Chapter 5 Lecture Notes Part 2

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29 Νοε 2013 (πριν από 3 χρόνια και 8 μήνες)

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Newtonian Forces


Tension is the pulling force away from an
object usually by a string, rope, or cable


Force is represented by mg


M stands for the mass of the object


G stands for the gravity constant on Earth (9.80
m/s
2
)



T

-
mg



If the mass is just hanging, then the
acceleration (ā) =
0


Σ
𝐹

̅

=
0


Σ
𝐹


=




Where m is the inertial mass




=
Σ
𝐹




= mg (down) + T (up)


T = mg (up)


M(
0
) = mg (down) + mg (up)


Where the mass of “mg (up)” is related to gravity


This is called the Gravitational Mass: this tells how
much weight the object has



In this class inertial mass ↔ gravitational mass





𝐹


=
(GM
1
M
2
)

𝑥






r
2


Where “r” is the distance between two objects with
different gravity











Tension must be the same if the rope is the same

𝐹


= (down) m
1
g

(down) m
2
g

|a|


|a|


Atwood

Machine

T

T


Σ
𝐹

1

= m
1
ā
1

= m
1
g (down) + T (up)


Σ
𝐹

2

= m
2
ā
2

= m
2
g (down) + T (up)


T (up) = [m
2
ā
2

= m
2
g (down) = m
1
ā
1

= m
1
g (down)]



ā
2
= a (down)


m
2
(a(down)


g (down)



ā
1
= a (up)


m
1
(a(up)


g(down) = m
1
(
-
a(down)


g(down))



m
2
a
-

m
2
g + m
1
a + m
1
g


= (m
2
+ m
1
)a


(m
2
+ m
1
)g =
0


A
=
(m
2

-

m
1
)g







(
m
2
+ m
1
)



The difference in the weights makes the
acceleration go up


The sum of the weights makes the
acceleration go down


(
m
2
+ m
1
)a = (m
2
+ m
1
)g


Inertial mass =
(
m
sys
)a


Gravitational mass =
(
m
sys
)g


Inertial
mass of the system ≠ gravitational
mass of the system



A new rope means you have two separate
tensions


T
1

T
1

T
2


Always acts against the motion


θ

-
mg



|

F
fr
|=

μ
s
N

or
|
F
fr
|=

μ
k
N


Where
μ
s
is the static friction constant and
μ
k


is the kinetic friction constant



μ
k

is like skating on ice


Pure slippage




μ
s
N

is like rubber on concrete


Very little motion




Σ
𝐹


=
mg
||

-

F
fr

= ma



mg
||
-


μ
k
N

= ma



mg
|

-

μ
k
N

= ma



g
||

-

μ
k
g
|

=
a





g
||
= g
cos
θ




g
|

= g sin
θ



g
sin
θ

-

μ
k
g
cos
θ

= a



g(sin
θ

-

μ
k

cos
θ
) = a




|
F
fr
|≤

μ
s
N


The static friction has a maximum balancing force

θ