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

Transcription

1 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 N, mg N Problem 3.2 he ring weighs 5 N and is in equilibrium. he force F 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 N cos 30 F N cos 30 cos 4.77 N ) 35.2 F 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

2 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? he free-bod diagram is shown below the picture. he equilibrium equations are: F : cos 35 cos 60 0 F : sin 35 sin 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: 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

3 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? he free-bod diagram is shown. he equilibrium equations are F : cos 40 cos 55 0 F : sin 40 sin 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 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

4 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 N he mass of is m j Lj jgj 24 N 2.45 kg 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 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

5 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 N/m m 60 N F 1 F N/m 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

6 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 kn 1 kn sin 45 0 F : F crane 1 kn cos kn 1 kn N crane kn,f crane kn (b) F : N caisson kn 1 kn sin 45 0 Fcrane F : 1 kn cos 45 F caisson 0 N crane N caisson 3.22 kn, F caisson kn 1 kn 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 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

7 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) kn F% : N kn cos 0 F- : F kn sin 0 N kn, F 4.03 kn (b) F 0.6 N F% : N kn cos 0 ) 31.0 α F F- : F 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 N Solving for S and using the given value for k we find S 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

8 Problem 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 N. For a tension of 400 N, (600 sin Solve for the unknown angle sin or 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

9 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 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: 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? F of the car being towed F- : cos kn sin 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

10 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 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 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 N 128 N (b) Force 98.1 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 93

11 Problem 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 (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 N 0 (b) 107 N Eamine one of the nails F : R cos N R F : R sin 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? F R cos 15 L cos 14 0 F R sin 15 L sin Solving, we get L 1490 N, R 1500 N 14 L R 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

12 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 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 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

13 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 ft/s ( ) 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

14 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) (b) he friction force F fr is given b F fr 0.35N N (a) For equilibrium we have F : cos N 0 F : sin 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

15 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 F : p p N Solving: 760 N, 838 N 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 : F : kn 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

16 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 F : p p p kn 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

17 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.) 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

18 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 ( ) ˇ tan Now from equilibrium we have F : cos cos ˇ 0 F : sin sin ˇ 981 N 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

19 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 ( ) , ( ) , ˇ 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 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

20 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 ( ) tan ( ) ˇ tan Now from equilibrium we have F : cos cos ˇ 0 F : sin sin ˇ 5800 N N 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? 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 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

21 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 p F : 8 p p 8 p N Solving: N, 68.2 N 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 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

22 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 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 and L s he stretch in the spring when in equilibrium is given b υ L s L u. arring out the calculations, we get L u m, L s m, and υ 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 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 N and 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

23 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 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

24 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 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 ( ) j j N ( ) j j N 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

25 Problem 3.38 onsider the 1000-N bank of lights in Problem 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 m. cos m 9 m β α he length is given b 9 m m. cos 30 β 19 α he altitude of the triangle for which is the hpotenuse is 9 tan m. he distance is given b m. he distance is given b β = α he new angles are given b the cosine law cos. β α Reduce and solve: ( ) cos , ( ) cos ˇ , ˇ 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 ( ) j j N, ( ) j j N 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 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

26 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 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 N and 70 α ( ) j j 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 j N, and 50, ˇ 70, ( ) j j N 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 ( ) Substitute: j j 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

27 Problem 3.40 truck dealer wants to suspend a 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 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 ( )( ) m 2 42 and S m. he interior angles are found from the cosine law: ( L b ) cos L b ( L b ) cos ˇ ˇ 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 N, and 13.97, ˇ 22.52, j j kn, j j 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 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

28 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 ( ) 4 tan ˇ ˇ 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

29 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 m 2 h m m tan h 0.75 m, tan ˇ h 0.25 m α 0.75 m 0.25 m β h ) h m, 31.39, ˇ 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

30 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 N e have 4 unknowns and 4 equations F : cos 0 F : sin N 0 F : cos F : sin N 0 Solving we find 79.7, N, 21.4 N, 62.6 N 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 p F : 2 p p 2 50 N F : 15 p F : 2 p N ) 2 25 N, 75.1 N 63.1 N, 70.8 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 113

31 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 p F : 2 p p F : 15 p F : 2 p , , is the critical cable N ) 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

32 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, , , 0.4 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

33 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

34 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 N k 1416 N/m 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 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, Now do the statics F F k 0.45 m 0.4 m F : cos F cos 0 θ φ F : sin F sin N 0 Solving: k 7560 N/m 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

35 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 F- :3 981 N sin N 283 N 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

36 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

37 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

38 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 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

39 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

40 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 : (c) Fupper :2 1 0 Fmiddle : Flower : (c) For three pulles (a) For one pulles 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

41 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 ,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, N he equilibrium equations are F cos sin 0 α γ L α = 4 γ = 10 F sin L cos 0 60 kn N Solving, we get 36.0 kn, L 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

42 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 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 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

43 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 ( ) 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 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

44 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 : kg F : N 0 Fz : 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 : N 0 F : N 0 Fz : 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

45 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 : F 0 ) F 2.72 F : F 0 ) F Fz : F z 0 ) F z e also have the constraint equation ) 363 N Solving, we find F 2 F 2 F z 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

46 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/ 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 : F : N 10 N 0 Fz : N 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

47 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 i j k, e i j k, e i j 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 j j j j j j2000 i0 F j j0.1961j j j j j 0 Fz j j j j0j j k 0 he commercial program K Solver Plus was used to solve these equations. he results are j j N, j j N, j j 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

48 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 , 8, 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 Fz 0.667F 0.684F 0.218F 0 Solving, we get F 64.8 N¾ 2 students F 99.8 N¾ 3 students F 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

49 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 j he equations of equilibrium are F 0.228F F 0 F 0.608F 0.625F 0.371F Fz 0.760F 0.781F 0 0 e have 3 eqns in 3 unknowns solving, we get F N F 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