Help with Newton's Law of Gravitation

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Homework Help Overview

The discussion revolves around applying Newton's Law of Gravitation to determine the net gravitational force on Mass A due to two other masses, B and C, in a specified arrangement. The participants are exploring the calculations involved with gravitational forces between the masses, all of which are 2.00 kg.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss different methods for calculating the gravitational force, including whether to consider all three masses together or to calculate the forces between A and each of the other masses separately. There is also a question about the relevance of the force between C and B in the context of the problem.

Discussion Status

The discussion is active, with participants providing feedback on each other's reasoning and calculations. Some guidance has been offered regarding the appropriate approach to take, particularly emphasizing the focus on the forces acting on Mass A.

Contextual Notes

There are indications of potential confusion regarding the distances used in calculations, as well as the importance of significant digits in the final results. Participants are encouraged to verify their calculations and reasoning as they proceed.

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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