F 65 35 F F he equilibrium equations are F : F F cos 65 F cos 35 F 0 F : F sin 65 F sin 35 0 Solving ields F 3680 N, F 2330 N 104 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.35 he collar slides on the smooth vertical bar. he masses m 20 kg and m 10 kg. hen h 0.1 m, the spring is unstretched. hen the sstem is in equilibrium, h 0.3 m. etermine the spring constant k. 0.25 m h k he triangles formed b the rope segments and the horizontal line level with can be used to determine the lengths L u and L s. he equations are L u 0.25 2 0.1 2 and L s 0.25 2 0.3 2. he stretch in the spring when in equilibrium is given b υ L s L u. arring out the calculations, we get L u 0.269 m, L s 0.391 m, and υ 0.121 m. he angle,, between the rope at and the horizontal when the sstem is in equilibrium is given b tan 0.3/0.25, or 50.2. From the free bod diagram for mass, we get two equilibrium equations. he are N m g F N cos 0 and F sin m g 0. e have two equations in two unknowns and can solve. e get N 163.5 N and 255.4 N. Now we go to the free bod diagram for, where the equation of equilibrium is m g kυ 0. his equation has onl one unknown. Solving, we get k 1297 N/m m g L u 0.25 m 0.1 m Kδ L s L u 0.3 m 0.25 m c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 105

Problem 3.36* Suppose that ou want to design a cable sstem to suspend an object of weight from the ceiling. he two wires must be identical, and the dimension b is fied. he ratio of the tension in each wire to its cross-sectional area must equal a specified value /. he cost of our design is the total volume of material in the two wires, V 2 p b 2 h 2. etermine the value of h that minimizes the cost. b b h From the equation F 2 sin 0, θ θ we obtain 2 sin p b 2 h 2. 2h Since /, p b 2 h 2 2h and the cost is V 2 p b 2 h 2 b2 h 2. h o determine the value of h that minimizes V, we set dv dh [ b2 h 2 ] h 2 2 0 and solve for h, obtaining h b. 106 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.37 he sstem of cables suspends a 1000-N bank of lights above a movie set. etermine the tensions in cables,, and E. Isolate juncture, and solve the equilibrium equations. Repeat for the cable juncture. he angle between the cable and the positive ais is. he tension in is j j i cos j sin he angle between the ais and is 180 ˇ. he tension is 10 m 9 m 45 30 E j j i cos 180 ˇ j sin 180 ˇ i cos ˇ j sin ˇ. he weight is 0i jjj. he equilibrium conditions are F 0 0. Substitute and collect like terms, F j j cos j j cos ˇ i 0 F j j sin ˇ j j sin jj j 0. Solve: j E jj j cos, j jj j sin ; for j j732 N and 30, j j896.6 N, Solving, we get ( ) ( ) cos jj cos ˇ j j j j and j j, cos ˇ sin ˇ jj 1000 N, and 30, ˇ 45 ( ) 0.7071 j j 1000 732.05 N 0.9659 ( ) 0.866 j j 732 896.5 N 0.7071 Isolate juncture. he angle between the positive ais and the cable is 180. he tension is j E j634 N, j j366 N β α j j i cos 180 j sin 180, or j j i cos j sin. 90 E he tension in the cable E is E ij E j0j. α he tension in the cable is 0i jj j. he equilibrium conditions are F 0 E 0 Substitute t and collect like terms, F j E j j j cos i 0, F j j j j sin j 0. c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 107

Problem 3.38 onsider the 1000-N bank of lights in Problem 3.37. technician changes the position of the lights b removing the cable E. hat is the tension in cable after the change? he original configuration in Problem 3.35 is used to solve for the dimensions and the angles. Isolate the juncture, and solve the equilibrium conditions. he lengths are calculated as follows: he vertical interior distance in the triangle is 10 m, since the angle is 45 deg. and the base and altitude of a 45 deg triangle are equal. he length is given b 10 m 14. 142 m. cos 45 10 m 9 m β α he length is given b 9 m 10. 392 m. cos 30 β 19 α he altitude of the triangle for which is the hpotenuse is 9 tan 30 5. 196 m. he distance is given b 10 5. 196 4. 804 m. he distance is given b 14.142 β 10.392 + 4.804 = 15.196 α 10. 392 4. 804 15. 196 he new angles are given b the cosine law 2 19 2 2 2 19 cos. β α Reduce and solve: ( 2. 2. 2 ) 19 15 196 14 142 cos 0.6787, 47.23. 2 19 15. 196 (. 2 2 2 ) 14 142 19 15.196 cos ˇ 0.6142, ˇ 52.1. 2 14.142 19 Isolate the juncture. he angle between the cable and the positive ais is. he tension is: j j i cos j sin. he angle between and the cable is 180 ˇ. he tension is Solve: and ( ) cos j j j j, cos ˇ ( ) jj cos ˇ j j. sin ˇ j j i cos ˇ j sin ˇ. he weight is 0i jjj he equilibrium conditions are F 0 0. For jj 1000 N, and 51.2, ˇ 47.2 ( ) 0.6142 j j 1000 621.03 N, 0.989 ( ) 0.6787 j j 622.3 687.9 N 0.6142 Substitute and collect like terms, F j j cos j j cos ˇ i 0, F j j sin ˇ j j sin jj j 0. 108 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.39 hile working on another ehibit, a curator at the Smithsonian Institution pulls the suspended Voager aircraft to one side b attaching three horizontal cables as shown. he mass of the aircraft is 1250 kg. etermine the tensions in the cable segments,, and. Isolate each cable juncture, beginning with and solve the equilibrium equations at each juncture. he angle between the cable and the positive ais is 70 ; the tension in cable is j j i cos j sin. he weight is 0i jjj. he tension in cable is jji 0j. he equilibrium conditions are 70 50 30 F 0. Substitute and collect like terms F j j cos jj i 0, F j j sin jj j 0. ( ) jj Solve: the tension in cable is j j. sin For jj 1250 kg (9.81 m ) s 2 12262.5 N and 70 α ( ) 12262.5 j j 13049.5 N 0.94 Isolate juncture. he angles are 50, ˇ 70, and the tension cable is j j i cos j sin. he angle between the cable and the positive ais is 180 ˇ ; the tension is j j i cos 180 ˇ j sin 180 ˇ j j i cos ˇ j sin ˇ α he tension in the left horizontal cable is jji 0j. he equilibrium conditions are F 0. β Substitute and collect like terms F j j cos j j cos ˇ jj i 0 F j j sin j j sin ˇ j 0. α Solve: j j ( ) sin ˇ j j. sin For j j13049.5 N, and 50, ˇ 70, ( ) 0.9397 j j 13049.5 16007.6 N 0.7660 Isolate the cable juncture. he angles are 30, ˇ 50. smmetr with the cable juncture above, the tension in cable is ( ) sin ˇ j j j j. sin ( ) 0.7660 Substitute: j j 16007.6 24525.0 N. 0.5 his completes the problem solution. β c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 109

