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Engineering Fundamentals

Engineering Fundamentals. Session 9. 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.

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Engineering Fundamentals

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  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 F1 + F2 + … + Fn = 0 Example F1 + F2 + F3= 0

  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 R mg R R

  8. Free Body Diagram

  9. Definitions • System of Particles or BodiesTwo or more bodies or particles connected together are referred to as a system of bodies or particles. • External ForceExternal 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 reactionsdue 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 E = - R Equilibrant E is the reaction of the barge to the ropes

  20. Moment and Couple • Moment of Force • Moment M of the force F about the point O is defined as:M = F dwhere 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 ComponentsRx = F1x + F2x + F3xRy = F1y + F2y + F3y

  23. Resultant Moment Resultant moment Mo= Sum of Moments Mo = F1 l1 + F2 l2 + F3 l3 = R l

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

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

  26. 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). Example 4 (Solution)

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

  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 =_____________ M L=20

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

  33. EXAMPLE 1 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 MA using scalar analysis.

  34. EXAMPLE 1 Solution +  Fx = -400 cos 20° N +  Fy = -400 sin 20° N + MA = {(400 cos 20°)(2) + (400 sin 20°)(3)} N·m = 1160 N·m

  35. Solution: +  Fy = - 40 cos 20° N +  Fx = - 40 sin 20° N + MO = {-(40 cos 20°)(200) + (40 sin 20°)(30)}N·mm = -7107 N·mm = - 7.11 N·m 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 MO using scalar analysis.

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