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SOLVED PROBLEMS CHAPTER- 2 : EQUILIBRIUM OF FORCE SYSTEMS Problem-2.1 Determine the magnitude of F1 and F2 so that particle P is in equilibrium. Figure for Problem-2.1 Solution: Soln. Fig. 2.1 Referring to Soln. Fig. 2.1 F x 0 yields F1 cos 45o F2 cos 30 o 500 0 0.707 F1 0.866 F2 500 0 F y ... ... ... ... ... (a) ... ... ... ... (b) 0 yields F1 sin 45o F2 sin 30 o 0 0.707 F1 0.5 F2 0 ... Subtract Eq. (b) from Eq. (a) to have 0.866F2 0.5F2 500 0 __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 1 F2 366.03 N Putting this value in Eq. (b) we get 0.707F1 0.5366.03 0 F1 258.86 N Problem-2.2 A weight W = 5000 N is suspended from a pin B which unites two inclined timbers AB and CB as shown in the figure. Determine the compressive force on each timber. Figure for Problem-2.2 Solution: Soln. Fig. 2.2 Referring to Soln. Fig. 2.2 F x 0 yields FAB cos 60 o FBC cos 60 o 0 0.5 FAB 0.5 FBC 0 FAB FBC F y 0 yields FAB cos 30 o FBC cos 30 o 5000 0 0.866 FAB 0.866 FBC 5000 0 0.866 FAB 0.866 FAB 5000 0 1.732 FAB 5000 0 FAB 2886.84 N __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 2 FBC 2886.84 N Problem-2.3 The permissible internal force on timber AB shown in the figure is 20 kN. If 30 o , what maximum safe load may be suspended at B? Figure for Problem-2.3 Solution: Soln. Fig. 2.3 Referring to Soln. Fig. 2.3 F x 0 yields 20 cos 45 o FBC cos 30 o 0 14.14 0.866 FBC 0 FBC 16.328 kN F y 0 yields 20 cos 45 o FBC cos 60 o W 0 200.707 16.3280.5 W 0 W 22.300 kN Problem-2.4 The derrick shown diagrammatically in the figure supports a load of W = 2 kips. Find the tension in the boom cable and the compression in the boom when the angle is 30o. __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 3 Figure for Problem-2.4 Solution: Soln Fig. 2.4 Referring to Soln. Fig. 2.4, apply cosine rule to have AB 2 40 2 70 2 2 4070cos 30 o AB 40.62 ft. Again applying sine rule, sin 30 o sin sin 40.62 40 70 o 120.50 and 29.50 o Now, Fx 0 gives FBC sin 30 o FAB cos90 o 29.50 o 30 o 0 __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 4 0.5FBC 0.862FAB 0 ... ... ... ... ... (a) ... (b) and Fy 0 gives FBC cos 30 o FAB cos29.50 o 30 o 2 0 0.866FBC 0.507 FAB 2 0 ... ... ... ... Multiply Eq. (a) by 0.866 and (b) by 0.5 and subtract to get 0.493FAB 1 0 FAB 2.03 kip Putting this value in Eq. (a) we get 0.5FBC 0.8622.03 0 FBC 3.50 kip Problem-2.5 Determine the magnitude and coordinate direction angles of F in the following figure, that are required for equilibrium of point O. Figure for Problem-2.5 Solution: Referring to figure for problem-2.5 above, F1 y 400 N, F1x 0 , F2 x 0 , F2 y 0 , F1z 0 F2 z 800 N __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 5 F3 x 700 2 200 N, 7 F3 y 700 3 300 N, 7 F3 z 700 6 600 N 7 Now, Fx 0 gives 0 0 200 Fx 0 Fx 200 N F F y 0 gives 400 0 300 Fy 0 Fy 100 N z 0 gives 0 800 600 Fz 0 Fz 200 N Hence, the magnitude of F is F F F F 2 x 2 y 2 z 2002 1002 2002 90000 300 N The coordinate direction angles are F 200 cos 1 x 48.19 o F 300 F 100 y cos 1 109.47 o F 300 F 200 cos 1 z 48.19 o F 300 Problem-2.6 Determine the tension in cords AB and AD for equilibrium of the 10-kg block shown in the figure. Figure for Problem- 2.6 Solution: __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 6 Soln Fig. 2.6 Referring to Soln. Fig. 2.6, Fx 0 gives TAB cos 30o TAD 0 0.866TAB TAD 0 ... ... ... ... ... ... (a) and, Fy 0 gives TAB sin 30o 98.1 0 TAB 196.2 N Putting this value in Eq. (a) we get 0.866196.2 TAD 0 TAD 169.91 N Problem-2.7 The ends of the 7-m-long cable AB are attached to the fixed walls. If a bucket and its contents have a mass of 10 kg and are suspended from the cable by means of a small pulley as shown, determine the location x of the pulley for equilibrium. __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 7 Figure for Problem-2.7 