Answer:
c. [tex]|F_T|=8\frac{Gm^2}{d^2}[/tex]
Explanation:
In order to calculate the gravitational force on the mass of the center, you take into account the following formula:
[tex]F=G\frac{m_1m_2}{r}[/tex] (1)
Furthermore, you take into account the components of the resultant vector.
By the illustration, you have that the force is given by:
[tex]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}][/tex]
where:
m1 = m
m2 = 2m
m3 = m
m4 = 4m
m: mass at the center of the system
The distance r is:
[tex]r=\sqrt{(\frac{d}{2})^2+(\frac{d}{2})^2}=\frac{d}{\sqrt{2}}[/tex]
You replace the values for all masses and sum the contributions of all forces:
[tex]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}][/tex]
and the magnitude is:
c. [tex]|F_T|=8\frac{Gm^2}{d^2}[/tex]
