5.43.
(a)
IDENTIFY:
Apply
m
=
∑
F
a
r
r
to the crate. Constant
v
implies
0.
a
=
Crate moving says that the friction is
kinetic friction. The target variable is the magnitude of the force applied by the woman.
SET UP:
The freebody diagram for the crate is sketched in Figure 5.43.
EXECUTE:
y
y
Fm
a
=
∑
sin
0
nm
gF
θ
−
−=
sin
=
+
kk
k
k
sin
fn
m
g
F
µµ
µ
==
+
Figure 5.43
x
x
a
=
∑
k
cos
0
Ff
cos
sin
0
g
F
θµ
−−
=
(cos
sin )
g
θµ θ µ
k
k
cos
sin
mg
F
=
−
(b)
IDENTIFY
and
SET UP:
“start the crate moving” means the same force diagram as in part (a), except that
k
is replaced by
s
.
Thus
s
s
.
cos
sin
mg
F
=
−
EXECUTE:
F
→∞
if
s
cos
sin
0.
θµ θ
This gives
s
cos
1
.
sin
tan
EVALUATE:
F
r
has a downward component so
.
g
>
If
0
=
(woman pushes horizontally),
g
=
and
.
m
g
5.45.
IDENTIFY:
Apply
m
∑
F
=a
r
r
to each block.
SET UP:
For block
B
use coordinates parallel and perpendicular to the incline. Since they are connected by
ropes, blocks
A
and
B
also move with constant speed.
EXECUTE:
(a)
The freebody diagrams are sketched in Figure 5.45.
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 Fall '07
 Evrard
 Force, Friction, Friction Force, Sin

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