90 2 = tan−1 1.2 0.7 = 59.74° iii. Force in member AB and AC. + → ∑ = 0 1.6 − 9.46° − 30.8° = 0 ----------------------- equation 1 + ↑ ∑ =0 − 9.46° − 30.8° = 0 = − 30.8 9.46 = −3.115 -------------------------- equation 2 Substitute = −3.115 to equation 1. 1.6 − (−3.115 ) 9.46° − 30.8° = 0
91 1.6 + 3.073 − 0.859 = 0 2.214 = −1.6 = − 1.6 2.214 = −0.723 (C) = −3.115 = −(3.115)(−0.723) = 2.252 ( )
92 6. The truss is subjected to the loading as shown in figure below. i. Find the value of 1 and 2 for the truss in figure below. ii. Illustrate free body diagram and label all the force acting for the truss in figure below. iii. Find the reaction force at supporter A and C in figure below. iv. Determine the force in all members and indicate if the member is in tension or compression. a. Member BD b. Member AD and AB c. Member BC, BD and DC Answer: i. Find the value of 1 and 2 for the truss in figure below. 1 = tan−1 4 3 = 53.13°
93 2 = tan−1 4 7 = 29.74° ii. Illustrate free body diagram and label all the force acting for the truss in figure below. iii. Find the reaction force at supporter A and C in figure below.
94 + → ∑ = 0 = 0 + ∑ = 0 (600)(3) − (10) = 0 1800 − 10 = 0 = −1800 −10 = 180 + ↑ ∑ =0 + = 600 + 180 = 600 = 600 − 180 = 420 iv. Determine the force in all members and indicate if the member is in tension or compression. a) Member BD Joint D + ↑ ∑ = 0 = 600 ()
95 b) Member AD and AB Joint A + → ∑ = 0 − 53.13° = 0 = 53.13° + ↑ ∑ = 0 53.13° − 420 = 0 = 420 53.13° = 525 () = 525 53.13° = 315 ()
96 c) Member BC, BD and DC Joint B + → ∑ = 0 525 53.13 − 29.74° = 0 = 525 53.13 29.74 = 362.785 + ↑ ∑ = 0 525 53.13° + 362.785 29.74° − = 0 = 420 + 180 = 600 ()
97 Joint C + → ∑ = 0 362.785 29.74° − = 0 = 315 (T)
98 7. The truss is subjected to the loading as shown in Figure below: i. Illustrate free body diagram for the truss as shown in Figure below. ii. Find the reaction force for each supporter. iii. Determine the force in each member of the truss and state the members are in tension or compression. Answer: i. Illustrate free body diagram for the truss as shown in Figure below. Free Body Diagram (FBD) ii. Find the reaction force for each supporter. + ∑ = 0 80(10) − (5) = 0 = 800 5 = 160 kN (C)
99 = 0 + = 80 + 160 = 80 = 80 − 160 = −80 () iii. Determine the force in each member of the truss and state the members are in tension or compression. Joint A 26.57° + = 0 ----------------------------- equation 1 26.57° − 80 = 0 = 80 26.57 = 178.85 () Substitute = 178.85 into equation 1. (178.85) 26.57° + = 0 159.96 + = 0 = −159.96 ()
100 Joint B 45° − (−159.96) = 0 = − 159.96 45° = −226.22 ()
101 8. The bridge in Figure below is subjected to the loading as shown below. By using method of section: i. Illustrate the free body diagram. ii. Find the reaction force at supporter A and E iii. Determine force in the part of frame HI, HB and BC of the truss. Answer: i. Ilustrate the free body diagram.