Problem 3.40 truck dealer wants to suspend a 4000- kg truck as shown for advertising. he distance b 15 m, and the sum of the lengths of the cables and is 42 m. Points and are at the same height. hat are the tensions in the cables? b 40 m etermine the dimensions and angles of the cables. Isolate the cable juncture, and solve the equilibrium conditions. he dimensions of the triangles formed b the cables: b 15 m, L 25 m, S 42 m. 15 m 25 m b L β α Subdivide into two right triangles with a common side of unknown length. Let the unknown length of this common side be d, then b the Pthagorean heorem b 2 d 2 2, L 2 d 2 2. Subtract the first equation from the second to eliminate the unknown d, L 2 b 2 2 2. Note that 2 2. Substitute and reduce to the pair of simultaneous equations in the unknowns ( L 2 b 2 ), S S Solve: ( )( 1 L 2 b 2 ) S 2 S ( )( 1 25 2 15 2 ) 42 25.762 m 2 42 and S 42 25.762 16.238 m. he interior angles are found from the cosine law: ( L b 2 2 2 ) cos 0.9704 13.97 2 L b ( L b 2 2 2 ) cos ˇ 0.9238 ˇ 22.52 2 L b Isolate cable juncture. he angle between and the positive ais is ; the tension is j j i cos j sin β Substitute and collect like terms F j j cos j j cos ˇ i 0, F j j sin j j sin ˇ jj j 0 ( ) cos ˇ Solve: j j j j, cos ( ) jj cos and j j. sin ˇ For jj 4000 9.81 39240 N, and 13.97, ˇ 22.52, j j64033 64 kn, j j60953 61 kn α he angle between and the positive ais is 180 ˇ ; the tension is j j i cos 180 ˇ j sin 180 ˇ j j i cos ˇ j sin ˇ. he weight is 0i jjj. he equilibrium conditions are F 0. 110 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.41 he distance h 12 cm, and the tension in cable is 200 N. hat are the tensions in cables and? 12 cm 8 cm 12 cm h 12 cm 8 cm Isolated the cable juncture. From the sketch, the angles are found from ( ) 8 tan 0.667 33.7 12 ( ) 4 tan ˇ 0.333 ˇ 18.4 12 he angle between the cable and the positive ais is 180, the tension in is: 12 cm α β 8 cm 4 cm j j i cos 180 j sin 180 j j i cos j sin. he angle between and the positive ais is 180 ˇ. he tension is j j i cos 180 ˇ j sin 180 ˇ j j i cos ˇ j sin ˇ. he tension in the cable is j ji 0j. he equilibrium conditions are F 0. Substitute and collect like terms, F j j cos j j cos ˇ j j i 0 F j j sin j j sin ˇ j 0. Solve: j j ( ) sin ˇ j j, sin ( ) sin and j j j j. sin ˇ For j j200 N, 33.7, ˇ 18.4 j j140.6 N, j j80.1 N c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 111

Problem 3.42 You are designing a cable sstem to support a suspended object of weight. ecause our design requires points and to be placed as shown, ou have no control over the angle, but ou can choose the angle ˇ b placing point wherever ou wish. Show that to minimize the tensions in cables and, ou must choose ˇ if the angle ½ 45. α β Strateg: raw a diagram of the sum of the forces eerted b the three cables at. raw the free bod diagram of the knot at point. hen draw the force triangle involving the three forces. Remember that is fied and the force has both fied magnitude and direction. From the force triangle, we see that the force can be smaller than for a large range of values for ˇ. inspection, we see that the minimum simultaneous values for and occur when the two forces are equal. his occurs when ˇ. Note: this does not happen when <45. α In this case, we solved the problem without writing the equations of equilibrium. For reference, these equations are: F cos cos ˇ 0 and F sin sin ˇ 0. Possible locations for lie on line?? α andidate β andidate values for Fied direction for line Problem 3.43* he length of the cable is 1.4 m. he 2-kN force is applied to a small pulle. he sstem is stationar. hat is the tension in the cable? 1 m 0.75 m 15 2 kn Eamine the geometr h 2 0.75 m 2 h 2 0.25 m 2 1.4 m tan h 0.75 m, tan ˇ h 0.25 m α 0.75 m 0.25 m β h ) h 0.458 m, 31.39, ˇ 61.35 Now draw a F and solve for the tension. e can use either of the equilibrium equations F : cos cos ˇ 2 kn sin 15 0 α β 1.38 kn 15 2 kn 112 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.44 he masses m 1 12 kg and m 2 6kg are suspended b the cable sstem shown. he cable is horizontal. etermine the angle and the tensions in the cables,, and. α 70 α m 1 m 2 117.7 N e have 4 unknowns and 4 equations F : cos 0 F : sin 117.7 N 0 F : cos 70 0 70 F : sin 70 58.86 N 0 Solving we find 79.7, 119.7 N, 21.4 N, 62.6 N 58.86 N Problem 3.45 he weights 1 50 N and 2 are suspended b the cable sstem shown. etermine the weight 2 and the tensions in the cables,, and. 2 m 3 m 3 m 3 m 1.6 m e have 4 unknowns and 4 equilibrium equations to use 1 2 F : 3 p 13 15 p 229 0 F : 2 p p 2 50 N 0 13 229 2 3 2 15 F : 15 p 229 15 17 0 F : 2 p 229 8 17 2 0 50 N ) 2 25 N, 75.1 N 63.1 N, 70.8 N 15 8 15 2 2 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 113

Problem 3.46 In the sstem shown in Problem 3.45, assume that 2 1 /2. If ou don t want the tension anwhere in the supporting cable to eceed 200 N, what is the largest acceptable value of 1? F : 3 p 13 15 p 229 0 F : 2 p p 2 1 0 13 229 2 3 2 15 F : 15 p 229 15 17 0 F : 2 p 229 8 17 1 2 0 1 1.502 1, 1.262 1, 1.417 1 is the critical cable 2 15 8 200 N 1.502 1 ) 1 133.2 N 15 2 = 1 /2 114 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.47 he hdraulic clinder is subjected to three forces. n 8-kN force is eerted on the clinder at that is parallel to the clinder and points from toward. he link eerts a force at that is parallel to the line from to. he link eerts a force at that is parallel to the line from to. (a) (b) raw the free-bod diagram of the clinder. (he clinder s weight is negligible). etermine the magnitudes of the forces eerted b the links and. Hdraulic clinder 0.6 m 0.15 m 1 m 0.6 m Scoop 1 m From the figure, if is at the origin, then points,, and are located at 0.15, 0.6 0.75, 0.6 1.00, 0.4 F and forces F, F, and F are parallel to,, and, respectivel. e need to write unit vectors in the three force directions and epress the forces in terms of magnitudes and unit vectors. he unit vectors are given b F F e r 0.243i 0.970j jr j e r 0.781i 0.625j jr j e r 0.928i 0.371j jr j Now we write the forces in terms of magnitudes and unit vectors. e can write F as F 8e kn or as F 8 e kn (because we were told it was directed from toward and had a magnitude of 8 kn. Either wa, we must end up with F 6.25i 5.00j kn Similarl, F 0.243F i 0.970F j F 0.928F i 0.371F j For equilibrium, F F F 0 In component form, this gives F 0.243F 0.928F 6.25 (kn) 0 F 0.970F 0.371F 5.00 (kn) 0 Solving, we get F 7.02 kn, F 4.89 kn c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 115

