4.37
(a)
Since the crate has constant velocity,
0
x
y
a
a
=
=
.
Applying Newton’s second law:
, or
cos20.0
0
x
k
x
Assignment 4 solutions (Fall term Physics 102)
F
F
f
ma
Σ
=
° −
=
=
(
)
300 N
cos20.0
282 N
k
f
=
° =
and
, or
sin 20.0
0
y
F
n
F
w
Σ
=
−
° −
=
(
)
3
300 N
sin 20.0
1000 N
1.10
10
N
n
=
° +
=
×
.
The coefficient of friction is then
3
282 N
1.10
10
N
k
k
f
n
µ
=
=
=
×
0.256
.
(b) In this case,
,
so
.
The friction force now becomes
sin 20.0
0
y
F
n
F
w
Σ
=
+
° −
=
sin 20.0
897 N
n
w
F
=
−
° =
(
)(
)
0.256
897 N
230 N
k
k
f
n
µ
=
=
=
.
Therefore,
cos20.0
x
k
x
w
x
F
F
f
ma
g
⎛
⎞
Σ
=
° −
=
=
⎜
⎟
⎝
⎠
a
and the acceleration is
(
)
(
)
[
]
(
)
2
300 N
cos20.0
230 N
9.80 m s
cos20.0
1000 N
k
F
f
g
a
w
° −
° −
=
=
=
2
0.509 m s
4.41
The normal force acting on the crate is given by
cos
n
F
mg
θ
=
+
. The net force tending to move the crate down
the incline is
sin
s
mg
f
θ
−
, where
s
f
is the force of static
friction between the crate and the incline. If the crate is in
equilibrium, then
sin
0
s
mg
f
θ
−
=
, so that
sin
s
f
mg
θ
=
.
But, we also know
(
)
cos
s
s
s
f
n
F
mg
µ
µ
θ
≤
=
+
.
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 Fall '08
 VANDENBERG
 Physics, Force, Friction, Sin, ΣFx

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