1. MEC 0011 Statics Lecture 3 Prof. Sanghee Kim Fall_ 2012

2. Chapter Objectives 1.Condition for the Equilibrium of a Particle 2.The Free-Body Diagram (FBD) 3.Coplanar Systems 4.Three-Dimensional Force Systems

3. 3.1 Condition for the Equilibrium of a Particle Particle at equilibrium if - remains at rest (static equilibrium) - moving at constant velocity Newton’s first law of motion ∑F = 0 where ∑F is the vector sum of all the forces acting on the particle Newton’s second law of motion ∑F = ma  ma = 0  a = 0 therefore, the particle is moving in constant velocity or at rest

4. 3.2 The Free-Body Diagram Best representation of all the unknown forces (∑F) which acts on a body A sketch showing the particle “free” from the surroundings with all the forces acting on it Consider two common connections in this subject – Spring – Cables and Pulleys

5. Spring – Linear elastic spring: change in length is directly proportional to the force acting on it – spring constant or stiffness k: defines the elasticity of the spring – Magnitude of force when spring is elongated or compressed  F = ks s (distance) =l -l 0 Positive (elongation) Negative (compression)

6. Cables and Pulley – Cables (or cords) are assumed negligible weight and cannot stretch – Tension always acts in the direction of the cable – Tension force must have a constant magnitude for equilibrium – For any angle θ, the cable is subjected to a constant tension T

7. Procedure for Drawing a FBD 1.Draw outlined shape 2. Show all the forces - Active forces: particle in motion - Reactive forces: constraints that prevent motion 3. Identify each forces - Known forces with proper magnitude and direction - Letters used to represent magnitude and directions

8. QUIZ Q. Select the correct FBD of particle A. A 40  100 kg 30  A) A 100 kg B) 40° A F 1 F 2 C) 30° A F 100 kg A 30°40° F1F1 F2F2 100 kg D)

9. Example 3.1 The sphere has a mass of 6kg and is supported. Draw a free-body diagram of the sphere, the cord CE and the knot at C.

10. Solution 1. FBD at Sphere Two forces acting, weight and the force on cord CE 2. Cord CE Two forces acting: sphere and knot Newton’s 3 rd Law: F CE is equal but opposite F CE and F EC pull the cord in tension for equilibrium, F CE = F EC 3. FBD at Knot 3 forces acting: 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

11. 3.3 Coplanar Systems A particle is subjected to coplanar forces in the x-y plane Resolve into i and j components for equilibrium ∑F x = 0 ∑F y = 0 Scalar equations of equilibrium require that the algebraic sum of the x and y components to equal to zero

12. Procedure for Analysis 1. Free-Body Diagram a. Establish the x, y axes b. Label all the unknown and known forces 2. Equations of Equilibrium a. Apply the equations of equilibrium ∑F x = 0∑F y = 0 b. Apply F = ks to find spring force c. When negative result force is the reverse

13. Example 3.4 Determine the required length of the cord AC so that the 8kg lamp is suspended. The undeformed length of the spring AB is l’ AB = 0.4m, and the spring has a stiffness of k AB = 300N/m.

14. Solution FBD at Point A +→∑F x = 0; T AB – T AC cos30º = 0 +↑ ∑F y = 0; T AB sin30º – 78.5N = 0 Solving, T AC = 157.0kN T AB = 136.0kN T AB = k AB s AB ; 136.0N = 300N/m(s AB ) s AB = 0.453N For stretched length, l AB = l’ AB + s AB l AB = 0.4m + 0.453m = 0.853m For horizontal distance BC, 2m = l AC cos30° + 0.853m l AC = 1.32m

15. Exercise The cord BCA and CD can each support a maximum hold of 0.5 kN. Determine the maximum weight of the crate that can be hoisted at a constant velocity and the angle θ for equilibrium. Neglect the size of the smooth pulley at C.

16. Solution FBD

17. 3.4 Three-Dimensional Force Systems For particle equilibrium ∑F = 0 Resolving into i, j, k components ∑F x i + ∑F y j + ∑F z k = 0 Three scalar equations representing algebraic sums of the x, y, z forces ∑F x i = 0 ∑F y j = 0 ∑F z k = 0

18. Procedure for Analysis Free-body Diagram -Establish the x, y, z axes -Label all known and unknown force Equations of Equilibrium -Apply scalar equation of ∑F x = 0, ∑F y = 0 and ∑F z = 0  resolve -Substitute vectors into ∑F = 0 and set i, j, k components = 0 -Negative results indicate that the sense of the force is opposite to that shown in the FBD.

19. Example 3.7 Determine the force developed in each cable used to support the 40kN crate.

20. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution 1. FBD at Point A To expose all three unknown forces in the cables. 2. Equations of Equilibrium Expressing each forces in Cartesian vectors, F B = F B (r B / r B ) = -0.318F B i – 0.424F B j + 0.848F B k F C = F C (r C / r C ) = -0.318F C i – 0.424F C j + 0.848F C k F D = F D i W = -40k

21. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution For equilibrium, ∑F = 0; F B + F C + F D + W = 0 -0.318F B i – 0.424F B j + 0.848F B k - 0.318F C i – 0.424F C j + 0.848F C k + F D i - 40k = 0 ∑F x = 0; -0.318F B - 0.318F C + F D = 0 ∑F y = 0; – 0.424F B – 0.424F C = 0 ∑F z = 0; 0.848F B + 0.848F C - 40 = 0 Solving, F B = F C = 23.6kN F D = 15.0kN