Answer:
[tex]F'=4.5\times 10^{-5}\ N[/tex]
Explanation:
The gravitational force between two pickups is given by :
[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]
It means,
[tex]F\propto m_1m_2[/tex]
So,
[tex]\dfrac{F}{F'}=\dfrac{m_1m_2}{m_1'm_2'}[/tex]
We have,
[tex]m_1=m_2=2000\ kg\\\\F=3\times 10^{-5}\ N\\\\m_1'=2000+1000 =3000\ kg\\\\m_2'\ \text{remains the same i.e. 1000 kg}[/tex]
F' is the new force
So, putting all the values,
[tex]\dfrac{3\times 10^{-5}}{F'}=\dfrac{2000\times 2000}{3000\times 2000}\\\\\dfrac{3\times 10^{-5}}{F'}=\dfrac{2}{3}\\\\F'=\dfrac{9\times 10^{-5}}{2}\\\\F'=4.5\times 10^{-5}\ N[/tex]
So, the new force between the two trucks is [tex]4.5\times 10^{-5}\ N[/tex].