102 ii. Find the reaction force at supporter A and E + ∑ = 0 (16) − (5)(16) − (2)(12) − (2)(8) − (4)(4) = 0 = 136 16 = 8.5 kN (C) = 0 6.5 − 5 − 2 − 2 − 4 + = 0 = 6.5 ( ) iii. Determine force in the part of frame HI, HB and BC of the truss. + ∑ = 0 (8.5)(8) − (5)(8) − (2)(4) − (4) = 0 = 20 4 = 5 ()
103 + ∑ = 0 −(4) − (5)(4) + (8.5)(4) = 0 = 14 4 = 3.5 () = tan−1 4 4 = 45° 8.5 − 5 − 2 − 45° = 0 = −1.5 − 45 = 1.5 ()
104 9. Calculate the force in each member of the truss in figure below. State whether it is in tension or compression. Answer: = 0 = + 80 -------------------------------------- equation 1 + ∑ = 0 80(10) − (5) = 0 = 800 5 = 160
105 Substitute = 160 into equation 1. 160 = + 80 = 80 Joint A = tan−1 5 10 = 26.57° cos 26.57° = --------------------------------------- equation 1 sin 26.57° = 80 = 80 sin 26.57° = 178.85 ()
106 Substitute = 178.85 into equation 1. (178.85)cos 26.57° = = 160 () Joint C 160 − sin 45° = 0 = 160 sin 45° = 226.27 ()
107 10. Calculate the forces of each member of the roof truss shown in Figure below and indicate if the members are in tension or compression. Answer: + ∑ = 0 (2) − (500)() = 0 = 500 2 = 250 + = 500 + 500 30° 250 + = 500 + 250 = 500
108 Joint A 250 − 30° = 0 = 250 30° = 500 (C) = 30° = (500) 30° = 433.01 (T)
109 Joint C = 500 () = 433.01 () Joint B 500 30° + 500 30° = 30° 433.01 + 433.01 = 0.866 = 866.02 0.866 = 1000 ( )
110 11. Calculate the force of each member at BC, CF and EF of the truss as shown in fugure below and state whether the members are in tension or compression. Answer: + ∑ = 0 14(1) + 22(3) − (6) = 0 14 + 66 − 6 = 0 = 80 6 = 13.33
111 + ∑ = 0 (6) − 14(5) − 22(3) = 0 6− 70 − 66 = 0 = 136 6 = 22.67 1 2 + 2 = 2.5 2 2 = 6.25 − 1 2 = 5.25 = √5.25 = 2.29 sin = 1 2.5 = sin−1 0.4 = 23.6°
112 + ∑ = 0 −22(1) − 14(3) + (4) + (2.29) = 0 −22 − 42 + 4(22.67) + 2.29 = 0 26.68 + 2.29 = 0 = − 26.68 2.29 = −11.65 + ∑ = 0 −14(2) + 22.67(3) − (2.29) = 0 40.01 − 2.29 = 0 = 40.01 2.29 = 17.47 −14 − 22 + 22.67 − 23.6 = 0 −13.33 − 23.6 = 0 = − 13.33 23.6 = −14.55
113 12. Based on figure below, determine the tension developed in cable CA and CB Answer: 3500 − cos 30 ° − cos 40° = 0 ---------------------- equation 1 sin 30° − sin 40° = 0 ---------------------------------- equation 2 = sin 40° sin 30°
114 = 1.286 --------------------------- equation 3 Substitute = 1.286 into equation 1. 3500 − (1.286 ) cos 30 ° − cos 40° = 0 −1.114 − 0.766 = −3500 −1.88 = −3500 = −3500 −1.88 = 1861.7 = 1.286 = 1.286 (1861.7) = 2394.15
115 13. Using the method of joints, determine : i. The force in each member of the truss and state whether the members are in tension or compression. ii. The magnitude of the external reactions at A and B and show the direction of the reaction. Answer: i. The force in each member of the truss and state whether the members are in tension or compression. Joint C 500 cos 20° − cos 55° = 0 = −500 cos 20 − cos 55 = 819.15 (T)
116 500 sin 20° − − sin 55° = 0 171.01 − − (819.15) sin 55° = 0 171.01 − − 671 = 0 = −500 ( ) Joint B = 0 + = 0 −500 + = 0 = 500
117 ii. The magnitude of the external reactions at A and B and show the direction of the reaction. + + 500 sin 20° = 0 + = −171.01 ------------------------------ equation 1 + 500 cos 20° = 0 = −469.85 (← ) Substitute = 500 into equation 1 + 500 = −171.01 = −171.01 − 500 = −671.01 (↓)
118 14. The bridge in figure below is subjected to the loading shown. Identity whether the members IH, BH and BC of the truss are in tension or compression form. Answer: Free Body Diagram + ∑ = 0 20(4) + 20(8) + 40(12) − (16) = 0 80 + 160 + 480 − 16 = 0 720 − 16 = 0 = −720 −16 = 45
119 = 0 + − 40 − 20 − 20 − 40 = 0 + 45 − 120 = 0 − 75 = 0 = 75 + ∑ = 0 −20(4) − 40(8) + 75(8) − (4) = 0 −80 − 320 + 600 − 4 = 0 200 − 4 = 0 = −200 −4 = 50 (T)
120 + ∑ = 0 75(4) − 40(4) − (4) = 0 300 − 160 − 4 = 0 140 − 4 = 0 = −140 −4 = 35 (C) 75 − 40 − 20 − sin 45° = 0 15 − sin 45° = 0 = −15 − sin 45° = 21.21 ()
121 15. The bridge in figure below is subjected to the loading shown. Calculate the force in the part of frame IH, BH and BC of the truss. Identity whether the parts of frame are in tension or compression. Answer: Free Body Diagram + ∑ = 0 2(4) + 2(8) + 4(12) − (16) = 0 8 + 16 + 48 − 16 = 0 72 − 16 = 0 = −72 −16 = 4.5
