THE CONCEPT OF

EQUILIBRIUM

According to Rene Descartes, the human body is a mechanical

system designed by the hands of God.

HUMAN LEVER

Borelli, a student of

Galileo, considered the human body as

a system of levers.

In our body, the bones act as the rigid bar and the joints the fulcrum. It is through

these so-called human levers that we are able to move, no

matter how slight our movement is, to talk, walk,

and even eat.

EQUILIBRIUM

Is the condition of force where it is acted but

simply cancelled out. These forces may be even

large enough to cause

permanent deformation.

STATICS

It is concerned with the

calculation of the forces

acting on and within

structures that are in

equilibrium.

FORCE SYSTEMS

Concurrent and Nonconcurrent

Concurrent system occur when the lines of actions of the forces acting on a body intersect at a common point.

Nonconcurrent system occurs when the forces are acting at different

points.

Nonconcurrent forces may also me parallel and non parallel. The lines of

action of parallel forces

do not intersect.

PARTICLE AND RIGID BODYDistinction has to made whether

the object being acted upon by a system of forces is a particle or a

rigid body.

A PARTICLE may be

considered as a point of

mass

A RIGID BODY is an extended body in

space that does not

change its size and shape

TRANSITIONAL EQUILIBRIUMEquilibrium is a condition where

there is no change in the state of motion of a body. An object in

equilibrium may be at rest. It can also be in motion, provided that it moves with constant speed in the

same direction.

An object that has no net force acting on it is said to be in

TRANSLATIONAL EQUILIBRIUM. This condition of translational

equilibrium is usually referred to as the first condition for equilibrium

and may be expressed in the form:

∑F = 0

SAMPLE PROBLEMYou hang your picture frame by means of vertical string. Two strings in turn

support this string. Each string makes 30° with an overhead horizontal beam.

Find the tension in the strings.T1 T2

T3

30°30°

SOLUTIONWe are given w = 55N. The forces

acting on the frame are shown. The tension T3 in the rope pulls on the frame upward, while the weight of the frame acts downward. The frame will be in equilibrium when ∑F = 0.

55N

T3

Free body part of the frame

T3 - 55 N = 0

T3 = 55 N

To determine T1 and T2 consider the point where the three ropes meet as a particle in equilibrium.

T1

T2

30° 30°

w

Applying the first condition for equilibrium, we have:

∑Fx = 0: T1 cos 30°- T2 cos 30° = 0 T1 cos = T2

∑Fy = 0: T1 sin 30°+ T2 sin 30° - 55N = 0Since T1 = T2 we may replace T2 by T1 and solve the above equation for T1.∑Fy = 0: T1 (.5) + T1 (.5) – 55 N = 0

T1 = 55 NT2 = 55 N

TORQUEIn physics, a TORQUE (τ) is a vector that

measures the tendency of a force to rotate an object about some axis The

magnitude of a torque is defined as force times its lever arm . Just as a force is a

push or a pull, a torque can be thought of as a twist.The SI unit for torque is newton meters

(N m). In U.S. customary units, it is measured in foot pounds (ft·lbf) (also

known as 'pounds feet'). The symbol for torque is τ, the Greek letter tau .

EXPLANATIONThe force applied to a lever, multiplied by its distance from the lever's fulcrum, is

the torque. For example, a force of three newtons applied two meters from the fulcrum exerts the same torque as one

newton applied six meters from the fulcrum. This assumes the force is in a direction at right angles to the straight lever. The direction of the torque can be determined by using the right hand rule: Using your right hand, curl your fingers in

the direction of rotation, and stick your thumb out so it is aligned with the axis of

rotation. Your thumb points in the direction of the torque vector.

Mathematically, the torque on a particle (which has the position r in some reference frame) can be defined as the

cross product:where T = r x F

r is the particle's position vector relative to the fulcrum

F is the force acting on the particles, or, more generally, torque can be defined as the rate of

change of angular momentum,where T = dL dt

L is the angular momentum vector t stands for time.

As a consequence of either of these definitions, torque is a vector, which points along the axis of the rotation it

would tend to cause.

PROBLEMFor a position a, the lever arm is

already given as 0.25m. The force will produce a clockwise rotation respect to O and therefore the torque is negative.

t = - (60 N)(.25 m) = -15 Nm

SOLUTIONThe resultant force = 7.0 – 7.0 N =

0

CENTER OF GRAVITY

The center of mass or center of gravity is a useful concept when dealing with

equilibrium problems.Each of the particles making up a body hast its own weight or mass. The mass or weight of the object is the sum of the

masses or weights of these particles. The center of gravity of a body is the point

where its entire weight maybe assumed concentrated.

CENTER of GRAVITY of a GROUP of BODIES

The center of gravity of a group of bodies whose centers of gravity are

known from a fixed point is defined as the sum of the products of weight of

individual body and it center of gravity divided by the total weight. In symbols:

W1X1 + W2X2 ….

W1 + W2 + …..

PROBLEMFind the center of gravity of a flat piece of wood shaped in the form of letter L.

All dimensions are in meters. The weight per square meter of the wood is

2.4 N/m2.y

0

(X1, Y1)

3.0

4.0

3.0

(X2, Y2)

x

SOLUTIONThe center of gravity of the wood will

specified by two coordinates: x and y. We divide the plate into regularly shaped parts. Let A1 be the area of the bigger rectangle and

A2 the area of the smaller rectangle. The center of gravity of A1 labeled as (x1,y1) is the

(1.5 m, 3.0 m). The center of gravity of A2 labeled as (x2, y2) is (4.5 m, 1.0 m). A1 = (3 m) (6 m) = 18 m2

W1 = (2.4 kg/m2) (18 m2) = 43.2 N

A2 = (3 m)(6 m) = 18 m2

W2 = (2.4 kg/m2) (6.m2) = 14.4 N

X = (43.2 N) (1.5 m) + (14.4 N)(4.5 m) 43.2 N + 14.4 N

= 2.3 m

y = (43.2 N) (3 m) + (14.4 N)(1 m)

43.2 N + 14.4 N 2.5 m

=

ROTATIONAL EQUILIBRIUMA necessary condition for a body to be in rotational equilibrium is that the sum of

the torques with their proper signs about point must be zero.

∑t = 0The condition is known as the second

condition for equilibrium.

SAMPLE PROBLEMA 120 N child and a 200 N child sit at the opposite ends of 4.00 m uniform seesaw pivoted at its center. Where

should a 140 N child sit to balance the seesaw?

2.0 m

120 N 140 N

R

W 200 N

FREE BODY DIAGRAM OF THE SEESAW

SOLUTIONLet w represent the weight of the seesaw. R is the upward reaction

at the pivotal point. The third child must sit on the same side of the seesaw as the 120 N child. Let x be the distance of the third child

from the center of the seesaw. Applying the second condition for

equilibrium and talking summation of torques about the center of the

seesaw will eliminate the rotational effect of R and w.

+ (120 N)(2.00 m) + (140)(x) – (200 N)(2.00 m) = 0

Solving for x, x = 1.14 m from the center on the side of the 120 N

child.

STABILITY

Three types of equilibrium:

UNSTABLE – the great example of this is a cone that is balance on its apex but when disturbed slightly, it will fall over

STABLE – the condition of an object to return it is original position when slightly disturbed.

NEUTRAL – the condition where an object is lying on its side and displace but manages to remains its equilibrium about its new position