Download FE3

27
FE3
EQUILIBRIUM
OBJECTIVES
Aims
In this chapter you will learn the concepts and principles needed to understand mechanical
equilibrium. You should be able to demonstrate your understanding by analysing simple
examples of equilibrium. You will also learn about the interactions between fluids and solid
bodies as well as the concepts - buoyant force and pressure - used to describe the interactions.
Minimum learning goals
When you have finished studying this chapter you should be able to do all of the following.
1.
Explain, interpret and use the terms
translational motion, rotational motion, rigid body, equilibrium, stable equilibrium,
unstable equilibrium, neutral equilibrium, axis, torque [moment of a force], centre of
gravity, buoyancy, buoyant force, Archimedes' principle, pressure, pascal, density,
barometer.
2.
State and apply the relation between force and torque.
3.
State the conditions for equilibrium and apply them to simple problems.
4.
Describe and explain how the centre of gravity of a body can be located.
5.
Describe and explain the forces acting on a body which is wholly or partly immersed in
fluid. Solve simple problems involving buoyancy.
PRE-LECTURE
Introduction
This chapter deals mainly with the equilibrium of rigid bodies. The conclusions about rigid
bodies can also be applied to some examples of non-rigid bodies, such as bodies of fluid at rest.
We start with two simple examples of objects in equilibrium: an object at rest and one moving
with constant velocity.
All the examples and principles discussed in this chapter are restricted to systems in which
all the forces can be described in two dimensions - a plane. The extension to general threedimensional systems uses the same concepts but is mathematically much more complex.
3-1
TRANSLATION AND ROTATION
The simple descriptions of motion that we have used so far implicitly treated the motion of only
one point in a body. That is a good kind of description provided that all points in the body follow
similar, parallel, paths. That kind of motion is called pure translational motion. As well as
translational motion, a body can also execute rotational motion (like that of a spinning wheel)
and vibration (like that of a shaking jelly). A body which cannot vibrate noticeably is said to be
rigid; its shape and size do not change significantly when it is acted on by a system of forces. The
most general kind of motion of a rigid body is, therefore, a combination of translation and
rotation; the flight of a boomerang is a good example.
28
FE3: Equilibrium
3-2
EQUILIBRIUM OF FORCES
Q3.1
A block is at rest on a table.
Figure 3.1
An object at rest. What are the forces?
a) On figure 3.1 draw in all the forces acting on the block.
The block is not accelerating, so the net force on it must be zero. What does this tell you about the
vertical forces and any horizontal forces that may be present?
b) Suppose that a horizontal force is applied in an attempt to push the block along the table, but the block
does not move. What is opposing the horizontal force?
Q3.2. Block moving at constant velocity
A block is being pulled across the table at constant velocity (figure 3.2).
pull
Figure 3.2
An object moving with constant velocity
Draw in the other forces.
A frictional force opposes the motion. Draw in this force and the other forces on the block.
Again, the block is not accelerating so the net force on it must be zero. What does this tell you about
the vertical and horizontal forces?
The meaning of equilibrium
The examples in questions 3.1 and 3.2 are simple illustrations of equilibrium. In both cases the
velocity of the block is constant, i.e. its acceleration is zero, and the total force on it is zero.
However it is not enough that the forces balance in order to have equilibrium. This
guarantees only that there is no change of translational motion, i.e. that the motion of the body as
a whole does not change. The definition of equilibrium needs to be extended to include the
requirement that the rotational motion of the body also remains constant. Thus, for example, a
body which is completely at rest is in equilibrium only if it does not start to move or rotate. As
another example, a wheel rotating about a fixed axle, is defined to be in equilibrium only if its
rotational speed does not change.
In order to consider this rotational aspect of equilibrium we need the concept of torque. We
shall then see that for equilibrium, torques, as well as forces, must balance.
FE3: Equilibrium
29
LECTURE
3-3
TORQUE
Example 3.1. Wheel on a fixed axle
Consider a wheel which can rotate about its axle. The axle remains in a fixed position. An
object is hung by a string from the rim of the wheel as shown in figure 3.3.
FA
W
F
Figure 3.3
An unbalanced wheel
When the object is released, the wheel starts to rotate so, clearly, it is not in equilibrium.
The forces acting on the wheel are shown. W is the wheel's weight, FA is the force exerted by
the axle on the wheel and F is the force exerted by the string attached to the falling object, which
causes the wheel to rotate. Note that these forces are not all acting through the same point.
In this example, rotation occurs about an axis - a line in space - which is perpendicular to
the plane of the forces involved. In such cases we define torque as follows.
