stretching of the wires allows us to find a relationship between
F
A
and
F
B
. If wire
A
originally had a length
L
A
and stretches by
Δ
L
A
, then
Δ
LF
L
A
E
AA
A
=
/
, where
A
is the
cross–sectional area of the wire and
E
is Young’s modulus for steel (200 × 10
9
N/m
2
).
Similarly,
Δ
L
A
E
BB
B
=
/
. If
A
is the amount by which
B
was originally longer than
A
then, since they have the same length after the log is attached,
AB
LL
Δ=
Δ+
A
. This means
FL
AE
AE
=+
A
.
We solve for
F
B
:
F
L
AE
L
B
=−
A
.
We substitute into
F
A
+ F
B
– mg
= 0 and obtain
F
mgL
AE
A
B
=
+
+
A
.
The cross–sectional area of a wire is
Ar
==
×
=
×
−−
ππ
23
2
6
120 10
4 52 10
..
mm
2
ch
.
Both
L
A
and
L
B
may be taken to be 2.50 m without loss of significance. Thus
26
2
9
2
3
(103kg)(9.8m/s )(2.50m)+(4.52 10 m )(200 10 N/m )(2.0 10 m)
2.50m+2.50m
866 N.
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 Spring '08
 Any
 Physics, Force, Mass

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