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net,
net,
0
cos
cos
0s
i
n
s
i
n
xC
A
yA
C
B
FF
F
F
F
φθ
θφ
==−
==+
−
¦
¦
To solve for
B
F
, we first compute
.
φ
With
220 N
A
F
=
,
170 N
C
F
=
and
47 ,
θ
=°
we get
cos
(220 N)cos47.0
cos
0.883
28.0
170 N
A
C
F
F
φφ
°
==
=
¡
Substituting the value into the second force equation, we find
sin
sin
(220 N)sin 47.0
(170 N)sin 28.0
241 N.
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Force

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