5.1.4 Force and moments

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5.1.4 Force and moments
Tuesday, 24 November 2009
You should be able to (a) understand that the
weight of a body may be taken as acting at a
single point known as its centre of gravity. (b)
understand a couple as a pair of equal parallel
forces tending to produce rotation only. (c)
define and use the moment of a force and the
torque of a couple. (d) show an understanding
that, when there is no resultant force and no
resultant torque, a system is in equilibrium. (e)
apply the principle of moments to solve problems
involving forces acting in two dimensions.
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The centre of mass is the point where all of the
mass of the object is concentrated. When an
object is supported at its centre of mass it will
remain in equilibrium. If the object is
uniform, for example a meter stick, the center of
mass will be at the exact geometric center if
the object is irregular in shape the center of
mass will be closer to the heavier end.  An easy
way to determine the location of the center of
mass of a rigid pole is to support the pole on
one finger from each hand. Gently slide your
fingers together. When your fingers meet, you
will be at the centre of mass. Try it with a
bat or a broom.
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To find the center of mass of an planar object
use a plumb line. Suspend the mass from each
vertex and trace the plumb line's location. Since
the center of mass will fall below the suspension
point the center of mass will be at the
intersection of all of the plumb lines.
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Where is the centre of mass here ?
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The Moment of a Force (also called torque) The
moment of a force is a measure of its turning
effect. The moment can be calculate using the
following equation.
Moment Force Perpendicular distance of force
from "pivot"

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Consider a person trying to open a door, by
applying a force, of magnitude, F, as shown below.
F
rsin?
The perpendicular distance, d, between the line
of action of the force and the pivot (the hinge
of the door) is rsin? Therefore, the moment of
the force is given by

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• The two obvious changes the person could make in
  order to open the door more easily are he/she
  could
• increase the distance "r
• push at 90 to the door.

If the angle between the line of action of the
force and the door is 90, we have
This is the maximum value of the turning effect.
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What happens when the pairs of forces shown below
act on the green block ?
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Two equal and oppositely directed parallel but
not collinear forces acting upon a body. The
moment of the couple (or torque) is given by the
product of one of the forces by the perpendicular
distance between them.
If the distance between the horses is d and the
force each one exerts is f. The moment of the
couple is given by Moment (f ? ½ d) (f ? ½
d) f ? d
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5.1.4 Pressure
Tuesday, 24 November 2009
You should be able to (f) define pressure. (g)
recall and use the equation p F/A.
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• Pressure
• Pressure is the force per unit area.This means
  that the pressure a solid object exerts on
  another solid surface is its weight in newtons
  divided by its area in square metres.Pressure
  can be found using the equation
• pressure force
• area
•               
• The unit for pressure is the pascal (Pa).
• One pascal is equivalent to one newton per
  square metre.
• Atmospheric pressure is approximately 1 105
  Pa.

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5.1.4 Work and Power
Tuesday, 24 November 2009
You should be able to (h) understand the concept
of work in terms of the product of force and
displacement in the direction of the force. (i)
define the joule. (j) recall and use the equation
W Fx, where F is a constant force along the
direction of motion. (k) recall and use equations
for kinetic energy ?Ek ½ mv2 and change in
gravitational potential energy ? Ep mg ? h. (l)
relate power to work done and time taken. (m)
define the watt. (n) recall and use the equation
W Pt.
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Work Work is defined as The product of force
and the distance moved in the direction of the
force. Work Force distance moved in the
direction of the force.

• Units are newton metres (Nm) or joules (J).
• Work is actually a scalar quantity despite being
  the product of a vector quantity.
• When work is done, there must be movement. This
  can result in acceleration, a rise in
  temperature, or deformation in shape.

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Example 1  These kids are trying to bump-start
this car, but they cannot get it to move. Sweat
is pouring off them. Explain why they have done
no work.
The car has not moved.  Therefore the
displacement is zero.   Unless you enjoy trouble,
don't mention that to them.          
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Example 2  A horse is pulling a barge along a
canal as shown in the diagram. It pulls the
barge with a force of 1000 N a distance of 75 m.
The angle the rope is at 15o to the direction of
travel. The situation is shown in the diagram
(a) Can you explain why the answer is NOT 75000
J? (b) What is the work done by the horse?
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a) The line of action of the force is not in the
same direction as the movement.  So we need the
horizontal (or  more correctly) forward component
of the force. b) What is the work done by the
horse? W Fs cos ?   1000 x 75 x cos 15
1000 x 75 x 0.966 72 400 J  
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Energy Energy and work are very closely related.
Energy is the ability to do work. When work is
done, energy is transferred. Energy comes in
many forms. Some kinds of energy can be stored,
while others cannot. Energy is always
conserved. Example 3 A box is pushed 5 m across
a room with a force of 30 N. What is the work
done and how much energy is used? 
Work done Force x displacement 30
x 5 150 J Energy used 150 J
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Power Power is the rate at which energy is
used. Power energy transferred (J)
time taken (s) work done (J)                
      time taken (s)

•  Units of power are watts (W).
• 1 watt 1 joule per second.
• Also kilowatt (kW). 1kW 1000 W.
• megawatt (MW). 1 MW 1 106 W.

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Example 4 It takes 20 seconds to push the box in
Example 3 across the room.  What is the power?
Energy used 150 J Power energy / time
150 20 7.5 W
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Kinetic Energy Kinetic energy is the ability to
do work through motion. If the motion is in a
straight line, we call the kinetic energy
translational. This is the only kinetic energy
we will consider.
Example 5 Calculate the kinetic energy of a 4 kg
shot-put thrown by an athlete at a speed of 15
m/s.
Kinetic energy 1/2 mv2 1/2 x 4 x (15) 2
1/2 x 4 x 225 450 J
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Potential Energy This term is often used in the
context of gravitational potential energy. If we
lift an object of mass m against gravity, we are
doing a job of work. Work done PE weight
distance moved against gravity.
Notice the term ?h ("delta h").  This means
"change in height".  So if we lifted an object
from 200 m above sea level to 300 m above sea
level, the change in height is 100 m, which we
would use in the equation. 
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6 What is the potential energy of a 12 kg mass
raised from the ground to a to a height of 25 m?
Potential Energy weight x height change Weight
12 x 10 120 N Height change
height at end - height at start
25 - 0 25
m Potential energy 120 x 25
3000 J