Three point masses are fixed in position in an xy plane Two

Three point masses are fixed in position in an xy plane. Two of them, particle A of mass 6.00 g and particle B of mass 12.0 g, are shown in the figure with a separationofdAB=0.500matanangleof ?=30!. ParticleC, with mass 8.00 g is not shown. The net gravitational force acting on particle A due to particles B and C is 2.77×10?14 N at an angle of -163.8 degrees from the positive direction of the x axis. What are (a) the x coordinate and (b) the y coordinate of the particle C?

Solution

force, F_AB = GMm/r² =[(6.67*10^-11)*(6*10^-3)*(12*10^-3)]/0.5² = 1.92*10^-14 N

Angle between Fnet and F_AB is, 30 + 16.2 = 46.2^o.

F_AB =1.92*10^-14 and F_net = 2.77*10^-14

force, F_AC = [ (F_AB)² + (F_net)² - 2(F_AB)(F_net)*cos(46.2^o) ] = 2.0*10^-14 N

By law of sine, we get

        F_AC/sin(46.2) = F_AB/sin()

                          =43.86^o

So we know that Fac would be 180 -43.86 - 16.2 = 119.94^o away from x axis

To find x, y coordinate, we need to know that actually distance,

magnitude of F_AC =2.0*10^-14 = GMm/r^2 = [(6.67*10^-11)*(6*10^-3)*(8*10^-3)]/d²

then, the distance d is,

                  d = 0.4 m

x = d*cos(119.94^o) = -0.199 m

y = - d*sin(119.94^o) = 0.3466 m