Engineering Mechanics: Statics in SI Units, 12e

Transcription

1 Engineering Mechanics: Statics in SI Units, 12e 3 Equilibrium of a Particle Chapter Objectives To introduce the concept of the free-body diagram for a particle To show how to solve particle equilibrium problems using the equations of equilibrium 1

2 Chapter Outline 1. Condition for the Equilibrium of a Particle 2. The Free-Body Diagram 3. Coplanar Systems 4. Three-Dimensional Force Systems 3.1 Condition for the Equilibrium of a Particle Particle at equilibrium if - At rest - Moving at a constant velocity Newton s first law of motion F = 0 where F is the vector sum of all the forces acting on the particle 2

3 3.1 Condition for the Equilibrium of a Particle Newton s second law of motion F = ma When the force fulfill Newton's first law of motion, ma = 0 a = 0 therefore, the particle is moving in constant velocity or at rest 3.2 The Free-Body Diagram (Continued) Best representation of all the unknown forces ( F) which acts on a body A sketch showing the particle free from the surroundings with all the forces acting on it Consider two common connections in this subject Spring Cables and Pulleys 3

4 3.2 The Free-Body Diagram (Continued) Spring Linear elastic spring: change in length is directly proportional to the force acting on it spring constant or stiffness; k: defines the elasticity of the spring Magnitude of force when spring is elongated or compressed F = k s 3.2 The Free-Body Diagram (Continued) Cables and Pulley Cables (or cords) are assumed negligible weight and cannot stretch Tension always acts in the direction of the cable Tension force must have a constant magnitude for equilibrium For any angle θ, the cable is subjected to a constant tension T 4

5 3.2 The Free-Body Diagram (Concluded) Procedure for Drawing a FBD 1. Draw outlined shape 2. Show all the forces -Active forces: particle in motion -Reactive forces: constraints that prevent motion 3. Identify each forces - Known forces with proper magnitude and direction -Letters used to represent magnitude and directions Example 3.1 The sphere has a mass of 6kg and is supported as. Draw a free-body diagram of the sphere, the cord CE and the knot at C. 5

6 FCBA FCBA FCBA FCBA FCD FCD FCE FCE= 58.9 N WE = 6kg 9.81 m/(s^2)= N 58.9 N FCBASin60 FCBACos60 FCBA FCD FCD FCE= 58.9 N WE = 6kg 9.81 m/(s^2)= N 58.9 N 6

7 FCBASin60 = FCE= 58.9 N ( 3)/2 Hence, FCBA= N FCD =FCBACos60 1/2 Hence, FCD = * 0.5 = N FCE= 58.9 N FCBA FCD Graphical Solution WE 7

8 Solution FBD at Sphere Two forces acting, weight and the force on cord CE. Weight of 6kg (9.81m/s 2 ) = 58.9N Cord CE Two forces acting: sphere and knot Newton s 3 rd Law: F CE is equal but opposite F CE and F EC pull the cord in tension For equilibrium, F CE = F EC Solution FBD at Knot 3 forces acting: cord CBA, cord CE and spring CD Important to know that the weight of the sphere does not act directly on the knot but subjected to by the cord CE 8

9 3.3 Coplanar Force Systems A particle is subjected to coplanar forces in the x-y plane Resolve into i and j components for equilibrium F x = 0 F y = 0 Scalar equations of equilibrium require that the algebraic sum of the x and y components to equal to zero 3.3 Coplanar Systems Procedure for Analysis 1. Free-Body Diagram - Establish the x, y axes - Label all the unknown and known forces 2. Equations of Equilibrium - Apply F = ks to find spring force - When negative result force is the reserve - Apply the equations of equilibrium F x = 0 F y = 0 9

10 Example 3.4 Determine the required length of the cord AC so that the 8kg lamp is suspended. The undeformed length of the spring AB is l AB = 0.4m, and the spring has a stiffness of k AB = 300N/m. Solution FBD at Point A Three forces acting, force by cable AC, force in spring AB and weight of the lamp. If force in spring AB is known, stretch of the spring can be found using F = ks. + F x = 0; T AB T AC cos30º = 0 + F y = 0; T AC sin30º 78.5N = 0 Solving, T AC = N T AB = N 10

11 Solution T AB = k AB s AB ; N = 300N/m*(s AB ) Hence, s AB = 0.453N For stretched length, ι AB =ι AB + s AB AB ι = 0.4m m = 0.853m For horizontal distance BC, 2m = ι AC cos m ι AC = 1.32m 1 11

12 Problem 3-5 y 8 kn 30 x 45 T F + ve 5 kn Fx = 0, T cos 30 8 kn -T cos kn + 5 kn sin 45 = 0, Hence, T = kn 13.3 kn + ve Fy = 0, - 5 kn cos 45 + F -T sin 30 = 0, Hence, F = 10.2 kn 5 kn sin 45 5 kn cos 45 F T sin 30 12

