Solution :
We know that the volume of sphere is given by :
[tex]$V=\frac{4}{3}\pi r ^3$[/tex]
Given r = 50 mm
The density of copper is :
[tex]$\rho_c = 8910 \ kg/m^3$[/tex]
The density of steel is :
[tex]$\rho_s = 7830\ kg/m^3$[/tex]
Therefore, mass of copper is :
[tex]$m_c= V . \rho$[/tex]
[tex]$m_c= \frac{4}{3} \pi (0.05)^3 \times 8910$[/tex]
= 4.665 kg
= 4665 g
Similarly, mass of steel is :
[tex]$m_s= V . \rho$[/tex]
[tex]$m_s= \frac{4}{3} \pi (0.025)^3 \times 7830$[/tex]
= 0.512 kg
= 512 g
Now the distance between the centers of the particles is by Pythagoras theorem,
[tex]$l^2=(4r)^2+(2r)^2$[/tex]
l = 4.472 x r
= 4.472 x 50
= 224 mm
= 0.224 m
From the law of gravitation,
We know, [tex]G = 6673 \times 10^{-11} \ m^3 / kg-s^2[/tex]
∴ [tex]$F=G\times \frac{m_c \times m_s}{l^2}$[/tex]
[tex]$F=6.673 \times 10^{-11}\times \frac{4.665 \times 0.512}{(0.224)^2}$[/tex]
[tex]$F = 3.176 \times 10^{-9} \ N$[/tex]
Now,
[tex]$\alpha = \tan^{-1} \left( \frac{2r}{4r}\right)$[/tex]
= [tex]$26.56^\circ$[/tex]
Therefore, we can write it in the vector form as
[tex]$\text{F}=F(-i . \cos \alpha - j. \sin \alpha)$[/tex]
[tex]$\text{F}=3.176 \times 10^{-9}(-i . \cos 26.56 - j. \sin 26.56)$[/tex]
[tex]$\text{F}=(-2.840\ i- 1.420 \ j) \times 10^{-9} \ N$[/tex]
