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Physics Lesson 8 Static Equilibrium

Physics Lesson 8 Static Equilibrium. Eleanor Roosevelt High School Chin -Sung Lin. Net Force. Net Force. Net Force – Sum of all Forces. The combination of all forces acting on an object is called the net force. Net Force. Calculate Net Force – Vector Addition.

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Physics Lesson 8 Static Equilibrium

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  1. Physics Lesson 8Static Equilibrium Eleanor Roosevelt High School Chin-Sung Lin

  2. Net Force

  3. Net Force Net Force – Sum of all Forces • The combination of all forces acting on an object is called the net force

  4. Net Force Calculate Net Force – Vector Addition • Vector addition can be used to calculate the net force

  5. Net Force Calculate Net Force – Vector Resolution/Addition • Since vector addition by graphical method is not precise, it is preferred to resolve all the force vectors into horizontal and vertical components, and then add them respectively y F Fy Fx = F cosθ Fy = F sin θ x Fx

  6. Net Force Calculate Net Force B A

  7. Net Force Calculate Net Force B A + B A

  8. Net Force Calculate Net Force B A

  9. Net Force Calculate Net Force A + B B A

  10. Net Force Calculate Net Force B A

  11. Net Force Calculate Net Force A – B B A

  12. Net Force Calculate Net Force B A

  13. Net Force Calculate Net Force A – B B A

  14. Net Force Calculate Net Force A A

  15. Net Force Calculate Net Force 0 A A

  16. Net Force Calculate Net Force A A

  17. Net Force Calculate Net Force 0 A A

  18. Gravitational Force (Fg) The gravitational force (Fg) is the attraction force exerted by Earth on the block Fg

  19. Normal Force (FN) The normal force (FN) is the reaction force exerted by ground on the block FN Fg

  20. Normal Force (FN) Normal force is always perpendicular to the surface FN • FN = Fg Fg

  21. Friction Force (Ff) The friction force (Ff) is the reaction force exerted by ground on the block Friction force is always against the motion F Ff

  22. Friction Force (Ff) When the net force is equal to zero, friction force (Ff) is equal to the the force (F) exerted on the block F Ff • Ff= FF

  23. Tension Force (FT) The tension force (FT) is the force exerted by the rope on the block FT Fg

  24. Tension Force (FT) When the net force is equal to zero,tension force (FT) is equal to the the gravitational force (Fg) exerted on the block FT • FT = Fg Fg

  25. Equilibrium

  26. Statics Staticsis the branch of mechanics concerned with the analysis of force on physical systems in static equilibrium, that is, in a state where the relative positions of subsystems are at rest, or where components and structures are at a constant velocity

  27. Equilibrium When an object is at rest, or at constant velocity, with the net force on it being zero, the object is in a state of equilibrium

  28. Free-Body Diagram A diagram showing all the forces acting on an object is called a free-body diagram Drawing a free-body diagram should always the first step in solving physics problems involving forces

  29. Free-Body Diagram We isolate, or free the object of concern from everything else, and represent that object by a dot. We then draw all the forces vectors acting on the object with its tail starting on the dot. We label each vector to indicate what type of force it represents – • W(or Fg) for a gravitational force • N (or FN) for a normal force • f(or Ff) for a friction force, and • T(or FT) for a tension force FT Fg

  30. Vector Addition & Vector Resolution

  31. Vector Addition An object of 10 N hangs from two spring scales When the spring scales hang at 60o from the vertical, their readings are 10 N each When the angle has increased to 75.5o, the readings are 20 N each

  32. Vector Addition As the angle between the scales increases, the tension in the scales must increase for the resultant to remain 10 N

  33. Vector Addition For any pair of scales, ropes, or wires supporting a load, the greater their angles from the vertical, the larger the tension force in them. The resultant of the tension forces, or the diagonal of the parallelogram they describe, must be equal and opposite to the load being supported

  34. Vector Resolution 2 FTcos = 2 x 10 N xcos 60o = 2 x 10 N x 1/2 = 10 N 2 FTcos = 2 x 20 N xcos 75.5o = 2 x 20 N x 1/4 = 10 N

  35. Q & A

  36. The End

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