In the figure, three 5.00kgspheres are located at distances${\mathbf{d}}_{\mathbf{1}}$=0.300 mand${\mathbf{d}}_{\mathbf{2}}$=0.400 m. What are the (a) magnitude and (b) direction (relative to the positive direction of thexaxis) of the net gravitational force on sphereBdue to spheresAandC?

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,$G=6.67\times {10}^{-11}N-m^2/k{g}^2$

This problem is based on Newton’s 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:

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

Force between B & C:

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

Net force on B due to A & C:

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

The net force is$2.13\times {10}^{-8}N$

Direction of force on B:

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

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