13 W = 2 * 9.81 N = N AB = [(3^2)+(4^2)]= 5 m 3 m C (1) F AC 3 m β θ A W AC = [(3^2)+(3^2)] = 3 2 m Sin θ = 4m/AB = 4m/5m = 0.8 Cos θ = 3m/AB = 3m/5m = 0.6 Sin β = 3m/AC = 3m/(3 2 m) = 1/ 2 Cos β = 3m/AC = 3m/(3 2 m) = 1/ 2 4 m F AB B (2) (3) (4) (5) + ve Fx = 0, F AB Sin θ -F AC Sin β = 0, + ve Fy = 0, F AC C F AC Sin β F AB Cos θ + F AC Cos β-w= 0, β θ F AB Cos θ F AC Cos β W F AB Substituting by the values of Sin θ & Sin β from Eqs. (2) and (4) in Eq. (6) to get y B F AB Sin θ 0.8 *F AB - F AC (1/ 2) = 0, Hence, F AC = 0.8 * 2 F AB, (6) (7) (8) x 13

14 Substituting by the values of W, Cos θ, Cos β & (F AC in terms of F AB ) from Eqs.(1), (3), (5) &(8) in Eq. (7) to get F AB * (0.6)+ (0.8 * 2 F AB )* 1/ N = 0, Hence, (1.4 * F AB ) = N, i.e., F AB = (19.62 /1.4) = N Hence, F AC = 0.8 * 2 F AB = 0.8 * 2 * N = N, i.e., & F AB N F AC N 14

15 A 30 B 45 FBD 45 F BD Sin 45 F C CD Sin F CD 30 F BD Cos 45 D F CD Cos 30 W Point D; Equation of Equilibrium + ve Fx = 0, F CD Cos 30 -F BD Cos 45 = 0, (9) F CD ( 3 /2) - F BD (1/ 2) = 0, Hence, F BD = ( 3/2) F CD = F CD (10) + ve Fy = 0, F BD Sin 45 + F CD Sin 30 - W = 0, Substituting by F BD in terms of F CD from Eq. (10) in Eq. (11) to get (1.225F CD ) (1/ 2)+0.5*F CD -W= 0, (11) [(1.225 / 2 )+ 0.5 ]*F CD -W= 0, Hence, F CD = *W, (12) 15

16 m = 50 kg, Hence, W = mg = 50*9.81= N = 0.49 kn, Hence, F CD = *W = * 0.49 kn, i.e., F CD = kn, (12 ) Substituting by F CD from Eq. (12) in Eq. (10) to get Hence, F BD = * *W i.e.,f BD = *W (13) i.e.,f BD = * 0.49 kn = kn (13 ) A F AB Sin 30 F AB Cos B F BD Cos F BD F BC 30 C F BD Sin 45 D Point B; Equation of Equilibrium + ve Fy = 0, F AB Sin 30 -F BD Sin 45 = 0, (14) 16

17 i.e., F AB = (Sin 45 / Sin 30 ) F BD = F BD (15) Substituting by F BD in terms of W from Eq. (13) in Eq. (15) to get: F AB = * ( *W )= W + ve (16) Fx = 0, F BD Cos 45 +F BC -F AB Cos 30 = 0, Hence, F BC = F AB Cos 30 - F BD Cos 45 From Eq. (16) From Eq. (13) Hence, F BC = ( W ) Cos 30 - ( *W) Cos 45 i.e., F BC = ( W ) Cos 30 - ( *W) Cos 45 i.e., F BC = W (17) i.e., F AB = W = * 0.49 kn = kn i.e., (16 ) i.e., F BC = W= = *(0.49 kn) = kn (17 ) 17

18 Examples and Problems Review the following examples: 3.2 and 3.3 Resolve the following problems for Homework: 3.27, 3.28 and Three-Dimensional Force Systems For particle equilibrium F = 0 Resolving into i, j, k components F x i + F y j + F z k = 0 Three scalar equations representing algebraic sums of the x, y, z forces F x i = 0 F y j = 0 F z k = 0 18

19 3.4 Three-Dimensional Force Systems Procedure for Analysis Free-body Diagram - Establish the x, y, z axes - Label all known and unknown force magnitudes and directions on the digram Equations of Equilibrium - Apply F x = 0, F y = 0 and F z = 0 - Substitute vectors into F = 0 and set i, j, k components = 0 - Negative results indicate that the sense of the force is opposite to that shown in the FBD. Example 3.7: Determine the force developed in each cable used to support the 40kN crate. F B F C F D 19

