Objects with masses of 120 kg and a 420 kg are separated by 0.310 m.?

Question

Objects with masses of 120 kg and a 420 kg are separated by 0.310 m.

(a) Find the net gravitational force exerted by these objects on a 50.0 kg object placed midway between them.
4.1642E-5 N

(b) At what position (other than infinitely remote ones) can the 50.0 kg object be placed so as to experience a net force of zero?




[That's not the correct answer, hmata3. stfu.]

Answer 1

Imagine this setup.

(a) For the Left Half:

(420 kg) ------0.155 m------ (50 kg)

F = G M m / d^2
= 5.8339180 E -5 N

For the Right Half:

(50 kg) ------0.155 m------ (120 kg)

F = G N m / d^2
= 1.6668337 E -5 N

The difference, or net force, is 4.167084 E -5 N.

--------------------------------------...

(b) The solution to this has infinite solutions. There is NOT one point in space where the net force will be zero. There are infinitely many.

You will need to place one mass at any distance from the 50 kg mass. Then solve for the distance required to balane the fores. Otherwise the solution will cancel out your distances, "d's."

Assign a distance for the 420 kg mass as a variable "k". Thus the force will be

F = G ( 420 ) ( 50 ) / k^2

Assign a distance for the 120 kg mass as a variable "d". Thus the force will be

f = G ( 120 ) ( 50 ) / d^2

Solve

F = f

21000 / k^2 = 6000 / d^2, Rearranging

d = k / sqrt (3.5)

Setting the 420 kg mass any distance, k, from the 50 kg mass, ...

...will be the equivalent of placing the 120 kg mass a distance, d, away from the 50 kg mass,...

...where d = k / sqrt (3.5), in meters, and k is any number.

Answer 2

given that the gravitational force is m1m2G/r^2, and the r is the same for the two masses
net force = G/.310^2 (420x50 - 120x50)

as for the net force = 0, intuitively the 50kg will have a net fprce of 0 when it's between the masses at a point where the force of each mass is equal. And also intuitive is that the 50kg mass will be closer to the smaller mass. you can easily set up to find the ratio of the distance from 50kg to 120kg over the distance from 50kg to 420kg.

f1 / f2 = 1 (because you're looking for when they are equal)
1= [Gx50x120/r1^2] / [Gx50x420/r2^2]
Gx50 cancels
1=[120/r1^2] / [420/r2^2]
420/r2^2 = 120/r1^2
420xr1^2 = 120xr2^2
sqrt420 x r1 = sqrt120 x r2
r1/r2 = sqrt 120/sqrt420
given that ratio and total the distance, 0.310m, you can find the point with balanced forces.