Physics question below?

Question

The drawing shows three particles far away from any other objects and located on a straight line. The masses and the position of these particles are ( in order) mA = 363 kg, mB = 517 kg, and mC = 154 kg. The distance from particle A to B is 0.500 and the distance from particle B to C is 0.250. Find the magnitude and direction of the net gravitational force acting on each of the three particles (let the direction to the right be positive).

Answer 1

F = G(m1*m2) / r^2 where G = 6.67 x 10^-11 Nm^2/kg^2 or


So between mA and mB:

F = 6.67 x 10^-11(363 x 517) / (.500m)^2

F = 5.00 x 10^-5 N (That's the magnitude)

The direction is opposite each body. mB is exerting a force that pulls mA to the right and mA has an attractive force to pull mB to the left.

Between B & C use the same formula and you get:

8.49 x 10^-5 N

Between A & C...note the distance is now .250 + .500

You get:

6.62 x 10^-6 N (which makes sense because as the distance increases, gravitational force will decrease.)

Regards,

Mysstere