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Gravitational Forces on three masses at the corners of an equilateral triangle
I solved the math using vector rule R= \sqrt{F^2 +F^2 +2F^2cos\frac{\pi}{3}} =\sqrt{3}\frac{Gm^2}{a^2} But the answer is showing: \sqrt{3}\frac{Gm^2}{a^2} (-\hat{j}) My question is: Why is (-\hat{j}) added here? Why is it negative?- hasibx
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- Gravitation
- Replies: 4
- Forum: Introductory Physics Homework Help