EQUILIBRIUM OF A PARTICLE IN 2-D Today’s Objectives:
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1 EQUILIBRIUM OF A PARTICLE IN 2-D 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.In-Class Activities:Reading quizApplicationsWhat, why and how of a FBDEquations of equilibriumAnalysis of spring and pulleysConcept quizGroup problem solvingAttention quizStatics:The Next Generation Mehta & Danielson Lecture Notes for Sections
2 READING QUIZ1) 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) zeroD) a negative number E) an integer.2) For a frictionless pulley and cable, tensions in the cable (T1 and T2) are related as _____ .A) T1 > T2B) T1 = T2C) T1 < T2D) T1 = T2 sin Answers:1. C2. BStatics:The Next Generation Mehta & Danielson Lecture Notes for Sections
3 APPLICATIONSFor a spool of given weight, what are the forces in cables AB and AC ?
4 APPLICATIONS (continued) For a given cable strength, what is the maximum weight that can be lifted ?
5 APPLICATIONS (continued) For a given weight of the lights, what are the forces in the cables? What size of cable must you use ?
6 EQUILIBRIUM OF PARTICLE IN 2-D (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 the engine, we need to learn how to draw a free body diagram and apply equations of equilibrium.Explain the following terms :Coplanar force system – The forces lie on a sheet of paper (plane).A particle – It has a mass, but a size that can be neglected.Equilibrium – A condition of rest or constant velocity.Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections
7 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.What ? - It is a drawing that shows all external forces acting on the particle.Why ? - It helps you write the equations of equilibrium used to solve for the unknowns (usually forces or angles).
8 2. Show all the forces that act on the particle. 1. Imagine the particle to be isolated or cut free from its surroundings.2. 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.3. Identify each force and show all known magnitudes and directions. Show all unknown magnitudes and / or directions as variables .Mainly explain the three steps using the example .ANote : Engine mass = 250 KgFBD at AStatics:The Next Generation Mehta & Danielson Lecture Notes for Sections
9 EQUATIONS OF 2-D EQUILIBRIUM Since particle A is in equilibrium, the net force at A is zero.So FAB + FAB + FAC = 0or F = 0AFBD at AIn general, for a particle in equilibrium, F = 0 orFx i + Fy j = = 0 i j (A vector equation)Or, written in a scalar form,Fx = 0 and Fy = 0These are two scalar equations of equilibrium (EofE). They can be used to solve for up to two unknowns.
10 Solving the second equation gives: TB = 4.90 kN EXAMPLENote : Engine mass = 250 KgFBD at AWrite the scalar EofE:+ Fx = TB cos 30º – TD = 0+ Fy = TB sin 30º – kN = 0You may tell students that they should know how to solve linear simultaneous equations by hand and also using calculator . If they need help they can see you .Solving the second equation gives: TB = 4.90 kNFrom the first equation, we get: TD = 4.25 kNStatics:The Next Generation Mehta & Danielson Lecture Notes for Sections
11 SPRINGS, CABLES, AND PULLEYS Spring Force = spring constant * deformation, orF = k * SWith a frictionless pulley, T1 = T2.
12 Given: Sack A weighs 20 lb. and geometry is as shown. EXAMPLEGiven: Sack A weighs 20 lb. and geometry is as shown.Find: Forces in the cables and weight of sack B.Plan:1. Draw a FBD for Point E.2. Apply EofE at Point E to solve for the unknowns (TEG & TEC).3. Repeat this process at C.You may ask the students to give a planYou may explain why analyze at E, first, and C, laterStatics:The Next Generation Mehta & Danielson Lecture Notes for Sections
13 + Fx = TEG sin 30º – TEC cos 45º = 0 EXAMPLE (continued)A FBD at E should look like the one to the left. Note the assumed directions for the two cable tensions.The scalar EofE are:+ Fx = TEG sin 30º – TEC cos 45º = 0+ Fy = TEG cos 30º – TEC sin 45º – 20 lbs = 0Solving these two simultaneous equations for the two unknowns yields:TEC = 38.6 lbTEG = 54.6 lbYou may ask students to give equations of equilibrium.students hand-outs should be without the equations.Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections
14 Fy = (3/5) TCD + 38.64 sin 45 – WB = 0 EXAMPLE (continued)Now move on to ring C. A FBD for C should look like the one to the left.The scalar EofE are: Fx = cos 45 – (4/5) TCD = 0 Fy = (3/5) TCD sin 45 – WB = 0You may ask the students to give equations of equilibrium.Students hand-outs should be without the equation. You may also explain about the direction of 38.6 lb force (see fig. 3.7 d, page 88)Solving the first equation and then the second yieldsTCD = 34.2 lb and WB = 47.8 lb .Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections
15 A) The weight is too heavy. B) The cables are too thin. CONCEPT QUESTIONS1000 lb1000 lb1000 lb( A ) ( B ) ( 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 lb weight.Answers:1. C2. CYou may ask the students to record their answers, first, without group discussion and later after the group discussion with their neighbors. Again, like after reading quiz, they should verbally indicate the answers. You can comment on the answers, emphasizing statically indeterminate condition.Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections
16 GROUP PROBLEM SOLVINGGiven: The car is towed at constant speed by the 600 lb force and the angle is 25°.Find: The forces in the ropes AB and AC.Plan:1. Draw a FBD for point A.2. Apply the EofE to solve for the forces in ropes AB and AC.
17 GROUP PROBLEM SOLVING (continued) 30°25°600 lbFABFACAFBD at point AApplying the scalar EofE at A, we get;+ Fx = FAC cos 30° – FAB cos 25° = 0+ Fy = -FAC sin 30° – FAB sin 25° = 0Solving the above equations, we get;FAB = 634 lbFAC = 664 lb
18 1. Select the correct FBD of particle A. ATTENTION QUIZ1. Select the correct FBD of particle A.A3040100 lbF F2AA)B)3040°Answers:1. D100 lbAF2FF1D)C)30°40°30°AA100 lb100 lbStatics:The Next Generation Mehta & Danielson Lecture Notes for Sections
19 ATTENTION QUIZ2. 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 = 0B) F2 cos 50° – 20 = 0C) F2 sin 50° – F1 = 0D) F2 cos 50° = 0F220 lb50°CF1Answers:2. BStatics:The Next Generation Mehta & Danielson Lecture Notes for Sections