Answers: 1
(1)![F_T=F_1+F_2+F_3+F_4\\\\F_1=\frac{Gm_1m}{r^2}[-cos45\°\hat{i}+sin45\°\hat{j}]\\\\F_2=\frac{Gm_2m}{r^2}[cos45\°\hat{i}+sin45\°\hat{j}]\\\\F_3=\frac{Gm_3m}{r^2}[cos45\°\hat{i}-sin45\°\hat{j}]\\\\F_4=\frac{Gm_4m}{r^2}[-cos45\°\hat{i}-sin45\°\hat{j}]](/tpl/images/0706/0697/e2190.png)
![F_1=\frac{\sqrt{2}}{2}\frac{Gm^2}{(\frac{d^2}{2})}[-\hat{i}+\hat{j}]=\sqrt{2}\frac{Gm^2}{d^2}[-\hat{i}+\hat{j}]\\\\F_2=\frac{\sqrt{2}}{2}\frac{2Gm^2}{(\frac{d^2}{2})}[\hat{i}+\hat{j}]=2\sqrt{2}\frac{Gm^2}{d^2}[\hat{i}+\hat{j}]\\\\F_3=\frac{\sqrt{2}}{2}\frac{Gm^2}{(\frac{d^2}{2})}[\hat{i}-\hat{j}]=\sqrt{2}\frac{Gm^2}{s^2}[\hat{i}-\hat{j}]\\\\F_4=\frac{\sqrt{2}}{2}\frac{4Gm^2}{(\frac{d^2}{2})}[-\hat{i}-\hat{j}]=4\sqrt{2}\frac{Gm^2}{d^2}[-\hat{i}-\hat{j}]\\\\F_T=-2\sqrt{2}\frac{Gm^2}{d^2}}[\hat{i}+\hat{j}]](/tpl/images/0706/0697/43367.png)
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