Problem 3.48 surfaces. (a) (b) he 50-N clinder rests on two smooth raw the free-bod diagram of the clinder. If 30, what are the magnitudes of the forces eerted on the clinder b the left and right surfaces? α 45 Isolate the clinder. (a) he free bod diagram of the isolated clinder is shown. (b) he forces acting are the weight and the normal forces eerted b the surfaces. he angle between the normal force on the right and the ais is 90 ˇ. he normal force is α β N R jn R j i cos 90 ˇ j sin 90 ˇ N L N R N R jn R j i sin ˇ j cos ˇ. he angle between the positive ais and the left hand force is normal 90 ; the normal force is N L jn L j i sin j cos. he weight is 0i jjj. he equilibrium conditions are F NR N L 0. Substitute and collect like terms, F jn R j sin ˇ jn L j sin i 0, Solve: jn R j ( ) sin jn L j, sin ˇ ( ) jj sin ˇ and jn L j. sin ˇ For jj 50 N, and 30, ˇ 45, the normal forces are jn L j36.6 N, jn R j25.9 N F jn R j cos ˇ jn L j cos jj j 0. Problem 3.49 For the 50-N clinder in Problem 3.48, obtain an equation for the force eerted on the clinder b the left surface in terms of the angle in two was: (a) using a coordinate sstem with the ais vertical, (b) using a coordinate sstem with the ais parallel to the right surface. he solution for Part (a) is given in Problem 3.48 (see free bod diagram). jn R j ( ) sin jn L j sin ˇ ( ) jj sin ˇ jn L j. sin ˇ Part (b): he isolated clinder with the coordinate sstem is shown. he angle between the right hand normal force and the positive ais is 180. he normal force: N R jn R ji 0j. he angle between the left hand normal force and the positive is 180 ˇ. he normal force is N L jn L j i cos ˇ j sin ˇ. he angle between the weight vector and the positive ais is ˇ. he weight vector is jj i cos ˇ j sin ˇ. he equilibrium conditions are β α N R N L Substitute and collect like terms, F jn R j jn L j cos ˇ jj cos ˇ i 0, F jn L j sin ˇ jj sin ˇ j 0. F NR N L 0. Solve: ( ) jj sin ˇ jn L j sin ˇ 116 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.50 he two springs are identical, with unstretched length 0.4 m. hen the 50-kg mass is suspended at, the length of each spring increases to 0.6 m. hat is the spring constant k? 0.6 m k k F k 0.6 m 0.4 m F F F :2F sin 60 490.5 N 0 60 60 k 1416 N/m 490.5 N Problem 3.51 he cable is 0.5 m in length. he unstretched length of the spring is 0.4 m. hen the 50-kg mass is suspended at, the length of the spring increases to 0.45 m. hat is the spring constant k? 0.7 m k he Geometr Law of osines and Law of Sines θ 0.7 m φ 0.7 2 0.5 2 0.45 2 2 0.5 0.45 cos ˇ sin 0.45 m sin 0.5 m sin ˇ 0.7 m 0.5 m β 0.45 m ˇ 94.8, 39.8 45.4 Now do the statics F F k 0.45 m 0.4 m F : cos F cos 0 θ φ F : sin F sin 490.5 N 0 Solving: k 7560 N/m 490.5 N c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 117

Problem 3.52 he small sphere of mass m is attached to a string of length L and rests on the smooth surface of a fied sphere of radius R. he center of the sphere is directl below the point where the string is attached. Obtain an equation for the tension in the string in terms of m, L, h, and R. h L m R From the geometr we have cos R h, sin L L cos R, sin R hus the equilibrium equations can be written F : L R N 0 F : R h L R N mg 0 Solving, we find mgl R h Problem 3.53 he inclined surface is smooth. etermine the force that must be eerted on the cable to hold the 100-kg crate in equilibrium and compare our answer to the answer of Problem 3.11. 60 F- :3 981 N sin 60 0 3 981 N 283 N N 60 118 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.54 In Eample 3.3, suppose that the mass of the suspended object is m and the masses of the pulles are m 0.3m, m 0.2m, and m 0.2m. Show that the force necessar for the sstem to be in equilibrium is 0.275m g. From the free-bod diagram of pulle 2 m g 0 ) 2 m g hen from the free-bod diagram of pulle 2 m g m g m g 0 mg hus 1 4 m m m g 1 4 m 0.3m 0.2m g 0.275m g (a) 2 mg 0.275m g mg m g (b) c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 119

Problem 3.55 he mass of each pulle of the sstem is m and the mass of the suspended object is m. etermine the force necessar for the sstem to be in equilibrium. raw free bod diagrams of each pulle and the object. Each pulle and the object must be in equilibrium. he weights of the pulles and object are mg and m g. he equilibrium equations for the weight, the lower pulle, second pulle, third pulle, and the top pulle are, respectivel, 0, 2 0, 2 0, 2 0, and F S 2 0. egin with the first equation and solve for, substitute for in the second equation and solve for, substitute for in the third equation and solve for, and substitute for in the fourth equation and solve for, to get in terms of and. he result is, 2 2, 4 3 4, and 8 7 8, F s or in terms of the masses, g 8 m 7m. 120 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.56 he suspended mass m 1 50 kg. Neglecting the masses of the pulles, determine the value of the mass m 2 necessar for the sstem to be in equilibrium. m 2 m 1 F : 1 2m 2 g m 1 g 0 1 F : 1 2m 2 g 0 m 2 m 1 4 12.5 kg 1 m 1 g = m 2 g c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 121

Problem 3.57 he bo is lifting himself using the block and tackle shown. If the weight of the block and tackle is negligible, and the combined weight of the bo and the beam he is sitting on is 600 N, what force does he have to eert on the rope to raise himself at a constant rate? (Neglect the deviation of the ropes from the vertical.) free-bod diagram can be obtained b cutting the four ropes between the two pulles of the block and tackle and the rope the bo is holding. he tension has the same value in all five of these ropes. So the upward force on the free-bod diagram is 5 and the downward force is the 600 N weight. herefore the force the bo must eert is 600 N /5 120 N 120 N 600 N 122 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.58 Pulle sstems containing one, two, and three pulles are shown. Neglecting the weights of the pulles, determine the force required to support the weight in each case. (a) One pulle (b) wo pulles (c) hree pulles (a) F :2 0 ) 2 (b) For two pulles (b) Fupper :2 1 0 Flower :2 1 0 1 1 4 (c) Fupper :2 1 0 Fmiddle :2 1 2 0 Flower :2 2 0 8 (c) For three pulles (a) For one pulles 1 1 2 2 2 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 123

Problem 3.59 Problem 3.58 shows pulle sstems containing one, two, and three pulles. he number of pulles in the tpe of sstem shown could obviousl be etended to an arbitrar number N. etrapolation of the previous problem (a) 2 N (a) (b) Neglecting the weights of the pulles, determine the force required to support the weight as a function of the number of pulles N in the sstem. Using the result of part (a), determine the force required to support the weight for a sstem with 10 pulles. (b) 1024 Problem 3.60 14,000-kg airplane is in stead flight in the vertical plane. he flight path angle is 10, the angle of attack is 4, and the thrust force eerted b the engine is 60 kn. hat are the magnitudes of the lift and drag forces acting on the airplane? (See Eample 3.4). Let us draw a more detailed free bod diagram to see the angles involved more clearl. hen we will write the equations of equilibrium and solve them. mg 14,000 9.81 N he equilibrium equations are F cos sin 0 α γ L α = 4 γ = 10 F sin L cos 0 60 kn 60000 N Solving, we get 36.0 kn, L 131.1 kn γ Path γ α L Horizon 124 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.61 n airplane is in stead flight, the angle of attack 0, the thrust-to-drag ratio / 2, and the lift-to-drag ratio L/ 4. hat is the flight path angle? (See Eample 3.4). Use the same strateg as in Problem 3.52. he angle between the thrust vector and the positive ais is, jj i cos j sin he lift vector: L 0i jljj he drag: jji 0j. he angle between the weight vector and the positive ais is 270 ; jj i sin j cos. he equilibrium conditions are F L 0. Substitute and collect like terms F jj cos jj jj sin i 0, and F jj sin jlj jj cos j 0 Solve the equations for the terms in : jj sin jj cos jj, and jj cos jj sin jlj. ake the ratio of the two equations tan ( ) jj cos jj. jj sin jlj ivide top and bottom on the right b jj. For 0, jj jj 2, jlj jj 4, ( ) 2 1 tan 1 4 4 or 14 Path L α γ Horizontal c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 125

