Chapter 3 Equilibrium of a Particle Engineering Mechanics

Chapter 3: Equilibrium of a Particle Engineering Mechanics: Statics Ahmad Faizal bin Salleh 2014/2015

Chapter Objectives • To introduce the concept of the free-body diagram for a particle. • To show to solve 2 D and 3 D particle equilibrium problems using the equations of equilibrium.

Chapter Outline • Condition for the Equilibrium of a Particle • The Free-Body Diagram • Coplanar Systems • Three-Dimensional Force Systems

WHY TREAT A BODY AS A PARTICLE In a real world there is no such things as particles, which occupy zero volume in space. However, when the body doesn’t rotate, (therefore no angular velocity, angular acceleration, angular kinetic energy or momentum), we can treat the body as a particle.

WHY TREAT A BODY AS A PARTICLE Even, when the body rotates but its angular velocity remain constant, we can still treat the body as particle, since it rotational quantities (angular velocity, angular acceleration, angular kinetic energy or momentum) remain unchanged during the motion. Keep it mind that the body is treat as a particle not because of its size, BUT because of its insignificancy of rotation.

WHY TREAT A BODY AS A PARTICLE Example Space airfcraft orbiting earth where the observer at earth did not care so much about rotational motion of the satelite. If the observer at earth want to consider the rotational motion of the satelite (if he start thinking about the effect of aircarft rotational motion on the passengers), he/she need to treat aircarft as BODY.

3. 1 Condition for the Equilibrium of a Particle • Particle is at equilibrium if - at rest - moving at a constant velocity

3. 1 Condition for the Equilibrium of a Particle • Newton’s first law of motion ∑F = 0 where ∑F is the vector sum of all the forces acting on the particle

3. 1 Condition for the Equilibrium of a Particle • Newton’s second law of motion ∑F = ma

3. 2 The Free-Body Diagram Two common connections in particle equilibrium – Spring – Cables and Pulleys

3. 2 THE FBD Two common connections in particle equilibrium – Spring – Cables and Pulleys

3. 2 The Free-Body Diagram Spring • Spring constant or stiffness k, defines the elasticity of the spring • Magnitude of force when spring is elongated or compressed F = ks

3. 2 The Free-Body Diagram • Spring deformed distance, s where s is determined from the difference in spring’s deformed length l and its undeformed length lo s = l - lo - If s is positive, F “pull” onto the spring - If s is negative, F “push” onto the spring

3. 2 The Free-Body Diagram Example Given lo = 0. 4 m and k = 500 N/m To stretch it until l = 0. 6 m, A force, F = ks =(500 N/m)(0. 6 m – 0. 4 m) = 100 N is needed To compress it until l = 0. 2 m, A force, F = ks =(500 N/m)(0. 2 m – 0. 4 m) = -100 N is needed

3. 2 The Free-Body Diagram Cables and Pulley (Assumptions) • Negligible weight and cannot stretch. • Only support tension or pulling force.

3. 2 The Free-Body Diagram Cables and Pulley (Assumptions) • Tension always acts in the direction of the cable • Tension force in a continuous cable must have a constant magnitude for equilibrium

3. 2 The Free-Body Diagram • Cables and Pulley For any angle θ, the cable is subjected to a constant tension T, throughout its length

3. 2 The Free-Body Diagram Procedure for Drawing a FBD 1. Draw outlined shape - Isolate particle from its surroundings 2. Show all the forces - Indicate all the forces - Active forces: set the particle in motion - Reactive forces: result of constraints and supports that tend to prevent motion

3. 2 The Free-Body Diagram Procedure for Drawing a FBD 3. Identify each forces - Known forces should be labeled with proper magnitude and direction - Letters are used to represent magnitude and directions of unknown forces

3. 2 The Free-Body Diagram Example 3. 1 The sphere has a mass of 6 kg and is supported. Draw a free-body diagram of i)the sphere, ii)the cord CE iii)and the knot at C.

