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Chapter 12
373
P12.57
From geometry, observe that
cos
θ
=
1
4
and
=°
75 5
.
For the left half of the ladder, we have
FTR
xx
∑
=− =
0
(1)
FRn
yyA
∑
=+−
=
686
0
N
(2)
τ
top
N
∑
+
°
686
1 00
75 5
2 00
75 5
.c
o
s.
.s
i
n.
af
T
−°
=
n
A
400
755
0
o
(3)
For the right half of the ladder we have
FRT
∑
=−
=
0
Fn R
yBy
∑
=−=
0
(4)
top
∑
−
°
=
nT
B
200
0
o
i
(5)
FIG. P12.57
Solving equations 1 through 5 simultaneously yields:
(a)
T
=
133 N
(b)
n
A
=
429 N
and
n
B
=
257 N
(c)
R
x
=
133 N
and
R
y
=
257 N
The force exerted by the left half of the ladder on the right half is to the right and
downward.
P12.58
(a)
x
mx
m
y
ii
i
CG
CG
kg
m
kg
kg
kg
m
1375 kg
m
kg
m
kg
m
kg
m
kg
kg
m
=
=
+++
=
=
=
∑
∑
1 000
10 0
125
0
125
0
125
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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

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