# STATICS: CE201 Chapter 3 Equilibrium of a Particle

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STATICS: CE201 Chapter 3 Equilibrium of a Particle

1/11/2013 STATICS: CE201 Chapter 3 Equilibrium of a Particle Notes are prepared based on: Engineering Mechanics, Statics by R. C. Hibbeler, 12E Pearson Dr M. Touahmia & Dr M. Boukendakdji Civil Engineering Department, University of Hail (2012/2013) 3. Equilibrium of a Particle ________________________________________________________________________________________________________________________________________________ Chapter Goals: 1. 2. Introduce the concept of the free-body diagram (FBD) for an object modeled as a particle. Show how to solve particle equilibrium problems using the equations of equilibrium. Contents: 1 2 3 4 Condition for the Equilibrium of a Particle The Free-Body Diagram (FBD) Coplanar Force Systems Three-Dimensional Force Systems 1 1/11/2013 Condition for the Equilibrium of a Particle A particle is said to be in equilibrium when the resultant of all forces acting on it is zero. Considering Newton’s first law of motion, equilibrium can mean that the particle is either at rest or moving at constant velocity. To maintain equilibrium it is necessary and sufficient that the resultant force acting on a particle be equal to zero. Equilibrium of a Particle In terms of Newton`s first law of motion, this can be expressed mathematically as: F 0 Equation of Equilibrium where F is the vector sum of all the forces acting on the particle. For Equilibrium: F x F y 0 0 2 1/11/2013 The Free-Body Diagram (FBD) To apply the equation of equilibrium, we must account for all the known and unknown forces which act on the particle. In order to account for all the forces that act on a particle, it is necessary to draw its Free-Body Diagram (FBD) The free-body diagram is simply a sketch which shows the particle “free” from its surroundings with all the forces that act on it. Types of Connections Two types of connections often encountered in particle equilibrium: Springs and Cable and Pulleys. Springs: The magnitude of force exerted on a linearly elastic spring is: F ks which has a stiffness k and is deformed (elongated or compressed) a distance s, measured from its unloaded position. s l lo where lo the original length and l the final length 3 1/11/2013 Types of Connections Cables and Pulleys: All cables are assumed to have negligible weight and they cannot stretch. A cable can support only a tension or “pulling” force, and this force always act in the direction of the cable. The tension in a cable is constant throughout its length. Cable in tension Procedure for Drawing a Free-Body Diagram: Draw Outlined Shape: Imagine the particle to be isolated or cut “free” from its surroundings with all the forces outlined shape. Show All Forces: Indicate on this sketch all the forces that act on the particle. Identify Each Force: The forces which are known should be labeled with their proper magnitudes and directions. Letters are used to represent the magnitudes and directions of forces that are unknown. 4 1/11/2013 Example 1: The sphere has a mass of 6 kg and is supported as shown. Draw a free-body diagram of the sphere, the cord CE, and the knot at C. Solution 1: Sphere: There are two forces acting on the sphere: ◦ The weight of the sphere W, W 6kg 9.81m/s 2 58.9 N ◦ The force FCE of the cord CE acting on the sphere The free body diagram: 5 1/11/2013 Solution 1: Cord CE: When the cord CE is isolated from its surroundings, its free-body diagram shows only two forces acting on it: ◦ The force of the sphere and FCE. ◦ The force of the knot FEC. FCE = FEC Solution 1: Knot: The knot at C is subjected to three forces. They are caused by the cords CBA and CE and the spring CD: 6 1/11/2013 Coplanar Force Systems If a particle is subjected to a system of coplanar forces that lie in the x-y plane, then each force can be resolved into i and j components. For equilibrium, these forces must sum to produce a zero resultant. Hence: F 0 F i F x y F x j0 0 F 0 y Example 2: Determine the tension in cables BA and BC necessary to support the 60-kg cylinder. 7 1/11/2013 Solution 2: Due to equilibrium, the weight of the cylinder causes the tension in cable BD to be: TBD 609.81 588.6 N The forces in cable BA and BC can be determined by investigating the equilibrium of the ring B. Its free body diagram is then: The magnitudes of TA and TC are unknown, but their directions are known. Solution 2: Equations of equilibrium along the x and y axes: F 0 x 4 T cos 45 T 0 5 o C A 3 F 0 TC sin 45o TA 588.6 0 y 5 TA 0.8839 TC 3 TC sin 45o (0.8839TC ) 588.6 0 5 TC 475.66 N TA 420 N 8 1/11/2013 Example 3: The 200-kg box is suspended using the ropes AB and AC. Each rope can withstand a maximum force of 10 kN before it breaks. If AB always remains horizontal, determine the smallest angle θ to which the box can be suspended before one of the ropes breaks. Solution 3: Free-Body Diagram of the ring: There are three forces acting on the ring: FC , FB , FD The magnitude of FD is equal to the weight of the box. FD = 200 (9.81) N = 1962 N < 10 kN 9 1/11/2013 Solution 3: Equations of Equilibrium: FC sin 1962 0 FC FC cos FB 0 FB cos FC 10 kN 10 10 3 sin 1962 0 sin 1 0.1962 11.31 The force developed in rope AB: FB FC cos 11.31 10 10 3 cos 11.31 FB 9.81 kN Three-Dimensional Force Systems The necessary and sufficient condition for particle equilibrium is: F 0 We can resolve the forces into their respective i, j, k components , so that: F x i Fy j Fz k 0 To satisfy this equation we require: F F x 0 y 0 F z 0 10 1/11/2013 Example 5: If the mass of cylinder C is 40 kg, determine the mass of cylinder A in order to hold the assembly in the position shown. 11