Question

The answer is required in terms of d, note there's no value for mA.

4. Three point particles are fixed in place in an xy plane. Particle A has mass ma, particle B has mass 2.00ma, and particle C has mass 3.00mA. A fourth particle D, with mass 4.00mA, is to be placed near the other three particles. In terms of distance d, at what (a) x coordinate and (b) y coordinate should particle D be placed so that the net gravitational force on particle A from particles B, C, and D is zero? 1.5d

Answer 1

a)Let the x coordinate be x and y coordinate be y,

x component of Gravitational field at A = 0

-G*3mA/(1.5d)^2 + G*4mA/(x^2+y^2) *x/sqrt(x^2+y^2) = 0

or   -3/(1.5d)^2 + 4x/(x^2+y^2)^1.5 =0

y component of Gravitational field at A = 0

G*2mA/(d)^2 + G*4mA/(x^2+y^2) *y/sqrt(x^2+y^2) = 0

    2/(d)^2 +4y/(x^2+y^2)^1.5 = 0

solving the two equations -3/(1.5d)^2 + 4x/(x^2+y^2)^1.5 =0 and  2/(d)^2 +4y/(x^2+y^2)^1.5 = 0

we get

x = 0.7156d answer of part A

y = -1.0734d

b) As found above , y = -1.0734d