122 = 0 + − 4 − 2 − 2 − 4 = 0 + 4.5 − 12 = 0 − 7.5 = 0 = 7.5 + ∑ = 0 −2(4) − 4(8) + 7.5(8) − (4) = 0 −8 − 32 + 60 − 4 = 0 20 − 4 = 0 = −20 −4 = 5 (T)
123 + ∑ = 0 7.5(4) − 4(4) − (4) = 0 30 − 16 − 4 = 0 14 − 4 = 0 = −14 −4 = 3.5 (C) 7.5 − 4 − 2 − sin 45° = 0 1. 5 − sin 45° = 0 = −1.5 − sin 45° = 2.121 ()
124 CHAPTER 3 KINEMATICS OF PARTICLES 1. State the following terms: i. Displacement of particle ii. Acceleration Answer: i. Displacement of a particle is defined as the change in its position during the interval time. ii. Acceleration is the rate of change of velocity of an object during the time interval. 2. Give the definition and S.I. unit for accelerations. Answer: Accelerations is the rate of change of velocity and S.I. unit for acceleration is m/s2 . 3. Kinematics is a branch of dynamics. i. Define kinematics. ii. Compare velocity and speed in kinematics. Answer: Kinematics is the branch of mechanics that concerned only on geometric aspects of motion for object without reference to the forces which cause the motion. Velocity is defined as the rate of change of displacement. Velocity is a movement in a known direction, so it is a vector quantity. Meanwhile, speed is the rate of change of distance. The direction is not known, so it is a scalar quantity.
125 4. Define the following terms: i. Velocity ii. Speed iii. Angular velocity Answer: Velocity is defined as the rate of change of displacement. Velocity is a movement in a known direction, so it is a vector quantity. Speed is the rate of change of distance to time Angular velocity is defined as the rate of change of angular displacement , with respect to time t, and it is for an object rotating about a fixed axis at a constant speed. = / where = angular velocity v= linear velocity r= radius 5. Define the following terms: i. Constant velocity ii. Constant acceleration Answer: Constant velocity - The object in motion travels the same every second where magnitude and direction of the velocity is remaining constant. Constant acceleration - The object in motion is changing the speed uniformly throughout the motion.
126 6. Define the terms below. i. Kinematic ii. Kinetics Answer: i. Kinematic which treats only the geometric aspect of the motion ii. Kinetics which is the analysis of the force causing the motion. Figure below shows the sketch of a circular motion, interpret , , , . Answer: – linear velocity (m/s) – angular acceleration (rad/s2 ) − angular velocity (rad/s) −rotation angle (rad) −radius (m)
127 7. Visualize using suitable velocity-time graph. i. When the object starts to rest. ii. When the object is in acceleration. iii. when the object is in deceleration iv. When the object is in uniform motion. Answer: i. When the object starts to rest. ii. When the object is in acceleration. iii. When the object is decelerated. iv. When the object is in uniform motion.
128 8. A car with wheels of 500 mm in diameter each is travelling at 64 km/h. Determine the angular velocity of the wheels in rad/s. Answer: 64 /ℎ = 64 ℎ × 1000 1 × 1ℎ 3600 = 17.78 / The radius of a wheel is 0.25 m = = 17.78 0.25 = 71.12 /
129 9. A motorcycle moves with a wheel speed of 250 rpm. If the diameter of motorcycle wheel is 60 cm. i. Convert the velocity of motorcycle in km/h unit. ii. If the motorcycle moves for 10 minutes from rest, express the value of acceleration for the motorcycle in m/s2 . Answer: i. N = 250 rpm D = 60 cm = 0.6 m r = 0.3 m = 2 60 = 2(250) 60 = 26.18 / = = (0.3)(26.18) = 7.854 / = 7.854 × 3600 1000 = 28.27 /ℎ ii. u = 0, t = 10 min = 600 s = + 7.854 = 0 + (600) = 7.854 600 = 0.01309 m/s2
130 10. The distance between building A and building B is 6.8 km. A car starts from rest at building A with a constant acceleration for 32 seconds, then a car travels with a constant velocity before it decelerates constantly and stops at building B in the last 15 seconds of the journey. If the total time taken is 7 minutes: i. Draw a velocity-time graph. ii. Calculate the constant velocity of the car in m/s iii. Calculate the acceleration of the car in m/s2 . iv. Calculate the distance traveled in the first 3 minutes of the journey. Answer: i. Draw a velocity-time graph. ii. Calculate the constant velocity of the car in m/s 1 = 1 2 × 32 × 2 = 373 × 3 = 1 2 × 15 ×
