Answer to Question #188841 in Physics for .jac

Question

An equilateral triangle 10.0m on a side has 1.00kg mass at one corner a 2.00kg mass at another corner and a 3.00kg mass at the third corner. Find the magnitude abd direction of the net force acting on the 2.00 kg mass

Answer 1

$F_1=G\frac{m_1\cdot m_2}{x^2}=6.67\cdot10^{-11}\cdot\frac{1\cdot2}{10^2}=1.334\cdot10^{-12} \ (N)$

$F_2=G\frac{m_3\cdot m_2}{x^2}=6.67\cdot10^{-11}\cdot\frac{3\cdot2}{10^2}=4.002\cdot10^{-12} \ (N)$

$F=\sqrt{F_1^2+F_2^2+2F_1F_2\cos60°}=$

$=\sqrt{(1.334\cdot10^{-12})^2+(4.002\cdot10^{-12})^2+2\cdot1.334\cdot10^{-12}\cdot 4.002\cdot10^{-12}\cdot \cos60°}=$

$=4.81\cdot10^{-12}\ (N)$

$F_x=F_1+F_2\cdot\cos60°=1.334\cdot10^{-12}+4.002\cdot10^{-12}\cdot\cos60°=3.335\cdot10^{-12}\ (N)$

$F_y=F_2\cdot\sin60°=4.002\cdot 10^{-12}\cdot \sin60°=3.46\cdot 10^{-12} \ (N)$

$\alpha=\tan^{-1}\frac{F_y}{F_x}=\tan^{-1}\frac{3.46\cdot 10^{-12}}{3.335\cdot 10^{-12}}=46°$ (relative to the base 1-2)