# Finding net gravitational attractions in a triangle

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1. Jul 31, 2017

### AfronPie

1. The problem is attached in a picture. I've done it five times and keep getting it wrong. The correct answer is also displayed in the picture. Thank you for any help.

2. Fg=Gm_1m_2/r^2

3. I calculated Fg between AB and BC (they are equal). So Fg=(6.67*10^-11)(4)(4)/(.10)^2=1.067*10^-10. Then I drew a triangle (also attached). I labeled the point in between B and C the center of my axis. So when I calculated the sum of the forces in the x direction I got 0. In the y direction the sum of the forces is 2*1.067*10^-10*sin60. I did that and got 2.13*10^-10. Why isn't that the right answer?

2. Jul 31, 2017

### Staff: Mentor

Hello AfronPie,

Welcome to Physics Forums!

Recheck your calculation of the force between two of the masses (your Fg). The value you're getting is too small, so check how you're handling the scientific notation exponents.

3. Jul 31, 2017

### AfronPie

Thank you very very much. I did the whole thing right but that one mistake messed it up completely.