131 1 + 2 + 3 = 16 + 373 + 7.5 = 6800 396.5 = 6800 = 6800 396.5 = 17.15 / iii. Calculate the acceleration of the car in m/s2 . = + 17.15 = 0 + (32) = 17.15 32 = 0.536 m/s2 iv. Calculate the distance traveled in the first 3 minutes of the journey. 32 = 1 2 ( + ) = 1 2 (17.15 + 0)32 = 274.4 m 148 = 1 2 ( + ) = 1 2 (17.15 + 17.15)148 = 2538.2 m 180 = 32 + 140 = 274.4 + 2538.2 = 2812.6 m
132 11. A car starts from rest and accelerates uniformly until it reaches a velocity of 130 m/s in 65 second. Its velocity is maintained for 30 seconds and then it decelerates to stops within 35 seconds. i. Sketch the graph of velocity versus time. ii. Calculate the total distance of the journey. iii. Calculate the average velocity for the whole journey. Answer: i. Sketch the graph of velocity versus time. ii. Calculate the total distance of the journey. 1 = 1 2 × 65 × 130 = 4225 m 2 = 130 × 30 = 3900 m 3 = 1 2 × 35 × 130 = 2275 m
133 = 1 + 2 + 3 = 4225 + 3900 + 2275 = 10400 m iii. Calculate the average velocity for the whole journey. = Total Distance/Total Time = 10400/130 = 80 m/s
134 12. A car starts from rest and accelerates uniformly for 20 seconds until it reaches 22 m/s at the end of acceleration. Then the car moves constantly for 40 seconds, after that it accelerates for 10 seconds until it reaches 30 m/s and stops in 30 seconds. i. Draw a velocity-time graph for this trip. ii. Calculate the second acceleration. iii. Calculate each of distance travelled by the car during first acceleration uniform velocity and deceleration. Answer: i. Draw a velocity-time graph for this trip. ii. Calculate the second acceleration. = + 30 = 22 + (10) 30 −22 = 10 = 8 10 = 0.8 m/s2
135 iii. Calculate each distance travelled by the car during first acceleration uniform velocity and deceleration. During first acceleration 1 = 1 2 × 20 × 22 = 220 During uniform velocity 2 = 40 × 20 = 800 m During deceleration 3 = 1 2 × 30 × 30 = 450
136 13. The movement of a particle is expressed in relation of = 3 − 5 2 + 15 + 40 where s and t are in meters and second respectively. When = 2, express the answer in mathematical method. i. Displacement ii. Velocity Answer: i. Displacement = (2) 3 − 5(2) 2 + 15(2) + 40 = 58 ii. Velocity = 3 − 5 2 + 15 + 40 = = 3 2 − 10 + 15 when = 2 = 3 2 − 10 + 15 = 3(2) 2 − (10)(2) + 15 = 7 /
137 14. A man starts his exercise buy running from rest and accelerates uniformly for 40 seconds and reaches a velocity of 15 m/s at the end of the acceleration. Its velocity is maintained for a while and then it stops within 60 seconds with a constant deceleration. The total distance travelled by that man is 2.3 km. i. Sketch a graph of velocity (v) against time (t). ii. Calculate the acceleration of that running man. iii. Calculate the total time taken for the whole exercise sessions. iv. Calculate the deceleration of that running man. Answer: i. Sketch a graph of velocity (v) against time (t). ii. Calculate the acceleration of that running man. = + 15 = 0 + 40 = 15 40 = 0.375 m/s2
138 iii. Calculate the total time taken for the whole exercise sessions. 1 = 1 2 × 40 × 15 = 300 m 2 = 15 3 = 1 2 × 60 × 15 = 450 m = 1 + 2 + 3 2300 = 300 + 15 + 450 15 = 1550 = 1550 15 = 103.33 iv. Calculate the deceleration of that running man. = + 0 = 15 + 60 = − 15 60 = 0.25 m/s2
139 15. A car with a velocity of 15 m/s accelerates to reach velocity of 80 m/s with an acceleration of 3.5 m/s2 . This velocity is maintained for 35 seconds before deceleration until it stops within 25 seconds i. Draw a velocity-time diagram for the trip. ii. Calculate the time taken during the acceleration. iii. Calculate the car deceleration. iv. Calculate total distance for the trip Answer: i. Draw a velocity-time diagram for the trip. ii. Calculate the time taken during the acceleration. = + = + . = − . = .