Question

masses of 2 kg each are placed at corners b and a

Answer 1

You are adding three vectors here. You need to cancel the lateral vector elements of the left and right masses and only add the remaining vector elements which point at the most distant mass.

Thus you get sin (45º) times vector left and right, plus the vector of the far mass.

Fl/r=Gm1⋅m2r2=6.674×10−11900⋅900.122=0.00375418125N

So for the two vectors, left and right, we get:

2⋅sin(45∘)⋅0.00375418125=0.0053N

And for the straight across vector, we get:

6.674×10−11900⋅900.16972=0.0019N

For a total of 0.0072N