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EQUILIBRIUM APPLICATION OF THE  FIRST LAW
STATIC EQUILIBRIUM A 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 know If the object is In equilibrium?
CENTER OF GRAVITY
Center of Gravity The location where all of the weight of an object seemed to be concentrated.
How do we locate the C.G? For a regularly-shaped object, it is at its geometric center.
 Center of Gravity C.G. C.G. Height, h 1/3 h 1/4 h Solid cone Triangle
Sometimes the C.G. is found outside the body. How do we locate the C.G? C.G
Locating the C.G. of Irregularly-shaped object For an irregularly-shaped object, it could be determined by balancing it by trial and error method or by the plumb bob method Assignment:   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.
STATES OF EQUILIBRIUM Any object at rest may be in one of three states of equilibrium. STABLE UNSTABLE and NEUTRAL
STABLE EQUILIBRIUM For 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.
UNSTABLE EQUILIBRIUM For 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.
NEUTRAL EQUILIBRIUM For objects with neutral equilibrium: the C.G. is neither lowered nor raised when the object is toppled. they roll from one side to another.
TOPPLING  Toppling the upright book requires only a slight raising of C.G. Physics by AlveaJoiSikat Physicsby AlveaJoiSikat Physicsby AlveaJoiSikat
TOPPLING  Physicsby AlveaJoiSikat Physicsby AlveaJoiSikat Toppling the flat book requires a relatively large raising of its C.G. Physicsby AlveaJoiSikat
TOPPLING Toppling the cylinder does not change the height of its C.G.
3 FACTORS FOR STABILITY Mass of the object Location of the center of gravity Area of the base of support
The First Condition Of Equilibrium
EQUILIBRIUM What force/s are acting on the block of wood? Draw a free-body diagram.
STATIC EQUILIBRIUM FN Normal force F FFFFFFFF ΣFTable ΣFWood Fg Weight of the wood,    W = mg Gravitational Force
STATIC EQUILIBRIUM A 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.
F3 The Ring Four forces are acting on the ring. If the ring is to remain at rest: F1 F2 F4
F3 +y Draw the free-body diagram. F1 F2 +x -x F4 -y
F3 +y F1 F2 +x -x F4 -y ΣF = ma =0 ΣF =  F1 + F2 + F3 + F4 = 0 ΣF x = (-F1) + F2 = 0 ΣF = F3 + (-F4) =  0
STATIC EQUILIBRIUM The 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.
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 =  0 ΣFx = 0 and ΣFy= 0
Example no.1 The chandelier has a mass of 3.0 kg. What is the tension in the cord? T Tension Fg Gravitational force W
Free-body Diagram Given: m= 3.8kg Find: T = ?  ΣF = 0  ΣFx = 0 ΣFy = T – W = 0 T – mg = 0 T = mg = (3.0 kg)(9.8m/s2) T = 29.4 N T W= mg
            Find F1 and F2 F2 F1 60° Jaztene’s Internet Café W = 600 N
ΣF = 0 ΣFX = -F1x + F2x = 0 -F1cos 60°+F2cos60°= 0  F1 = F2   ----- eq. 1 Free-body diagram F1 y F2 F1y F2y ΣFy = F1y + F2x = 0 F1 sin60°+F2sin60°-W = 0  2F1 sin60°= 600N  60° 60° 60° x F1X F2X 600N 2 sin60° 600N 1.73 F1 =  =  W = 600 N F1 = F2 = 347 N
3. Determine the tension in the cords supporting the 2000-N load? ΣF = 0 ΣFX = -T1x + T2 = 0 -T1cos 30°+ T2= 0  T2 = T1cos 30°   ---- eq. 1 60° T1 T2 ΣFy = T1y- W = 0 T1 sin30°-W = 0 T1 sin30°= 2000N T1 = 4000 N T2 = 2000 N  W = 2000N

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Static Equilibrium and Center of Gravity

  • 2. STATIC EQUILIBRIUM A 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 know If the object is In equilibrium?
  • 4. Center of Gravity The 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 Gravity C.G. C.G. Height, h 1/3 h 1/4 h Solid cone Triangle
  • 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 object For an irregularly-shaped object, it could be determined by balancing it by trial and error method or by the plumb bob method Assignment: 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 EQUILIBRIUM Any object at rest may be in one of three states of equilibrium. STABLE UNSTABLE and NEUTRAL
  • 10. STABLE EQUILIBRIUM For 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 EQUILIBRIUM For 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 EQUILIBRIUM For 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 AlveaJoiSikat Physicsby AlveaJoiSikat Physicsby AlveaJoiSikat
  • 14. TOPPLING Physicsby AlveaJoiSikat Physicsby AlveaJoiSikat Toppling the flat book requires a relatively large raising of its C.G. Physicsby AlveaJoiSikat
  • 15. TOPPLING Toppling the cylinder does not change the height of its C.G.
  • 16. 3 FACTORS FOR STABILITY Mass of the object Location of the center of gravity Area of the base of support
  • 17.
  • 18. The First Condition Of Equilibrium
  • 19. EQUILIBRIUM What force/s are acting on the block of wood? Draw a free-body diagram.
  • 20. STATIC EQUILIBRIUM FN Normal force F FFFFFFFF ΣFTable ΣFWood Fg Weight of the wood, W = mg Gravitational Force
  • 21. STATIC EQUILIBRIUM A 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. F3 The Ring Four forces are acting on the ring. If the ring is to remain at rest: F1 F2 F4
  • 23. F3 +y Draw the free-body diagram. F1 F2 +x -x F4 -y
  • 24. F3 +y F1 F2 +x -x F4 -y ΣF = ma =0 ΣF = F1 + F2 + F3 + F4 = 0 ΣF x = (-F1) + F2 = 0 ΣF = F3 + (-F4) = 0
  • 25. STATIC EQUILIBRIUM The 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 = 0 ΣFx = 0 and ΣFy= 0
  • 27. Example no.1 The chandelier has a mass of 3.0 kg. What is the tension in the cord? T Tension Fg Gravitational force W
  • 28. Free-body Diagram Given: m= 3.8kg Find: T = ? ΣF = 0 ΣFx = 0 ΣFy = T – W = 0 T – mg = 0 T = mg = (3.0 kg)(9.8m/s2) T = 29.4 N T W= mg
  • 29. Find F1 and F2 F2 F1 60° Jaztene’s Internet Café W = 600 N
  • 30. ΣF = 0 ΣFX = -F1x + F2x = 0 -F1cos 60°+F2cos60°= 0 F1 = F2 ----- eq. 1 Free-body diagram F1 y F2 F1y F2y ΣFy = F1y + F2x = 0 F1 sin60°+F2sin60°-W = 0 2F1 sin60°= 600N 60° 60° 60° x F1X F2X 600N 2 sin60° 600N 1.73 F1 = = W = 600 N F1 = F2 = 347 N
  • 31. 3. Determine the tension in the cords supporting the 2000-N load? ΣF = 0 ΣFX = -T1x + T2 = 0 -T1cos 30°+ T2= 0 T2 = T1cos 30° ---- eq. 1 60° T1 T2 ΣFy = T1y- W = 0 T1 sin30°-W = 0 T1 sin30°= 2000N T1 = 4000 N T2 = 2000 N W = 2000N