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 FAD is equal in
magnitude to the net force of FAB and
FAC.
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:
rx = 0.157m
ry = -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.
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
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