nnouncements Equilibrium of a Particle in 2-D Today s Objectives Draw a free body diagram (FBD) pply equations of equilibrium to solve a 2-D problem Class ctivities pplications What, why, and how of a Free Body Diagram Equations of equilibrium in 2-D Springs and pulleys Examples Engr221 Chapter 3 1
pplications For a spool of given weight, what are the forces in cables B and C? pplications - continued For a given cable strength, what is the maximum weight that can be lifted? Engr221 Chapter 3 2
pplications - continued For a given engine weight, what are the forces in the cables? What size of cable should you use? Equilibrium of a Particle in 2-D This is an example of a 2-D, or coplanar force system. If the whole assembly is in equilibrium, then particle 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. Engr221 Chapter 3 3
The What, Why, and How of a Free Body Diagram Free Body Diagrams are one of the most important things for you to know how to draw, and how to use. What is it? - drawing that shows all external forces acting on a particle. Why draw it? - It helps you write the equations of equilibrium used to solve for the unknowns (usually forces or angles). How? Imagine the particle to be isolated or cut free from its surroundings Show all the forces that act on the particle - ctive forces: They tend to move the particle - Reactive forces: They tend to resist the motion Identify each force and show all known magnitudes and directions as variables Engine mass = 250 Kg FBD at Engr221 Chapter 3 4
Equations of 2-D Equilibrium FBD at Since particle is in equilibrium, the net force at is zero. ΣF = 0 (vector equation) or T B + T D + W = 0 In general, for a particle in equilibrium, ΣF = 0 or ΣF x i + ΣF y j = 0 = 0 i + 0 j ( vector equation) Or written in scalar form, ΣF x = 0 and ΣF y = 0 These are two scalar equations of equilibrium (E of E). They can be used to solve for up to two unknowns. Example Engine mass = 250 Kg Write the scalar Equations of Equilibrium (E of E): + ΣF x = T B cos 30º T D = 0 + ΣF y = T B sin 30º 2.452 kn = 0 FBD at Solving the second equation gives: T B = 4.90 kn From the first equation, we get: T D = 4.25 kn Engr221 Chapter 3 5
Springs Spring Force = spring constant deformation F = k s Cables and Pulleys With a frictionless pulley, tensions are equal Engr221 Chapter 3 6
Example B Given: Sack weighs 20 lb with geometry as shown Find: Forces in the cables and weight of sack B Plan: 1. Draw a FBD at point E 2. pply E of E at point E to solve for the unknowns (T EG & T EC ) 3. Repeat this process at point C Example B - continued FBD at E should look like the one to the left. Note the assumed directions for the two cable tensions. The scalar E of E are: + ΣF x = T EG sin 30º T EC cos 45º = 0 + ΣF y = T EG cos 30º T EC sin 45º 20 lbs = 0 Solving these two simultaneous equations for the two unknowns yields: T EC = 38.6 lb T EG = 54.6 lb Engr221 Chapter 3 7
Example B - continued Now, move on to ring C. FBD for C should look like the one to the left. The scalar E of E are: + Σ F x = 38.64 cos 45 (4/5) T CD = 0 + Σ F y = (3/5) T CD + 38.64 sin 45 W B = 0 Solving the first equation and then the second yields: T CD = 34.2 lb and W B = 47.8 lb Example C Given: The car is towed at constant speed by the 600 lb force The angle θ is 25 Find: The forces in the ropes B and C Plan: 1. Draw a FBD at point 2. pply the E of E to solve for the forces in ropes B and C Engr221 Chapter 3 8
Example C - continued 600 lb 25 30 FBD at point F B F C pplying the E of E at point, we get: + F x = F C cos 30 F B cos 25 = 0 + F y = -F C sin 30 F B sin 25 + 600 = 0 Solving the above equations, we get: F B = 634 lb F C = 664 lb Questions 1) When a particle is in equilibrium, the sum of forces acting on it equals ) 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 (T 1 and T 2 ) are related as ) T 1 > T 2 B) T 1 = T 2 C) T 1 < T 2 D) T 1 = T 2 sin θ Engr221 Chapter 3 9
Question Select the correct FBD of particle. 30 40 100 lb ) 100 lb F C) 30 100 lb B) D) F1 F2 30 40 F1 30 40 100 lb F2 Question Using the FBD for point C, the sum of forces in the x-direction (Σ F X ) is. Use a sign convention of + ) F 2 sin 50 20 = 0 B) F 2 cos 50 20 = 0 C) F 2 sin 50 F 1 = 0 D) F 2 cos 50 + 20 = 0 20 lb 50 C F 1 F 2 Engr221 Chapter 3 10