Problem 3.62 n airplane glides in stead flight ( 0), and its lift-to-drag ratio is L/ 4. (a) hat is the flight path angle? (b) If the airplane glides from an altitude of 1000 m to zero altitude, what horizontal distance does it travel? (See Eample 3.4.) See Eample 3.4. he angle between the thrust vector and the positive ais is : Path jj i cos j sin. he lift vector: L 0i jljj. L α γ he drag: jji 0j. he angle between the weight vector and the positive ais is 270 : jj i sin j cos. Horizontal he equilibrium conditions are F L 0. Substitute and collect like terms: F jj cos jj jj sin i 0 1 km γ F jj sin jlj jj cos j 0 Solve the equations for the terms in, h γ jj sin jj cos jj, and jj cos jj sin jlj Part (a): ake the ratio of the two equilibrium equations: tan ( ) jj cos jj. jj sin jlj ivide top and bottom on the right b jj. For 0, jj 0, jlj jj 4, tan ( ) 1 14 4 Part (b): he flight path angle is a negative angle measured from the horizontal, hence from the equalit of opposite interior angles the angle is also the positive elevation angle of the airplane measured at the point of landing. tan 1 h, h 1 tan ( 1 ) 4km 1 4 126 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.63 In ctive Eample 3.5, suppose that the attachment point is moved to the point (5,0,0) m. hat are the tensions in cables,, and? he position vector from point to point can be used to write the force. r 5i 4j m ( 2, 0, 2) m ( 3, 0, 3) m z (4, 0, 2) m (0, 4, 0) m r jr j 0.781i 0.625j Using the other forces from ctive Eample 3.5, we have F :0.781 0.408 0.514 0 100 kg F :0.625 0.816 0.686 981 N 0 Fz : 0.408 0.514 0 Solving ields 509 N, 487 N, 386 N Problem 3.64 he force F 800i 200j (N) acts at point where the cables,, and are joined. hat are the tensions in the three cables? (0, 6, 0) m F (12, 4, 2) m (0, 4, 6) m (6, 0, 0) m z e first write the position vectors r 6i 4j 2k m r 12i 6k m r 12i 2j 2k m Now we can use these vectors to define the force vectors r jr j 0.802i 0.535j 0.267k r jr j 0.949i 0.316k r jr j 0.973i 0.162j 0.162k he equilibrium equations are then F : 0.802 0.949 0.973 800 N 0 F : 0.535 0.162 200 N 0 Fz : 0.267 0.316 0.162 0 Solving, we find 405 N, 395 N, 103 N c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 127

Problem 3.65 Suppose that ou want to appl a 1000-N force F at point in a direction such that the resulting tensions in cables,, and are equal. etermine the components of F. (0, 4, 6) m (0, 6, 0) m F (12, 4, 2) m (6, 0, 0) m z e first write the position vectors r 6i 4j 2k m r 12i 6k m r 12i 2j 2k m Now we can use these vectors to define the force vectors r 0.802i 0.535j 0.267k jr j r 0.949i 0.316k jr j r 0.973i 0.162j 0.162k jr j he force F can be written F F i F j F z k he equilibrium equations are then F : 0.802 0.949 0.973 F 0 ) F 2.72 F : 0.535 0.162 F 0 ) F 0.732 Fz : 0.267 0.316 0.162 F z 0 ) F z 0.113 e also have the constraint equation ) 363 N Solving, we find F 2 F 2 F z 2 1000 F 990 N, F 135 N, F z 41.2 N N 128 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.66 he 10-N metal disk is supported b the smooth inclined surface and the strings and. he disk is located at coordinates (5,1,4) m. hat are the tensions in the strings? (0, 6, 0) m (8, 4, 0) m he position vectors are r 5i 5j 4k m r 3i 3j 4k m he angle between the inclined surface the horizontal is z 10 m 8 m 2 m tan 1 2/8 14.0 e identif the following force: r jr j 0.615i 0.615j 0.492k r jr j 0.514i 0.514j 0.686k N N cos j sin k N 0.970j 0.243k 10 N j he equilibrium equations are then F : 0.615 0.514 0 F :0.615 0.514 0.970N 10 N 0 Fz : 0.492 0.686 0.243N 0 Solving, we find N 8.35 N 1.54 N, 1.85 N c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 129

Problem 3.67 he bulldozer eerts a force F 2i (kn) at. hat are the tensions in cables,, and? 6 m Isolate the cable juncture. Epress the tensions in terms of unit vectors. Solve the equilibrium equations. he coordinates of points,,, are: 8, 0, 0, 0, 3, 8, 0, 2, 6, 0, 4, 0. he radius vectors for these points are 3 m z 4 m 8 m 2 m 8 m r 8i 0j 0k, r 0i 3j 8k, r 0i 2j 6k, r 0i 4j 0k. definition, the unit vector parallel to the tension in cable is e r r jr r j. arring out the operations for each of the cables, the results are: e 0.6835i 0.2563j 0.6835k, e 0.7845i 0.1961j 0.5883k, e 0.8944i 0.4472j 0k. he tensions in the cables are epressed in terms of the unit vectors, j je, j je, j je. he eternal force acting on the juncture is F 2000i 0j 0k. he equilibrium conditions are F 0 F 0. Substitute the vectors into the equilibrium conditions: F 0.6835j j 0.7845j j 0.8944j j2000 i0 F 0.2563j j0.1961j j 0.4472j j j 0 Fz 0.6835j j 0.5883j j0j j k 0 he commercial program K Solver Plus was used to solve these equations. he results are j j780.31 N, j j906.49 N, j j844.74 N. 130 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.68 Prior to its launch, a balloon carring a set of eperiments to high altitude is held in place b groups of student volunteers holding the tethers at,, and. he mass of the balloon, eperiments package, and the gas it contains is 90 kg, and the buoanc force on the balloon is 1000 N. he supervising professor conservativel estimates that each student can eert at least a 40-N tension on the tether for the necessar length of time. ased on this estimate, what minimum numbers of students are needed at,, and? (0, 8, 0) m ( 16, 0, 4) m z (10,0, 12) m (16, 0, 16) m 1000 N F 1000 90 9.81 0 0, 8, 0 117.1 N (90) g 16, 0, 16 10, 0, 12 16, 0, 4 e need to write unit vectors e, e, and e. e 0.667i 0.333j 0.667k (0, 8, 0) e 0.570i 0.456j 0.684k F e 0.873i 0.436j 0.218k e now write the forces in terms of magnitudes and unit vectors F 0.667F i 0.333F j 0.667F k F 0.570F i 0.456F j 0.684F k F ( 16, 0, 4) (10, 0, 12) m F 0.873F i 0.436F j 0.218F k 117.1j (N) z (16, 0, 16) m he equations of equilibrium are F 0.667F 0.570F 0.873F 0 F 0.333F 0.456F 0.436F 117.1 0 Fz 0.667F 0.684F 0.218F 0 Solving, we get F 64.8 N¾ 2 students F 99.8 N¾ 3 students F 114.6 N¾ 3 students c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 131

Problem 3.69 he 20-kg mass is suspended b cables attached to three vertical 2-m posts. Point is at (0, 1.2, 0) m. etermine the tensions in cables,, and. 1 m 1 m 0.3 m z 2 m Points,,, and are located at 0, 1.2, 0, 0.3, 2, 1, 0, 2, 1, 2, 2, 0 F F F rite the unit vectors e, e, e e 0.228i 0.608j 0.760k e 0i 0.625j 0.781k e 0.928i 0.371j 0k z (20) (9.81) N he forces are F 0.228F i 0.608F j 0.760F k F 0F i 0.625F j 0.781F k F 0.928F i 0.371F j 0k 20 9.81 j he equations of equilibrium are F 0.228F 0 0.928F 0 F 0.608F 0.625F 0.371F 20 9.81 0 Fz 0.760F 0.781F 0 0 e have 3 eqns in 3 unknowns solving, we get F 150.0 N F 146.1 N F 36.9 N 132 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.70 he weight of the horizontal wall section is 20,000 N. etermine the tensions in the cables,, and. Set the coordinate origin at with aes as shown. he upward force,, at point will be equal to the weight,, since the cable at supports the entire wall. he upward force at is k. From the figure, the coordinates of the points in metre are 4, 6, 10, 0, 0, 0, 12, 0, 0, and 4, 14, 0. he three unit vectors are of the form e I I i I j z I z k I 2 I 2 z I z 2, where I takes on the values,, and. he denominators of the unit vectors are the distances,, and, respectivel. Substitution of the coordinates of the points ields the following unit vectors: e 0.324i 0.487j 0.811k, e 0.566i 0.424j 0.707k, and e 0i 0.625j 0.781k. he forces are e, e, and e. he equilibrium equation for the knot at point is 10 m 6 m 4 m 8 m z 7 m 10 m 14 m 6 m X 4 m 8 m 7 m 14 m 0. From the vector equilibrium equation, write the scalar equilibrium equations in the,, and z directions. e get three linear equations in three unknowns. Solving these equations simultaneousl, we get 9393 N, 5387 N, and 10,977 N 10 m 7 m 6 m 14 m 4 m 8 m c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 133

