11. We use m
1
for the 20 kg of the sphere at (x
1
, y
1
) = (0.5, 1.0) (SI units understood), m
2
for the 40 kg
of the sphere at (x
2
, y
2
) = (
−1.0, −1.0), and m
3
for the 60 kg of the sphere at (x
3
, y
3
) = (0,
−0.5). The
mass of the 20 kg object at the origin is simply denoted m. We note that r
1
=
√
1.25, r
2
=
√
2, and
r
3
= 0.5 (again, with SI units understood). The force
F
n
that the n
th
sphere exerts on m has magnitude
Gm
n
m/r
2
n
and is directed from the origin towards m
n
, so that it is conveniently written as
F
n
=
Gm
n
m
r
2
n
x
n
r
n
ˆi+
y
n
r
n
ˆj
=
Gm
n
m
r
3
n
x
n
ˆi+ y
n
ˆj
.
Consequently, the vector addition to obtain the net force on m becomes
F
net
=
3
n=1
F
n
=
Gm
3
n=1
m
n
x
n
r
3
n
ˆi+
3
n=1
m
n
y
n
r
3
n
ˆj
=
−9.3 × 10
−9
ˆi
− 3.2 × 10
−7
ˆj
in SI units. Therefore, we find the net force magnitude is
| F
net
| = 3.2 × 10
−7
N.