Solution: Soln Fig. 2.7 The cable is a continuous one, which passes over a frictionless pulley, so the tension-force developed, T has a constant magnitude. Now referring to Soln. Fig. 2.7, Fx 0 yields T cos1 T cos 2 0 1 2 (say). Now, cos x AC AC x cos ... ... ... ... ... ... ... (a) __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 8 Again, 4 x BC 4 x BC cos cos ... ... ... ... ... ... ... (b) Adding Eqs. (a) and (b) we get x 4 x AC BC cos cos x4 x 7 cos 4 cos 7 4 cos 1 55.15 o 7 Now, h 4 x h 4 x tan 1.436 4 x tan h 1 x 1.436 4 x 1 1.436 x 1.436x 1.4364 1.436 x 1 2.872 x 6.744 6.744 x 2.35 ft 2.872 Again, tan Problem-2.8 If the cylinder at A in the figure has a weight of 20 lb, determine the weight of B and the force in each cord needed to hold the system in the equilibrium position shown. __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 9 Figure for Problem- 2.8 Solution: Soln. Fig. 2.8a Soln. Fig. 2.8b Referring to Soln. Fig. 2.8a, Fx 0 yields TEG cos 60 o TEC cos 45o 0 0.5TEG 0.70 TEC 0 F y ... ... ... ... ... ... (a) ... ... ... (b) 0 yields TEG cos 30 o TEC cos 45 o 20 0 0.866 TEG 0.707 TEC 20 0 ... ... Subtract Eq. b from a we get 0.366 TEG 20 0 TEG 54.64 lb Putting this value in Eq. a we get 0.554.64 0.707 TEC 0 __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 10 TEC 38.64 lb 3 36.87 o , and, referring to Soln. Fig. 2.8b, 4 Fx 0 yields TEC cos 45o TCD cos 0 Now, tan 1 38.640.707 TCD cos 36.87 o 0 TCD 34.150 lb F y 0 yields TEC sin 45 o TCD sin 36.87 o WB 0 38.640.707 34.150.6 WB 0 27.318 20.49 WB 0 WB 47.81 lb Problem-2.9 Two weights are suspended from a flexible cable as shown in the figure. Determine the internal forces in the various parts of the cable and the weight W. Figure for Problem- 2.9 Solution: Soln. Fig. 2.9b __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 11 Soln. Fig. 2.9a Referring to Soln. Fig. 2.9a, Fx 0 yields TAB sin 30 o TBC cos15o 0 0.5TAB 0.966 TBC 0 F y ... ... ... ... ... ... (a) ... ... ... (b) 0 yields TAB cos 30 o TBC cos 75 o 10 0 0.866 TAB 0.259 TBC 10 0 ... ... Multiply Eq. a by 0.866 and Eq. b by 0.5 and subtract to get 0.707 TBC 5 0 TBC 7.07 kN Putting this value in Eq a, 0.5TAB 0.9667.07 0 TAB 13.66 kN Again referring to Soln. Fig. b Fx 0 gives TCD cos 45o TBC cos15o 0 0.707 TCD 7.07 0.966 0 TCD 9.66 kN F y 0 gives TCD sin 45 o TBC sin 15 o W 0 9.660.707 7.070.259 W 0 6.83 1.83 W 0 W 8.66 kN Problem-2.10 Determine the required length of cord AC in the figure, so that the 8-kg lamp is suspended in the position shown. The unstretched length of the spring AB is l AB 0.4 m, and the spring has a stiffness of k AB 300 N/m. __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 12 Figure for Problem- 2.10 Solution: Soln Fig. 2.10 Referring to Soln. Fig. 2.10, Fx 0 gives TAB TAC cos 30 o 0 TAB 0.866TAC 0 F y ... ... ... ... ... (a) 0 gives T AC cos 60 o W 0 0.5TAC 78.48 0 TAC 156.96 N Putting this value in Eq. a we get TAB 0.866156.96 0 TAB 135.93 N The stretch in spring AB is TAB k x __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 13 135.93 300 x x 0.453 m The stretched length of the spring AB is l AB l AB x l AB 0.4 0.453 0.853 m The horizontal distance from C to B requires 2 l AC cos 30 o 0.853 0.866 l AC 0.853 2 0 l AC 1.324 m Problem-2.11 A 25-kN sphere rests on a smooth plane inclined at an angle 45o with the horizontal and against a smooth vertical wall. What are the reactions at the contact surfaces A and B? Figure for Problem- 2.11 Solution: Soln Fig. 2.11 Referring to Soln. Fig. 2.11, __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 14 F x 0 yields FA FB cos 45 o 0 FA 0.707FB 0 ... ... ... ... ... (a) F 0 yields FB sin 45o 25 0 25 35.36 kN FB sin 45 o y Putting this value in Eq. a FA 0.70735.36 0 FA 25 kN Problem-2.12 A 12.5-kN wheel with a radius of 90 cm is acted upon by a force F, as shown in the figure, which