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Statics free body diagram



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Statics free body diagram

  1. 1. Statics (MET 2214) Statics of Particles MET 2214
  2. 2. Statics (MET 2214) Static Equilibrium for a Particle Objective: To introduce the concept of the free-body diagram for a particle and to show how to solve particle equilibrium problems using the equations of equilibrium. A particle: An object with inertia (mass) but of negligible dimensions. A particle at rest: A particle is at rest if originally at rest or has a constant velocity if originally in motion.
  3. 3. Statics (MET 2214) Equilibrium equations for a particle A particle is in equilibrium if the resultant of ALL forces acting on the particle is equal to zero. (Newton’s first law is that a body at rest is not subjected to any unbalanced forces). Sum of all forces acting on a particle = 0F =∑
  4. 4. Statics (MET 2214) Equilibrium equations in component form In a rectangular coordinate system the equilibrium equations can be represented by three scalar equations: 0 0 0 x y z F F F = = = ∑ ∑ ∑
  5. 5. Statics (MET 2214) Free-Body Diagram (FBD): To apply equilibrium equations we must account for all known and unknown forces acting on the particle. The best way to do this is to draw a free-body diagram of the particle. FBD: A diagram showing the particle under consideration and all the forces and moments acting on this particle. This is a sketch that shows the particle “free” from its surroundings with all the forces acting on it.
  6. 6. Statics (MET 2214) Parallelogram Law Copyright of Ohio University Two forces on a body can be replaced by a single force called the resultant by drawing the diagonal of the parallelogram with sides equivalent to the two forces.
  7. 7. Statics (MET 2214) Principal of Transmissibility The conditions of equilibrium or motion of a body remain unchanged if a force on the body is replaced by a force of the same magnitude and direction along the line of action of the original force.
  8. 8. Statics (MET 2214) Mechanical components String or cable: A mechanical device that can only transmit a tensile force along itself.
  9. 9. Statics (MET 2214) Mechanical components Linear spring: A mechanical device which exerts a force along its line of action and proportional to its extension (F = kX). K is constant of proportionality which is a measure of stiffness or strength.
  10. 10. Statics (MET 2214) Mechanical components Cables: Cables are assumed to have negligible weight and they cannot stretch. They can only support tension or pulling (you can’t push on a rope!). Frictionless pulleys: Pulleys are assumed to be frictionless.
  11. 11. Statics (MET 2214) Mechanical components A continuous cable passing over a frictionless pulley must have tension force of a constant magnitude. The tension force is always directed in the direction of the cable. For a frictionless pulley in static equilibrium, the tension in the cable is the same on both sides of the pulley.
  12. 12. Statics (MET 2214) Force types Force types: Active Forces - tend to set the particle in motion. Reactive Forces - result from constraints or supports and tend to prevent motion. Active force Reactive force Active force Reactive force
  13. 13. Statics (MET 2214) Free Body Diagram (FBD) How to draw a Free Body Diagram: Draw outlined shape - Imagine the particle isolated or cut “free” from its surroundings Show all forces and moments - Include “active forces” and “reactive forces”. Place each force and couple at the point that it is applied. Identify each force: Known forces labeled with proper magnitude and direction. Letters used for unknown quantities. Add any relevant dimensions onto your picture.
  14. 14. Statics (MET 2214) FBD
  15. 15. Statics (MET 2214) FBD F.B.D of the ring A:
  16. 16. Statics (MET 2214) Example 1 The sphere has a mass of 6 kg and is supported as shown. Draw a free-body diagram of the sphere, cord CEsphere, cord CE, and the knot at C.the knot at C.
  17. 17. Statics (MET 2214) Sphere There are two forces acting on the sphere. These are its weight and the force of cord CE. The weight is: W = 6 kg (9.81 m/s2 ) = 58.9 N.
  18. 18. Statics (MET 2214) FBD of sphere This is the way we show the FBD of the sphere: FCE 58.9 N
  19. 19. Statics (MET 2214) Cord CE There are two forces acting on the cord. These are the force of the sphere, and the force of the knot. A cord is a tension only member. Newton’s third law applies.
  20. 20. Statics (MET 2214) FBD of the cord CE FCE FEC C E
  21. 21. Statics (MET 2214) Knot at C There are three forces acting on the knot at C. These are the force of the cord CBA, and the force of the cord CE, and the force of the spring CD.
  22. 22. Statics (MET 2214) FBD of the knot at C FCE FCBA FCD C 60o
  23. 23. Statics (MET 2214) Example 2 Draw the FBD diagram of the ring A: W= 2.452 KN
  24. 24. Statics (MET 2214) FBD of the ring A Is this the FBD of A? No! this is not the free body diagram of A!
  25. 25. Statics (MET 2214) FBD of the ring A
  26. 26. Statics (MET 2214) Example 3 Draw the free body diagrams of C and E and the cable CE:
  27. 27. Statics (MET 2214) FBD of E
  28. 28. Statics (MET 2214) FBD of C
  29. 29. Statics (MET 2214) FBD of cable EC
  30. 30. Statics (MET 2214) Example 4 Draw the FBD of ring A. W=78.5 N
  31. 31. Statics (MET 2214) FBD of A
  32. 32. Statics (MET 2214) Part 2 Applying the Equilibrium Equations
  33. 33. Statics (MET 2214) FBD Draw the free body diagrams: W N W N f Normal force = The force you have when there is a contact between surfaces (the ball is in contact with the ground). Friction force = You have this when the surface in contact is not frictionless and the friction prevents the motion of the object. 30
  34. 34. Statics (MET 2214) FBD 0 0 x y F F = = ∑ ∑ W N W N f 30 0 0 x y F F = = ∑ ∑ x y x y
  35. 35. Statics (MET 2214) yFBD W N N f N = W N - W.cos 30 = 0 f - W.sin 30 = 0 W.cos30 W.sin30 x x y
  36. 36. Statics (MET 2214) Example 2: Determine the tension in cables AB and AD for equilibrium of the 250 kg engine. FBD of the ring A
  37. 37. Statics (MET 2214) B 0, cos30 0 0, sin30 2.452 0 Solving for T : sin30 2.452 , 4.90 Subsituting into the first equation: 4.25 x B D y B B B D F T T F T kN T kN T kN T kN = − = = − = = = = ∑ ∑ Solution of Example 2 According to the free body diagram of the ring A, we have three forces acting on the ring. The forces TB and TD have unknown magnitudes but known directions. Cable AC exerts a downward force on A equal to: W = (250kg)(9.81m/s2 ) = 2452N = 2.245KN TBcos30 TBsin30


  • Ok. Lets get started.
    My name is Dr Simin Nasseri, a faculty member here at MET and I am going to teach this course MET ????, Course name.
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