40 15. Refer to figure below, calculate: i. Force in component x and y axis and resultant force in terms of cartesian vector. ii. Magnitude of the resultant force, FR. Answer: i. Force in component x and y axis and resultant force in terms of cartesian vector. = −1 ( 4 4 ) = 45° 1 = −750 cos 45° + 750 45° = −530.33 + 530.33 2 = −650 53.13° − 650 53.13° = −390 − 520 3 = 450 30° − 450 30° = 389.71 − 225
41 = 1 + 2 + 3 = (−530.33 + 530.33 ) + (−390 − 520) + (389.71 − 225 ) = (−530.62 − 214.67 ) ii. Magnitude of the resultant force, FR. Magnitude of Resultant Force, = √(−530.62) 2 + (−214.67) 2 = 572.4
42 16. Express the value of the internal forces in cable CB, CE, and spring CD in figure below, if the mass of the ball is 70 kg. Answer: Free Body Diagram at joint E W = mg = 70 × 9.81 = 686.7 +↑ ∑ = 0 − = 0 = = 686.7
43 Free Body Diagram at joint C sin 90° = sin 150° = sin 120° sin 150° = 686.7 sin 120° = 686.7 sin 120 × sin 150 = 396.47 sin 90° = sin 120° = 686.7 sin 120° × sin 90° = 792.93 Or CE = 686.7 N CE = 686.7 N
44 +↑ ∑ = 0 sin 60° − 686.7 = 0 = 686.7 sin 60° = 792.93 + → ∑ = 0 − cos 60° + = 0 (−792.93 cos 60°) + = 0 (−396.47) + = 0 = 396.47
45 17. Figure below shows the magnitude and direction of two forces acting on a ring bracket. Calculate: i. Magnitude of the resultant force. ii. Direction of the resultant force. Answer: i. Magnitude of the resultant force. FRD + → ∑ = 0 ∑ = 100 20° + 200 65° = 93.97 + 84.52 = 178.49 kN
46 +↑ ∑ = 0 ∑ = 100 20° + 200 65° = 34.2 + 181.26 = 215.46 kN = √( ) 2 + () 2 = √(178.49) 2 + (215.46) 2 =√78281.69 = 279.79 kN ii. Direction of the resultant force. = tan−1 ( ) = tan−1 ( 215.46 178.49) = 50.36 °
47 18. Figure below shows a load of 40 kg hanging in equilibrium on cable AO and BO. If the system is in equilibrium, calculate: i. Angle ii. Force at cable AO and BO Answer: i. Angle = tan−1 = tan−1 1.5 1 = 56.31° ii. + → ∑ = 0 56.31° − 40°= 0 = 40° 56.31° = 1.38 ----------------------------------------------- equation 1
48 +↑ ∑ = 0 40° + 56.31 = 392.4 -------------------------------- equation 2 Substitute = 1.38 to equation 2. ∴ 40° + (1.38 ) 56.31 = 392.4 0.643 + 1.148 = 392.4 1.791 = 392.4 = 392.4 1.791 = 219.1 = 1.38 = 1.38 (219.1) = 302.358 N
49 19. Based on figure below, if 2 = 35 and = 55°: i. Calculate each force into component -x () and component -y (). ii. Calculate the magnitude of the resultant in cartesian vector form. Answer: i. Calculate each force into component -x () and component -y (). 1 = 40 ∶ 1 = 40 36.87 = 32 1 = 40 36.87 = 24 2 = 35 : 2 = −35 55° = −20.08 2 = −35 55° = −28.67 3 = 50 : 3 = 50 60° = 25 3 = −50 60 = −43.30
50 ii. Calculate the magnitude of the resultant in cartesian vector form. = () + () = (32 − 20.08 + 25) + (24 − 28.67 − 43.30) = (36.92 − 47.97) = √( ) 2 + () 2 = √(36.92) 2 + (47.97) 2 =√3662.21 = 60.52 kN
51 20. Figure below shows a plane force system at O. Calculate the magnitude of the forces, 1 and 2 if the system is in equilibrium. Answer: = tan−1 3 4 = 36.9° + → ∑ = 0 1 60° − 2 70° + 5 60° − 10 36.9° = 0 0.871 − 0.942 + 2.5 − 8 = 0 0.871 − 0.942 = 5.5 -------------------------equation 1 +↑ ∑ = 0 1 60° + 2 70° − 5 60° − 10 36.9° = 0 (0.5)1 + (0.342)2 – (4.33) − 6 = 0 (0.5)1 + (0.342)2 = 10.33 ------------------equation 2