tends to pull the wheel over the obstruction at A. At the instant the wheel is about to move, the pressure between the wheel and the ground is zero. What is the magnitude of the force F at this instant if 30o? Figure for Problem- 2.12 Solution: Soln Fig. 2.12 Referring to Soln. Fig. 2.12, __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 15 cos 1 F x 45 60 o 90 0 yields F cos 30 o FA sin 60 o 0 0.866F 0.866FA 0 FA F F y ... ... ... ... ... ... (a) 0 yields F sin 30 o FA cos 60 o 12.5 0 0.5F 0.5 FA 12.5 0 0.5F 0.5 F 12.5 0 F 12.5 kN Problem-2.13 A 2.5-kN cylinder A rests on a smooth inclined plane, as shown in the figure. For a tension in the rope of 1.25 kN find the inclination of the plane and the plane reaction. Figure for Problem- 2.13 Solution: __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 16 Soln Fig. 2.13 Referring to Soln. Fig. 2.13, Fx 0 gives 1.25 2.5sin 0 1.25 0.5 2.5 sin 1 0.5 30 o sin F y 0 giveds FR 2.5 cos 0 FR 2.5 cos 30 o 2.16 kN Problem-2.14 Two smooth spheres, each of radius 25 cm and weight 500 N rest in a horizontal channel having vertical walls, the distance between which is 90 cm. Find the pressure exerted on the walls and floor at the points of contact A, B and D. Figure for Problem- 2.14 __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 17 Solution: Soln. Fig. 2.14b Soln. Fig. 2.14a Soln. Fig. 2.14d Soln. Fig. 2.14c cos 1 40 36.87 o [Soln. Fig. 2.14b] 50 Referring to Soln. Fig. 2.14c, Fx 0 gives RC cos 36.87 o R A 0 0.8RC R A 0 ... ... ... ... ... ... (a) F 0 gives RC sin 36.87 o 500 0 500 833.33 N RC sin 36.87 o y Putting this value in Eq. a we get, 0.8833.33 RA 0 __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 18 RA 666.66 N Now referring to Fig. 2.14d, Fx 0 gives RB RC cos 36.87 o 0 RB 833.330.8 0 RB 666.66 N F y 0 gives RD 500 RC sin 36.87 o 0 RD 1000 N Problem-2.15 Two spheres are at rest against smooth surfaces, as shown in the figure. Sphere A weighs 3200 Ib and sphere B weighs 400 Ib. Let F = 1000 Ib and 75o, and find the reactions at C, D and E. Figure for Problem- 2.15 Solution: Soln. Fig. 2.15b __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 19 Soln. Fig. 2.15a Referring to Soln. Fig. 2.15a, Fx 0 gives 1000 RC cos15o RO cos 60 o 0 1000 0.966RC 0.5RO 0 F y ... ... ... ... ... (a) ... ... ... (b) 0 gives RC sin 15 o RO sin 60 o 3200 0 0.259 RC 0.866 RO 3200 0 ... ... Multiply Eq. a by 0.259 and Eq. b by 0.966 and subtract to get 3350 0.965RO 0 RO 3471.50 lb Putting this value in Eq a, we get 1000 0.966RC 0.53471.50 0 0.966RC 735.75 0 RC 761.65 lb Now referring to Soln. Fig. 2.15b Fx 0 gives RO cos 60 o RE 0 3471.500.5 RE 0 RE 1735.75 lb F y 0 gives 400 RO sin 60 o RD 0 400 3471.500.866 RD 0 RD 3406.32 lb Problem-2.16 Determine the moment of force F about points A and B of the beam shown in the figure. Figure for Problem-2.16 Solution: __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 20 M A F d 2002 400 N.m By the right-hand rule, the moment is directed into the page, since the force tends to rotate the beam in a clockwise direction about an axis passing through A. M B F d 2000.5 100 N.m This moment is clockwise and hence directed out of the page. Problem-2.17 The crank shown in the figure has acting on it a 1 kN and a 1.5 kN force as shown. What is the moment of these forces about the axis A and in what direction will the crank turn? Figure for Problem-2.17 Solution: Compute the components of 1 kN force as Fx 1 cos 45 o 0.707 kN, and Fy 1 sin 45o 0.707 kN. Letting the clockwise moment be positive and summing moments about A, we get M A 0.70750 1.525 2.15 kN.cm Since the answer is negative, the net moment is in the counterclockwise sense, the direction in which the crank will turn if it moves. Problem-2.18 Two coplanar forces act on the beam shown in the figure. Determine the moment of each force about point