Question

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles are mA = 363 kg, mB = 517 kg, and mC = 154 kg. 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).

Particle A is 0.5m from B and B is .25m from C... All in a astraight line

Answer 1

A: 5.67*10^-5 N

B: 3.49*10^-5 N

C: -9.16*10^-5 N

Step-by-step explanation:

`\\A\\F_(A) = F_(BA) + F_(CA) =G*(m_(b)*m_(a))/(r^2_(AB)) + G*(m_(c)*m_(a))/(r^2_(AC)) \\\\= (6.67*10^(-11))*((363*517)/(0.5^2) + (154*363)/(0.75^2))\\\\= 5.67*10^-5 N \\\\B\\\\F_(B) = -F_(BA) + F_(CB) =-G*(m_(b)*m_(a))/(r^2_(AB)) + G*(m_(c)*m_(b))/(r^2_(BC)) \\\\= (6.67*10^(-11))*(-(517*363)/(0.5^2) + (517*154)/(0.25^2))\\\\= 3.49*10^-5\\\\`

`F_(C) = -F_(CA) - F_(CB) =-G*(m_(b)*m_(c))/(r^2_(BC)) - G*(m_(c)*m_(a))/(r^2_(AC)) \\\\= (6.67*10^(-11))*(-(517*154)/(0.25^2) - (154*363)/(0.75^2))\\\\= -9.16*10^-5`