Posted by **Anonymous** on Sunday, January 2, 2011 at 5:38pm.

three point particles are fixed in place in an xy plane. Particle A has mass mA = 3 g, 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. 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 (d = 15 cm)?

In the diagram, particle A is at the origin. Particle is on the y axis with the point (0,d). Particle C is on the x axis with the point (-1.5d,0).

When i solved for the force from particle b to a i got 5.336E-14 and when i solved for the force from particle c to a i got -3.557E-14. Now i just don't know how to find the place to put particle D

- physics -
**drwls**, Sunday, January 2, 2011 at 6:41pm
Place particle D where the force due to that particle, upon A, cancels the vector sum of the B and C particle's forces on A.

- physics -
**Anonymous**, Sunday, January 2, 2011 at 6:47pm
when i tired that i get the x to be .26m and y to be -.212m is that correct?

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