O. __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 21 Figure for Problem-2.18 Solution: Soln. Fig. 2.18 Each force is resolved into its x and y components. Now For the force 200 N: Taking counterclockwise positive, M O 200 cos 60 o 0 200 sin 60 o 0.4 69.28 N.m For the force 400 N: Taking counterclockwise positive, M O 400 sin 30 o 0.2 400 cos 30 o 0.4 40 138.56 98.56 N.m The negative sign indicates M O acts clockwise. __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 22 Problem-2.19 Four wheels A, B, C and D apply loads of 15, 30, 50 and 75 kN respectively, to a beam GH as shown in the figure. Find the resultant of these loads. Figure for Problem-2.19 Solution: The magnitude of the resultant is obtained as R 15 30 50 75 170 kN acting downward. To define R completely, we locate its line of action by using the principle of moment. Choosing any point for center of moments, say point A, we get (clockwise positive). M A 150 301.8 504.2 757.2 54 210 540 804 kN.m Since, R r 804 r 804 170 4.73 m. Therefore the resultant R 170 kN acts vertically downward through a point 4.73 m from A. Problem-2.20 Three parallel forces A, B and C are acting as shown in the figure. Determine their resultant. __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 23 Figure for Problem-2.20 Solution: The magnitude of the resultant is R F 3.0 0.5 105 2.0 kN acting in the same direction as the 3.0 kN and 0.5 kN forces. Now taking clockwise moments as positive and choosing C as the moment center, we get M C 1.537.5 0.512.5 56.25 6.25 50.5 kN.cm (+ve) where the positive sign indicates that the resultant must act in such a manner as to produce a clockwise moment about C. Then the length of the moment arm r of R is obtained from R r 50 r 50 2 25 cm [Note: 1. that the line of action of R is 25 cm from C, 2. that R acts downward, and 3. that R must produce a clockwise moment about C.] Problem-2.21 Determine the magnitude and direction of the couple moment. __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 24 Figure for Problem-2.21 Solution: Soln. Fig. 2.21 The couple forces are resolved. Now there are two couples. M C x 4 sin 30 o 4 8 kN.m (counterclockwise) MC y 4 cos 30 4 13.86 kN.m (counterclockwise) o MC MCx MC y 8.00 13.86 21.86 kN.m (counterclockwise) Problem-2.22 A couple acts at the end of the beam shown. Replace it by an equivalent one having a pair of forces that act through (a) points A and B; (b) points D and E. __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 25 Figure for Problem-2.22 Solution: The couple has a magnitude of M C F d 4000.2 80 N.m. Here the forces tend to rotate clockwise. (a) To preserve the direction of M C horizontal forces acting through points A and B must be directed as shown. Soln. Fig. 2.22a The magnitude of each force is M C F d 80 F 0.25 F 320 N (b) To generate the required clockwise rotation, forces acting through points D and E must be vertical and directed as shown. Soln. Fig. 2.22b The magnitude of each force is M C P d 80 P 0.1 __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 26 P 800 N Problem-2.23 Replace the force F = 300 Ib in the figure with a force through the axis and a couple. Figure for Problem-2.23 Solution: Introduce at the axis two equal, opposite and collinear forces F 300 lb, shown dotted, Soln. Fig. 2.23(a). We now have a couple F d 30014.1 4230 lb-in and a force F 300 lb acting through the axis downward toward the left. Soln. Fig. 2.23a Soln. Fig. 2.23b As suggested by Soln. Fig. 2.23(b), we may say that the original force F produced a force at the shaft of 300 lb and a turning moment of C 4030 lb-in. __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 27 Problem-2.24 Determine the horizontal and vertical components of reaction for the beam loaded as shown in the figure. Neglect the weight of the beam in the calculations. Figure for Problem- 2.24 