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374
Static Equilibrium and Elasticity
P12.60
When it is on the verge of slipping, the cylinder is in equilibrium.
F
x
∑
=
0:
fn
n
s
12
1
==
µ
and
s
22
=
F
y
∑
=
Pn f F
g
++=
τ
∑
=
Pf f
=+
As
P
grows so do
f
1
and
f
2
Therefore, since
s
=
1
2
,
f
n
1
1
2
=
and
f
nn
2
21
24
FIG. P12.60
then
Pn
n
F
g
++ =
1
1
4
(1)
and
P
n
=+=
11
1
3
4
(2)
So
F
g
+=
5
4
1
becomes
PP
F
g
+
F
H
G
I
K
J
=
5
4
4
3
or
8
3
PF
g
=
Therefore,
g
=
3
8
P12.61
(a)
FkL
=∆
af
, Young’s modulus is
Y
FL
AL
F
A
L
L
i
i
∆
∆
Thus,
Y
kL
A
i
=
and
k
YA
L
i
=
(b)
W
Fdx
kx dx
YA
L
xdx
YA
L
L
LL
i
L
i
=−
−
=
=
zz
z
00
0
2
2
∆∆
∆
∆
P12.62
(a)
Take both balls together. Their weight is 3.33 N
and their CG is at their contact point.
F
x
∑
=
0:
+−=
31
0
F
y
∑
=
+−
=
P
2
333
0
.
N
P
2
=
N
A
∑
=
−+
−
+
°
PR PR
R R
32
450
.c
o
s
.
N
++
°
=
PR R
1
5
0
0
cos
.
Substituting,
−
+
°
°
=
°=
°
P
RRR
PR
P
1
1
1
13
1
12 4
5
0 0
2
167
..c
o
s
.
cos
.
.
cos
.
cos
.
..
N
N
N
s
o
N
a
f
F
g
P
1
P
2
P
3
3.33 N
FIG. P12.62(a)
(b)
Take the upper ball. The lines of action of its weight, of
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics, Static Equilibrium

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