Problem 3.71 he car in Fig. a and the pallet supporting it weigh 13500 N. he are supported b four cables,,, and E. he locations of the attachment points on the pallet are shown in Fig. b. he tensions in cables and are equal. etermine the tensions in the cables. Isolate the knot at. Let,, and E be the forces eerted b the tensions in the cables. he force eerted b the vertical cable is (13500 N)j. efirst find the position vectors and then epress all of the forces as vectors. r 2i 6j 2k m (0, 6, 0) m r 2i 6j 2k m r 3i 6j 1.5 k m E r E 2.5 i 6j 2k m z ( ) r jr j 2 i 6 j 2 k p 44 ( ) r jr j 2 i 6 j 2 k p 44 ( ) r jr j 3 i 6 j 1.5 k p 47.25 ( ) r E E E jr E j 2.5 i 6 j 2 k E p 46.25 he equilibrium equations are 1.5 m (a) 3 m 2 m 2 m F : p 2 p 2 3 2.5 p p E 0 44 44 47.25 46.25 6 6 6 6 F : p p p p E 13500 N 0 44 44 47.25 46.25 Fz : p 2 p 2 1.5 2 p p E 0 44 44 47.25 46.25 E 2.5 m 2 m z (b) 2 m Solving, we find 4344 N, 4344 N, 3693 N, E 2741 N 134 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.72 he 680-kg load suspended from the helicopter is in equilibrium. he aerodnamic drag force on the load is horizontal. he ais is vertical, and cable O lies in the - plane. etermine the magnitude of the drag force and the tension in cable O. 10 O F O sin 10 0, O F O cos 10 680 9.81 0. 10 Solving, we obtain 1176 N, O 6774 N. (680) (9.81) N c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 135

Problem 3.73 In Problem 3.72, the coordinates of the three cable attachment points,, and are ( 3.3, 4.5, 0) m, (1.1, 5.3, 1) m, and (1.6, 5.4, 1) m, respectivel. hat are the tensions in cables O, O, and O? he position vectors from O to pts,, and are r O 3.3i 4.5j (m), r O 1.1i 5.3j k (m), r O 1.6i 5.4j k (m). ividing b the magnitudes, we obtain the unit vectors e O 0.591i 0.806j, e O 0.200i 0.963j 0.182k, e O 0.280i 0.944j 0.175k. Using these unit vectors, we obtain the equilibrium equations F O sin 10 0.591 O 0.200 O 0.280 O 0, F O cos 10 0.806 O 0.963 O 0.944 O 0, Fz 0.182 O 0.175 O 0. From the solution of Problem 3.72, O 6774 N. Solving these equations, we obtain O 3.60 kn, O 1.94 kn, O 2.02 kn. O 10 O O O 136 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.74 If the mass of the bar is negligible compared to the mass of the suspended object E, the bar eerts a force on the ball at that points from toward. he mass of the object E is 200 kg. he -ais points upward. etermine the tensions in the cables and. Strateg: raw a free-bod diagram of the ball at. (he weight of the ball is negligible.) (0, 4, 3) m (0, 5, 5) m (4, 3, 1) m E z ( ) ( ) 4i 3j k 4i j 4k F F p, p, 26 33 he forces ( 4i 2j 4k 6 ), 200 kg 9.81 m/s 2 j he equilibrium equations F : 4 p 26 F 4 p 33 4 6 0 F : 3 p 26 F 1 p 33 2 6 1962 N 0 Fz : 1 p 26 F 4 p 33 4 6 0 ) 1610 N 1009 N c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 137

Problem 3.75* he 14000 N car is at rest on the plane surface. he unit vector e n 0.456i 0.570j 0.684k is perpendicular to the surface. etermine the magnitudes of the total normal force N and the total friction force f eerted on the surface b the car s wheels. he forces on the car are its weight, the normal force, and the friction force. e n he normal force is in the direction of the unit vector, so it can be written N Ne n N 0.456i 0.570j 0.684k he equilibrium equation is z Ne n f 14000 N j 0 he friction force f is perpendicular to N, so we can eliminate the friction force from the equilibrium equation b taking the dot product of the equation with e n. Ne n f 14000 N j Ð e n N 14000 N j Ð e n 0 N 14000 N 0.57 7980 N Now we can solve for the friction force f. f 14000 N j Ne n 14000 N j 7980 N 0.456i 0.570j 0.684k f 3639 i 9451j 5458 k N jfj 3639 N 2 9451 N 2 5458 N 2 11504 N jnj 7980 N, jfj 11504 N 138 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.76 he sstem shown anchors a stanchion of a cable-suspended roof. If the tension in cable is 900 kn, what are the tensions in cables EF and EG? Using the coordinates for the points we find r [ 3.4 2 i 1 1 j 0 0 k] m r 1.4 m i e r jr j i Using the same procedure we find the other unit vectors that we need. e 0.140i 0.700j 0.700k e 0.140i 0.700j 0.700k G (0, 1.4, 1.2) m F E (2, 1, 0) m (3.4, 1, 0) m (1, 1.2, 0) m (0, 1.4, 1.2) m (2.2, 0, 1) m z (2.2, 0, 1) m e E 0.981i 0.196j e EG 0.635i 0.127j 0.762k e EF 0.635i 0.127j 0.762k e can now write the equilibrium equations for the connections at and E. 900 kn e e e E e E 0, E e E EF e EF EG e EG 0 reaking these equations into components, we have the following si equations to solve for five unknows (one of the equations is redundant). 900 kn 0.140 0.140 E 0.981 0 0.700 0.700 E 0.196 0 0.700 0.700 0 E 0.981 EG 0.635 EF 0.635 0 E 0.196 EG 0.127 EF 0.127 0 EG 0.726 EF 0.726 0 Solving, we find 134 kn, E 956 kn EF EG 738 kn Problem 3.77* he cables of the sstem will each safel support a tension of 1500 kn. ased on this criterion, what is the largest safe value of the tension in cable? G (0, 1.4, 1.2) m From Problem 3.76 we know that if the tension in is 900 kn, then the largest force in the sstem occurs in cable E and that tension is 956 kn. o solve this problem, we can just scale the results from Problem 3.76 ( ) 1500 kn 900 kn 1410 kn 956 kn F (0, 1.4, 1.2) m E (1, 1.2, 0) m (2, 1, 0) m (3.4, 1, 0) m (2.2, 0, 1) m z (2.2, 0, 1) m c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 139

