# Help with Newton's Law of Gravitation

1. Feb 25, 2010

### SoulInNeed

1. Find the magnitude and direction of the net gravitational force on Mass A due to masses B and C in Figure 6.27. Each mass is 2.00 kg.

The figure looks like this:
A<----------->(10 cm distance)C<-------------------------->(40 cm distance)B

2.F(g)=G * (m(1)m(2))/(r^2)

3.I'm thinking of two ways to do this, one would be just include all 3 masses and the final distance between A & B, which would be 6.674e^-11 * (2 *2 *2)/(.50 m^2), which is 2.14e-9 to the right.

However, I could just do the two separately between A & C, and A & B, and just add them, or I could do A & C, and C & B, and just add them. Not too sure here, thanks for any help!

2. Feb 25, 2010

### Staff: Mentor

Now you're cooking.

Why C & B? You only care about the force on A, not between C and B.

3. Feb 25, 2010

### collinsmark

No, don't go that way. That's not the way the math works.

There ya go! try that one.

Yell, that last approach won't help you either. The problem statement is asking you the gravitational force on A, caused by the others. The gravitational force between C and B isn't relevant to the particular problem.

[Edit:] Looks like Doc Al beat me to it.

4. Feb 25, 2010

### SoulInNeed

OK, so would it be, A & C 6.674e-11 * (2 *2)/.1^2, which would be 2.7e-8.

Then, I would do A & B 6.674e-11 * (2 *2)/.5^2, which would be 1.07e-9.

Add them together and it's 2.807e-8. Can someone double check this? For the second part of the problem, I used .5m as the distance between A & B.

5. Feb 25, 2010

### collinsmark

You might want want to redo you final result (i.e. plugging in the numbers), keeping better track of significant digits, But yes, I believe you are essentially correct.

6. Feb 25, 2010

### SoulInNeed

Yeah, I was in a bit of a hurry, lol.