TAM 212 Spring, 2004, Exam 1A
Name:
solutions
1.
R
mv
F
s
m
F
ma
F
F
normal
normal
normal
normal
horizontal
/
sin
/
sin
2
2
=
θ
⇒
ρ
=
θ
⇒
=
=
∑
∑
&
(1)
mg
F
mg
F
ma
F
normal
normal
Vertical
Vertical
=
θ
⇒
=
−
θ
⇒
=
∑
cos
0
cos
(2)
( ) ( )
⇒
2
1
(
)
Rg
v
Rg
v
/
tan
/
tan
2
1
2
−
=
θ
⇒
=
θ
2. Cable / tractor problem:
The total length of the cable is constant and given by
(
)
(
)
l
y
h
const
L
+
−
=
+
2
.
, i.e.
(
)
(
)
2
2
2
.
x
h
y
h
const
L
+
+
−
=
+
(1)
The velocity of bale B (up) is given by
y
&
, so we should differentiate (1) w.r.t. time:
2
2
2
0
x
h
x
x
y
L
+
+
−
=
=
&
&
&
Therefore, the vertical velocity of the bale is
2
2
2
1
x
h
x
x
y
+
=
&
&
(3)
Now we have to write (3) in terms of time, so we have to determine
x
and
x
&
.
We are given
a
A
=
c
, which is the
same as
c
x
=
&
&
.
Integration w.r.t. time gives
D
ct
x
+
=
&
.
But the tractor starts from rest, so
D
=0.
Integration
one more time gives
E
t
c
x
+
=
2
2
.
We can say that the tractor starts at
x
=0 when
t
=0, so
E
=0.
If we plug these relations into (3), we get the vertical velocity of bale B as a function of time:
4
2
1
4
2
2
3
t
c
h
ct
y
+
=
&
.
3.
Vertical bead problem:
x
x
ma
N
F
F
=
−
=
∑
θ
sin
.
a
x
=0, so
θ
sin
F
N
=
.
(1)
dt
dv
m
ma
mg
N
F
F
y
k
y
=
=
−
−
=
∑
µ
θ
cos
(2)
where
v
is the vertical velocity of the collar.
We know that
θ
=
kt
(problem statement) and from (1) that
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 Spring '08
 Keane
 Trigraph, vertical velocity, 0k, MDV, w.r.t. time

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