52 From equation 1 0.871 − 0.942 = 5.5 1 = 5.5 + 0.94 2 0.87 = 6.32 + 1.08 2-------------------------------------------------------------- equation 3 Substitute 1 = 6.32 + 1.08 2 into equation 2. (0.5)(6.32 + 1.082) + 0.342 2 = 10.33 3.16 + 0.542 + 0.3422 = 10.33 0.542 + 0.3422 = 10.33 − 3.16 0.882 = 7.17 = 7.17 0.88 = 8.15 kN 1 = 6.32 + 1.08 2 = 6.32 + 1.08(8.15) = 15.122 kN
53 21. Two cables AB and BC are tied together at B and a load of 25 kg is hung at B as shown in Figure below. Calculate the tension developed in cable AB and BC. Answer: Free Body Diagram (FBD) 55° − 54.37° = 0 = 54.37° 55° = 1.0156 --------------------------------- equation 1
54 55° + 54.37° − 294.3 = 0 ---------------------- equation 2 Substitute equation 1 into equation 2. (1.0156)55° + 54.37° = 294.3 0.832 + 0.813 = 294.3 1.645 = 294.3 = 294.3 1.645 = 178.91 Substitute = 178.91 into equation 1. = 1.0156 = 1.0156 (178.91) = 181.76 N
55 22. Referring to force diagram in figure below, calculate: i. Resultant force of each component ii. Magnitude of the resultant force. iii. Determine the direction of the resultant force. iv. Illustrate that direction using force diagram for resultant force. Answer: i. Resultant force of each component. + → ∑ = 0 ∑ = 60 cos 45° − 120 cos 75° = 42.43 − 31.06 = 11.37 kN +↑ ∑ = 0 ∑ = 60 sin 45° − 120 sin 75° = 42.43 − 115.91 = −73.48
56 ii. Magnitude of the resultant force. = √( ) 2 + () 2 = √(11.37) 2 + (−73.48) 2 =√5528.5873 = 74.35 kN iii. Determine the direction of the resultant force. = tan−1 ( ) = tan−1 ( −73.48 11.37 ) = − 81.2° iv. Illustrate that direction using force diagram for resultant force.
57 23. Figure below shows the system in equilibrium. Represent the system in the form of free body diagram at: i. Point B ii. Point D Answer: i. Point B = tan−1 3 4 = 36.87°
58 ii. Point D
59 24. Determine the x and y components of each force acting on the screw eye shown in figure below. Answer: + → ∑ = 0 ∑ = 4 45° − 8 60° = 2.828 − 4 = −1.172 +↑ ∑ = 0 ∑ = −4 45° − 8 60° = −2.828 − 6.928 = −9.756
60 25. Based on diagram below, if 1 = 700 and = 25°, calculate the magnitude and direction measured counterclockwise from the positive x axis of the resultant force produced from the three forces acting on the A ring. Answer: Directian angle 600 N, = tan−1 3 4 = 36.87° + → ∑ = 0 ∑ = 400 cos 30° + 700 sin 25° − 600 cos 36.87° ∑ = 346.4 + 295.8 − 479.99 ∑ = 162.21 +↑ ∑ = 0 ∑ = 400 30° + 700 cos 25° + 600 sin 36.87° ∑ = 200 + 634.42 + 360 ∑ = 1194.42
61 = √( ) 2 + () 2 = √(162.21) 2 + (1194.42) 2 =√26312.0841 + 1426639.1364 = √1452951.22 = 1205.38 N = tan−1 ( ) = tan−1 ( 1194.42 162.21 ) = 82.27°
62 26. By using parallelogram, calculate: i. The magnitude of the component force F in figure below and the magnitude of the resultant force if is directed along the positive y axis. ii. Predict what happen to the resultant force if the value of F is 400 N Answer: i. The magnitude of the component force F in figure below and the magnitude of the resultant force if is directed along the positive y axis. sin 60° = sin 75° = 300 sin 45° sin 60° = 300 sin 45° = 300 × sin 60° sin 45°
63 = 259.81 0.707 = 367.48 sin 75° = 300 sin 45° = 300 × sin 75° sin 45° = 289.78 0.707 = 409.81 ii. Predict what happen to the resultant force if the value of F is 400 N sin 60° = sin 75° 400 sin 60° = 75° = 400×sin75° sin60° = 386.37 0.866 = 446.15 Therefore, the resultant force is increased when the value of F is 400 N.
