Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Chapter Objectives Concept of the free-body diagram for a particle Solve particle equilibrium problems using the equations of equilibrium Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Chapter Outline Condition for the Equilibrium of a Particle The Free-Body Diagram Coplanar Systems Three-Dimensional Force Systems Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Example 3.1 The sphere has a mass of 6kg and is supported. Draw a free-body diagram of the sphere, the cord CE and the knot at C. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd FCBA FCBA FCD FCBA FCD FCBA FCE= 58.9 N FCE WE = 6kg ×9.81 m/(s^2)= 58.86 N 58.9 N Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd FCBA Sin 60 FCBA FCD FCD FCBA Cos 60 FCE= 58.9 N WE = 6kg ×9.81 m/(s^2)= 58.86 N 58.9 N Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd FCBA Sin 60 = FCE= 58.9 N (3)/2 Hence, FCBA = 67.96 N FCD =FCBA Cos 60 1/2 Hence, FCD = 67.96 * 0.5 = 33.98 N Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd FCBA FCE= 58.9 N FCD WE Graphical Solution Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution FBD at Sphere Two forces acting, weight and the force on cord CE. Weight of 6kg (9.81m/s2) = 58.9N Cord CE Two forces acting: sphere and knot Newton’s 3rd Law: FCE is equal but opposite FCE and FEC pull the cord in tension For equilibrium, FCE = FEC Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 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 Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 3.3 Coplanar Systems A particle is subjected to coplanar forces in the x-y plane Resolve into i and j components for equilibrium ∑Fx = 0 ∑Fy = 0 Scalar equations of equilibrium require that the algebraic sum of the x and y components to equal to zero Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 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 ∑Fx = 0 ∑Fy = 0 Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 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 kAB = 300N/m. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution FBD at Point A Three forces acting, force by cable AC, force in spring AB and weight of the lamp. If force on cable AB is known, stretch of the spring is found by F = ks. +→ ∑Fx = 0; TAB – TAC cos30º = 0 +↑ ∑Fy = 0; TAC sin30º – 78.5N = 0 Solving, TAC = 157.0 N TAB = 135.9 N Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution TAB = kAB sAB ; 135.9 N = 300N/m*(sAB) Hence, sAB = 0.453N For stretched length, AB = ′AB+ sAB AB = 0.4m + 0.453m = 0.853m For horizontal distance BC, 2m = AC cos30°+ 0.853m AC = 1.32m Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Problem 3-5 y 8 kN 30 30 x 45 T F 5 kN + ve Fx = 0, 8 kN T cos 30 5 kN sin 45 -T cos 30+8 kN + 5 kN sin 45 = 0, Hence, T = 13.32 kN  13.3 kN 5 kN cos 45 + ve F Fy = 0, - 5 kN cos 45 + F -T sin 30 = 0, Hence, F = 10.2 kN T sin 30 Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 3 m 4 m C B W = 2 * 9.81 N = 19.62 N 3 m (1)   FAC FAB AB = [(3^2)+(4^2)]= 5 m A W AC = [(3^2)+(3^2)] = 32 m Sin  = 4m/AB = 4m/5m = 0.8 (2) Cos  = 3m/AB = 3m/5m = 0.6 (3) Sin  = 3m/AC = 3m/(32 m) = 1/2 (4) Cos  = 3m/AC = 3m/(32 m) = 1/2 (5) Copyright © 2010 Pearson Education South Asia Pte Ltd

