Physics Equilibrium

this is a sample powerpoint presentation on the concept of EQUILIBRIUM...

1. THE CONCEPT
OF
EQUILIBRIUM

2. According to Rene
Descartes, the
human body is a
mechanical system
designed by the
hands of God.
HUMAN LEVER

3. Borelli, a student of
Galileo, considered
the human body as
a system of levers.

4. 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.

5. 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.

6. STATICS
It is concerned with
the calculation of
the forces acting
on and within
structures that are
in equilibrium.

7. 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.

8. Nonconcurrent
forces may also me
parallel and non
parallel. The lines
of action of parallel
forces do not
intersect.

9. PARTICLE AND RIGID BODY
Distinction 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

10. TRANSITIONAL EQUILIBRIUM
Equilibrium 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.

11. 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

12. SAMPLE PROBLEM
You 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.
T
1
T
2
T
3
30°30°

13. SOLUTION
We are given w = 55N. The forces acting
on the frame are shown. The tension T
3
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
T
3
Free body part of the
frame
T
3
- 55 N = 0
T
3
= 55 N

14. To determine T
1
and T
2
consider the point where
the three ropes meet as a particle in
equilibrium.
T
1
T
2
30°
30°
w
Applying the first
condition for equilibrium,
we have:
∑F
x
= 0: T
1
cos 30°- T
2
cos 30° = 0
T
1
cos = T
2
∑F
y
= 0: T
1
sin 30°+ T
2
sin 30° - 55N = 0
Since T
1
= T
2
we may replace T
2
by T
1

and solve the above equation for T
1
.
∑F
y
= 0: T
1
(.5) + T
1
(.5) – 55 N = 0
T
1
= 55 N
T
2
= 55 N

15. TORQUE
In 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 .

16. EXPLANATION
The 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.

17. 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.

18. PROBLEM
For 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
SOLUTION
The resultant force = 7.0 – 7.0 N = 0

19. 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.

20. 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:
W
1
X
1
+ W
2
X
2
….
W
1
+ W
2
+ …..

21. PROBLEM
Find 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/m
2
.
y
0
(X
1
, Y
1
)
3.0
4.0
3.0
(X
2
, Y
2
)
x

22. SOLUTION
The center of gravity of the wood will specified by two
coordinates: x and y. We divide the plate into regularly
shaped parts. Let A
1
be the area of the bigger rectangle
and A
2 the area of the smaller rectangle. The center of
gravity of A
1
labeled as (x
1
,y
1
) is the (1.5 m, 3.0 m). The
center of gravity of A
2
labeled as (x2, y2) is (4.5 m, 1.0 m).
A
1
= (3 m) (6 m) = 18 m2
W
1
= (2.4 kg/m2) (18 m2) = 43.2 N
A
2
= (3 m)(6 m) = 18 m2
W
2
= (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=

23. ROTATIONAL EQUILIBRIUM
A 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 = 0
The condition is known as the second condition
for equilibrium.

24. SAMPLE PROBLEM
A 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

25. SOLUTION
Let 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.

26. + (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.

27. 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