11.59:
a)
,
0
=
∑
τ
axis at lower end of beam
Let the length of the beam be
L
.
N
2700
20
sin
40
cos
0
40
cos
2
)
20
(sin
2
1
=
°
°
=
=
°

=
°
mg
T
L
mg
L
T
b) Take +
y
upward.
N
1372
60
cos
gives
0
N
6
.
73
so
0
60
sin
gives
0
s
x
=
°
=
=
∑
=
=
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 mg cos, lower end, beam Let

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