64 27. A 10 kg block is suspended from pulley B and the sag of the cord is = 0.2 . Determine the force is cord ABC. Neglect the size of the pulley. Answer: +↑ ∑ = 0 sin 45° + sin 45° = 98.1 2 sin 45° = 98.1 = 98.1 2 sin 45° = 69.37
65 28. Calculate the magnitude of force T acting on the eyebolt and its angle , if the magnitude of the resultant force is 10 kN to be directed along the positive x axis. Answer: Cosinus law: 2 = 8 2 + 102 − 2(8)(10) 45° 2 = 50.863 = √50.863 = 7.13 Sinus law: 45 = 8 45 7.13 × 8 =
66 = 0.79 = sin−1 0.79 = 52.19° ∴ = 90 − 52.19° = 37.81 °
67 29. Solve each force acting into x and y components. Express the force as a Cartesian vector. Answer: = 4 3 = tan−1 1.333 = 53.13° 30 = 30 53.13° + 30 53.13° = 18 + 24 40 = −40 40° + 40 40 = −30.64 + 25.71 = 30 + 40 = (18 + 24) + (−30.64 + 25.71) = (−12.64 + 49.71) N
68 30. Based on figure below, if 1 = 100 and ∅ = 30°, calculate the magnitude of the resultant force acting on the bracket and its direction measured clockwise from positive x-axis. Answer: 3 = tan−1 5 12 3 = 22.62° + → ∑ = 0
69 ∑ = 100 sin 30° + 200 + 260 sin 22.62 ∑ = 50 + 200 + 100 ∑ = 350 +↑ ∑ = 0 ∑ = 100 30° − 260 cos 22.62° ∑ = 86.6 − 240 ∑ = −153.4 = √( ) 2 + () 2 = √(350) 2 + (−153.4) 2 = √122500 + 23531.56 = √146031.56 = 382.14 = tan−1 ( ) = tan−1 ( −153.4 350 ) = −23.67°
70 31. Based on figure below: i. Calculate the magnitude of the resultant of 1 and 2 in term of Cartesian vector. a. ii. Predict what will happen to the resultant force if the angle of 1 is increased to 60°. Answer: i. Calculate the magnitude of the resultant of 1 and 2 in term of Cartesian vector. = −1 ( 4 3 ) = 53.13° 1 = −150 sin 30 ° + 150 cos 30° = −75 + 129.9 2 = 200 sin 53.13° − 200 cos 53.13° = 160 − 120
71 = 1 + 2 = (−75 + 129.9) + (160 − 120) = (85 + 9.9) Magnitude of Resultant Force, = √(85) 2 + (9.9) 2 = √7225 + 98.01 = √7323.01 = 85.57 ii. Predict what will happen to the resultant force if the angle of 1 is increased to 60°. 1 = −150 sin 60 ° + 150 cos 60° = −129.9 + 75 = 1 + 2 = (−129.9 + 75) + (160 − 120) = (30.1 − 45) Magnitude of Resultant Force, = √(30) 2 + (−45) 2 = √900 + 2025 = √2925 = 54.08 By comparing the magnitude of , if angle of 1 is increased, the resultant force will decrease.
