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## EQUILIBRIUM OF A PARTICLE IN 2-D

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**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 quiz • Applications • What, why and how of a FBD • Equations of equilibrium • Analysis of spring and pulleys • Concept quiz • Group problem solving • Attention quiz**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 B) a positive number C) zero D) 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 > T2 B) T1 = T2 C) T1 < T2 D) T1 = T2 sin **APPLICATIONS**For a spool of given weight, what are the forces in cables AB and AC ?**APPLICATIONS(continued)**For a given cable strength, what is the maximum weight that can be lifted ?**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.**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).**How ?**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 . A Note : Engine mass = 250 Kg FBD at A**EQUATIONS OF 2-D EQUILIBRIUM**Since particle A is in equilibrium, the net force at A is zero. So FAB + FAD +FAC = 0 or F = 0 A FBD at A In general, for a particle in equilibrium, F = 0or Fxi + Fy j = 0 = 0 i +0 j (A vector equation) • Or, written in a scalar form, • These are two scalar equations of equilibrium (EofE). They can be used to solve for up to two unknowns. • Fx= 0and Fy= 0**EXAMPLE**Note : Engine mass = 250 Kg FBD at A Write the scalar EofE: + Fx = TB cos 30º – TD = 0 + Fy = TB sin 30º – 2.452 kN = 0 Solving the second equation gives: TB = 4.90 kN From the first equation, we get: TD = 4.25 kN**SPRINGS, CABLES, AND PULLEYS**Spring Force = spring constant * deformation, or F = k * S With a frictionless pulley, T1 = T2.**EXAMPLE**Given: Sack A weighs 20 N. 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.**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 = TEGsin 30º – TEC cos 45º = 0 + Fy = TEGcos 30º – TEC sin 45º – 20 N = 0 Solving these two simultaneous equations for the two unknowns yields: TEC = 38.6 N TEG = 54.6 N**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 = 38.64 cos 45 – (4/5) TCD = 0 Fy = (3/5) TCD + 38.64 sin 45 – WB = 0 Solving the first equation and then the second yields TCD = 34.2 N and WB = 47.8 N .**CONCEPT QUESTIONS**1000 N 1000 N 1000 N ( 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 N weight.**GROUP PROBLEM SOLVING**Given: The car is towed at constant speed by the 600 N 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.**600 N**FBD at point A A 30° 25° FAC FAB GROUP PROBLEM SOLVING (continued) Applying the scalar EofE at A, we get; + Fx = FAC cos 30° – FAB cos 25° = 0 + Fy = -FAC sin 30° – FAB sin 25° + 600 = 0 Solving the above equations, we get; FAB = 634 N FAC = 664 N**ATTENTION QUIZ**1. Select the correct FBD of particle A. A 30 40 100 N F1F2 A A) B) 30 40° 100 N A F2 F F1 D) C) 30° 40° 30° A A 100 N 100 N**ATTENTION QUIZ**2. 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 F2 20 N 50° C F1**EQUATIONS OF 3-D EQUILIBRIUM**Since particle A is in equilibrium, the net force at A is zero. So FB + FD +FC - W = 0 or F = 0 A FBD at A In general, for a particle in equilibrium, F = 0or Fxi + Fy j + Fzk = 0 i +0 j + 0 k (A vector equation) • Or, written in a scalar form, • Fx= 0, Fy= 0and Fz= 0 • These are three scalar equations of equilibrium (EofE). They can be used to solve for up to three unknowns.**EXAMPLE**Given: Crate mass 100 kg. and geometry is as shown. Find: Forces in the cords AC, AD and the stretch of the spring. Plan: 1. Draw a FBD for Point A. 2. Apply EofE at Point A to solve for the unknowns (FB, FC & FD). 3. Evaluate s = FB/k.**EXAMPLE (continued)**A FBD at A should look like the one to the left. Note the assumed directions for the three forces. From the FBD, one can write FB = FB i FC = FC cos 1200i + FC cos 1350j + FC cos 600k = -0.5FCi – 0.707 FCj + 0.5FCk FD= -0.333FDi + 0.667 FDj + 0.667FDk W = -981 k N**EXAMPLE (continued)**• Equilibrium requires • F = 0; FB + FC+ FD+ W = 0 • FBi+ -0.5FCi – 0.707 FCj + 0.5FCk + • -0.333FDi + 0.667 FDj + 0.667FDk - 981 k = 0 The scalar EofE are: + Fx = FB– 0.5FC -0.333FD = 0 (1) + Fy = -0.707FC– 0.667FD = 0 (2) + z Fz = 0.5FC + 0.667FD - 981 = 0 (3)**EXAMPLE (continued)**Solving these three simultaneous equations for the three unknowns yields: FB = 693.7 N FC = 813 N FD = 862 N The stretch of the spring is therefore F = ks; (or FB = ks) 693.7 =1500 s s = 0.462 m**End of the Lecture**Let Learning Continue