Download Equilibrium of Forces

STATICS
1STDIPS&M
Mass
The mass of a body is defined as the quantity of matter that it
contains. The SI unit of mass is the kilogram (kg).
Force
Force cannot be observed; only its effect can be seen, such as the
distortion of an object. Force is a vector quantity as it has size and
direction.
The Newton is the SI unit of force and is defined as:
The force required to accelerate a mass of one kilogram with an
acceleration of one metre per second per second.
One Newton is approximately one tenth of a kilogram.
Weight
The force, which attracts a body towards the earth, known as the force
of gravity, is the weight of a body.
This force is the product of the mass of the body and the acceleration
due to gravity (9.81 m/s2)
If W is the weight of a body, m its mass (kg) and g is the acceleration
due to gravity, then:
W = mg
newtons (N)
EXAMPLE 1
Determine the gravitational force acting on a body of mass 160 kg.
m = 160 kg and g = 9.81 m/s2
Force of gravity, weight, W = mg = 160  9.81 = 1570 N.
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1STDIPS&MSTATICS
Scalar and Vector quantities
Quantities that have both magnitude and direction are a vector
quantities e.g. force and velocity.
Quantities that possess magnitude only are scalar quantities e.g. mass
and speed.
Vectors
A vector quantity can be represented by a straight line whose length,
drawn to a scale, represents the magnitude of the quantity, and where
the direction of the line is parallel to the line of action of the quantity.
Fig 1 shows a gravitational force of 700 N represented by a vector
Fig 1
a
Gravitational force of 700 N
represented as a vector.
ab = 3.5 cm
cm
Scale 1 cm = 200 N
b
Fig 2 shows the resultant vector for two 5 tonne forces acting at 900.
5T
Fig 2
90o
5T
45o
Scale 1 cm = 1T
Resultant is 7.07 T
acting at 450 to the
horizontal
The resultant vector of 7.07 cm = 7.07 tonne acting at 45 0 to the
horizontal may be found by the parallelogram of forces rule that is
covered later.
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1STDIPS&MSTATICS
Equilibrium
A body is said to be in stable equilibrium if it tends to return to the
same position when acted upon by a force.
A body is said to be in an unstable equilibrium if it moves away from
its equilibrium position to a different position when acted upon by a
force.
A body is said to be in neutral equilibrium if it moves from one
equilibrium position to another equilibrium position, which is
essentially the same position, when acted upon by a force.
Fig 3 shows examples of bodies in stable, unstable and neutral
equilibrium.
Fig 3
Unstable equilibrium
Stable equilibrium
Neutral equilibrium
The pyramid resting on its base will return to its equilibrium position
if displaced slightly by a force.
The pyramid balanced on its apex will topple at the slightest force and
move away from its original equilibrium position and take up a
different stable equilibrium position on its side.
The sphere can rest in equilibrium at any position on the surface. If it
is moved slightly it will rest in equilibrium in essentially the same
position. It is said to be in neutral equilibrium.
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1STDIPS&MSTATICS
Moments
The moment of a force is a measure of the turning effect of the force
about a fixed point.
Force F Newtons acts at a perpendicular distance d metres from point
O as shown in Fig 4.
A moment is the product of the force
F
and the perpendicular distance from
d
O
a point to the line of action of the force.
Fig 4
Moment of force = Fd (Nm)
Beams
A beam is a rigid structural member subjected to external forces along
its length.
Beams are usually supported horizontally with external vertical forces
acting at points along its length and reactions at the supports as shown
in Fig 5.
Load
Load
Support
Fig 5
Support
A uniform beam is one that has the same cross-sectional area
throughout its length, whose mass is considered to act at the midpoint.
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1STDIPS&MSTATICS
Equilibrium of Forces
A body is said to be in equilibrium when:
(a)
The resultant force acting on the body is zero
(b)
The algebraic sum of the moments of the forces about any
point is zero, i.e. the clockwise moments equal the
anticlockwise moments
EXAMPLE 1
Show that the beam in Fig 6 is in equilibrium.
4 kN
Fulcrum
3m
3 kN
4m
7 kN
Fig 6
For equilibrium, the clockwise moments must equal the anticlockwise
moment.
Taking moments about the fulcrum.
Clockwise moments = 3  4 = 12 kNm
Anticlockwise moments = 4  3 = 12 kN
For equilibrium, the vertical forces must be zero:
7 kN = 3 + 4 = 7 kN
The beam fulfils both conditions for static equilibrium.
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1STDIPS&MSTATICS
Simply Supported Beams
A simply supported beam is one that rests on two supports.
EXAMPLE 2
Find the value of the forces acting at the two supports RA and RB.
16 kN
6m
8m
RA
Fig 5
RB
To find RA, take moments about RB.
Equating clockwise and anticlockwise moments, for equilibrium,
gives:
RA  8 = 16  2 = 32 kNm
RA = 4 kN
To find RB, take moments about RA.
Equating clockwise and anticlockwise moments, for equilibrium,
gives:
16  6 = RB  8 = 96
RB = 12 kN
Check for equilibrium of vertical forces.
RA + RB = 4 + 12 = 16 kN
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1STDIPS&MSTATICS
Centre of gravity
The centre of gravity (c.g.) of a body is the point at which the weight
of a body is considered to be concentrated and to act vertically.
All of the gravitational forces acting on the individual particles that
make up the body are replaced by a single resultant force passing
through the centre of gravity.
Finding the centre of gravity of an irregular laminar
Suspend the laminar by a thin thread alternately from two points a
and b as shown in fig 6. The tension in the thread must equal the
weight of the lamina, and hence the centre of gravity must lie along
the dotted lines. The intersection of the two lines is the centre of
gravity.
Centre of gravity
a
b
a
Fig 6
If the lamina is a regular shape then the centre of gravity can usually
be found by construction as shown in fig 7 for a rectangle, triangle
and circle.
c.g.

