Three Particles - Gravitational Force

[Student3.41]

Homework Statement

The masses of these particles are mA = 379 kg, mB = 500 kg, and mC = 140 kg. d1 = 0.558 m and d2 = 0.279 m. Calculate the magnitude of the net gravitational force acting on particle A

(Attached Diagram of Particles.)

Homework Equations

F=Gm1m2/r^2

G=6.673x10^-11Nm^2/kg^2

The Attempt at a Solution

The question asks what is the magnitude of the g force acting on particle A. So, I added the masses of particles b and c.

Particle B and C = 640kg

F = (6.673x10^-11)(379kg)(640kg)/(0.558m+0.279m) = (0.0000162)/(0.701)= 0.0000231N (WRONG)

Then I thought particle C has no effect on the force of particle A so I just used the mass of particle A and B.

F=(6.673x10^-11)(379)(500)/(0.558)^2 = (0.0000126)/(0.346) = 0.0000364N

Then I added the Force from A-B and Force B-C...

Fa-b=0.0000364N

Fb-c=(6.673x10^-11)(500)(140)/(0.279)^2 = 0.0000600N

Therefore, 0.0000364N+0.0000600N=Fnet=0.0000964N

Can anyone help?

 

[Dadface]

Find the force on A due to B on on its own then A due to C on its own.Both forces on A act in the same direction so to find the total force just add them.

 

[tiny-tim]

Hi Student3.41!

The question asks what is the magnitude of the g force acting on particle A. So, I added the masses of particles b and c.

Particle B and C = 640kg

F = (6.673x10^-11)(379kg)(640kg)/(0.558m+0.279m) = (0.0000162)/(0.701)= 0.0000231N (WRONG)

You can't do that!

You must add the force from B to the force from C (ie, use the formula twice, separately) …

you can't just add the masses, and add the distances, and then chuck them all into the same formula!

Try again. ​

 

[Student3.41]

Thank you!, got to analyze the problem more