3. 2 The Free-Body Diagram Solution FBD at Sphere • Two forces acting, weight and the force on cord CE. • Weight of 6 kg (9. 81 m/s 2) = 58. 9 N

3. 2 The Free-Body Diagram Solution Cord CE • Two forces acting, force of the sphere and force of the knot • Newton’s Third Law: FCE is equal but opposite • FCE and FEC pull the cord in tension • For equilibrium, FCE = FEC

3. 2 The Free-Body Diagram Solution FBD at Knot • Three forces acting, force by cord CBA, cord CE and spring CD • Important to know that the weight of the sphere does not act directly on the knot but subjected to by the cord CE

FBD & Coordinate System In solving equilibrium equation, it is good to resolve forces into a component of their coordinate system. The problems could be solved in either 1 D, 2 D or 3 D depends on the movement, required accuracies and etc.

WED 13 Sept 2017 Re explain about the concept of FBD in Ex. 3. 1

3. 3 Equilibrium in Coplanar (2 D) Systems A particle is subjected to coplanar forces in the x-y plane

3. 3 Equilibrium in Coplanar (2 D) Systems Example 3. 2 Determine the tension in cables AB and AD for equilibrium of the 250 kg engine.

3. 3 Equilibrium in Coplanar (2 D) Systems To solve for tension in AB and AD, you need to establish equilibrium equation. You need to establish the Cartesian coordinate system. Where to locate the origin of the coordinate system? Locating the origin of coordinate system at point D, make it difficult for determining the point A in later stage (distance of A from D were not given).

3. 3 Equilibrium in Coplanar (2 D) Systems The best is to locate origin of the coordinate system at point A. By doing so, it is easy for you to resolve the component of tension force in AD and AB in x and y direction of 2 D cartesian coordinate system. Always try to solve the equilibrium AT A POINT where all the forces of the system acts.

3. 3 Equilibrium in Coplanar (2 D) Systems Solution FBD at Point A - Initially, two forces acting, forces of cables AB and AD - Engine Weight = (250 kg)(9. 81 m/s 2) = 2. 452 k. N supported by cable CA - Finally, three forces acting, forces TB and TD and engine weight on cable CA

3. 3 Equilibrium in Coplanar (2 D) Systems Solution +→ ∑Fx = 0; +↑ ∑Fy = 0; Solving, TBcos 30° - TD = 0 TBsin 30° - 2. 452 k. N = 0 TB = 4. 90 k. N TD = 4. 25 k. N *Note: Neglect the weights of the cables since they are small compared to the weight of the engine

3. 3 Equilibrium in Coplanar (2 D) Systems Example 3. 3 If the sack at A has a weight of 20 N (≈ 2 kg), determine the weight of the sack at B and the force in each cord needed to hold the system in the equilibrium position shown.

3. 3 Equilibrium in Coplanar (2 D) Systems You can locate the origin of the coordinate system at either point C or E. However, it is better to locate it at point E since the weight of sack A is given and it helps to solve for tension EC and EG. Once tension in EC and EG were obtained, then proceed with point C.

3. 3 Equilibrium in Coplanar (2 D) Systems Solution FBD at Point E - Three forces acting, forces of cables EG and EC and the weight of the sack on cable EA

3. 3 Equilibrium in Coplanar (2 D) Systems Solution +→ ∑Fx = 0; +↑ ∑Fy = 0; Solving, TEGsin 30° - TECcos 45° = 0 TEGcos 30° - TECsin 45° - 20 N = 0 TEC = 38. 6 k. N TEG = 54. 6 k. N *Note: use equilibrium at the ring to determine tension in CD and weight of B with TEC known

3. 3 Equilibrium in Coplanar (2 D) Systems Solution FBD at Point C - Three forces acting, forces by cable CD and EC (known) and weight of sack B on cable CB