The torque of a force F about a specified axis is defined as

 = Fx
... (3.1)
where x is the perpendicular distance from the axis to the force's line of action. Torque is also
known as the moment of a force. Note that we can, in principle, define many torques for each
force, one for every possible choice of axis. It is not necessary for the axis to be a possible axis of
rotation.
Example 3.1 - continued
x
F
Figure 3.4
Torque on a pivoted wheel
In this example, we consider torques about the wheel's axis of rotation, its axle. Only the
force F has a non-zero torque about the axle so there is a net torque, due to F, acting on the
wheel. This net torque produces an angular acceleration, i.e. it changes the angular speed of
rotation about the axle.
30
FE3: Equilibrium
Demonstrations
•
Consider two wheels which have the same shape and the same total mass. One has a dense
metal rim, the other has a dense metal axle. Identical loads are hung from their rims. This
demonstration shows that the angular acceleration depends on the distribution of mass in the
object.
metal rim
metal
axle
Figure 3.5
Effect of mass distribution on angular acceleration
Identical loads were attached to the wheels and then released. The angular acceleration
was greater for the wheel whose mass is more concentrated near the axle.
•
3-4
Other demonstrations show rotational motion when the axis is not fixed.
EQUILIBRIUM OF TORQUES
Example 3.1 - continued
The wheel has no translational motion so the total force on it must be zero.
Forces
Torques
FA
Torque
Fx
W
F
Figure 3.6
Forces and torque on the pivoted wheel
The vertically upward component of the total force = FA - W - F = 0 , but the net torque, Fx, is
unbalanced. Note that this torque produces a clockwise rotation.
Example 3.2: Equilibrium of the wheel
To bring our wheel back into equilibrium, a torque of the same magnitude but in the opposite
(anti-clockwise) sense would have to be provided. Another object with the same mass could be
attached to the opposite side of the wheel in order to achieve this. See figure 3.7. (Note that the
supporting force FA will take on a new value.)
Forces
Torques
FA
-Fx
Fx
W
F
F
Figure 3.7
Forces and torques at equilibrium
FE3: Equilibrium
3-5
31
CONDITIONS FOR EQUILIBRIUM
In general, then, the conditions for equilibrium of an object which is free to rotate about a fixed
axis are:
(i) total force acting on the object = 0;
(ii) total torque about the axis = 0 .
Note that, since force is a vector quantity, the calculation of the net force must take account
of directions. This can be done using the method of components introduced in chapter FE2.
Torque, as defined here, is a scalar quantity whose values need to be associated with either
clockwise or anti-clockwise rotation. We can assign positive values to one of these two senses,
and negative values to the other. (The usual convention makes anti-clockwise values positive.)
Example 3.2 - continued
When we apply these conditions to the wheel with two objects hanging from its rim, we can
choose vertically down as a component direction for the forces (all horizontal force components
are equal to zero).
(i)
F + W + F - FA = 0 .
(FA takes on a new value when the second object is attached.)
The condition for balancing the torques is satisfied because the two torques have the
same magnitudes but opposite senses:
(ii)
Fx - Fx = 0.
It is obviously inconvenient to have to draw two diagrams for each example - one showing
forces and one showing torques. Henceforth both forces and torques will be shown on the same
diagram. Remember, however, that forces and torques are quite different entities and must be
combined separately.
3-6
CENTRE OF GRAVITY
Example 3.3. Centre of gravity of a flat object
A flat object is pivoted at the point P. See figure 3.8. Imagine that the object is divided up into
little pieces. The weight of each piece provides a torque about the pivot. The object will be in
equilibrium only if the torques due to all these pieces add up to zero.
P
Figure 3.8
Distribution of gravitational forces on an object
It is obviously inconvenient to have to consider the weights of all the little pieces of the
object separately. Fortunately, their total can be represented by the total weight, W, acting
through a point called the centre of gravity.
32
FE3: Equilibrium
Torque
Equilibrium
P
Centre
of gravity
P
W
Figure 3.9
W
Centre of gravity and the total weight force
Then W provides a torque about the pivot (equal to the sum of the many small torques).
The object will be in equilibrium if its centre of gravity lies vertically below the fixed
pivot point. In this position the length of the perpendicular from the pivot to the line of action of
the weight is zero, and so is that from the pivot to the supporting force at the pivot. Therefore
the torques of the weight and the supporting force about the pivot are both zero.