Solution: Soln. Fig. 2.24 F x 0 yields 600 cos 45 o B x 0 Bx 424.26 N F y 0 yields Ay By 600 sin 45o 100 0 Ay B y 524.26 M A ... ... ... ... ... ... (a) 0 yields 600 sin sin 45o 2 100 5 By 7 0 B y 192.65 N Substitute this value of B y in Eq. a Ay 524.26 192.65 331.61 N Problem-2.25 Determine the reactions at the fixed support A for the loaded frame shown in the figure. __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 28 Figure for Problem- 2.25 Solution: Soln. Fig. 2.25 F y 0 yields Ay 200 400 0 Ay 600 N __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 29 M A 0 yields M A 400 1 200 2 0 M A 800 N-m Problem-2.26 The uniform beam shown in the figure is subjected to a force and couple. If the beam is supported at A by a smooth wall and at B and C either at the top or bottom by smooth contacts, determine the reactions at these supports. Neglect the weight of the beam. Figure for Problem- 2.26 Solution: Soln. Fig. 2.26 F x 0 yields FA cos 30 o 300 sin 30 o 0 __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 30 FA 173.20 N F y 0 yields FC FB FA sin 30 o 300 cos 30 o 0 FC FB 0.5FA 259.81 FC FB 0.5 173.20 259.81 ... ... FC FB 346.41 M A 0 yields ... ... ... ... (a) ... (b) 4000 FB 2 FA sin 30 o 4 FC 2 300 cos 30 o 4 0 2 FB 2 FC 4000 173.20 sin 30 o 4 1039.23 0 ... ... ... ... FB FC 2346.41 ... Add Eq. a and b to get 2FB 2000 FB 1000 N Put this value of FB in Eq. a 1000 FC 346.41 FC 1346.41 N Problem-2.27 The beam shown in the figure is pin-connected at A and rests against a roller at B. Compute the horizontal and vertical components of reaction at the pin A. Figure for Problem- 2.27 Solution: __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 31 Soln. Fig. 2.27 F x 0 yields Ax FB sin 30 o 0 Ax 0.5FB 0 F y ... ... ... ... (a) ... ... ... ... (b) 0 yields Ay 60 FB cos 30o 0 Ay 0.866 FB 60 M ... A ... 0 gives FB 0.75 60 1 60 0 FB 160 N Put this value in Eq. a Ax 0.5 160 0 Ax 80 N Now put FB 160 N in Eq. b Ay 0.866 160 60 Ay 198.56 N Problem-2.28 The man has a weight of 150 lb and stands at the end of the beam shown. Determine the magnitude and direction of the reaction at the pin A and the tension in the cable. __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 32 Figure for Problem- 2.28 Solution: Soln. Fig. 2.28 3 36.87 o 4 Fx 0 yields T cos 36.87 o Ax 0 tan 1 0.8T Ax 0 F y ... ... ... ... ... (a) ... ... ... (b) 0 yields Ay T sin 36.87 o 150 0 0.6T Ay 150 ... ... 0 gives 150 10 T sin 36.87 o 2 T cos 36.87 o 3 0 150 1.2 T 2.4 T 0 3.6 T 1500 T 416.67 lb M A Put this value of T in Eq. a __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 33 0.8 416.67 Ax 0 Ax 333.33 lb Now put T 416.67 lb in Eq. b 0.6 416.67 Ay 150 Ay 100.00 lb FA Ax2 Ay2 333.332 100.00 2 348.00 lb Direction of FA , Ay 100.00 tan 1 tan 1 16.70 o Ax 333.33 Problem-2.29 The 100-kg uniform beam AB shown in the figure is supported at A by a pin and at B and C by a continuous cable which wraps around a frictionless pulley located at D. If a maximum tension force of 800 N can be developed in the cable before it breaks, determine the greatest length b of a uniform 2.5 kN/m distributed load that can be placed on the beam. What are the horizontal and vertical components of reaction at A just before the cable breaks? Figure for Problem- 2.29 Solution: __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 34 Soln. Fig. 2.29 Referring to Soln. Fig. 2.29 Fx 0 yields Ax 800 cos 60 o 0 Ax 400 N F y 0 yields Ay 2500 b 981 800 sin 60o 800 0 Ay 2500 b 511.82 0 ... ... ... ... ... (a) and, b 0 gives 80010 800 sin 60 o 8 9815 2500 b 0 2 2 800 5542.56 4905 1250 b 0 b 2 6.91 b 2.63 m M A Put this value in Eq. (a) to get Ay 25002.63 511.82 6063.18 N __________________________________________________________________________________________________ SOLVED PROBLEMS ON ENGINEERING MECHANICS Chapter- 2 Equilibrium of Force Systems Page 35