20 Solution FBD at Point A To expose all three unknown forces in the cables. Equations of Equilibrium Expressing each forces in Cartesian vectors, F B = F B (r B / r B ) r B = (3m) 2 +(8m) 2 +(4m) 2 = m r B = AB x i + AB y j + AB z k = (-3m) i + (-4m) j + (8m) k Hence, F B = F B [{(-3m) i + (-4m) j + (8m) k} / m ] 20

21 i.e. F B = F B i 0.424F B j F B k Similarly, F C = F C (r C / r C ) r C = (3m) 2 +(8m) 2 +(4m) 2 = m r C = AC x i + AC y j + AC z k = (-3m) i + (4m) j + (8m) k Hence, F C = F C [{(-3m) i + (4m) j + (8m) k} / m ] i.e. F C = F C i F C j F C k F D = F D i W = -40 k For equilibrium, F = 0; F B + F C + F D + W = 0 (-0.318F B i 0.424F B j F B k) + ( F C i F C j F C k) + F D i 40 k = 0 F x = 0; F B F C + F D = 0 F y = 0; 0.424F B F C = 0 F z = 0; 0.848F B F C - 40 = 0 (18) (19) (20) From Eq. (19), F C =F B (21) Substituting by F C in terms of F B from Eq. (21) in Eq. (20) to get : 21

22 2* 0.848F B - 40 kn = 0 Hence, F B = F C = 20/0.848 kn = 23.6kN (22) Substituting by F B & F C from Eq. (22) in Eq. (18) to get : *23.6kN * 23.6kN + F D = 0 i.e. F D = 2* 0.318* 23.6kN =15.0kN (23) 22

23 Solution FBD at Point A To expose all three unknown forces in the cables. Equations of Equilibrium Expressing each forces in Cartesian vectors, F B = F B (r B / r B ) r B = (2 ft) 2 +(1 ft) 2 +(2 ft) 2 = 3.0 ft r B = AB x i + AB y j + AB z k = (- 2 ft) i + (1 ft) j + (2 ft) k Hence, F B = F B [{(- 2 ft) i + 1 ft) j + (2 ft) k} / 3.0 ft ] 23

24 i.e., F B = F B i F B j F B k Similarly, F C = F C (r C / r C ) r C = (2 ft) 2 +(2 ft) 2 +(1 ft) 2 = 3.0 ft r C = AC x i + AC y j + AC z k = (- 2 ft) i + (-2 ft) j + (1 ft) k Hence, F C = F C [{(- 2 ft) i + (-2 ft) j + (1 ft) k} / 3.0 ft ] i.e. F C = F C i F C j F C k F D = F D i W = -300 k For equilibrium, F = 0; [F B ] + (F C ) + F D + W = 0 [ F B i F B j F B k] + ( F C i F C j F C k) + F D i 300 k = 0 F x = 0; F B F C + F D = 0 (24) F y = 0; F B F C = 0 (25) F z = 0; F B F C = 0 (26) From Eq. (25), F C = 0.5 F B (27) Substituting by F C in terms of F B from Eq. (27) in Eq. (26) to get : F B + (0.333 * 0.5 F B ) = 0 24

25 i.e., F B + (0.333 * 0.5 F B ) = 0, Hence, F B = Ib, (28) Hence, From Eq. (27), F C = Ib (29) Substituting by F B &F C from Eqs. (28) & (29) in Eq. (24) to get : * Ib * Ib + F D = 0, Hence, F D = Ib, (30) Example 3.5: 25

26 Solution FBD at Point A To expose all three unknown forces in the cables. Equations of Equilibrium Expressing each forces in Cartesian vectors, F C = F C (r C / r C ) F C = F C [Cos α i + Cos β j + Cos γ k ] C =F C [(- 4/5) i + (0) j + (3/5) k] =F C [- 0.8 i k] =-0.8F C i + 0.6F C k F D = F D (r D / r D ) F D = F D [Cos α i + Cos β j + Cos γ k ] D =F D [Cos 60 i - Cos 30 j + 0 k] =F D [(1/2) i - ( 3/2) j] = 0.5F D i F D j W = -90 k F B = F B j For equilibrium, F = 0; F B + (F C ) + [F D ]+ W = 0 F B j + (- 0.8F C i + 0.6F C k) + [0.5F D i F D j ] 90 k = 0 F x = 0; - 0.8F C +0.5 F D = 0 (31) F y = 0; F B F D = 0 (32) F z = 0; 0.6 F C - 90 = 0 (33) 26

27 Fron Eq. (33), F C = 90/0.6=150 Ib (34) Fron Eq. (31), F D =1.6 F C = 240 Ib (35) Fron Eq. (32), F B =0.866 F D = Ib (36) But, F B =k AB x AB Hence, Ib = (500 Ib/ft) x AB, i.e. x AB = /500 = ft. 27

