Equilibrium
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Transcript of Equilibrium
![Page 1: Equilibrium](https://reader033.fdocuments.us/reader033/viewer/2022061119/546afa72af795920668b68dd/html5/thumbnails/1.jpg)
EQUILIBRIUM
APPLICATION OF THE FIRST LAW
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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 KNOWIF THE OBJECT IS
IN EQUILIBRIUM?
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CENTER OF GRAVITY
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Center of Gravity
•The location where all of the weight of an object seemed to be concentrated.
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How do we locate the C.G?
• For a regularly-shaped object, it is at its geometric center.
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Center of Gravity
Height, h
C.G.
1/3 h1/4 h
C.G.
Solid cone Triangle
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• Sometimes the C.G. is found outside the body.
How do we locate the C.G?
C.G
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• For an irregularly-shaped object,• it could be determined by balancing it by
trial and error method or by the plumb bob method
Locating the C.G. of Irregularly-shaped object
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.
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STATES OF EQUILIBRIUM
Any object at rest may be in one of three states of equilibrium.
• STABLE• UNSTABLE and• NEUTRAL
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STABLE EQUILIBRIUM
• 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.
For stable objects:
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UNSTABLE EQUILIBRIUM
• 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.
For unstable objects:
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NEUTRAL EQUILIBRIUM
• the C.G. is neither lowered nor raised when the object is toppled.
• they roll from one side to another.
For objects with neutral equilibrium:
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TOPPLING
Physics
by A
lvea Jo
i Sika
t
Physics by Alvea Joi Sikat
Physics
by A
lvea Jo
i Sika
t
Toppling the upright book requires only a slight raising of C.G.
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TOPPLING
Physi
cs b
y A
lvea Joi Sik
at
Physics by Alvea Joi Sikat Phys
ics
by
Alv
ea Joi Sik
atToppling the flat book
requires a relatively large raising of its C.G.
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TOPPLINGToppling the cylinder does not change the height of its C.G.
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3 FACTORS FOR STABILITY1. Mass of the object2. Location of the center of gravity3. Area of the base of support
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The First ConditionOf Equilibrium
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EQUILIBRIUM
• What force/s are acting on the block of wood? Draw a free-body diagram.
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STATIC EQUILIBRIUM
Fg
FNNormal force
Gravitational Force
F F F F F F F F FΣFTable
ΣFWood
Weight of the wood, W = mg
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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.
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F1 F2
F3
F4
The Ring Four forces are acting on the ring.
If the ring is to remain at rest:
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F1 F2
F3
F4
+x
+y
-y
-x
Draw the free-body diagram.
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ΣF = ma =0ΣF = F1 + F2 + F3 + F4 = 0ΣF x = (-F1) + F2 = 0ΣF = F3 + (-F4) = 0
F1 F2
F3
F4
+x
+y
-y
-x
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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.
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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
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T
Fg
Tension
Gravitational force
The chandelier has a mass of 3.0 kg. What is the tension in the cord?
Example no.1
W
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Free-body Diagram
T
W= mg
Given: m= 3.8kgFind: T = ?
ΣF = 0 ΣFx = 0ΣFy = T – W = 0T – mg = 0T = mg = (3.0 kg)(9.8m/s2)T = 29.4 N
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Find F1 and F2
Jaztene’s Internet Café
F1F2
W = 600 N
60°
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Free-body diagramΣF = 0ΣFX = -F1x + F2x = 0
-F1 cos 60°+F2cos60°= 0
F1 = F2 ----- eq. 1
x
y F2F1
W = 600 N
60°60°60°
F1X F2X
F1y F2y ΣFy = F1y + F2x = 0
F1 sin60°+F2sin60°-W = 0
2F1 sin60°= 600N F1 = 600N
2 sin60°600N1.73
=
F1 = F2 = 347 N
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60°
T1
T2
W = 2000N
3. Determine the tension in the cords supporting the 2000-N load?
ΣF = 0ΣFX = -T1x + T2 = 0
-T1 cos 30°+ T2= 0
T2 = T1 cos 30° ---- eq. 1ΣFy = T1y - W = 0
T1 sin30°-W = 0
T1 sin30°= 2000N
T1 = 4000 N
T2 = 2000 N