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.