Physics E1a
Homework Assignment #4 Solutions
Fall 2011
1.
(a)
(b)
The velocity is constant so we know that the acceleration is zero
a = 0
F
y
=
F
N
!
"
F
push
sin
# !
F
gravity
=
ma
=
0
Fx
=
!
F
cos
"#
F
fr
=
=
0
we can find F
push
from this second equation.
F
=
F
cos
!
or
F
=
60N
cos 35
0
( )
=
73.2N
2.
F
y
=
F
!
N
+
F
pull
sin
"
#
F
gr
=
0
and
=
F
!
"
F
pull
cos
#
=
0
$
F
=
F
cos
#
We also know that
F
= μ
F
N
!
F
N
=
F
μ
Now substitute this expression for F
N
into the
F
y
!
equation.
Which gives us
F
fr
μ
+
F
pull
sin
!
"
F
gr
=
0
then substitute for F
fr
F
=
F
cos
!
F
cos
!
μ
+
F
sin
!
=
F
gr
"
F
=
F
gr
cos
!
μ
+
sin
!
#
$
%
&
'
F
=
mg
cos 35
0
( )
μ
+
sin 35
0
( )
!
"
#
$
%
&
We can determine
μ
using information from problem 1.
We know that
F
= μ
F
N
=
60N
where
F
N
=
F
sin
!
+
F
gr
plugging in numbers we find that
F
N
=
73.2N
( )
sin 35
0
( ) +
28kg
( )
9.8m s
2
( ) =
42.0N
+
274.4N
=
316N
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then,
μ
=
F
fr
F
N
=
60
N
316
N
=
0.19
Finally solve for F
pull
F
pull
=
274.4N
4.31
+
0.57
=
56.2N
Pulling the cart requires less force.
3.
(a) The distance the wheel travels in one revolution is the circumference of the outer edge of
the wheel.
C
=
!
D
=
!
0.850m
( ) =
2.67m
, if the plane is travelling at 60m/s then the wheel is
revolving at
60m
s
!
1rev
2.67m
!
60
min
=
1348rpm
rev
"
#
$
%
(b)
a
centripetal
=
v
2
R
=
60m s
( )
2
0.850m 2
=
8471m s
2
(c)
F
centripetal
=
ma
centripetal
=
10
!
15
kg
( )
2
( ) =
8.47
"
10
!
12
N
(d)The weight of the bacterium is
mg
=
10
!
kg
( )
9.8m s
2
( ) =
9.8
"
10
!
N
The ratio of the force
to the weight of the bacterium is
8.47
!
10
"
9.8
!
10
"
=
864
1
4
. (a)
The horizontal component of the normal force F
N
supplies the centripetal
force.
F
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 Fall '11
 WolfgangRueckner
 mechanics, Acceleration, Force, Friction, Gravity, Work, Sin, Cos, FN, Fpull

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