Example 3.4. Locating the centre of gravity
To locate the centre of gravity of a flat object, first mark a vertical line showing the line of
action of the weight. The centre of gravity must be somewhere on this line. Then choose a
different pivot point and repeat the process. The centre of gravity must be also be somewhere
on the new line; so it must be at the intersection of the lines.
A wheel is symmetric about its axis and the centre of gravity is at the centre of the wheel.
This is easily verified.
3-7
EQUILIBRIUM OF A SYSTEM OF OBJECTS
Example 3.5: Two children balancing on a seesaw.
This example is analysed in terms of forces and torques acting on a system which consists of the
two children and the plank. The forces acting on this system are the weights of the two children,
the weight of the seesaw's plank and a vertical supporting force at the pivot.
The weights of the children, W1 and W2 act at distances x1 and x2 from the pivot, as
shown on the figure. These forces give torques W1x1 (anticlockwise) and W2x2 (clockwise)
respectively about the pivot.
Suppose that the centre of gravity of the plank is directly above the pivot so that the
weight WS of the seesaw plank acts downwards at the pivot. N is the upward supporting force
exerted by the pivot on the plank. Each of these two forces gives zero torque about the pivot.
N
x2
x1
W1 x 1
W 2x 2
W1
Figure 3.10
W
S
W2
Balancing on a seesaw
For equilibrium the following conditions must be satisfied.
(i) Total force acting on the system = 0.
FE3: Equilibrium
33
Taking force components in the vertically downward direction:
W1 + W2 + WS - N = 0.
(ii) Total torque about an axis through the pivot = 0.
Taking clockwise as the positive sense:
W2 x2 - W1x1 = 0.
This analysis is essentially the same as that for a beam balance.
3-8
EQUILIBRIUM OF A FREE OBJECT
A free object is one that is not pivoted. This is a more general situation than the case of a fixed
axis.
Demonstration
moves forward
spins anticlockwise
push
push
moves forward
spins clockwise
moves forward
push
no spin
Figure 3.11
Motion of a free object
The centre of gravity is shown as a heavy dot.
A net force not acting through the centre of gravity of a rigid body will cause translational
acceleration of the object as well as change in its rotational motion. The resulting motion can be
described as a combination of translational motion of the centre of gravity and rotational motion
about the centre of gravity.
3-9
GENERAL CONDITIONS FOR EQUILIBRIUM
The general conditions for equilibrium are as follows
(i) The total force must be zero (as before).
(ii) The total torque about any axis must be zero.
In many cases it is convenient to consider torques about axes through the centre of gravity.
3-10
BUOYANCY
When a solid object is wholly or partly immersed in a fluid, the fluid molecules are continually
striking the submerged surface of the object. The forces due to these impacts (which are
sometimes called pressure forces) can be combined into a single force, the buoyant force.
FE3: Equilibrium
Figure 3.12
Forces exerted by a fluid
Note that, for clarity, we show only the forces exerted by the surrounding fluid in this and
the following diagram.
We want to find the magnitude of the buoyant force and the point through which it acts.
Buoyant force on a completely submerged object
Figure 3.13
Buoyant force on a submerged object
To work out how big this buoyant force is, and where it acts, we use the trick of thinking
about a 'block' of fluid, which has exactly the same shape and size as the solid object. This
imaginary portion of fluid is often called the 'displaced fluid'.
Figure 3.14
Fluid "displaced" by the submerged object
The pressure forces on this imaginary displaced fluid are exactly the same as the pressure
forces on the solid object. So their total effect, the buoyant force, will be the same, irrespective of
what is inside the broken outline. The buoyant forces on the solid object and the 'displaced fluid'
are identical.
Now the displaced fluid must be in equilibrium. Since the only other force on it is its
weight, which acts through its centre of gravity, the buoyant force must be equal to its weight, and
it must act vertically upward through its centre of gravity. Hence the buoyant force on the
submerged block must be equal to the weight of the displaced fluid and it must act vertically up
through the centre of gravity of the 'displaced' fluid body.
34
35
FE3: Equilibrium
Buoyant force
Weight of displaced fluid
Figure 3.15
Forces on the displaced fluid
This conclusion is known as Archimedes' principle.
Buoyant force on a partly submerged object
Air
Buoyant force
Liquid
Figure 3.16
Buoyant force on a partly submerged object
Note that there is also a gravitational force (weight).
In this case the object is immersed in two fluids one of which is the air. The diagram shows
only the total force exerted by these fluids on the object.
Consider an imaginary block composed of the two fluids 'displaced' by the object.
