91影视

Three spheres of equal mass=5kg

Distance between sphere A and B,$r_{AB}=0.3m$

Distance between sphere B and C,$r_{BC}=0.4m$

Gravitational constant,$G=6.67脳{10}^{-11}N-m^2/k{g}^2$

This problem is based on Newton鈥檚 law of gravitation. Net force due to two forces can be calculated by using vector addition.

Formula:

Gravitational force of attraction, $F=\frac{GMm}{r^2}$ (i)

Using equation (i), Force between spheres A & B can be given by:

$$\begin{gathered} F_{AB}=\frac{G{m}_A{m}_B}{r_{AB}^2} \\ =\frac{6.67脳{10}^{-11}脳5脳5}{0.3^2} \\ =1.853脳{10}^{-8}N \end{gathered}$$

Force between B & C:

$$\begin{gathered} F_{BC}=\frac{G{m}_B{m}_B}{r_{BC}^2} \\ =\frac{6.67脳{10}^{-11}脳5脳5}{0.4^2} \\ =1.042脳{10}^{-8}N \end{gathered}$$

Net force on B due to A & C:

$$\begin{gathered} F=\sqrt{\left(F_{AB}^2+F_{BC}^2\right)} \\ =\sqrt{\left({\left(1.853脳{10}^{-8}\right)}^2+{\left(1.042脳{10}^{-8}\right)}^2\right)} \\ =2.126脳{10}^{-10}N \\ 鈮�2.13脳{10}^{-8}N \end{gathered}$$

The net force is$2.13脳{10}^{-8}N$

Direction of force on B:

$$\begin{gathered} 胃={\tan }^{-1}\left(\frac{F_{BA}}{F_{BC}}\right) \\ ={\tan }^{-1}\left(\frac{1.853脳{10}^{-8}}{1.042脳{10}^{-8}}\right) \\ =60.6掳 \end{gathered}$$

The net gravitational force makes an angle of $60.6掳$with positive x axis.