1. Engineering Fundamentals Session 9

2. Equilibrium A body is in Equilibrium if it moves with constant velocity. A body at rest is a special case of constant velocity i.e. v = 0 = constant. For a body to be in Equilibrium the resultant force (meaning the vector addition of all the forces) acting on the body must be zero. Resulting force = vector addition of force vectors A Force can be defined as 'that which tends to cause a particle to accelerate ‘.

3. Equilibrium of Concurrent Forces Equilibrant E are equal and opposite to Resultant R E = -R

4. Particle Vs Rigid Body A particle has dimension = 0 A Rigid body is a non-particle body and it does not deform (change shape). Concurrent forces: all forces acting a the same point Coplanar forces: all forces lie on the same plane

5. Conditions for Equilibrium Explanation: Sum of forces = 0, Or F 1 + F 2 + … + F n = 0 F 1 + F 2 + F 3 = 0 Example

6. Conditions for Equilibrium Breaking down into x and y components Example: For three forces acting on a particle

7. Free Body Diagram Free body diagram isolates a rigid body to describe the system of forces acting on it. R mg R R R

8. Free Body Diagram

9. Definitions System of Particles or Bodies Two or more bodies or particles connected together are referred to as a system of bodies or particles. External Force External forces are all the forces acting on a body defined as a free body or free system of bodies, including the actions due to other bodies and the reactions due to supports.

10. Transmissibility of Force

11. Load and Reaction Loads are forces that are applied to bodies or systems of bodies. Reactions at points supporting bodies are a consequence of the loads applied to a body and the equilibrium of a body.

12. Tensile and Compressive Forces Pushing force on the body -- compressive force Pulling force on a body -- a tensile force

13. Procedure for drawing a free body diagram Step 1: Draw or sketch the body to be isolated Step 2: Indicate all the forces that act on the particle. Step 3: Label the forces with their proper magnitudes and directions

14. Example 1

15. Example 2

16. Example 3

17. Solution Resultant R of the two forces in two ropes:

18. Solution Equilibrant E = - R

19. Solution Resultant R is the sum of the actions of the tow ropes on the barge Equilibrant E is the reaction of the barge to the ropes E = - R

20. Moment and Couple Moment of Force Moment M of the force F about the point O is defined as: M = F d where d is the perpendicular distance from O to F Moment is directional

21. Moment and Couple Moment = Force x Perpendicular Distance

22. Resultant of a system of forces An arbitrary body subjected to a number of forces F1, F2 & F3. Resultant R = F1 + F2 + F3 Components Rx = F1x + F2x + F3x Ry = F1y + F2y + F3y

23. Resultant Moment Resultant moment Mo = Sum of Moments Mo = F 1 l 1 + F 2 l 2 + F 3 l 3 = R l

24. Couple For a Couple R =  F = 0 But Mo  0 lMo = F(d+l) - Fl = Fd lMoment of couple is the same about every point in its plane Mo = F d

25. Example 4 Calculate the total (resultant) moment on the body.

26. Example 4 (Solution) Taking moments about the corner A Note that the forces form two couples or pure moments 3.6 Nm and 3.0 Nm(resultant force =0, moment is the same about any point).

27. Exercise 1. What is the moment of the 10 N force about point A (M A )? A) 10 N·m B) 30 N·m C) 13 N·m D) (10/3) N·m E) 7 N·m A d = 3 m F = 10 N

28. APPLICATIONS What is the net effect of the two forces on the wheel?

29. APPLICATIONS What is the effect of the 30 N force on the lug nut?

30. MOMENT IN 2-D The moment of a force about a point provides a measure of the tendency for rotation (sometimes called a torque).

31. Moment F=100 L=20 M =_____________

32. Moment F=55N F=32N L=50cm L=300mm +M=27.5N M=-9.6N +

33. Given: A 400 N force is applied to the frame and  = 20°. Find: The moment of the force at A. Plan: 1) Resolve the force along x and y axes. 2) Determine M A using scalar analysis. EXAMPLE 1

34. Solution +  F x = -400 cos 20° N +  F y = -400 sin 20° N + M A = {(400 cos 20°)(2) + (400 sin 20°)(3)} N·m = 1160 N·m

35. GROUP PROBLEM SOLVING Given: A 40 N force is applied to the wrench. Find: The moment of the force at O. Plan: 1) Resolve the force along x and y axes. 2) Determine M O using scalar analysis. Solution: +  F y = - 40 cos 20° N +  F x = - 40 sin 20° N + M O = {-(40 cos 20°)(200) + (40 sin 20°)(30)}N·mm = -7107 N·mm = - 7.11 N·m