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# Equilibrium

## by Nestor Enriquez on Sep 25, 2010

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## EquilibriumPresentation Transcript

• 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