Problem 3.78 he 200-kg slider at is held in place on the smooth vertical bar b the cable. (a) (b) etermine the tension in the cable. etermine the force eerted on the slider b the bar. 2 m 5 m 2 m 2 m he coordinates of the points, are 2, 2, 0, 0, 5, 2. he vector positions z r 2i 2j 0k, r 0i 5j 2k he equilibrium conditions are: N F N 0. Eliminate the slider bar normal force as follows: he bar is parallel to the ais, hence the unit vector parallel to the bar is e 0i 1j 0k. he dot product of the unit vector and the normal force vanishes: e Ð N 0. ake the dot product of e with the equilibrium conditions: e Ð N 0. e Ð F e Ð e Ð 0. he weight is e Ð 1j Ð jjj jj 200 9.81 1962 N. Note: For this specific configuration, the problem can be solved without eliminating the slider bar normal force, since it does not appear in the -component of the equilibrium equation (the slider bar is parallel to the -ais). However, in the general case, the slider bar will not be parallel to an ais, and the unknown normal force will be projected onto all components of the equilibrium equations (see Problem 3.79 below). In this general situation, it will be necessar to eliminate the slider bar normal force b some procedure equivalent to that used above. End Note. he unit vector parallel to the cable is b definition, e r r jr r j. Substitute the vectors and carr out the operation: e 0.4851i 0.7278j 0.4851k. (a) he tension in the cable is jje. Substitute into the modified equilibrium condition e F 0.7276jj 1962 0. Solve: jj 2696.5 N from which the tension vector is jje 1308i 1962j 1308k. (b) he equilibrium conditions are F 0 N 1308i 1308k N 0. Solve for the normal force: N 1308i 1308k. he magnitude is jnj 1850 N. 140 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.79 In Eample 3.6, suppose that the cable is replaced b a longer one so that the distance from point to the slider increases from 6 m to 8 m. etermine the tension in the cable. 4 m r 8 m e he vector from to is now 6 m ( 4 r 8 m 9 i 7 9 j 4 ) 9 k r 3.56i 6.22j 3.56k m e can now find the unit vector form to. r r O r O r [ 7j 4k z 7 m 4 m O 4 m f 7j 3.56i 6.22j 3.56k g] m r 3.56i 6.22j 0.444k m e r 0.495i 0.867j 0.0619k jr j Using N to stand for the normal force between the bar and the slider, we can write the equilibrium equation: e N 100 N j 0 e can use the dot product to eliminate N from the equation [e N 100 N j] Ð e e Ð e 100 N j Ð e 0 ([ ] [ 4 [ 0.495] 7 ] [ ] ) 4 [0.867] [0.0619] 100 N 0.778 0 9 9 9 0.867 77.8 N 0 ) 89.8 N 89.8 N c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 141

Problem 3.80 he cable keeps the 8-kg collar in place on the smooth bar. he ais points upward. hat is the tension in the cable? 0.15 m vectors e develop the following position vectors and unit 0.4 m r 0.2i 0.3j 0.25k m e r 0.456i 0.684j 0.570k jr j r 0.2 m e 0.091i 0.137j 0.114k m r r O r O r z 0.5 m 0.2 m O 0.25 m 0.2 m 0.3 m r [ 0.5j 0.15k f0.4i 0.3jgf 0.091i 0.137j 0.114kg ] m r 0.309i 0.337j 0.036k m e r 0.674i 0.735j 0.079k jr j e can now write the equilibrium equation for the slider using N to stand for the normal force between the slider and the bar. e N 8 kg 9.81 m/s 2 j 0 o eliminate the normal force N we take a dot product with e. [e N 8 kg 9.81 m/s 2 j] Ð e 0 e Ð e 78.5 N j Ð e 0 [ 0.674][ 0.456] [0.735][ 0.684] [0.079][0.570] 78.5 N 0.684 0 0.150 53.6 N 0 357 N Problem 3.81 etermine the magnitude of the normal force eerted on the collar b the smooth bar. 0.15 m From Problem 3.81 we have e 0.674i 0.735j 0.079k 0.4 m 357 N he equilibrium equation is e N 78.5 N j 0 e can now solve for the normal force N. N 78.5 N j 357 N 0.674i 0.735j 0.079k z 0.5 m 0.2 m O 0.25 m 0.2 m 0.3 m N 240i 184j 28.1k N he magnitude of N is jnj 240 N 2 184 N 2 28.1 N 2 jnj 304 N 142 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.82* he 10-kg collar and 20-kg collar are held in place on the smooth bars b the 3-m cable from to and the force F acting on. he force F is parallel to the bar. etermine F. (0, 5, 0) m he geometr is the first part of the Problem. o ease our work, let us name the points,, E, and G as shown in the figure. he unit vectors from to and from E to G are essential to the location of points and. he diagram shown contains two free bodies plus the pertinent geometr. he unit vectors from to and from E to G are given b e er i e j e z k, z (0, 3, 0) m (0, 0, 4) m F 3 m (4, 0, 0) m and e EG er EG i e EG j e EGz k. Using the coordinates of points,, E, and G from the picture, the unit vectors are e 0.625i 0.781j 0k, and e EG 0i 0.6j 0.8k. he location of point is given b N (0, 5, 0) m G (0, 3, 0) m F N 3 m e, e, and z z e z, where 3 m. From these equations, we find that the location of point is given b (2.13, 2.34, 0) m. Once we know the location of point, we can proceed to find the location of point. e have two was to determine the location of. First, is 3 m from point along the line (which we do not know). lso, lies on the line EG. he equations for the location of point based on line are: e, e, and z z e z. he equations based on line EG are: E Ee EG, E Ee EG, and z z E Ee EGz. e have si new equations in the three coordinates of and the distance E. Some of the information in the equations is redundant. However, we can solve for E (and the coordinates of ). e get that the length E is 2.56 m and that point is located at (0, 1.53, 1.96) m. e net write equilibrium equations for bodies and. From the free bod diagram for, weget N e Fe 0, z m g E (0, 0, 4) m m g (4, 0, 0) m e now have two fewer equation than unknowns. Fortunatel, there are two conditions we have not et invoked. he bars at and are smooth. his means that the normal force on each bar can have no component along that bar. his can be epressed b using the dot product of the normal force and the unit vector along the bar. he two conditions are N Ð e N e N e N z e z 0 for slider and N Ð e EG N e EG N e EG N z e EGz 0. Solving the eight equations in the eight unknowns, we obtain F 36.6 N. Other values obtained in the solution are E 2.56 m, N 145 N, N 116 N, N z 112 N, N 122 N, N 150 N, and N z 112 N. N e Fe m g 0, and N z e z Fe z 0. From the free bod diagram for, we get N e 0, N b e m g 0, and N z e z 0. c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 143

Problem 3.83 he 100-N crate is held in place on the smooth surface b the rope. etermine the tension in the rope and the magnitude of the normal force eerted on the crate b the surface. 45 30 he free-bod diagram is sketched. he equilibrium equations are F - : cos 45 30 100 N sin 30 0 N F% : sin 45 30 100 N cos 30 N 0 Solving, we find 51.8 N,N 73.2 N Problem 3.84 he sstem shown is called Russell s traction. If the sum of the downward forces eerted at and b the patient s leg is 32.2 N, what is the weight? 20 60 25 he force in the cable is everwhere. he free-bod diagram of the leg is shown. he downward force is given, but the horizontal force F H is unknown. he equilibrium equation in the vertical direction is F : sin 25 sin 60 32.2 N 0 hus 32.2 N sin 25 sin 60 25.0 N N 144 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.85 he 1600-N engine block is suspended b the cables and. If ou don t want either or to eceed 1600 N, what is the smallest acceptable value of the angle? a a 1600 N he equilibrium equations are F : cos cos 0 F : sin sin 1600 N 0 Solving, we find 1600 N 2 sin If we limit the tensions to 1600 N, we have 1600 N 1600 N ) sin 1 2 sin 2 ) 30 Problem 3.86 he cable is horizontal, and the bo on the right weighs 100 N. he surface are smooth. (a) hat is the tension in the cable? (b) hat is the weight of the bo on the left? 20 40 e have the following equilibrium equations F : N cos 40 100 N 0 N F : N sin 40 0 F : N sin 20 0 F : N cos 20 0 Solving these equations sequentiall, we find N 131 N, 83.9 N N 245 N, 230.5 N hus we have 83.9 N, 230.5 N c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 145