72 32. If the mass of chandelier in figure below is 50 kg, determine the tension developed in BD and CD cable used to support the chandelier. Answer: Free Body Diagram + → ∑ = 0 cos 30° − cos 45° = 0 = cos 45° cos 30° = 0.816 -------------------------- equation 1 +↑ ∑ = 0 sin 30° + sin 45° = 490.5 0.5 + 0.707 = 490.5 --------------------------- equation 2
73 Substitute = 0.816 into equation 2 0.5(0.816) + 0.707 = 490.5 0.408 + 0.707 = 490.5 1.115 = 490.5 = 490.5 1.115 = 439.9 = 0.816(439.9) = 358.95
74 33. Figure below shown the three forces acting on the particle O. Calculate: i. Magnitude resultant force for these forces. ii. The direction of the resultant force counterclockwise from the positive x-axis. Answer: i. Magnitude resultant force for these forces. + → ∑ = 0 ∑ = 30 cos 45° + 40 cos 15° + 25 sin 15° = 21.21 + 38.64 + 6.47 = 66.32 N +↑ ∑ = 0 ∑ = 30 sin 45° − 40 sin 15° − 25 15° = 21.21 − 10.35 − 24.15 = −13.29 = √( ) 2 + () 2 = √(66.32) 2 + (−13.29) 2 = √4574.9665 = 67.64
75 ii. The direction of the resultant force counterclockwise from the positive x-axis. = tan−1 = tan−1 −13.29 66.32 = −11.33° ∴ = 360° − 11.33° = 348.7
76 34. Two forces 1 = 60 and 2 = 40 acting on the particle O as shown in figure below: i. Calculate magnitude of the resultant force of the forces as Cartesian vector form. ii. Predict what will happen to the resultant force if the angle of 2 is increased to 30°. Answer: i. Calculate magnitude of the resultant force of the forces as Cartesian vector form. 1 = 60 cos 35 ° + 60 sin 35° = 49.15 + 34.41 2 = 40 sin 15° − 40 cos 15° = 10.35 − 38.64 = 1 + 2 = ( 49.15 + 34.41) + (10.35 − 38.64) = (59.5 − 4.23) Magnitude of Resultant Force, = √(59.5) 2 + (−4.23) 2 = √3540.25 + 17.89 = √3468.14
77 = 58.89 ii. Predict what will happen to the resultant force if the angle of 2 is increased to 30°. 2 = 40 sin 30° − 40 cos 30° = 20 − 34.64 = 1 + 2 = (49.15 + 34.41) + (20 − 34.64) = (69.15 − 0.23) Magnitude of Resultant Force, = √(69.15) 2 + (−0.23) 2 = √4781.72 + 0.053 = √4781.743 = 69.15 By comparing the magnitude of , if angle of 2 is increased, the resultant force will increase.
78 35. If the mass of bucket in figure below is 10 kg, determine the tension developed in EB and ED cable used to support the bucket. Answer: Analysis at point E × . = . + → ∑ = 0 cos 30° − cos 45° = 0 = cos 45° cos 30° = 0.816 -------------------------- equation 1
79 +↑ ∑ = 0 sin 30° + sin 45° = 98.1 0.5 + 0.707 = 98.1 --------------------------- equation 2 Substitute = 0.816 into equation 2. 0.5(0.816) + 0.707 = 98.1 0.408 + 0.707 = 98.1 1.115 = 98.1 = 98.1 1.115 = 87.98 = 0.816(87.98) = 71.79
80 CHAPTER 2 STRUCTURES 1. Describe the following term: i. The conditions for equilibrium ii. Plane truss Answer: i. The conditions for equilibrium A particle is in equilibrium provided it is at rest if originally at rest, or has a constant velocity if originally in motion. To maintain a state of equilibrium, it is necessary to satisfy Newton’s first law of motion, which states that if the resultant force acting on a particle is zero, then the particle is in equilibrium. ii. Plane truss A planar truss is one where all members and nodes lie within a two dimensional plane.
