Linear Motion Newtons 2nd Law

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Linear Motion Newtons 2nd Law

• Engineering 1h
• Dynamics

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Velocity
Velocity is defined as the rate of change of
distance with time. For a constant velocity, we
can write
where s is the position at time, t, and s0 and t0
are the starting position and time respectively
More generally, if the velocity is not constant,
then for any instant,
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Acceleration
Acceleration is defined as the rate of change of
velocity with time. For a constant acceleration,
we can write
If we start at time t0 0, then this can be
re-written as
To find the distance travelled, we integrate the
velocity
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Acceleration
Velocity is rate of change of distance, u is the
starting velocity and doesnt change, and we are
assuming constant acceleration, so
We can remove t if we know v
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Newton II

• The resultant force on a particle is equal to the
  rate of change of momentum of the particle

Units 1 Newton accelerates 1kg at 1ms-2. The
form Fma is only valid if the mass is constant.
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Example
A crate of mass 800kg is on a slipway inclined at
25o to the horizontal. Will the crate slip a)
by itself (?s0.65) b) when prodded (?k0.45)
Steps 1. draw the physical system 2. draw the
FBD 3. Calculate R and hence Fmax 4. Calculate F
and compare Fmax
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Example - sketch and FBD
sketch
F.B.D
Axes
F
mg
25o
R
25o
mgcos25o
mg
Components of g
mgsin25o
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Example - solution
? R 7110N
To find R
R - mgcos25o 0
Therefore, for the static case, the maximum
friction force available to stop the crate
slipping is
Fmax ?sR 0.65 7110 4620N
The force needed to stop the crate slipping can
be found from summing the forces up and down the
slope
(zero, since the crate is not moving)
mgsin25o-F 0
? F 3310N
Since Fmax is greater than the force required, F,
the crate will not slip and so the static
assumption was correct
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Example - crate nudged
For this case, we assume that the crate has been
nudged in some way and examine whether it will
continue to accelerate, or come to a halt.
As the crate is slipping, we use the coefficient
of kinetic friction, ?k0.45.
F ?kR 0.457110 3200N
This is less than the force required, so the
crate will continue to slip.
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Acceleration of crate
The crate will now move according to Newton II.
mgsin25o-F ma
a gsin25o - ?kR/m
9.8 sin 25o - 0.457710/800
0.14ms-2
After being nudged, the crate will accelerate
down the slope at 0.14ms-2.
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Systems of connected bodies
Consider two blocks in contact with each other on
a surface
P
A
B
What is the acceleration of the blocks?
FBD
P
mAg
mBg
FA
FB
RB
RA
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Acceleration of connected bodies
Forces are in balance vertically
RA mAg
RB mBg
FA?ARA
FB?BRB
friction forces
FA?A mAg
FB?B mBg
?
The body is moving horizontally, so Newton II
applies
P - FA - FB (mA mB)a
P - ?A mAg - ?B mBg (mA mB)a
?
(?A mA- ?B mB)g
?
a
-
(mA mB)
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Forces between connected bodies
To get the interactive force, treat as two
separate systems
P
s
s
mAg
mBg
FA
FB
RA
RB
Vertical forces are as before. The horizontal
forces are
P - FA - s mA a
s - FB mB a
adding these equations gives the same answer for
the acceleration as before
and s can be found from either equation
s mB a - FB