Problem 3.87 ssume that the forces eerted on the 700 N climber b the slanted walls of the chimne are perpendicular to the walls. If he is in equilibrium and is eerting a 650-N force on the rope, what are the magnitudes of the forces eerted on him b the left and right walls? 10 he forces in the free-bod diagram are in the directions shown on the figure. he equilibrium equations are: F : sin 10 N L cos 4 N R cos 3 0 4 3 F : cos 10 700 N N L sin 40 N R sin 3 0 where 650 N. Solving we find N L 94.7 N, N R 18.4 N Left all: 94.7 N Right all: 18.4 N 700 N 146 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.88 he mass of the suspended object is m and the masses of the pulles are negligible. etermine the force necessar for the sstem to be in equilibrium. reak the sstem into four free bod diagrams as shown. arefull label the forces to ensure that the tension in an single cord is uniform. he equations of equilibrium for the four objects, starting with the leftmost pulle and moving clockwise, are: F S 3 0, R 3S 0, F 3R 0, R and 2 2S 2R m g 0. e want to eliminate S, R, and F from our result and find in terms of m and g. From the first two equations, we get S 3, and R 3S 9. Substituting these into the last equilibrium equation results in 2 2 3 2 9 m g. Solving, we get m g/26. S S S S R S S R R R R m g Note: e did not have to solve for F to find the appropriate value of. he final equation would give us the value of F in terms of m and g. e would get F 27m g/26. If we then drew a free bod diagram of the entire assembl, the equation of equilibrium would be F m g 0. Substituting in the known values for and F, we see that this equation is also satisfied. hecking the equilibrium solution b using the etra free bod diagram is often a good procedure. c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 147

Problem 3.89 he assembl, including the pulle, weighs 60 N. hat force F is necessar for the sstem to be in equilibrium? F From the free bod diagram of the assembl, we have 3F 60 0, or F 20 N F F F F F F F 60 N. 148 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.90 he mass of block is 42 kg, and the mass of block is 50 kg. he surfaces are smooth. If the blocks are in equilibrium, what is the force F? F 45 20 Isolate the top block. Solve the equilibrium equations. he weight is. he angle between the normal force N 1 and the positive ais is. he normal force is. he force N 2 is. he equilibrium conditions are F N1 N 2 0 N 2 from which F 0.7071jN 1 j jn 2 j i 0 N 1 α F 0.7071jN 1 j 490.5 j 0. Solve: N 1 693.7 N, jn 2 j490.5 N N 1 Isolate the bottom block. he weight is 0i jjj 0i 42 9.81 j 0i 412.02j (N). F β α he angle between the normal force N 1 and the positive ais is 270 45 225. N 3 he normal force: N 1 jn 1 j i cos 225 j sin 225 jn 1 j 0.7071i 0.7071j. he angle between the normal force N 3 and the positive -ais is 90 20 70. he normal force is N 1 jn 3 j i cos 70 j sin 70 jn 3 j 0.3420i 0.9397j. he force is...f jfji 0j. he equilibrium conditions are F N1 N 3 F 0, from which: F 0.7071jN 1 j0.3420jn 3 jjfj i 0 F 0.7071jN 1 j0.9397jn 3 j 412 j 0 For jn 1 j693.7 N from above: jfj 162 N c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 149

Problem 3.91 he climber is being helped up an ic slope b two friends. His mass is 80 kg, and the direction cosines of the force eerted on him b the slope are cos 0.286, cos 0.429, cos z 0.857. he ais is vertical. If the climber is in equilibrium in the position shown, what are the tensions in the ropes and and the magnitude of the force eerted on him b the slope? (2, 2, 0) m (5, 2, 1) m z (3, 0, 4) m Get the unit vectors parallel to the ropes using the coordinates of the end points. Epress the tensions in terms of these unit vectors, and solve the equilibrium conditions. he rope tensions, the normal force, and the weight act on the climber. he coordinates of points,, are given b the problem, 3, 0, 4, 2, 2, 0, 5, 2, 1. he vector locations of the points,, are: N r 3i 0j 4k, r 2i 2j 0k, r 5i 2j 1k. he unit vector parallel to the tension acting between the points, in the direction of is e r r jr r j he unit vectors are e 0.2182i 0.4364j 0.8729k, e 0.3482i 0.3482j 0.8704k, and e N 0.286i 0.429j 0.857k. Substitute and collect like terms, F 0.2182j j0.3482j j 0.286jNj i 0 F 0.4364j j0.3482j j0.429jnj 784.8 j 0 Fz 0.8729j j0.8704j j 0.857jNj k 0 e have three linear equations in three unknowns. he solution is: j j100.7 N, j j889.0 N, jnj 1005.5 N. where the last was given b the problem statement. he forces are epressed in terms of the unit vectors, j je, j je, N jnje N. he weight is 0i jjj 0k 0i 80 9.81 j 0k 0i 784.8j 0k. he equilibrium conditions are F 0 N 0. 150 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.92 onsider the climber being helped b his friends in Problem 3.91. o tr to make the tensions in the ropes more equal, the friend at moves to the position (4, 2, 0) m. hat are the new tensions in the ropes and and the magnitude of the force eerted on the climber b the slope? Get the unit vectors parallel to the ropes using the coordinates of the end points. Epress the tensions in terms of these unit vectors, and solve the equilibrium conditions. he coordinates of points,, are 3, 0, 4, 4, 2, 0, 5, 2, 1. he vector locations of the points,, are: r 3i 0j 4k, r 4i 2j 0k, r 5i 2j 1k. he unit vectors are e 0.2182i 0.4364j 0.8729k, e 0.3482i 0.3482j 0.8704k, e N 0.286i 0.429j 0.857k. where the last was given b the problem statement. he forces are epressed in terms of the unit vectors, j je, j je, N jnje N. he weight is 0i jjj 0k 0i 80 9.81 j 0k 0i 784.8j 0k. he equilibrium conditions are F 0 N 0. Substitute and collect like terms, F 0.281j j0.3482j j 0.286jNj i 0 F 0.4364j j0.3482j j0.429jnj 784.8 j 0 Fz 0.8729j j0.8704j j 0.857jNj k 0 he HP-28S hand held calculator was used to solve these simultaneous equations. he solution is: j j420.5 N, j j532.7 N, jnj 969.3 N. c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 151

Problem 3.93 climber helps his friend up an ic slope. His friend is hauling a bo of supplies. If the mass of the friend is 90 kg and the mass of the supplies is 22 kg, what are the tensions in the ropes and? ssume that the slope is smooth. hat is, onl normal forces are eerted on the man and the bo b the slope. 20 Isolate the bo. he weight vector is 40 75 2 22 9.81 j 215.8j (N). he angle between the normal force and the positive ais is 90 60 30. 60 he normal force is N jn j 0.866i 0.5j. he angle between the rope and the positive ais is 180 75 105 ; the tension is: β 2 j 2 j i cos 105 j sin 105 j 2 j 0.2588i 0.9659j he equilibrium conditions are F 0.866jN j0.2588j 2 j i 0, F 0.5jN j0.9659j 2 j 215.8 j 0. N α Solve: N 57.8 N, j 2 j193.5 N. Isolate the friend. he weight is 1 20 90 9.81 j 882.9j (N). he angle between the normal force and the positive ais is 90 40 50. he normal force is: N 40 75 2 N F jn F j 0.6428i 0.7660j. he angle between the lower rope and the ais is 75 ; the tension is 2 j 2 j 0.2588i 0.9659j. he angle between the tension in the upper rope and the positive ais is 180 20 160, the tension is 1 j 1 j 0.9397i 0.3420j. he equilibrium conditions are F 1 2 N F 0. From which: F 0.6428jN F j0.2588j 2 j 0.9397j 1 j i 0 F 0.7660jN F j 0.9659j 2 j0.3420j 1 j 882.9 j 0 Solve, for j 2 j193.5 N. he result: jn F j1051.6 N, j 1 j772.6 N. 152 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.94 he 1440 kg car is moving at constant speed on a road with the slope shown. he aerodnamic forces on the car the drag 530 N, which is parallel to the road, and the lift L 360 N, which is perpendicular to the road. etermine the magnitudes of the total normal and friction forces eerted on the car b the road. 15 L he free-bod diagram is shown. If we write the equilibrium equations parallel and perpendicular to the road, we have: 360 N F- : N 1440 9.81 N cos 15 360 N 0 F% : f 530 N 1440 9.81 N sin 15 0 530 N Solving, we find N 13.29 kn, f 4.19 kn 1440 g c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 153

