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

Answer:

A: 5.67*10^-5 N

B: 3.49*10^-5 N

C: -9.16*10^-5 N

Explanation:

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

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