Help with Newton's Law of Gravitation

AI Thread Summary
To find the net gravitational force on Mass A from masses B and C, the gravitational force formula is applied, F(g) = G * (m1 * m2) / (r^2). The discussion suggests calculating the forces separately between A & C and A & B, rather than including the forces between C and B, which are irrelevant to the problem. The calculations yield forces of approximately 2.7e-8 N from A & C and 1.07e-9 N from A & B, leading to a total of about 2.807e-8 N. Participants emphasize the importance of tracking significant digits in the final results.
SoulInNeed
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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!
 
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SoulInNeed said:
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.
:bugeye:

However, I could just do the two separately between A & C, and A & B, and just add them,
Now you're cooking.

or I could do A & C, and C & B, and just add them.
Why C & B? You only care about the force on A, not between C and B.
 
SoulInNeed said:
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.

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

However, I could just do the two separately between A & C, and A & B, and just add them,

There you go! try that one. :approve:

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

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.
 
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.
 
SoulInNeed said:
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.

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. :cool:
 
collinsmark said:
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. :cool:
Yeah, I was in a bit of a hurry, lol.
 
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