28 Solution FBD at Point A To expose all three unknown forces in the cords. Equations of Equilibrium Expressing each forces in Cartesian vectors, F C = F C (r C / r C ) F C = F C [Cos α i + Cos β j + Cos γ k ] C =F C [Cos 120 i + Cos 135 j + Cos 60 k ] =F C [- 0.5 I j k] =-0.5 F C i F C j + 0.5F C k Similarly, F D = F D (r D / r D ) r D = (1 m) 2 +(2 m) 2 +(2 m) 2 = 3.0 m r D = AD x i + AD y j + AD z k = (- 1 m) i + (2 m) j + (2 m) k Hence, F D = F D [{(- 1 ft) i + (2 ft) j + (2 ft) k} / 3.0 ft ] i.e. F D = F D i F D j F D k F B = F B i W = -mg = -100*9.81 k = kn k For equilibrium, F = 0; F B + (F C ) + [F D ]+ W = 0 [F B i] + (- 0.5 F C i F C j + 0.5F C k) + [ F D i F D j F D k ] kn k = 0 F x = 0; F B F C F D = 0 (37) F y = 0; F C F D = 0 (38) F z = 0; 0.5 F C F D kn = 0 (39) 28

29 From Eq. (38), F D = (0.708/ 0.666)F C = F C (38 ) Substituting by F D in terms of F C from Eq. (38 ) in Eq. (39) to get, 0.5 F C F D =0.981 kn i.e., 0.5 F C * F C =0.981 kn i.e., F C = kn (40) Substituting by F C from Eq. (40) in Eq. (38 ) to get, F D = * kn= kn (41) Substituting by F C & F D from Eqs. (40) & (41) in Eq. (37) to get F B = 0.5 F C F D = 0.5* kn * kn i.e., F B = kn (42) Examples and Problems Review the following examples: 3.5 and 3.8 Resolve the following problems for Homework: 3.45, 3.52 and 3.63 Test No (1) Tuesday 4/1/1436 h at AM. Chapters, 1,2 and 3. 29

30 QUIZ 1. When a particle is in equilibrium, the sum of forces acting on it equals. (Choose the most appropriate answer) A) A constant B) A positive number C) Zero D) A negative number E) An integer 2. For a frictionless pulley and cable, tensions in the cables are related as A) T 1 > T 2 B) T 1 = T 2 C) T 1 < T 2 D) T 1 = T 2 sin θ T 1 T 2 QUIZ 100 N 100 N 100 N ( A ) ( B ) ( C ) 3. Assuming you know the geometry of the ropes, you cannot determine the forces in the cables in which system above? 4. Why? A) The weight is too heavy. B) The cables are too thin. C) There are more unknowns than equations. D) There are too few cables for a 100 kg weight. 30

31 QUIZ 5. Select the correct FBD of particle A. 30 A kg A) C) 30 A 100 kg F A B) D) F1 F A A F2 40 F kg 100 kg QUIZ 6. Using this FBD of Point C, the sum of forces in the x- direction (Σ FX) is. Use a sign convention of +. A) F 2 sin = 0 B) F 2 cos = 0 C) F 2 sin 50 F 1 = 0 D) F 2 cos = 0 31

32 QUIZ 7. Particle P is in equilibrium with five (5) forces acting on it in 3-D space. How many scalar equations of equilibrium can be written for point P? A)2 B) 3 C) 4 D) 5 E) 6 8. In 3-D, when a particle is in equilibrium, which of the following equations apply? A) (Σ F x ) i + (Σ F y ) j + (Σ F z ) k = 0 B) ΣF = 0 C) Σ F x = ΣF y = ΣF z = 0 D) All of the above. E) None of the above. QUIZ 9. In 3-D, when you know the direction of a force but not its magnitude, how many unknowns corresponding to that force remain? A) One B) Two C) Three D) Four 10. If a particle has 3-D forces acting on it and is in static equilibrium, the components of the resultant force. A) have to sum to zero, e.g., -5i+ 3 j + 2 k B) have to equal zero, e.g., 0i+ 0 j + 0 k C) have to be positive, e.g., 5i+ 5j + 5 k D) have to be negative, e.g., -5 i - 5 j - 5 k 32

33 QUIZ 11. Four forces act at point A and point A z is in equilibrium. Select the correct force vector P. P A) {-20 i + 10 j 10 k}lb A B) {-10 i 20 j 10 k} lb C) {+ 20 i 10 j 10 k}lb D) None of the above. x 12. In 3-D, when you don t know the direction or the magnitude of a force, how many unknowns do you have corresponding to that force? A) One B) Two C) Three D) Four F 3 = 10 N F 2 = 10 N y 33