EQUILIBRIUM OF A PARTICLE, THE FREE-BODY DIAGRAM & COPLANAR FORCE SYSTEMS Today’s Objectives: Students will be able to : a) Draw a free-body diagram (FBD), and, b) Apply equations of equilibrium to solve a 2-D problem Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  In-Class Activities: ‡ Reading Quiz ‡ Applications ‡ What, Why, and How of a FBD ‡ Equations of Equilibrium ‡ Analysis of Spring and Pulleys ‡ Concept Quiz ‡ Group Problem Solving ‡ Attention Quiz © Pearson Education South Asia Pte Ltd 2013 All rights reserved READING QUIZ 1) When a particle is in equilibrium, the sum of forces acting on it equals _ (Choose the most appropriate answer) A) A constant D) A negative number B) A positive number E) An integer C) Zero 2) For a frictionless pulley and cable, tensions in the cable (T1 and T2) are related as _ A) T1 > T2 B) T1 = T2 C) T1 < T2 T1 D) T1 = T2 sin T T2 Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved APPLICATIONS The crane is lifting a load To decide if the straps holding the load to the crane hook will fail, you need to know the force in the straps How could you find the forces? Straps Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved APPLICATIONS (continued) For a spool of given weight, how would you find the forces in cables AB and AC? If designing a spreader bar like this one, you need to know the forces to make sure the rigging doesn’t fail Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved APPLICATIONS (continued) For a given force exerted on the boat’s towing pendant, what are the forces in the bridle cables? What size of cable must you use? Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved COPLANAR FORCE SYSTEMS (Section 3.3) This is an example of a 2-D or coplanar force system If the whole assembly is in equilibrium, then particle A is also in equilibrium To determine the tensions in the cables for a given weight of cylinder, you need to learn how to draw a freebody diagram and apply the equations of equilibrium Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved THE WHAT, WHY, AND HOW OF A FREE-BODY DIAGRAM (FBD) Free-body diagrams are one of the most important things for you to know how to draw and use for statics and other subjects! What? - It is a drawing that shows all external forces acting on the particle Why? - It is key to being able to write the equations of equilibrium—which are used to solve for the unknowns (usually forces or angles) Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved How? Imagine the particle to be isolated or cut free from its surroundings Show all the forces that act on the particle Active forces: They want to move the particle Reactive forces: They tend to resist the motion Identify each force and show all known magnitudes and directions Show all unknown magnitudes and / or directions as variables y FBD at A FB A FD A  x FC = 392.4 N (What is this?) Note : Cylinder mass = 40 Kg Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap 30˚ © Pearson Education South Asia Pte Ltd 2013 All rights reserved EQUATIONS OF 2-D EQUILIBRIUM FBD at A FD A y A FB Since particle A is in equilibrium, the net force at A is zero x So FB + FC + FD = 30˚ or A 6F=0 FC = 392.4 N FBDat atA A FBD In general, for a particle in equilibrium, F = or Fx i + Fy j = = 0i + 0j (a vector equation) Or, written in a scalar form, Fx = and Fy = These are two scalar equations of equilibrium (E-of-E) They can be used to solve for up to two unknowns Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved EQUATIONS OF 2-D EQUILIBRIUM (continued) y FBD at A A FDA 30˚ FB x FC = 392.4 N Note : Cylinder mass = 40 Kg Write the scalar E-of-E: + o Fx = FB cos 30º – FD = + n Fy = FB sin 30º – 392.4 N = Solving the second equation gives: FB = 785 N ĺ From the first equation, we get: FD = 680 N ĸ Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved SIMPLE SPRINGS Spring Force = spring constant * deformation of spring or F = k * s Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved CABLES AND PULLEYS With a frictionless pulley and cable T1 = T2 T1 T2 Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved EXAMPLE Given: The box weighs 550 N and geometry is as shown Find: The forces in the ropes AB and AC Plan: Draw a FBD for point A Apply the E-of-E to solve for the forces in ropes AB and AC Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved EXAMPLE (continued) y FB FC 30˚ FBD at point A A x FD = 550 N Applying the scalar E-of-E at A, we get; + o ¦ F x = FB cos 30° – FC (4/5) = + o ¦ F y = FB sin 30° + FC (3/5) - 550 N = Solving the above equations, we get; FB = 478 N and FC = 518 N Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved CONCEPT QUESTIONS 1000 N 1000 N (A) (B) 1000 N (C) 1) Assuming you know the geometry of the ropes, you cannot determine the forces in the cables in which system above? 