Displaced air
Displaced liquid
Figure 3.17
Fluids displaced by the partly submerged object
This fluid block is in equilibrium and, just as before,
buoyant force = total weight of the displaced fluids,
and acts vertically upward through the centre of gravity of the displaced fluids.
Note that, for an object floating in a liquid, the buoyant force due to the displaced air is
usually negligibly small.
FE3: Equilibrium
Example 3.6: A heavy object supported in water by a string
String
Force exerted by string
Water
Buoyant force
Weight
Figure 3.18
A submerged object in equilibrium
The force exerted by the string adjusts to balance the weight and the buoyant force.
Here all the forces on the submerged object are shown.
For equilibrium of this object:
force exerted by string = weight - buoyant force.
We might call the weight of the block minus the buoyant force the "effective gravitational
force".
This situation is demonstrated using a spring balance.
If the buoyant force were greater than the weight, a force would have to be applied to
hold the object down. This happens, for example with a helium-filled balloon.
Example 3.7: Floating object
Buoyant force
Air
Liquid
Figure 3.19
Weight
A floating object in equilibrium
The displaced liquid adjusts so that the forces balance.
For equilibrium, taking components in the direction vertically down:
weight - buoyant force = 0 .
The total torque will be zero if the buoyant force and the weight act along the same line.
This means that the centres of gravity of the floating object and the displaced fluids must lie in
the same vertical line.
POST-LECTURE
3-11
MOMENT OF INERTIA
The physical quantity which describes the distribution of matter about the axis of rotation is the
moment of inertia. Objects with their masses concentrated about the axis of rotation (or axle)
have smaller moments of inertia about that axis.
More precisely, if we divide the object up into small pieces each with mass m and at some
distance r from the axle, then the moment of inertia is
I = m r 2
where the sum is taken over all the small pieces.
36
37
FE3: Equilibrium
m
Axle
r
Figure 3.20
Defining the moment of inertia
For objects with a fixed axis of rotation, total torque about the axis equals moment of inertia
about the axis times angular acceleration. This is the reason why the wheel with the metal axle in
the lecture demonstration had the larger angular acceleration.
3-12
QUESTIONS AND PROBLEMS ON EQUILIBRIUM
Q3.3 a)
A tall block sitting on a table is being pulled.
F
Figure 3.21
Tipping an object
If the pull is applied near the top, the object will tip over instead of sliding along the table. Why?
Hint: to get the block moving, the force F must be greater than the maximum frictional force acting at
the bottom on the block. F cannot be any smaller than this. Now consider torques. What point is the
block going to rotate about?
How could you pull the block along the table without tipping it?
b)
Pulling trees down with a tractor can be a dangerous occupation. Which of the methods shown
below is the less dangerous way to tie the rope to the tractor? Why?
Figure 3.22
Uprooting a tree. Which way is safer?
Equilibrium of pivoted objects
Q3.4
Two children are balancing on a seesaw. Their weights are 200 N and 300 N . The smaller child is 1.80 m
from the centre.
Suppose that the weight of the seesaw is 500 N and the centre of gravity is directly above the pivot.
Draw all the forces acting on the seesaw.
a) Calculate the magnitude of the force that the pivot exerts on the seesaw.
b) How far from the centre is the larger child at equilibrium?
Q3.5
Explain why a beam balance gives the same value for the mass of an object on the Moon as on the Earth.
38
FE3: Equilibrium
3-13
STABLE, UNSTABLE AND NEUTRAL EQUILIBRIUM
We sometimes distinguish between stable and unstable equilibrium. For example, consider a
cone.
Stable
Stable
Figure 3.23
Unstable
Unstable
Neutral
Neutral
Stable and unstable equilibrium
In both cases the cone is in equilibrium because the total force is zero and the total torque is
zero. But the first case is stable, a slight displacement has no effect, while the second case is
unstable, a slight displacement causes the cone to tip over. When the cone is lying on its side it is
in neutral equilibrium.
Q3.6
What forces are acting and where are they acting in each case? What torques are responsible for restoring
the cone to its original position or otherwise?
Centre of Gravity
Q3.7 a)
b)
3-14
Does the centre of gravity always lie within an object? If not, give examples.
Suggest a way of locating the centre of gravity of a "lumpy" object (not a flat object).
FLUIDS
Pressure
When a solid object is immersed in a fluid, the force exerted on the object by the fluid is
distributed over the contact surface. For a complete description we need to look at the force
acting on each small part of the surface. We can define the average pressure on a flat surface to
be the component, Fn, of the force perpendicular to the surface divided by the area, A, of the
surface. The limit of this quotient as we take smaller and smaller pieces of the contact surface
(and hence also smaller and smaller forces) is the pressure, P, at a point on the surface:
F 
... (3.2)
P  lim  n  .