Fx = 0, Fy = 0, FAB Cos  FAC Cos  FAB FAC FAC Sin  FAB Sin  W + ve x Fx = 0, FAC Sin  FAB Sin  W FAB Sin  - FAC Sin  = 0, (6) + ve Fy = 0, FAB Cos  + FAC Cos  - W = 0, (7) Substituting by the values of Sin  & Sin  from Eqs. (2) and (4) in Eq. (6) to get 0.8 *FAB - FAC (1/2) = 0, Hence, FAC = 0.8 * 2 FAB , (8) Copyright © 2010 Pearson Education South Asia Pte Ltd Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Substituting by the values of W, Cos  , Cos  & (FAC in terms of FAB ) from Eqs. (1), (3), (5) & (8) in Eq. (7) to get FAB * (0.6) + (0.8 * 2 FAB )* 1/2 - 19.62 N = 0, Hence, (1.4 * FAB ) = 19.62 N, i.e., FAB = (19.62 /1.4) = 14.014 N Hence, FAC = 0.8 * 2 FAB = 0.8 * 2 * 14.014 N = 12.2815 N, i.e., FAB  14.01 N & FAC  12.28 N Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd FBDSin 45 A 30 C FCDSin 30 30 B 45 FCD FBD 30 45 D FBDCos 45 FCDCos 30 W Point D; Equation of Equilibrium + ve Fx = 0, FCD Cos 30 - FBD Cos 45 = 0, (9) Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd FCD (3 /2) - FBD (1/2) = 0, (10) Hence, FBD = (3/2) FCD = 1.225 FCD + ve Fy = 0, FBD Sin 45+ FCD Sin 30- W = 0, (11) Substituting by FBD in terms of FCD from Eq. (10) in Eq. (11) to get (1.225 FCD ) (1/2)+0.5 *FCD - W = 0, [(1.225 /2 )+ 0.5 ]*FCD - W = 0, FCD = 0.731954 *W , (12) Hence, Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd m = 50 kg, Hence, W = mg = 50*9.81= 490.5 N = 0.49 kN , FCD = 0.731954*W = 0.731954* 0.49 kN , Hence, FCD = 0.359 kN , (12′) i.e., Substituting by FCD from Eq. (12) in Eq. (10) to get Hence, FBD = 1.225 * 0.731954*W i.e., FBD = 0.896644*W (13) i.e., FBD = 0.896644* 0.49 kN = 0.439 kN (13′) Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd FAB Sin 30 A FBC 30 FBD Cos 45 30 B 45 FAB Cos 30 C FBD FBD Sin 45 D Point B; Equation of Equilibrium + ve Fy = 0, FAB Sin 30 - FBD Sin 45 = 0, (14) Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd i.e., FAB = (Sin 45 / Sin 30) FBD = 1.414214 FBD (15) Substituting by FBD in terms of W from Eq. (13) in Eq. (15) to get: FAB = 1.414214 * (0.896644*W )= 1.268046 W (16) + ve Fx = 0, FBD Cos 45 + FBC - FAB Cos 30 = 0, Hence, FBC = FAB Cos 30 - FBD Cos 45 From Eq. (16) From Eq. (13) Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Hence, FBC = (1.268046 W ) Cos 30 - (0.896644*W) Cos 45 i.e., FBC = (1.268046 W ) Cos 30 - (0.896644*W) Cos 45 i.e., FBC = 0.464137 W (17) i.e., FAB = 1.268046 W = 1.268046 * 0.49 kN = 0.621 kN (16′) i.e., i.e., FBC = 0.464137 W= = 0.464137*(0.49 kN) = 0.2274 kN (17′) Copyright © 2010 Pearson Education South Asia Pte Ltd

3.4 Three-Dimensional Force Systems For particle equilibrium ∑F = 0 Resolving into i, j, k components ∑Fx i + ∑Fy j + ∑Fz k = 0 Three scalar equations representing algebraic sums of the x, y, z forces ∑Fx i = 0 ∑Fy j = 0 ∑Fz k = 0 Copyright © 2010 Pearson Education South Asia Pte Ltd

3.4 Three-Dimensional Force Systems Procedure for Analysis Free-body Diagram Establish the z, y, z axes Label all known and unknown force Equations of Equilibrium Apply ∑Fx = 0, ∑Fy = 0 and ∑Fz = 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. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Example 3.7: Determine the force developed in each cable used to support the 40kN crate. FB FC FD Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution FBD at Point A To expose all three unknown forces in the cables. Equations of Equilibrium Expressing each forces in Cartesian vectors, FB = FB (rB / rB) rB = √(3m) 2 +(8m) 2 +(4m) 2 = 9.434 m rB = ABx i + ABy j + ABz k = (-3m) i + (-4m) j + (8m) k Hence, FB = FB [{(-3m) i + (-4m) j + (8m) k} / 9.434 m ] Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd i.e. FB = -0.318FB i – 0.424FB j + 0.848FB k Similarly, FC = FC (rC / rC) rC = √(3m) 2 +(8m) 2 +(4m) 2 = 9.434 m rC = ACx i + ACy j + ACz k = (-3m) i + (4m) j + (8m) k Hence, FC = FC [{(-3m) i + (4m) j + (8m) k} / 9.434 m ] i.e. FC = -0.318FC i + 0.424FC j + 0.848FC k FD = FD i W = -40 k Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd For equilibrium, ∑F = 0; FB + FC + FD + W = 0 (-0.318FB i – 0.424FB j + 0.848FB k) + (- 0.318FC i + 0.424FC j + 0.848FC k) + FD i – 40 k = 0 ∑Fx = 0; -0.318FB - 0.318FC + FD = 0 ∑Fy = 0; – 0.424FB + 0.424FC = 0 ∑Fz = 0; 0.848FB + 0.848FC - 40 = 0 From Eq. (19), FC =FB (21) Substituting by FC in terms of FB from Eq. (21) in Eq. (20) to get: (18) (19) (20) Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 2* 0.848FB - 40 kN = 0 Hence, FB = FC = 20/0.848 kN = 23.6kN (22) Substituting by FB & FC from Eq. (22) in Eq. (18) to get: -0.318*23.6kN - 0.318* 23.6kN + FD = 0 i.e. FD = 2* 0.318* 23.6kN =15.0kN (23) Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution FBD at Point A To expose all three unknown forces in the cables. Equations of Equilibrium Expressing each forces in Cartesian vectors, FB = FB (rB / rB) rB = √(2 ft) 2 +(1 ft) 2 +(2 ft) 2 = 3.0 ft rB = ABx i + ABy j + ABz k = (- 2 ft) i + (1 ft) j + (2 ft) k Hence, FB = FB [{(- 2 ft) i + 1 ft) j + (2 ft) k} / 3.0 ft ] Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd i.e., FB = -0.666 FB i + 0.333 FB j + 0.666 FB k Similarly, FC = FC (rC / rC) rC = √(2 ft) 2 +(2 ft) 2 +(1 ft) 2 = 3.0 ft rC = ACx i + ACy j + ACz k = (- 2 ft) i + (-2 ft) j + (1 ft) k Hence, FC = FC [{(- 2 ft) i + (-2 ft) j + (1 ft) k} / 3.0 ft ] i.e. FC = -0.666 FC i - 0.666 FC j + 0.333 FC k FD = FD i W = -300 k Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd For equilibrium, ∑F = 0; [FB] + (FC) + FD + W = 0 [-0.666 FB i + 0.333 FB j + 0.666 FB k] + (-0.666 FC i - 0.666 FC j + 0.333 FC k) + FD i – 300 k = 0 ∑Fx = 0; -0.666FB -0.666 FC + FD = 0 (24) ∑Fy = 0; 0.333 FB - 0.666 FC = 0 (25) ∑Fz = 0; 0.666 FB + 0.333 FC - 300 = 0 (26) From Eq. (25), FC = 0.5 FB (27) Substituting by FC in terms of FB from Eq. (27) in Eq. (26) to get: 0.666 FB + (0.333 * 0.5 FB ) - 300 = 0 Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd i.e., 0.666 FB + (0.333 * 0.5 FB ) - 300 = 0, Hence, FB = 360.36 Ib, (28) Hence, From Eq. (27), FC = 180.18 Ib (29) Substituting by FB &FC from Eqs. (28) & (29) in Eq. (24) to get: -0.666 * 360.36 Ib -0.666 * 180.18 Ib + FD = 0, Hence, FD = 360.36 Ib, (30) Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Example 3.5: Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution FBD at Point A To expose all three unknown forces in the cables. Equations of Equilibrium Expressing each forces in Cartesian vectors, FC = FC (rC / rC) FC = FC [Cos  i + Cos  j + Cos  k ]C = FC [(- 4/5) i + (0) j + (3/5) k] = FC [- 0.8 i + 0.6 k] = - 0.8 FC i + 0.6 FC k Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd FD = FD (rD / rD) FD = FD [Cos  i + Cos  j + Cos  k ]D = FD [Cos 60 i - Cos 30 j + 0 k] = FD [(1/2) i - (√3/2) j] = 0.5 FD i - 0.866 FD j W = -90 k FB = FB j For equilibrium, ∑F = 0; FB + (FC) + [FD ]+ W = 0 FB j + (- 0.8 FC i + 0.6 FC k) + [0.5 FD i - 0.866 FD j ] – 90 k = 0 ∑Fx = 0; - 0.8 FC +0.5 FD = 0 (31) ∑Fy = 0; FB - 0.866 FD = 0 (32) ∑Fz = 0; 0.6 FC - 90 = 0 (33) Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Fron Eq. (33), FC = 90/0.6=150 Ib (34) Fron Eq. (31), FD = 1.6 FC = 240 Ib (35) Fron Eq. (32), FB = 0.866 FD = 207.84 Ib (36) But, FB = kAB xAB Hence, 207.84 Ib = (500 Ib/ft) xAB , i.e. xAB = 207.84/500 = 0.416 ft. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Solution FBD at Point A To expose all three unknown forces in the cords. Equations of Equilibrium Expressing each forces in Cartesian vectors, FC = FC (rC / rC) FC = FC [Cos  i + Cos  j + Cos  k ]C = FC [Cos 120 i + Cos 135 j + Cos 60 k ] = FC [- 0.5 I -0.708 j + 0.5 k] = - 0.5 FC i - 0.708 FC j + 0.5 FC k Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd Similarly, FD = FD (rD / rD) rD = √(1 m) 2 +(2 m) 2 +(2 m) 2 = 3.0 m rD = ADx i + ADy j + ADz k = (- 1 m) i + (2 m) j + (2 m) k Hence, FD = FD [{(- 1 ft) i + (2 ft) j + (2 ft) k} / 3.0 ft ] i.e. FD = -0.333 FD i + 0.666 FD j + 0.666 FD k FB = FB i W = -mg = -100*9.81 k = -0.981 kN k For equilibrium, ∑F = 0; FB + (FC) + [FD ]+ W = 0 [FB i] + (- 0.5 FC i - 0.708 FC j + 0.5 FC k) + [-0.333 FD i + 0.666 FD j + 0.666 FD k ] -0.981 kN k = 0 ∑Fx = 0; FB - 0.5 FC- 0.333 FD = 0 (37) ∑Fy = 0; - 0.708 FC + 0.666 FD = 0 (38) ∑Fz = 0; 0.5 FC + 0.666 FD -0.981 kN = 0 (39) Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd From Eq. (38), FD = (0.708/ 0.666) FC = 1.063 FC (38′) Substituting by FD in terms of FC from Eq. (38′) in Eq. (39) to get, 0.5 FC + 0.666 FD =0.981 kN i.e., 0.5 FC + 0.666 * 1.063 FC =0.981 kN i.e., FC = 0.812 kN (40) Substituting by FC from Eq. (40) in Eq. (38′) to get, FD = 1.063 * 0.812 kN= 0.863 kN (41) Substituting by FC & FD from Eqs. (40) & (41) in Eq. (37) to get FB = 0.5 FC + 0.333 FD = 0.5* 0.812 kN + 0.333* 0.863 kN i.e., FB = 0.6934 kN (42) Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 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) T1 > T2 B) T1 = T2 C) T1 < T2 D) T1 = T2 sin  T1 T2 Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 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. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ A 40 100 kg 30 5. Select the correct FBD of particle A. 30 A) A 100 kg B) 40° F1 F2 C) 30° F F1 F2 D) Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 6. Using this FBD of Point C, the sum of forces in the x-direction ( FX) is ___ . Use a sign convention of +  . A) F2 sin 50° – 20 = 0 B) F2 cos 50° – 20 = 0 C) F2 sin 50° – F1 = 0 D) F2 cos 50° + 20 = 0 Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 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) ( Fx) i + ( Fy) j + ( Fz) k = 0 B)  F = 0 C)  Fx =  Fy =  Fz = 0 D) All of the above. E) None of the above. Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd 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., -5 i + 3 j + 2 k B) have to equal zero, e.g., 0 i + 0 j + 0 k C) have to be positive, e.g., 5 i + 5 j + 5 k D) have to be negative, e.g., -5 i - 5 j - 5 k Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 11. Four forces act at point A and point A is in equilibrium. Select the correct force vector P. A) {-20 i + 10 j – 10 k}lb B) {-10 i – 20 j – 10 k} lb C) {+ 20 i – 10 j – 10 k}lb D) None of the above. 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 z F3 = 10 N P x A F2 = 10 N y Copyright © 2010 Pearson Education South Asia Pte Ltd