2) 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 1000 N weight Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved GROUP PROBLEM SOLVING Given: The mass of lamp is 20 kg and geometry is as shown Find: The force in each cable Plan: Draw a FBD for Point D Apply E-of-E at Point D to solve for the unknowns (FCD & FDE) Knowing FCD, repeat this process at point C Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved GROUP PROBLEM SOLVING (continued) FBD at point D y FDE FCD 30˚ D x W = 20 (9.81) N Applying the scalar E-of-E at D, we get; +n ¦ Fy = FDE sin 30° – 20 (9.81) = +o ¦ Fx = FDE cos 30° – FCD = Solving the above equations, we get: FDE = 392 N and FCD = 340 N Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved GROUP PROBLEM SOLVING (continued) FBD at point C y FAC FCD =340 N C FBC x 45˚ Applying the scalar E-of-E at C, we get; +o ¦ Fx = 340 – FBC sin 45° – FAC (3/5) = + n ¦ Fy = FAC (4/5) – FBC cos 45° = Solving the above equations, we get; FBC = 275 N and FAC = 243 N Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved ATTENTION QUIZ Select the correct FBD of particle A 30q A 40q 100 N F1 A) A B) 100 N F2 30q 40ƒ A C) F 30ƒ D) A F2 F1 30ƒ 40ƒ A 100 N 100 N Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved EXAMPLE Given: Loads as shown on the truss Find: The force in members KJ, KD, and CD Plan: a) Take a cut through members KJ, KD and CD b) Work with the left part of the cut section Why? c) Determine the support reactions at A What are they? d) Apply the E-of-E to find the forces in KJ, KD and CD Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved EXAMPLE (continued) 56.7 kN Analyzing the entire truss for the reactions at A, we get FX = AX = A moment equation about G to find AY results in: ∑MG = AY (18) – 20 (15) – 30 (12) – 40 (9) = 0; AY = 56.7 kN Now take moments about point D Why this? + MD = – 56.7 (9) + 20 (6) + 30 (3) – FKJ (4) = FKJ = − 75.1 kN or 75.1 kN ( C ) Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved EXAMPLE (continued) 56.7 kN Now use the x and y-directions equations of equilibrium Ĺ + FY = 56.7 – 20 – 30 – (4/5) FKD = 0; FKD = 8.38 kN (T) ĺ + FX = (– 75.1) + (3/5) (8.38) + FCD = 0; FCD = 70.1 kN (T) Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved CONCEPT QUIZ Can you determine the force in member ED by making the cut at section a-a? Explain your answer A) No, there are unknowns B) Yes, using MD = C) Yes, using ME = D) Yes, using MB = Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved CONCEPT QUIZ (continued) If you know FED, how will you determine FEB? A) By taking section b-b and using ME = B) By taking section b-b, and using FX = and FY = C) By taking section a-a and using MB = D) By taking section a-a and using MD = Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved GROUP PROBLEM SOLVING Given: Loads as shown on the truss Find: The force in members GB and GF Plan: a) Take the cut through members GF, GB, and AB b) Analyze the left section Determine the support reactions at A Why? c) Draw the FBD of the left section d) Apply the equations of equilibrium (if possible, try to it so that every equation yields an answer to one unknown Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved GROUP PROBLEM SOLVING (continued) 1) Determine the support reactions at A by drawing the FBD of the entire truss +ĺ FX = AX = + MD = – AY (28) + 600 (18) + 800 (10) = 0; AY = 671.4 kN Why is Ax equal zero by inspection? Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved GROUP PROBLEM SOLVING (continued) 2) Analyze the left section + MB = – 671.4 (10) + FGF (10) = 0; FGF = 671 kN (C) Ĺ + FY = 671.4 – FGB = 0; FGB = 671 kN (T) Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved ATTENTION QUIZ As shown, a cut is made through members GH, BG, and BC to determine the forces in them Which section will you choose for analysis and why? A) Right, fewer calculations B) Left, fewer calculations C) Either right or left, same amount of work D) None of the above, too many unknowns Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved ATTENTION QUIZ When determining the force in member HG in the previous question, which one equation of equilibrium is the best one to use? A) MH = B) MG = C) MB = D) MC = Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap   © Pearson Education South Asia Pte Ltd 2013 All rights reserved FRAMES AND MACHINES Today’s Objectives: Students will be able to: a) Draw the free-body diagram of a frame or machine and its members In-Class Activities: ‡ Check Homework, if any b) Determine the forces acting at the joints and supports of a frame or machine ‡ Reading Quiz ‡ Applications ‡ Analysis of a Frame/Machine ‡ Concept Quiz ‡ Group Problem Solving ‡ Attention Quiz Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved READING QUIZ Frames and machines are different as compared to trusses since they have _ A) Only two-force members B) Only multiforce members C) At least one multiforce member D) At least one twoforce member Forces common to any two contacting members act with _ on the other member A) Equal magnitudes but opposite sense B) Equal magnitudes and the same sense C) Different magnitudes and the opposite sense D) Different magnitudes and the same sense Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved APPLICATIONS Frames are commonly used to support various external loads How is a frame different than a truss? To be able to design a frame, you need to determine the forces at the joints and supports Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved APPLICATIONS (continued) “Machines,” like those above, are used in a variety of applications How are they different from trusses and frames? How can you determine the loads at the joints and supports? These forces and moments are required when designing the machine’s members Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved FRAMES AND MACHINES: DEFINITIONS Frame Machine Frames and machines are two common types of structures that have at least one multi-force member (Recall that trusses have nothing but two-force members) Frames are generally stationary and support external loads Machines contain moving parts and are designed to alter the effect of forces Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved STEPS FOR ANALYZING A FRAME OR MACHINE Draw a FBD of the frame or machine and its members, as necessary Hints: a) Identify any two-force members, b) Note that forces on contacting surfaces (usually between a pin and a member) are equal and opposite, and FAB c) For a joint with more than two members or an external force, it is advisable to draw a FBD of the pin Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved STEPS FOR ANALYZING A FRAME OR MACHINE Develop a strategy to apply the equations of equilibrium to solve for the unknowns Look for ways to form single equations and single unknowns FAB Problems are going to be challenging since there are usually several unknowns A lot of practice is needed to develop good strategies and ease of solving these problems Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved EXAMPLE Given: The frame supports an external load and moment as shown Find: The horizontal and vertical components of the pin reactions at C and the magnitude of reaction at B Plan: a) Draw FBDs of the frame member BC Why pick this part of the frame? b) Apply the equations of equilibrium and solve for the unknowns at C and B Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved EXAMPLE (continued) 800 N.m 400 N CX CY 1m 1m B 2m FBD of member BC (Note AB is a 2-force member!) 45° FAB Please note that member AB is a two-force member Equations of Equilibrium: Start with ¦ MC since it yields one unknown + ¦ MC = FAB sin45° (1) – FAB cos45° (3) + 800 N m + 400 (2) = FAB = 1131 N Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved EXAMPLE (continued) 800 N.m 400 N CX CY 1m 1m 2m B 45° FAB FBD of member BC o + ¦ FX = – CX + 1131 sin 45° = CX = 800 N n + ¦ FY = – CY + 1131 cos 45° – 400 = CY = 400 N Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved CONCEPT QUIZ The figures show a frame and its FBDs If an additional couple moment is applied at C, how will you change the FBD of member BC at B? A) B) C) D) No change, still just one force (FAB) at B Will have two forces, BX and BY, at B Will have two forces and a moment at B Will add one moment at B Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved CONCEPT QUIZ (continued) xD The figures show a frame and its FBDs If an additional force is applied at D, then how will you change the FBD of member BC at B? A) B) C) D) No change, still just one force (FAB) at B Will have two forces, BX and BY, at B Will have two forces and a moment at B Will add one moment at B Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved GROUP PROBLEM SOLVING Given: A frame