A0  A 
Provided that the body and the fluid are not moving, the force on each small part of the
contact surface is perpendicular to the surface (see figure 3.12) so the interaction can be described
completely in terms of pressure. (On the other hand, if there is relative motion between fluid and
solid object, the force has components parallel to the surface, not described by the pressure.)
This idea of pressure can be used also to describe what goes on inside the fluid; just imagine
the fluid divided into two parts as in the argument about buoyancy. Wherever we draw the fluid
boundary, we can define a pressure exerted by one part of the fluid on the other part. So we can
say that pressure exists within the fluid.
The following are important statements about pressure.
39
FE3: Equilibrium
•
Pressure is a scalar quantity - it has no direction.
•
The pressure within a uniform stationary fluid is the same at all points in the same
horizontal plane.
•
The SI unit of pressure is the pascal, symbol Pa; 1 Pa = 1 N.m-2 .
Density
Pressure variations within a fluid are affected by its density. The density, , of a uniform
substance is defined as the quotient: mass volume:
m

 = V .
... (3.3)
The barometer
h
S
S'
Figure 3.24
A mercury barometer
A tube, closed at one end, is filled with mercury and is then inverted over a container of
mercury as shown. If the tube is sufficiently long, the mercury falls from the top leaving an
evacuated region there.
By considering the equilibrium of a part of the surface SS' of the mercury outside the tube,
we can show that the pressure in the mercury just under that surface is equal to the atmospheric
pressure. So too is the pressure at the same level inside the tube.
We consider the equilibrium of a body consisting of the column of mercury in the tube
which is above the level SS'. The forces on this column of mercury are its weight, W, downward
and the upward force, F, exerted by the mercury below the column. The magnitude of the upward
force must be equal to atmospheric pressure multiplied by the cross-sectional area, A, of the tube:
F = PA.
The weight of the mercury column is equal to the product of its density, itsvolume andg.
Since the volume is equal to the product of the column's height and cross-sectional area,
W = hAg .
These two forces, must have equal magnitudes, so
P = gh .
...(3.4)
The density of mercury is 13.6  103 kg.m-3 and normal atmospheric pressure is 1.01  105 Pa
(101 kPa).
Q3.8 a)
Estimate the height of a column of mercury in a barometer.
3
-3
b) The density of water is 1.00  10 kg.m . Estimate the height of the column of water in a water
barometer.
3-15
Q3.9
QUESTIONS ON BUOYANT FORCES
-3
Density of air
= 1.29 kg.m .
Density of helium
= 0.18 kg.m .
-3
40
FE3: Equilibrium
3
A balloon is filled with helium. Its volume is 1.00 m . What downward force must be applied to stop
the balloon rising up into the sky?
Q3.10
Estimate the buoyant force exerted on you by the atmosphere.
Q3.11
An ice cube is floating in a glass of water. The water and the ice are at about 0°C.
Density of ice at 0°C
-3
=
917 kg.m .
Density of water at 0°C =
1000 kg.m .
-3
a) What fraction of the ice cube is submerged?
b) Explain what happens to the level of the water as the ice melts.
Q3.12
Steel is about eight times as dense as water. How can ships float?
Q3.13
The density of water varies with temperature as shown in the graph below. The curve has a maximum at
about 4°C.
1000
Density/kg.m
-3
999
998
997
996
995
0
Figure 3.25
10
20
Temperature/°C
30
40
How the density of water changes with temperature
a) When a beaker of water is heated from below, why does the warm water at the bottom rise?
b) When a lake ices over, why does freezing occur only at the top of the lake? Is there any biological
significance in that?
Interlude 3: Masses
INTERLUDE 3 - THE RANGE OF MASSES IN THE UNIVERSE
Mass of the Universe
Mass of the Sun
Mass of the Moon
Person
Postage stamp
Protein molecule
Proton
mass/kilograms
1050 __
_
_
_
_
40
10 __ Mass of our Galaxy
_
_
_
_
30
10 __
_
_
_
Mass of the Earth
_
20
10 __
_
_
_
_
10
10 __
_
_
Ocean liner
_
_
1
__
_
_
Sugar cube
_
_
10-10__
_
_
Red blood corpuscle
_
_
-20
10 __
_
_
_
Oxygen molecule
_
-30
10 __ Electron
41