3. 3 Equilibrium in Coplanar (2 D) Systems Solution +→ ∑Fx = 0; 38. 6 cos 30° - (4/5)TCD = 0 +↑ ∑Fy = 0; (3/5)TCD – 38. 6 sin 45°N – WB = 0 Solving, TCD = 34. 1 k. N WB = 47. 8 k. N *Note: components of TCD are proportional to the slope of the cord by the 3 -45 triangle

3. 3 Equilibrium in Coplanar (2 D) Systems • Procedure for Analysis 1. Free-Body Diagram - Establish the x, y axes in any suitable orientation - Label all the unknown and known forces magnitudes and directions - Sense of the unknown force can be assumed

3. 3 Equilibrium in Coplanar (2 D) Systems • Procedure for Analysis 2) Equations of Equilibrium - If given a spring, apply F = ks to find spring force using deformation of spring - If the solution yields a negative result, the sense of force is the reserve of that shown in the free-body diagram

3. 3 Equilibrium in Coplanar (2 D) Systems • Procedure for Analysis 2) Equations of Equilibrium - Apply the equations of equilibrium ∑Fx = 0 ∑Fy = 0 - Components are positive if they are directed along the positive negative axis and negative, if directed along the negative axis

3. 4 Equilibrium in 3 D Force Systems • Make use of the three scalar equations to solve for unknowns such as angles or magnitudes of forces

3. 4 Equilibrium in 3 D Force Systems • Ring at A subjected to force from hook and forces from each of the three chains • Hook force = weight of the electromagnet and the load, denoted as W • Three scalars equations applied to FBD to determine FB, FC and FD

3. 4 Equilibrium in 3 D Force Systems Example 3. 4 A 90 N load is suspended from the hook. The load is supported by two cables and a spring having a stiffness k = 500 N/m. Determine the force in the cables and the stretch of the spring for equilibrium. Cable AD lies in the x-y plane and cable AC lies in the x-z plane.

3. 4 Equilibrium in 3 D Force Systems Establish the coordinate system where the origin is located at point A.

3. 4 Equilibrium in 3 D Force Systems Solution FBD at Point A - Point A chosen as the forces are concurrent at this point

3. 4 Equilibrium in 3 D Force Systems Solution Equations of Equilibrium, ∑Fx = 0; FDsin 30° - (4/5)FC = 0 ∑Fy = 0; -FDcos 30° + FB = 0 ∑Fz = 0; (3/5)FC – 90 N = 0 Solving, FC = 150 N FD = 240 N FB = 208 N

3. 4 Equilibrium in 3 D Force Systems Solution For the stretch of the spring, FB = ks. AB 208 N = 500 N/m(s. AB) s. AB = 0. 416 m

3. 4 Equilibrium in 3 D Force Systems Example 3. 6 Determine the magnitude and coordinate direction angles of force F that are required for equilibrium of the particle O.

3. 4 Equilibrium in 3 D Force Systems Vector F 3 need to be established using unit vector. Unit vector could be established using position vector.

3. 4 Equilibrium in 3 D Force Systems Solution FBD at Point O - Four forces acting on particle O

3. 4 Equilibrium in 3 D Force Systems Solution Equations of Equilibrium Expressing each forces in Cartesian vectors, F 1 = {400 j} N F 2 = {-800 k} N F 3 = F 3(r. B / r. B) = {-200 i – 300 j + 600 k } N F = Fxi + Fyj + Fzk

3. 4 Equilibrium in 3 D Force Systems Solution For equilibrium, ∑F = 0; F 1 + F 2 + F 3 + F = 0 400 j - 800 k - 200 i – 300 j + 600 k + Fxi + Fyj + Fzk = 0 ∑Fx = 0; - 200 + Fx = 0 Fx = 200 N ∑Fy = 0; 400 – 300 + Fy = 0 Fy = -100 N ∑Fz = 0; - 800 + 600 + Fz = 0 Fz = 200 N

3. 4 Equilibrium in 3 D Force Systems Solution

3. 4 Equilibrium in 3 D Force Systems Example 3. 7 Determine the force developed in each cable used to support the 40 k. N (≈ 4 tonne) crate.