Textbook Problem 3-10 The 500-lb crate is hoisted using the ropes B and C. Each rope can withstand a maximum tension of 2500 lb before it breaks. If B always remains horizontal, determine the smallest angle θ to which the crate can be hoisted. For T B = 2500 lb, θ = 11.31º For T C = 2500 lb, θ = 11.54º Summary Draw a free body diagram (FBD) pply equations of equilibrium to solve a 2-D problem Engr221 Chapter 3 11
nnouncements F = mg F = ks Test Monday Equations of 2-D Equilibrium - Review FBD at Since particle is in equilibrium, the net force at is zero. ΣF = 0 (vector equation) or T B + T D + W = 0 In general, for a particle in equilibrium, ΣF = 0 or ΣF x i + ΣF y j = 0 = 0 i + 0 j ( vector equation) Or written in scalar form, ΣF x = 0 and ΣF y = 0 These are two scalar equations of equilibrium (E of E). They can be used to solve for up to two unknowns. Engr221 Chapter 3 12
Free Body Diagrams - Review Imagine the particle to be isolated or cut free from its surroundings Show all the forces that act on the particle - ctive forces: They tend to move the particle - Reactive forces: They tend to resist the motion Identify each force and show all known magnitudes and directions as variables Engine mass = 250 Kg FBD at Textbook Problem 3-14 The unstretched length of spring B is 2 m. If the block is held in the equilibrium position shown, determine the mass of the block at D. F B F C W = 90 N = 102 N = 126 N Mass = 12.8 kg Engr221 Chapter 3 13
Textbook Problem 3-32 Determine the un-stretched length of spring C if a force P = 80 lb causes the angle θ = 60º for equilibrium. Cord B is 2 ft long. Take k = 50 lb/ft. F s L uns = 40 lb = 2.66 ft Equations of 3-D Equilibrium FBD at Since the particle is in equilibrium, the net force at the origin is zero. So F 1 + F 2 + F 3 = 0 or ΣF = 0 In general, for a particle in equilibrium in 3-D, ΣF = 0 or ΣF x i + ΣF y j + ΣF z k = 0 = 0 i + 0 j + 0 k Or written in scalar form: ΣF x = 0, ΣF y = 0, and ΣF z = 0 These are the three scalar equations of equilibrium. They can be used to solve for up to three unknowns. Engr221 Chapter 3 14
Procedure for nalysis Free-Body Diagram Establish the x, y, and z axes in any suitable orientation Label all known and unknown force magnitudes and directions on the diagram ssume senses for unknown magnitudes Equations of Equilibrium Use scalar equations of equilibrium when it is easy to resolve each force into its x, y, and z components (not often) If the 3-D geometry is difficult, express the forces in Cartesian form, use ΣF = 0, and equate the i, j, and k components to 0 If the answer is negative, the assumed sense should be reversed Example 3-D Problem The 50-kg pot is supported from by the three cables. Determine the force acting in each cable for equilibrium. Take d = 2.5 meters. F B = 580 N F C = 312 N F D = 312 N Engr221 Chapter 3 15
nnouncements Homework notes Test Monday One crib sheet (given) Calculator Pencil Eraser FBD s, units, process are grading considerations Procedure for nalysis - Review Free-Body Diagram Establish the x, y, and z axes in any suitable orientation. Label all known and unknown force magnitudes and directions on the diagram. ssume senses for unknown magnitudes. Equations of Equilibrium Use scalar equations of equilibrium when it is easy to resolve each force into its x, y, and z components. If the 3-D geometry is difficult, express the forces in Cartesian form, use ΣF = 0, and set the i, j, and k components to 0. If the answer is negative, the assumed sense should be reversed. Engr221 Chapter 3 16
Textbook Example 3.8 Textbook Example 3.8 - continued Engr221 Chapter 3 17
Example Problem small peg P rests on a spring that is contained inside the smooth pipe. When the spring is compressed so that s = 0.15 m, the spring exerts an upward force of 60 N on the peg. Determine the point of attachment (x,y,0) of cord P so that the tension in cords PB and PC equals 30 N and 50 N, respectively. x = 0.1904 m y = 0.0123 m Example Problem Determine the tension developed in cables OD and OB and the strut OC, required to support the 50-kg crate. The spring O has an unstretched length of 0.8 m and a stiffness of k O = 1.2 kn/m. The force in the strut acts along the axis of the strut. nswers TBD Engr221 Chapter 3 18
Chapter 3 Summary Draw a free body diagram (FBD) pply equations of equilibrium to solve a 2-D problem pply equations of equilibrium to solve a 3-D problem Engr221 Chapter 3 19