This document discusses static equilibrium and concepts related to determining if an object is in equilibrium. It defines static equilibrium as a state of balance where the forces acting on an object are balanced, causing it to remain at rest. It also discusses the center of gravity and how to locate it for different shapes, as well as the three states of equilibrium - stable, unstable, and neutral. Factors that determine an object's stability like mass, center of gravity location, and base of support are also covered.

1. EQUILIBRIUMAPPLICATION OF THE FIRST LAW

2. STATIC EQUILIBRIUMA state of balance of an object at rest.A condition in which all forces acting on the body are balanced, causing the body to remain at rest.What is equilibrium?How do we knowIf the object isIn equilibrium?

3. CENTER OF GRAVITY

4. Center of GravityThe location where all of the weight of an object seemed to be concentrated.

5. How do we locate the C.G?For a regularly-shaped object, it is at its geometric center.

6. Center of GravityC.G.C.G.Height, h1/3 h1/4 hSolid coneTriangle

7. Sometimes the C.G. is found outside the body.How do we locate the C.G?C.G

8. Locating the C.G. of Irregularly-shaped objectFor an irregularly-shaped object,it could be determined by balancing it by trial and error method or by the plumb bob methodAssignment: In one long folder, draw the island of Luzon. Cut it out and locate its center of gravity. Mark it with a pen. At the back of the cut-out folder, write how you determined the location of the C.G.

9. STATES OF EQUILIBRIUMAny object at rest may be in one of three states of equilibrium.STABLEUNSTABLE andNEUTRAL

10. STABLE EQUILIBRIUMFor stable objects:the C.G. is at lowest possible position.the C.G. needs to be raised in order to topple the object.they are difficult to topple over.

11. UNSTABLE EQUILIBRIUMFor unstable objects:the C.G. is at the highest possible position.the C.G. is lowered in order to topple the object.They are easy to topple down.

12. NEUTRAL EQUILIBRIUMFor objects with neutral equilibrium:the C.G. is neither lowered nor raised when the object is toppled.they roll from one side to another.

13. TOPPLING Toppling the upright book requires only a slight raising of C.G.Physics by AlveaJoiSikatPhysicsby AlveaJoiSikatPhysicsby AlveaJoiSikat

14. TOPPLING Physicsby AlveaJoiSikatPhysicsby AlveaJoiSikatToppling the flat book requires a relatively large raising of its C.G.Physicsby AlveaJoiSikat

15. TOPPLINGToppling the cylinder does not change the height of its C.G.

16. 3 FACTORS FOR STABILITYMass of the objectLocation of the center of gravityArea of the base of support

18. The First ConditionOf Equilibrium

19. EQUILIBRIUMWhat force/s are acting on the block of wood? Draw a free-body diagram.

20. STATIC EQUILIBRIUMFNNormal forceF FFFFFFFFΣFTableΣFWoodFgWeight of the wood, W = mgGravitational Force

21. STATIC EQUILIBRIUMA state of balance of an object at rest.A condition in which all forces acting on the body are balanced, causing the body to remain at rest.

22. F3The RingFour forces are acting on the ring.If the ring is to remain at rest:F1F2F4

23. F3+yDraw the free-body diagram.F1F2+x-xF4-y

24. F3+yF1F2+x-xF4-yΣF = ma =0ΣF = F1 + F2 + F3 + F4 = 0ΣF x = (-F1) + F2 = 0ΣF = F3 + (-F4) = 0

25. STATIC EQUILIBRIUMThe body must be in translational equilibrium or the body does not accelerate along any line.If the acceleration is zero, then the resultant of the forces acting on the body is also zero.

26. The First Condition of Equilibrium:If the sum of all forces acting concurrently on a body is equal to zero, then the body must be in static equilibrium. Mathematically:ΣF = Fnet = 0ΣFx = 0 and ΣFy= 0

27. Example no.1The chandelier has a mass of 3.0 kg. What is the tension in the cord?TTensionFgGravitational forceW

28. Free-body DiagramGiven: m= 3.8kgFind: T = ? ΣF = 0 ΣFx = 0ΣFy = T – W = 0T – mg = 0T = mg = (3.0 kg)(9.8m/s2)T = 29.4 NTW= mg

29. Find F1 and F2F2F160°Jaztene’sInternet CaféW = 600 N

30. ΣF = 0ΣFX = -F1x + F2x = 0-F1cos 60°+F2cos60°= 0 F1 = F2 ----- eq. 1Free-body diagramF1yF2F1yF2yΣFy = F1y + F2x = 0F1 sin60°+F2sin60°-W = 0 2F1 sin60°= 600N 60°60°60°xF1XF2X600N2 sin60°600N1.73F1 = = W = 600 NF1 = F2 = 347 N

31. 3. Determine the tension in the cords supporting the 2000-N load?ΣF = 0ΣFX = -T1x + T2 = 0-T1cos 30°+ T2= 0 T2 = T1cos 30° ---- eq. 160°T1T2ΣFy = T1y- W = 0T1 sin30°-W = 0T1 sin30°= 2000NT1 = 4000 NT2 = 2000 N W = 2000N