Transcript Mechanics - akamdiplomaphysics

Mechanics
Topic 2.2
Forces and Dynamics
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G4 Physics Presentation
Forces and Free-body
Diagrams
• To a physicist a force is
recognised by the effect or
effects that it produces
• A force is something that can
cause an object to




Deform (i.e. change its shape)
Speed up
Slow Down
Change direction
• The last three of these can be
summarised by stating that a
force produces a change in
velocity
• Or an acceleration
Gioko, A. (2007). Eds.
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G4 Physics Presentation
Free-body Diagrams
• A free-body diagram is a
•
•
•
•
diagram in which the forces
acting on the body are
represented by lines with
arrows.
The length of the lines
represent the relative
magnitude of the forces.
The lines point in the direction
of the force.
The forces act from the centre
of mass of the body
The arrows should come from
the centre of mass of the body
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Example 1
Normal/Contact Force
A block resting on a worktop
Weight/Force due to
Gravity
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Example 2
A car moving with a constant velocity
Normal/Contact Force
Resistance
Motor Force
Weight/Force due to
Gravity
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Example 3
A plane accelerating horizontally
Upthrust/Lift
Motor Force
Air Resistance
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Weight/Force due to
Gravity
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Resolving Forces
Q. A force of 50N is
applied to a block on
a worktop at an
angle of 30o to the
horizontal.

What are the vertical
and horizontal
components of this
force?
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Answer
First we need to draw a freebody diagram
50N
30o
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Gioko, A. (2007). Eds.
G4 Physics Presentation
We can then resolve the
force into the 2 components
50N
Vertical = 50 sin 30o
30o
Horizontal = 50 cos 30o
Therefore


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Vertical = 50 sin 30o = 25N
Horizontal = 50 cos 30o = 43.3
= 43N Gioko, A. (2007). Eds.
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Determining the
Resultant Force
Two forces act on a body P
as shown in the diagram
Find the resultant force on
the body.
50N
30N
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30o
Gioko, A. (2007). Eds.
G4 Physics Presentation
Solution
Resolve the forces into the
vertical and horizontal
componenets (where
applicable)
50 sin 30o
30N
50N
30o
50 cos 30o
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G4 Physics Presentation
Add horizontal components
and add vertical components.
50 sin 30o = 25N
50 cos 30o – 30N = 13.3N
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Now combine these 2
components
25N
R
13.3N
R2 = 252 + 13.32
R = 28.3 = 28N
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G4 Physics Presentation
Finally to Find the Angle
R
25N

13.3N
tan  = 25/13.3
 = 61.987
 = 62o
The answer is therefore 28N at 62o upwards from
the horizontal to the right
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Springs
• The extension of a spring
which obeys Hooke´s law is
directly proportional to the
extending tension
• A mass m attached to the
end of a spring exerts a
downward tension mg on it
and if it is stretched by an
amount x, then if k is the
tension required to produce
unit extension (called the
spring constant and
measured in Nm-1) the
stretching tension is also kx
and so
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mg = kx
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Spring Diagram
x
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Newton´s Laws
The First Law
Every object
continues in a state
of rest or uniform
motion in a straight
line unless acted
upon by an external
force
examples
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Examples
Any stationary object!
Difficult to find
examples of moving
objects here on the
earth due to friction
Possible example
could be a puck on ice
where it is a near
frictionless surface
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Equilibrium
If a body is acted upon by a
number of coplanar forces
and is in equilibrium ( i.e.
there is rest (static
equilibrium) or
unaccelerated motion
(dynamic equilibrium))
then the following condition
must apply
The components of the
forces in both of any two
directions (usually taken at
right angles) must balance.
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Newton´s Laws
The Second Law
There are 2 versions of this
law
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Newton´s Second Law
1st version
The rate of change of
momentum of a body is
proportional to the resultant
force and occurs in the
direction of the force.
F = mv – mu
t
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F =
t
Newton´s Second Law
2nd version
The acceleration of a
body is proportional
to the resultant force
and occurs in the
direction of the force.

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F = ma
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Linear Momentum
The momentum p of a body
of constant mass m moving
with velocity v is, by
definition mv
Momentum of a body is
defined as the mass of the
body multiplied by its
velocity
Momentum = mass x velocity
 p = mv
 It is a vector quantity
 Its units are kg m s-1 or Ns
 It is the property of a moving
body. Gioko, A. (2007). Eds.

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Impulse
From Newtons second
law
F = mv – mu
t

F
=
t

Ft = mv – mu
 This quantity Ft is called the

impulse of the force on the
body and it is equal to the
change in momentum of a
body.
 It is a vector quantity
 Its units are kg m s-1or Ns
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Law of Conservation
of Linear Momentum
When bodies in a
system interact the
total momentum
remains constant
provided no
external force acts
on the system.
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Deriving This Law
To derive this law we apply
Newton´s 2nd law to each
body and Newton´s 3rd law
to the system
i.e. Imagine 2 bodies A and B
interacting
 If A has a mass of mA and B
has a mass mB If A has a
velocity change of uA to vA
and B has a velocity change
of uB to vB during the time of
the interaction t

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Deriving This Law
Then the force on A given by Newton 2 is
FA = mAvA – mAuA
t
And the force on B is
FB = mBvB – mBuB
t
But Newton 3 says that these 2 forces are
equal and opposite in direction
Therefore
mAvA – mAuA = -(mBvB – mBuB)
t
t
Therefore
mAvA – mAuA = mBuB – mBvB
Rearranging
mAvA + mBvB = mAuA + mBuB
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Total Momentum after
=
Total Momentum
before
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G4 Physics Presentation
Newton´s Law
• The Third Law
When two bodies A
and B interact, the
force that A exerts
on B is equal and
opposite to the force
that B exerts on A.
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Example of Newton´s 3rd
Q. According to
Newton’s third Law
what is the opposite
force to your weight?
A. As your weight is
the pull of the Earth
on you, then the
opposite is the pull
of you on the Earth!
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G4 Physics Presentation
Newton´s 3rd Law
The law is stating that forces
never occur singularly but always
in pairs as a result of the
interaction between two bodies.
For example, when you step
forward from rest, your foot
pushes backwards on the Earth
and the Earth exerts an equal
and opposite force forward on
you.
Two bodies and two forces are
involved.
Important
The equal and opposite forces
do not
Gioko, A. (2007). Eds.
act
on
the same body!
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