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32 CHAPTER 1 First-Order Differential Equations

N
mg sin(u)

u mg cos(u)
mg

π
6

FIGURE 1.7 Forces acting on the sliding block.

1 1
F = 48 − 36 − v = 12 − v.
2 2
Since the block weighs 96 pounds, its mass is 96/32 = 3 slugs. From Newton’s second law,
dv 1
3 = 12 − v.
dt 2
This is the linear equation
1

v + v = 4.
6

t/6
Compute the integrating factor e (1/6)dt = e . Multiply the differential equation by e t/6 to obtain
1 t/6 t/6
t/6
v e + ve = 4e
6
or
t/6
(ve ) = 4e .
t/6
Integrate to get
ve t/6 = 24e t/6 + c
so
v(t) = 24 + ce −t/6 .
Assuming that the block starts from rest at time zero, then v(0) = 0 = 24 + c,so c =−24 and
−t/6
v(t) = 24 1 − e .
This gives the block’s velocity at any time. We can also determine its position. Let x(t) be the
position of the block at time t measured from the top. Since v(t) = x (t) and x(0) = 0, then

t t
−τ/6
x(t) = v(τ)dτ = 24 1 − e dτ
0 0
−t/6
= 24t + 144 e − 1 .
Suppose we want to know when the block reaches the bottom of the ramp. This occurs at a time
T such that x(T ) = 50. We must solve for T in
−T/6
24T + 144 e − 1 = 50.
This equation cannot be solved algebraically for T , but a computer approximation yields T ≈5.8
seconds.
Notice that
lim v(t) = 24.
t→∞

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October 14, 2010 14:9 THM/NEIL Page-32 27410_01_ch01_p01-42