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 straight line.

Answer 1

`F_a=5.67* 10^(-5)\ N`

`F_b=3.49* 10^(-5)\ N`

`F_c=9.16* 10^(-5)\ N`

Step-by-step explanation:

Given:

• mass of particle A,
  `m_a=363\ kg`

• mass of particle B,
  `m_b=517\ kg`

• mass of particle C,
  `m_c=154\ kg`

• All the three particles lie on a straight line.

• Distance between particle A and B,
  `x_(ab)=0.5\ m`

• Distance between particle B and C,
  `x_(bc)=0.25\ m`

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

Force on particle A due to particles B & C:

`F_a=G. (m_a.m_b)/(x_(ab)^2) +G. (m_a.m_c)/((x_(ab)+x_(bc))^2)`

`F_a=6.67* 10^(-11)* ((363* 517)/(0.5^2)+(363* 154)/((0.5+0.25)^2))`

`F_a=5.67* 10^(-5)\ N`

Force on particle C due to particles B & A:

`F_c=G.(m_c.m_b)/(x_(bc)^2) +G.(m_c.m_a)/((x_(ab)+x_(bc))^2)`

`F_c=6.67* 10^(-11)* ((154* 517)/(0.25^2)+(154* 363)/((0.25+0.5)^2) )`

`F_c=9.16* 10^(-5)\ N`

Force on particle B due to particles C & A:

`F_b=G.(m_b.m_c)/(x_(bc)^2) -G.(m_b.m_a)/(x_(ab)^2)`

`F_b=6.67* 10^(-11)* ((517* 154)/(0.25^2)-(517* 363)/(0.5^2) )`

`F_b=3.49* 10^(-5)\ N`