3. 4 Equilibrium in 3 D Force Systems Consider, • Position vector • Unit vector • Force equilibrium

3. 4 Equilibrium in 3 D Force Systems Solution FBD at Point A - To expose all three unknown forces in the cables

3. 4 Equilibrium in 3 D Force Systems Solution Equations of Equilibrium Expressing each forces in Cartesian vectors, FB = FB(r. B / r. B) = -0. 318 FBi – 0. 424 FBj + 0. 848 FBk FC = FC (r. C / r. C) = -0. 318 FCi + 0. 424 FCj + 0. 848 FCk F D = F Di W = -40 k

3. 4 Equilibrium in 3 D Force Systems Solution For equilibrium, ∑F = 0; FB + FC + FD + W = 0 -0. 318 FBi – 0. 424 FBj + 0. 848 FBk - 0. 318 FCi – 0. 424 FCj + 0. 848 FCk + FDi - 40 k = 0 ∑Fx = 0; -0. 318 FB - 0. 318 FC + FD = 0 ∑Fy = 0; – 0. 424 FB – 0. 424 FC = 0 ∑Fz = 0; 0. 848 FB + 0. 848 FC - 40 = 0

3. 4 Equilibrium in 3 D Force Systems Solution Solving, FB = FC = 23. 6 k. N FD = 15. 0 k. N

3. 4 Equilibrium in 3 D Force Systems Example 3. 8 The 100 kg crate is supported by three cords, one of which is connected to a spring. Determine the tension in cords AC and AD and stretch of the spring.

3. 4 Equilibrium in 3 D Force Systems Consider, Position vector, Unit vector, Coordinate direction angles Force vector, Force equilibrium

3. 4 Equilibrium in 3 D Force Systems View Free Body Diagram Solution FBD at Point A - Weight of the crate = 100 (9. 81) = 981 N - To expose all three unknown forces in the cables

3. 4 Equilibrium in 3 D Force Systems Solution Equations of Equilibrium Expressing each forces in Cartesian vectors, FB = FBi FC = FCcos 120°i + FCcos 135°j – FCcos 60°k FD = -0. 333 FDi + 0. 667 FDj + 0. 667 FDk W = -981 k

3. 4 Equilibrium in 3 D Force Systems Solution For equilibrium, ∑F = 0; FB + FC + FD + W = 0 FBi + FCcos 120°i + FCcos 135°j – FCcos 60°k -0. 333 FDi + 0. 667 FDj + 0. 667 FDk - 981 k = 0 ∑Fx = 0; FB + FCcos 120° - 0. 333 FD = 0 ∑Fy = 0; FCcos 135° + 0. 667 FD = 0 ∑Fz = 0; FCcos 60° + 0. 667 FD - 981 = 0

3. 4 Equilibrium in 3 D Force Systems Solution Solving, FB = 693. 7 N FC = 813 N FD = 693. 7 N For the stretch of the spring, FB = ks 693. 7 N = 1500 s s = 0. 462 m

3. 4 Equilibrium in 3 D Force Systems • Procedure for Analysis Free-body Diagram - Establish the z, y, z axes in any suitable orientation - Label all known and unknown force magnitudes and directions - Sense of a force with unknown magnitude can be assumed

3. 4 Equilibrium in 3 D Force Systems • For particle equilibrium ∑F = 0 • Resolving into i, j, k components ∑Fxi + ∑Fyj + ∑Fzk = 0 • Three scalar equations representing algebraic sums of the x, y, z forces ∑Fxi = 0 ∑Fyj = 0 ∑Fzk = 0

IMPORTANT POINTS In solving equilibrium of a particle involving cable, pulley and spring, always solve the equilibrium AT A POINT where all the forces of the system acts.