Problem 3.95 n engineer doing preliminar design studies for a new radio telescope envisions a triangular receiving platform suspended b cables from three equall spaced 40-m towers. he receiving platform has a mass of 20 Mg (megagrams) and is 10 m below the tops of the towers. hat tension would the cables be subjected to? OP VIE Isolate the platform. hoose a coordinate sstem with the origin at the center of the platform, with the z ais vertical, and the, aes as shown. Epress the tensions in terms of unit vectors, and solve the equilibrium conditions. he cable connections at the platform are labeled a, b, c, and the cable connections at the towers are labeled,,. he horizontal distance from the origin (center of the platform) to an tower is given b 20 m 65 m L 65 37.5 m. 2 sin 60 he coordinates of points,, are 37.5, 0, 10, 37.5 cos 120,37.5 sin 120.10, a c z b 37.5 cos 240, 37.5 sin 240, 10, he vector locations are: r 37.5i 0j 10k, r 18.764i 32.5j 10k, r 18.764i 32.5j 10k. he distance from the origin to an cable connection on the platform is d 20 11.547 m. 2 sin 60 he coordinates of the cable connections are he tensions in the cables are epressed in terms of the unit vectors, a j a je a, b j b je b, c j c je c. he weight is 0i 0j 20000 9.81 k 0i 0j 196200k. he equilibrium conditions are F 0 a b c 0, from which: a 11.547, 0, 0, b 11.547 cos 120,11547 sin 120,0, F 0.9333j a j 0.4666j b j 0.4666j c j i 0 c 11.547 cos 240, 11.547 sin 240, 0. he vector locations of these points are, r a 11.547i 0j 0k, r b 5.774i 10j 0k, r c 5.774i 10j 0k. he unit vector parallel to the tension acting between the points, a in the direction of is b definition e a r r a jr r a. Perform this operation for each of the unit vectors to obtain e a 0.9333i 0j 0.3592k F 0j a j0.8082j b j 0.8082j c j j 0 Fz 0.3592j a j 0.3592j b j 0.3592j c 196200j k 0 he commercial package K Solver Plus was used to solve these equations. he results: j a j182.1 kn, j b j182.1 kn, j c j182.1 kn. heck: For this geometr, where from smmetr all cable tensions ma be assumed to be the same, onl the z-component of the equilibrium equations is required: Fz 3jj sin 196200 0, e b 0.4667i 0.8082j 0.3592k e c 0.4667i 0.8082j 0.3592k ( ) where tan 1 10 21.07, 37.5 11.547 from which each tension is jj 182.1 kn. check. 154 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.96 o support the tent, the tension in the rope must be 400 N. hat are the tensions in the ropes,, and E? e develop the following position vectors r 2i m r 6i j 3k m r 6i 2j 3k m r E 3i 4j m If we divide b the respective magnitudes we can develop the unit vectors that are parallel to these position vectors. e 1.00i (0, 6, 6) m (0, 5, 0) m (6, 4, 3) m (8, 4, 3) m E (3, 0, 3) m e 0.885i 0.147j 0.442k z e 0.857i 0.286j 0.429k e E 6.00i 0.800j he equilibrium equation is e e e E e E 0. If we break this up into components, we have F : 0.885 0.857 0.600 E 0 F :0.147 0.286 0.800 E 0 Fz : 0.442 0.429 0 If we set 400 N, we cans solve for the other tensions. he result is 206 N, 214 N, E 117 N c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 155

Problem 3.97 able is attached to the top of the vertical 3-m post, and its tension is 50 kn. hat are the tensions in cables O,, and? 5 m 5 m Get the unit vectors parallel to the cables using the coordinates of the end points. Epress the tensions in terms of these unit vectors, and solve the equilibrium conditions. he coordinates of points,,,, O are found from the problem sketch: he coordinates of the points are 6, 2, 0, 12, 3, 0, 0, 8, 5, 0, 4, 5, O 0, 0, 0. he vector locations of these points are: 8 m O 4 m (6, 2, 0) m r 6i 2j 0k, r 12i 3j 0k, r 0i 8j 5k, r 0i 4j 5k, r O 0i 0j 0k. z 12 m 3 m he unit vector parallel to the tension acting between the points, in the direction of is b definition e r r jr r j. 5 m 5 m Perform this for each of the unit vectors e 0.9864i 0.1644j 0k 4 m e 0.6092i 0.6092j 0.5077k e 0.7442i 0.2481j 0.6202k 8 m O (6, 2, 0) m e O 0.9487i 0.3162j 0k he tensions in the cables are epressed in terms of the unit vectors, j je 50e, j je, j je, O j O je O. he equilibrium conditions are F 0 O 0. Substitute and collect like terms, F 0.9864 50 0.6092j j 0.7422j j 0.9487j O j i 0 F 0.1644 50 0.6092j j0.2481j j 0.3162j O j j 0 Fz 0.5077j j 0.6202j j k 0. his set of simultaneous equations in the unknown forces ma be solved using an of several standard algorithms. he results are: j O j43.3 kn, j j6.8 kn, j j5.5 kn. 156 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior

Problem 3.98* he 1350-kg car is at rest on a plane surface with its brakes locked. he unit vector e n 0.231i 0.923j 0.308k is perpendicular to the surface. he ais points upward. he direction cosines of the cable from to are cos 0.816, cos 0.408, cos z 0.408, and the tension in the cable is 1.2 kn. etermine the magnitudes of the normal and friction forces the car s wheels eert on the surface. e p e n z ssume that all forces act at the center of mass of the car. he vector equation of equilibrium for the car is "car" e n F S 0. riting these forces in terms of components, we have F F N FS mgj 1350 9.81 13240j N, F S F S i F S j F Sz k, z and e, where e cos i cos j cos z k 0.816i 0.408j 0.408k. Substituting these values into the equations of equilibrium and solving for the unknown components of F S, we get three scalar equations of equilibrium. hese are: F S 0, F S 0, and F Sz z 0. Substituting in the numbers and solving, we get F S 979.2 N, F S 12, 754 N, and F Sz 489.6 N. he net step is to find the component of F S normal to the surface. his component is given b F N F N Ð e n F S e n F S e n F Sz e nz. Substitution ields F N 12149 N. From its components, the magnitude of F S is F S 12800 N. Using the Pthagorean theorem, the friction force is f F 2 S F2 N 4033 N. c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior 157

Problem 3.99* he brakes of the car in Problem 3.98 are released, and the car is held in place on the plane surface b the cable. he car s front wheels are aligned so that the tires eert no friction forces parallel to the car s longitudinal ais. he unit vector e p 0.941i 0.131j 0.314k is parallel to the plane surface and aligned with the car s longitudinal ais. hat is the tension in the cable? Onl the cable and the car s weight eert forces in the direction parallel to e p. herefore e p Ð mgj 0: 0.941i 0.131j 0.314k Ð [ 0.816i 0.408j 0.408k mgj] 0, 0.941 0.816 0.131 0.408 mg 0.314 0.408 0. Solving, we obtain 2.50 kn. 158 c 2008 Pearson Education South sia Pte Ltd. ll rights reserved. his publication is protected b opright and permission should be obtained from the publisher prior