c.g.

c.g.

Fig 7
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1STDIPS&MSTATICS
Exercise 1
1.
Find the value of the two reactions for the simply supported
beam in fig 8.
20 kN
3m
8m
RA
RB
Fig 8
2.
Find the value of the two reactions for the simply supported
beam in fig 9.
30 kN
40 kN
3m
6m
12 m
RA
Fig 9
RB
3.
A force is applied and then removed from a body in a state of
static equilibrium. State which of the following statements are true or
false.
(a) If the body is in unstable equilibrium, the displacement caused
by the force tends to increase.
TRUE/FALSE
(b) If the body is in stable equilibrium, the displacement caused by
the force tends to reduce to zero.
TRUE/FALSE
(c) If the body is in neutral equilibrium, the displacement caused by
the force remains essentially constant.
TRUE/FALSE
4.
Three vertical forces of 10 kN, 20 kN and 15 kN act vertically
downwards on a block of steel. The steel block is resting on a spring
that exerts a force of 22.5 kN vertically upwards. What is the resultant
force acting on the block?
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1STDIPS&MSTATICS
Fluids
Gases and liquids are both fluids but they have different
characteristics. Liquids are practically incompressible and have a
fixed size whereas gasses are compressible and completely adopt the
size and shape of the containing vessel.
Pressure
Pressure is defined as the force per unit area exerted by a fluid.
If F  force in newtons, and A  area in square metres, then
F
A
The unit of pressure is the Pascal (Pa), where 1 Pa = 1 N/m2
Pressure, p 
Gauge Pressure
Gauge pressure is the pressure above atmospheric pressure.
Atmospheric Pressure
Atmospheric pressure is the pressure exerted by the mass of air
covering the earth’s surface. It will support a 760 mm column of
mercury in a glass tube, sealed at one end and inverted in a bowl of
mercury at sea level as shown in Fig 1.
Vacuum
760 mm
Atmospheric
pressure
Fig 1
he space in the tube above the mercury is a vacuum, and the pressure
of air on the exposed surface of the mercury supports the column of
mercury.
Atmospheric pressure = 101 292.8 Pa = 101.3 kPa
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1STDIPS&MSTATICS
Absolute pressure
The absolute pressure is the pressure above a perfect vacuum, i.e. the
sum of the gauge pressure and atmospheric pressure.
Absolute pressure = Gauge Pressure + Atmospheric pressure
Properties of pressure in a fluid
The pressure in a liquid at rest always acts at right angles to its
containing surfaces.
The pressure at any point in a liquid acts equally in all directions.
The pressure in a liquid at any point is independent of the shape of the
container.
The pressure at any point in a fluid is proportional to the density of
the liquid and the depth of the point as shown in fig1.
h
Density of
Liquid,
 (kg/m3)
The gauge pressure on base area
Base area,
A (m2)
column h (m) of fluid above it
A is due to the weight of the
divided by the area.
Fig 1
The weight is mg, where the mass m is the volume multiplied by the
density of the fluid  i.e.
m  hA  
Pressure, p 
F
mg
hAg