81 2. The truss is subjected to the loading as shown in figure below: i. Draw the free body diagram for the truss as show in figure ii. Calculate the reaction force for each supporter iii. Calculate the force in each member by using the method of joint and state the members are in tension or compression. Answer: i. Draw the free body diagram for the truss as show in figure. ii. Calculate the reaction force for each supporter + → ∑ = 0, + 20 = 0 = −20 + ∑ = 0, (20)3 − (5) = 0
82 60 − 5 = 0 = 60 5 = 12 kN + ↑ ∑ =0 + = 0 + 12 = 0 = −12 iii. Calculate the force in each member by using the method of joint and state the members are in tension or compression. Free body diagram of joint C = tan−1 5 3 = 59.04° + → ∑ = 0, 20 − 59.04° = 0
83 = 20 59.04° = 23.32 (T) + ↑ ∑ =0, − 59.04 − = 0 −(23.32) 59.04 − = 0 −12 − = 0 − = 12 = −12 (C) At joint B + → ∑ = 0, − = 0 = 0
84 3. Figure below shows a truss on the floor with a supporter pin at A and supporter a roller at D. By using the method of section, i. Illustrate the free body diagram. ii. Find the reaction force at supporter A and D iii. Determine force in member BC, GF of the truss and state whether the member are in tension or compression. Answer: i. Illustrate the free body diagram. ii. Find the reaction force at supporter A and D + ∑ = 0, (20)(4) + (10)(8) − (12) = 0 80 + 80 − 12 = 0 20 KN
85 = 160 12 = 13.33 kN + ↑ ∑ =0 + − 20 − 10 − 50 = 0 + 13.33 − 20 − 10 − 50 = 0 − 66.67 = 0 = 66.67 + → ∑ = 0, = 0 iii. Determine force in member BC, GF of the truss and state whether the member are in tension or compression. Moment of G + ∑ = 0, − (4) − (50)(4) + (4) = 0
86 − (4) − 200 + (66.67)(4) = 0 − (4) − 200 + 266.68 = − (4) + 66.68 = 0 = 66.68 4 = 16.67 () Moment of C + ∑ = 0, (4) − (50)(8) − (20)(4) + (8) = 0 4 − 400 − 80 + (66.67)(8) = 0 4 − 400 − 80 + 533.36 = 0 4 + 53.36 = 0 = −53.36 4 = −53.36 () + ↑ ∑ =0 − 50 − 20 − 45° = 0 66.67 − 50 − 20 − 45° = 0 −3.33 − 45° = 0 = 3.33 − 45° = −4.71 ()
87 4. A truss is a structure composed of member joined together at their end points. Figure below shows a truss member with load F = 500 N acting at joint C. i. Classify 2 method of analysing a truss. ii. Find internal forces in member AC. iii. Find external reaction at support B Answer: i. Classify 2 method of analysing a truss. • Method of joint • Method of section ii. Find internal forces in member AC. FBD at joint C
88 + → ∑ = 0 500 − 45° = 0 = 500 45° = 707.11 (T) + ↑ ∑ =0 − − 45° = 0 − − (707.11) 45° = 0 − − 500 = 0 = − 500 () iii. Find external reaction at support B FBD at joint B + → ∑ = 0 = 0 + ↑ ∑ =0 + = 0 + (−500) = 0 = 500 ( )
89 5. Figure below shows a structure with load P = 1.6 kN acting at point A. From the structure given, find : i. The number of members in the structure. ii. Angle 1 2 . iii. Force in member AB and AC. Answer: i. The number of members in the structure. - 3 (member AB, member BC and member AC) ii. Angle 1 2 . 1 = tan−1 0.3 1.8 = 9.46°