This preview shows page 1. Sign up to view the full content.
12.71. Model:
Assume the string does not slip on the pulley.
Visualize:
The freebody diagrams for the two blocks and the pulley are shown. The tension in the string exerts an upward
force on the block
2
,
m
but a downward force on the outer edge of the pulley. Similarly the string exerts a force
on block
1
m
to the right, but a leftward force on the outer edge of the pulley.
Solve:
(a)
Newton’s second law for
1
m
and
2
m
is
11
Tm
a
=
and
22
2
.
gm
a
−
=
Using the constraint
21
,
aa
a
−=
+=
we have
1
a
=
and
.
a
−+
=
Adding these equations, we get
2
()
,
mg
m m a
=+
or
2
1
12
mmg
aT
m
a
mm
=⇒
=
=
++
(b)
When the pulley has mass
m
, the tensions
(
a
n
d
)
TT
in the upper and lower portions of the string are
different. Newton’s second law for
1
m
and the pulley are:
1
2
and
a
T
RT
R I
α
=
−
We are using the minus sign with
because the pulley accelerates clockwise. Also,
.
aR
=
Thus,
a
=
and
2
Ia a
I
R
RR
=
Adding these two equations gives
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '09
 geller
 Force

Click to edit the document details