30.0º = 150º (measured ccw from the +
x
axis).
The component along, say, the
x
axis of
one of the forcevectors
F
→
is simply
Fx/r
in this situation (where
F
is the magnitude of
F
→
).
Since the force itself (see Eq. 131) is inversely proportional to
r
2
then the
aforementioned
x
component would have the form
GmMx/r
3
; similarly for the other
components. With
m
A
= 0.0060 kg,
m
B
= 0.0120 kg, and
m
C
= 0.0080 kg, we therefore
have
F
net
x
=
Gm
A
m
B
x
B
r
B
3
+
Gm
A
m
C
x
C
r
C
3
= (2.77
×
10
−
14
N)cos(
−
163.8º)
and
F
net
y
=
Gm
A
m
B
y
B
r
B
3
+
Gm
A
m
C
y
C
r
C
3
= (2.77
×
10
−
14
N)sin(
−
163.8º)
where
r
B
=
d
AB
= 0.50 m, and (
x
B
,
y
B
) = (
r
B
cos(150º),
r
B
sin(150º)) (with SI units
understood).
A fairly quick way to solve for
r
C
is to consider the vector difference
This is the end of the preview. Sign up
to
access the rest of the document.
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

Click to edit the document details