A
A
A
p  hg
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1STDIPS&MSTATICS
EXAMPLE 1
Find the pressure in a column of mercury at a depth of 0.76 m if the
density of mercury is 13600 kg/m3.
 = 13600 kg/m3, h = 0.76 m,
g = 9.81 m/s2
p = hg = 13600  0.76  9.81 = 101396 Pa
Measurement of Pressure
Bourdon gauge
A Bourdon gauge works on the principle that pressure will tend to
straighten a curved flattened flexible tube.
Scale
Linkage
Gear segment
Flattened
flexible tube
Pressure
Bourdon gauge Fig 2
The pressure tries to straighten the curved flexible tube, which causes
the linkage mechanism to rotate the gear segment, which turns the
gear wheel and moves the pointer.
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1STDIPS&MSTATICS
Manometer
Pressures may be measured by means of a manometer (fig 3) which is
a U tube with one end connected to the vessel containing the gas and
the other end open to the atmosphere.
The U tube contains a fluid and the pressure in the vessel, which may
be greater or less than atmospheric, will cause the fluid to rise in one
leg of the tube and fall in the other.
Atmospheric
pressure
Gas
Pressure
h
Liquid
Manometer Fig 3
The difference in levels h is used to calculated the pressure using the
formula,
p = hg.
Where:
p = Pressure (Pa),
 = Density (kg/m3),
h = Height (m)
g = Gravity (9.81 m/s2)
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1STDIPS&MSTATICS
Exercise 2
1.
The pressure at any point in a liquid is:
a) the same at any depth.
b) acts only at right angles.
c) is independent of liquid density
d) is directly proportional to depth
2.
The pressure at any point in a liquid:
a) depends upon the shape of the container.
b) is greater in the vertical than horizontal direction.
c) acts equally in all directions.
d) is independent of depth
3.
A force of 2000N acting on a piston of area 2 m2 will transmit a
pressure into a closed cylinder of liquid of:
a)
c)
4000 Pa
1000 Pa
b)
d)
2000 Pa
1000 N
4.
If an object is 150 m below the surface of the sea and sea water
has a density of 1025  103 kg/m3, the pressure acting on the body is:
a)
b)
c)
d)
15.083  105 Pa
1508.3  105 Pa
1.5083  109 Pa
15.083  105 Nm
5.
If a force of 5 kN acts on a piston of diameter 50 cm, the
pressure exerted on the surface of the liquid inside the cylinder is:
a)
b)
c)
d)
254.65 bar
25.465 kPa
1000 kPa
25 MPa
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1STDIPS&MSTATICS
Answers to Exercise 1
1.
Find the value of the two reactions for the simply supported beam in fig 8.
20 kN
3m
8m
RA
RB
Fig 8
To find RA, take moments about RB.
Equating clockwise and anticlockwise moments, for equilibrium, gives:
RA  8 = 20  5 = 100
RA = 12.5 kNm
To find RB, take moments about RA.
Equating clockwise and anticlockwise moments, for equilibrium, gives:
20  3 = RB  8 = 60
RB = 7.5 kN
Check for equilibrium of vertical forces.
RA + RB = 12.5 + 7.5 = 20 kN
2.
Find the value of the two reactions for the simply supported beam in fig 9.
30 kN
40 kN
3m
6m
12 m
RA
Fig 9
RB
To find RA, take moments about RB.
Equating clockwise and anticlockwise moments, for equilibrium, gives:
RA  12 = 30  3 + 40  6 = 90 + 240 = 330
RA = 27.5 kNm
To find RB, take moments about RA.
Equating clockwise and anticlockwise moments, for equilibrium, gives:
40  6 + 30  = RB  12 = 240 + 270 = 510
RB = 42.5 kN
Check for equilibrium of vertical forces.
RA + RB = 27.5 + 42.5 = 70 kN
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1STDIPS&MSTATICS
3.
A force is applied and then removed from a body in a state of static equilibrium. State
which of the following statements are true or false.
(a)
If the body is in unstable equilibrium, the displacement caused by the force tends to
increase.
TRUE
(b)
If the body is in stable equilibrium, the displacement caused by the force tends to
reduce to zero.
TRUE
(c)
If the body is in neutral equilibrium, the displacement caused by the force remains
essentially constant.
TRUE
4.
Three vertical forces of 10 kN, 20 kN and 15 kN act vertically downwards on a block
of steel. The steel block is resting on a spring that exerts a force of 22.5 kN vertically
upwards. What is the resultant force acting on the block?
Total downward force = 10 + 20 + 15 = 45 kN
Upward force = 22.5 kN
Resultant downward force = 45 – 22.5 = 22.5 kN.
Answers to Exercise 2
1.
The pressure at any point in a liquid is:
e)
is directly proportional to depth
2.
The pressure at any point in a liquid:
c)
acts equally in all directions.
3.
A force of 2000N acting on a piston of area 2 m2 will transmit a pressure into a closed
cylinder of liquid of:
F
2000
Pressure, p 

 1000 Pa
A
2
c)
1000 Pa
4.
If an object is 150 m below the surface of the sea and sea water has a density of 1025
3
 10 kg/m3, the pressure acting on the body is:
p = hg = 1025  103 150  9.81
= 1 508 300  103 = 1.5083  109 Pa
c)
1.5083  109 Pa
5.
If a force of 5 kN acts on a piston of diameter 50 cm, the pressure exerted on the
surface of the liquid inside the cylinder is:
Pressure, p 
b)
F
5000

 25465 Pa  25.456 kPa
A
  0.252
25.465 kPa
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1STDIPS&MSTATICS