
While in motion, the friction force f
m
is a constant dependent on the coefficient of sliding friction m
s
and the weight mg of the object as seen in Equation 2.140. Note that f
m
is also equal to m
s
mg when
f f
B
and v > 0.
f
m
¼
f when f f
B
(v ¼ 0)
m
s
mg when f > f
B
(v > 0)
(
(2:140)
Example 2.10
The applied force f(t) is shown in Figure 2.35. Find the velocity of the object.
f(t) ¼
2f
B
t
t
1
0 t < t
1
2f
B
2
t
t
1
t
1
t < 2t
1
0 t 2t
1
8
>
>
>
>
>
<
>
>
>
>
>
:
(2:141)
The difference equation resulting from the substitution of the divided difference [v
A
(n þ1)
v
A
(n)]=T for the first derivative dv=dt in Equation 2.138 is
v
A
(n þ 1) ¼ v
A
(n) þ
T
m
[f (n) f
m
(n)] (2:142)
A recursive solution ...