32a. If the straight wire illustrated in Figure 3.2a has a uniform
mass per unit length
equal to
ξ
,
the total mass of the wire is given by
o
o
mass
L
=
ξ
If the mass per unit length is given by ()
x
ξ
,the total mass is determined by the following
line
integral
:
0
()
xL
x
mass
x dx
=
=
=ξ
∫
For the following conditions
2
1
o
2
3
o
0.0065 kg/m,
0.0017 kg/m ,
1.4 m
xx
L
L
ξ=
ξ
+
α
−
α=
=
determine the total mass of the wire.
Figure 3.2a
. Wire having a uniform or nonuniform mass density
32a. The mass contained in the wire can be expressed as
2
00
(
/2
)
o
mass
x dx
x
L
dx
==
= ξ
+
α
−
∫∫
(1)
and evaluation of the integral leads to
333
324
1
o
LLL
L
mass
L
L
3
2
o
α
=ξ +
α
−
+
(2)
Making use of the data that are given provides
(3)
23
(0.0065 kg/m ) (1.4m)
(0.0017 kg/m ) (1.4m) /12
mass
=+
3
and the mass is calculated to be
0.0095 kg
mass
=
(4)
3.2b. If the flat plate illustrated in Figure 3.2b has a uniform
mass per unit area
equal to
ψ
, the
total mass of the plate is given by
o
oo
mass
A
L L
1
2
=
ψ=
ψ
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 Winter '10
 B.G.Higgins
 ρ, 0.0017 kg, 0.00017 kg, 0.0095 kg, 0.02518 kg

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