Equilibrium

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 KNOWIF 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

Height, h

C.G.

1/3 h1/4 h

C.G.

Solid cone Triangle

• Sometimes the C.G. is found outside the body.

How do we locate the C.G?

C.G

• 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.

STATES OF EQUILIBRIUM

Any object at rest may be in one of three states of equilibrium.

• STABLE• UNSTABLE and• NEUTRAL

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:

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:

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:

TOPPLING

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Toppling the upright book requires only a slight raising of C.G.

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requires a relatively large raising of its C.G.

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

3 FACTORS FOR STABILITY1. Mass of the object2. Location of the center of gravity3. Area of the base of support

The First ConditionOf Equilibrium

EQUILIBRIUM

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

STATIC EQUILIBRIUM

Fg

FNNormal force

Gravitational Force

F F F F F F F F FΣFTable

ΣFWood

Weight of the wood, W = mg

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.

F1 F2

F3

F4

The Ring Four forces are acting on the ring.

If the ring is to remain at rest:

F1 F2

F3

F4

+x

+y

-y

-x

Draw the free-body diagram.

ΣF = ma =0ΣF = F1 + F2 + F3 + F4 = 0ΣF x = (-F1) + F2 = 0ΣF = F3 + (-F4) = 0

F1 F2

F3

F4

+x

+y

-y

-x

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

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

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

Find F1 and F2

Jaztene’s Internet Café

F1F2

W = 600 N

60°

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

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