Question

The figure shows a system of five objects. Determine the magnitude of the gravitational force acting on the

Answer 1

Answer:

c.  $|F_T|=8\frac{Gm^2}{d^2}$

   

Explanation:

In order to calculate the gravitational force on the mass of the center, you take into account the following formula:

$F=G\frac{m_1m_2}{r}$         (1)

Furthermore, you take into account the components of the resultant vector.

By the illustration, you have that the force is given by:

$F_T=F_1+F_2+F_3+F_4\\\\F_1=\frac{Gm_1m}{r^2}[-cos45\°\hat{i}+sin45\°\hat{j}]\\\\F_2=\frac{Gm_2m}{r^2}[cos45\°\hat{i}+sin45\°\hat{j}]\\\\F_3=\frac{Gm_3m}{r^2}[cos45\°\hat{i}-sin45\°\hat{j}]\\\\F_4=\frac{Gm_4m}{r^2}[-cos45\°\hat{i}-sin45\°\hat{j}]$

where:

m1 = m

m2 = 2m

m3 = m

m4 = 4m

m: mass at the center of the system

The distance r is:

$r=\sqrt{(\frac{d}{2})^2+(\frac{d}{2})^2}=\frac{d}{\sqrt{2}}$

You replace the values for all masses and sum the contributions of all forces:

$F_1=\frac{\sqrt{2}}{2}\frac{Gm^2}{(\frac{d^2}{2})}[-\hat{i}+\hat{j}]=\sqrt{2}\frac{Gm^2}{d^2}[-\hat{i}+\hat{j}]\\\\F_2=\frac{\sqrt{2}}{2}\frac{2Gm^2}{(\frac{d^2}{2})}[\hat{i}+\hat{j}]=2\sqrt{2}\frac{Gm^2}{d^2}[\hat{i}+\hat{j}]\\\\F_3=\frac{\sqrt{2}}{2}\frac{Gm^2}{(\frac{d^2}{2})}[\hat{i}-\hat{j}]=\sqrt{2}\frac{Gm^2}{s^2}[\hat{i}-\hat{j}]\\\\F_4=\frac{\sqrt{2}}{2}\frac{4Gm^2}{(\frac{d^2}{2})}[-\hat{i}-\hat{j}]=4\sqrt{2}\frac{Gm^2}{d^2}[-\hat{i}-\hat{j}]\\\\F_T=-2\sqrt{2}\frac{Gm^2}{d^2}}[\hat{i}+\hat{j}]$

and the magnitude is:

c.  $|F_T|=8\frac{Gm^2}{d^2}$