Question

Three particles are fixed in positions on the xy plane. Particle A, with a mass of 6.00 g, is located at the origin. Particle B, with a mass of 12.0 g, is located in the second quadrant a distance 0.500 m from the origin at an angle of 30◦ from the x axis. Particle C has a mass of 8.00 g. 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 ◦ from the +x axis. What are the x and y coordinates of particle C?

Answer 1

net force = 2.77*10^-14 at -163.8 from positive x

x-component= 2.77*10^-14 cos(-163.8)= -2.66*10^-14 N

y-component = 2.77*10^-14 sin(-163.8)= -0.77*10^-44 N

F by B= 6.67*10^-11*0.006*0.012/0.500^2 = 1.92*10^-14 N

x component = -1.92*10^-14 cos30 = -1.66*10^-14 N

y component = 1.92*10^-14 sin30 = 0.96*10^-14 N

Cx= -2.66*10^-14 N- (-1.66*10^-14 N )= -1*10^-14 N

Cy= -0.77*10^-44 N-0.96*10^-14 N=-1.73*10^-14 N

Net force by C= 2.05*10^-14 = 6.67*10^-11*0.006*0.008/r^2

r=0.395 m

angle = 57.55 degress from negative x

so rx= -0.21m

ry=-0.33m

so answer = (-0.21,-0.33)

kindly check the calculations once.