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4.65.
θ
φ
G
O
P
d
a
Figure 1
We first draw a freebody diagram of the half cylinder as shown in Figure 2.
θ
φ
G
O
P
d
a
W
N
F
φ
Figure 2
The normal reaction force
N
must be perpendicular to the plane and hence to the tangent at
the point
P
. This force must therefore pass through the center of the half cylinder at
O
.
Summing the forces along the direction of
N
, we have
N
=
W
cos
θ
1
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View Full Document and summing along the direction of the friction force
F
,
F
=
W
sin
θ
.
For there to be no slip, we require
F
<
μ
N
, with
=
0
.
3, so
W
sin
<
0
.
3
W
cos
,
or
tan
<
0
.
3
and
<
16
.
7
o
.
To find the corresponding value of
φ
we sum the moments about
O
, obtaining
Wd
sin

Fr
=
0
(remember that
N
passes through
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This note was uploaded on 11/22/2011 for the course ME ME211 taught by Professor Barber during the Fall '10 term at University of Michigan.
 Fall '10
 Barber

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