Section 3.1-3.3 Condition for the Equilibrium of a Particle

Engineering Mechanics: Statics Fifteenth Edition Chapter 3 Equilibrium of a Particle Section 3.1 Condition for the Equilibrium of a Particle Section 3.2 The Free-Body Diagram Section 3.3 Coplanar Force Systems Copyright © 2022 Pearson Education, Inc. All Rights Reserved Equilibrium of A Particle, The Freebody Diagram & Coplanar Force Systems Today’s Objectives: a. Draw a free body diagram (FBD), and, b. Apply equations of equilibrium to solve a 2-D problem. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 3.3: Coplanar Force Systems 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 free-body diagram and apply the equations of equilibrium. Copyright © 2022 Pearson Education, Inc. All Rights Reserved The What, Why and How of A Free Body Diagram (FBD) (1 Of 2) 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). Copyright © 2022 Pearson Education, Inc. All Rights Reserved The What, Why and How of A Free Body Diagram (FBD) (2 of 2) 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: Cylinder mass = 40Kg Copyright © 2022 Pearson Education, Inc. All Rights Reserved Equations of 2-D Equilibrium (1 of 2) Since particle A is in equilibrium, the net force at A is zero. So FB + FC + FD = 0 F =0 or In general, for a particle in equilibrium,  F = 0 or F i + F X y j = 0 = 0i +0 j (a vector equation) Or, written in a scalar form, F X = 0 and  FY = 0 These are two scalar equations of equilibrium (E-of-E). They can be used to solve for up to two unknowns. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Equations of 2-D Equilibrium (2 of 2) Note: Cylinder mass = 40 Kg Write the scalar E-of-E: + →  FX = FB cos 30o − FD = 0 +   FY = FB Sin30o − 392.4N = 0 Solving the second equation gives: FB = 785N → From the first equation, we get: FD = 680N  Copyright © 2022 Pearson Education, Inc. All Rights Reserved Simple Springs Spring Force = spring constant * deformation of spring or F = k*s Copyright © 2022 Pearson Education, Inc. All Rights Reserved Cables and Pulleys With a frictionless pulley and cable T1 = T2  Cable can support only a tension or “pulling” force, and this force always acts in the direction of the cable. Cable is in tension Copyright © 2022 Pearson Education, Inc. All Rights Reserved Smooth Contact If an object rests on a smooth surface, then the surface will exert a force on the object that is normal to the surface at the point of contact. In addition to this normal force N, the cylinder is also subjected to its weight W and the force T of the cord. Since these three forces are concurrent at the center of the cylinder, we can apply the equation of equilibrium to this “particle,” which is the same as applying it to the cylinder. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example I (1 of 2) Given: The box weighs 550 l b and geometry is as shown. Find: The forces in the ropes A B and AC. Plan: 1. Draw a FBD for point A. 2. Apply the E-of-E to solve for the forces in ropes AB and A C. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example I (2 of 2) Applying the scalar E-of-E at A, we get; 4 + →  FX = − FB cos30o + FC   = 0 5 3 +   F Y = FB sin30 + FC   − 550 lb = 0 5 o Solving the above equations, we get; FB = 478 lb and FC = 518 lb Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅱ (1 of 2) Given: The mass of cylinder C is 40 kgrand geometry is as shown. Find: The tensions in cables DE, EA, and EB. Plan: 1. Draw a FBD for point E. 2. Apply the E-of-E to solve for the forces in cables D E, EA, and EB. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅱ (2 of 2) Applying the scalar E-of-E at E, we get; + →  F X = − TED + ( 40  9.81) cos30o = 0 +   F Y = ( 40  9.81) sin30o − TEA = 0 and TE A = 196 N ↓ TE D =340 N ← Solving the above equations, we get; Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving (1 of 3) Given: The mass of lamp is 20 kgrand geometry is as shown. Find: The force in each cable. Plan: 1. Draw a FBD for Point D. 2. Apply E-of-E at Point D to solve for the unknowns (FCD & F D E). 3. Knowing FCD, repeat this process at point C. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving (2 of 3) Applying the scalar E-of-E at D, we get; +   FY = FDE sin30o − 20 ( 9.81) = 0 + →  FX = FDE cos30o − FCD = 0 Solving the above equations, we get: FDE = 392N and FCD =340N  Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving (3 of 3) Applying the scalar E-of-E at C, we get; 3 + →  FX = 340 − FBC sin 45o − FAC   = 0 5 4 +   FY = FAC   − FBC cos 45o = 0 5 Solving the above equations, we get; FBC = 275N and FAC = 243N Copyright © 2022 Pearson Education, Inc. All Rights Reserved Attention Quiz (1 of 2) 1. Select the correct FBD of particle A. A. C. B. D. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Attention Quiz (2 of 2) 2. Using this FBD of Point C, the sum of forces in the x-direction (F ) X Is ______. Use a sign convention of + → . A. F2 sin50o − 20 = 0 B. F2 cos50o − 20 = 0 o C. F2 sin50 − F1 = 0 D. F2 cos50o + 20 = 0 Copyright © 2022 Pearson Education, Inc. All Rights Reserved Reading Quiz (1 of 2) 1. When a particle is in equilibrium, the sum of forces acting on it equals ______ . (Choose the most appropriate answer) A. B. C. D. E. A constant A positive number Zero A negative number An integer Copyright © 2022 Pearson Education, Inc. All Rights Reserved Reading Quiz (2 of 2) 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  Copyright © 2022 Pearson Education, Inc. All Rights Reserved