Question

Two dimensions. In Figure 13-34, 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 = 22 cm)?

MY WORK

      (1)
     (2)
     (3)

because F_AD is equal in magnitude to the net force of F_AB and F_AC.

Plug in the forces from (1), (2), and (3), and solve for r, which is the length from D to A.





Solve...



But these are wrong, I know that the correct answer is:
r_x = 0.157m
r_y = -0.236m

I am guessing that something with my trig and angle is wrong. I have tried this problem so many times I am obviously overlooking something easy and I need an outsider to look in. What did I do wrong?

Thanks, I appreciate it.

Answer 1

I think you substituted incorrectly for the other masses

mA = 3g
mB = 2.00mA = 6g
mC = 3.00mA = 9g
mD = 4.00mA = 12g

these must be converted to kg to work with the units of G