Page 58 - Advanced engineering mathematics

38 CHAPTER 1 First-Order Differential Equations

The length of chain that is actually in motion varies with time. Let x(t) be the length of that
part of the chain that has left the support by time t and is currently in motion. The equation of
motion is
dv dm
m + v = F,
dt dt
where F is the magnitude of the total external force acting on the chain. Now F = ρx = mg,so
m = ρx/g = ρx/32. Then
dm ρ dx ρ
= = v.
dt 32 dt 32
Furthermore,
dv dv dx dv
= = v .
dt dx dt dx
Substituting this into the previous equation gives us

ρxv dv ρ 2
+ v = ρx.
32 dx 32
Multiply by 32/ρxv to obtain
dv 1 32
+ v = . (1.7)
dx x v
2
This is a Bernoulli equation with α =−1. Make the change of variable w = v 2−α = v . Then
1/2
v = w , and
dv 1 −1/2 dw
= w .
dx 2 dx
Substitute this into equation (1.7) to obtain
1 dw 1
w −1/2 + w 1/2 = 32w −1/2 .
2 dx x
Multiply by 2w 1/2 to obtain the linear equation
2
w + w = 64.

x
Solve this to obtain
64 c
2
w(x) = v(x) = x + .
3 x 2
Since v = 0 when x = 10,
64 c
(10) + = 0,
3 100
so c =−64,000/3. Therefore,

64
1000
2
v(x) = x − .
3 x 2
The chain leaves the support when x = 40, so at this time,
64
1000
2
v = 40 − = 4(210).
3 1600
√
The velocity at this time is v = 2 210, which is about 29 feet per second.

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October 14, 2010 14:9 THM/NEIL Page-38 27410_01_ch01_p01-42