supports a 50-kN load as shown Find: The reactions exerted by the pins on the frame members at B and C Plan: a) b) Draw a FBD of member BC and another one for AC Apply the equations of equilibrium to each FBD to solve for the four unknowns Think about a strategy to easily solve for the unknowns Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  GROUP PROBLEM © Pearson Education South Asia Pte Ltd 2013 All rights reserved SOLVING (continued) FBDs of members BC and AC CY CX 50 kN 3.5 m 6m 8m AX Applying E-of-E to member AC: AY + ¦ MA = – CY (8) + CX (6) + 50 (3.5) = o+ ¦ FX = CX – AX = n+ ¦ FY = 50 – AY – CY = (1) 0 Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved GROUP PROBLEM SOLVING (continued) FBDs of members BC and AC CY CX 50 kN 3.5 m 6m 8m AX Applying E-of-E to member BC: AY + ¦ MB = – 50 (2) – 50 (3.5) + CY (8) = ; CY = 34.38 = 34.4 kN From Eq (1), CX can be determined; CX = 16.67 = 16.7 kN o+ ¦ FX = 16.67 + 50 – BX = ; BX = 66.7 kN n+ ¦ FY = BY – 50 + 34.38 = ; BY = 15.6 kN Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved ATTENTION QUIZ When determining the reactions at joints A, B, and C, what is the minimum number of unknowns in solving this problem? A) B) C) D) For the above problem, imagine that you have drawn a FBD of member BC What will be the easiest way to write an equation involving unknowns at B? A) ¦ MC = B) ¦ MB C) ¦ MA = D) ¦ FY = Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  = 0 © Pearson Education South Asia Pte Ltd 2013 All rights reserved Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved [...]... C A) F2 sin 50ƒ – 20 = 0 F1 B) F2 cos 50ƒ – 20 = 0 C) F2 sin 50ƒ – F1 20 N =0 D) F2 cos 50ƒ + 20 = 0 Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap © Pearson Education South Asia Pte Ltd 2013 All rights reserved Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap © Pearson Education South Asia Pte Ltd 2013 All rights reserved ... the above Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved APPLICATIONS You know the weight of the electromagnet and its load But, you need to know the forces in the chains to see if it is a safe assembly How would you do this? Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler. .. of a force, how many unknowns do you have corresponding to that force? A) One B) Two C) Three D) Four Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap © Pearson Education South Asia Pte Ltd 2013 All rights reserved Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap © Pearson Education South Asia Pte Ltd 2013 All rights reserved ... moments together Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved EXAMPLE I (continued) Solution + n Fy = – 100 (3/5) N + o Fx = 100 (4/5) N + MO = {– 100 (3/5)N (5 m) – (100)(4/5)N (2 m)} N·m = – 460 N·m or 460 N·m CW Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai... F = { 10 k } N, the moment r x F equals { _ } N·m A) 50 i B) 50 j D) – 50 j E) 0 C) –50 i Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap   © Pearson Education South Asia Pte... i u i = 0 Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved CROSS PRODUCT (continued) Also, the cross product can be written as a determinant Each component can be determined using 2 u 2 determinants Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap ... number Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved APPLICATIONS With the force P, a person is creating a moment MA using this flex-handle socket wrench Does all of MA act to turn the socket? How would you calculate an answer to this question? Mechanics for Engineers: Statics, 13th SI Edition. .. F Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved APPLICATIONS Beams are often used to bridge gaps in walls We have to know what the effect of the force on the beam will have on the supports of the beam What do you think is happening at points A and B? Mechanics for Engineers: Statics, 13th. .. analysis Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved GROUP PROBLEM SOLVING I (continued) y x Solution: + n Fy = 100 sin 30° N + o Fx = 100 cos 30° N + MA = {–(100 cos 30°)N (450 mm) – (100 sin 20°)N (125 mm)} = – 43.2464 N·mm = 43.2 N·m (clockwise or CW) Mechanics for Engineers: Statics, 13th. .. Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap  © Pearson Education South Asia Pte Ltd 2013 All rights reserved GROUP PROBLEM SOLVING II Given: The force and geometry shown Find: Moment of F about point A Plan: 1) Find F and rAC 2) Determine MA = rAC u F Mechanics for Engineers: Statics, 13th SI Edition R C Hibbeler and Kai Beng Yap