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Spring, 2004
2.136 Given:
For the image shown, F(t) = 5 + 2t
lb
(where
t
is in seconds).
The coefficient of friction (static
and kinetic) = 0.2.
The box is at rest when t = 0.
Its weight is 50lb.
Find:
The velocity of the box when t = 10 s
Solution:
Use Impulse = change in momentum
()
0
1
2
2
1
−
=
−
=
−
=
=
∫
x
m
x
x
m
dt
i
f
t
t
&
&
&
L
L
F
I
In the horizontal direction,
( ) ( )
x
m
f
t
F
f
t
F
x
m
F
x
&
&
&
&
=
−
=
−
⇒
=
∑
2
/
3
30
cos
Until the force F(t) is larger than the frictional force,
0
=
x
&
&
, i.e.,
2
/
3
2
5
t
f
N
+
=
≥
µ
This happens at t*, where t* is found by finding where
( )
2
/
3
*
2
5
t
N
+
=
(1)
We find
N
by looking in the ydirection:
() ( )
t
mg
N
mg
N
t
mg
N
t
F
y
m
F
y
+
+
=
⇒
=
−
+
+
−
⇒
=
−
+
⇒
=
=
∑
5
.
2
0
2
/
2
5
0
30
sin
0
&
&
t
N
+
=
⇒
5
.
52
(2)
Using this in (1)
( )
2
/
3
*
2
5
*
5
.
52
2
.
0
t
t
+
=
+
⇒
, i.e., t*~4.03 s
So, until 4.03s, the net force is zero, and after 4.03 s (once the block starts sliding,
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This note was uploaded on 04/14/2010 for the course TAM 212 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Staff

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