Question

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?

Short answer

a) The magnitude of the net gravitational force on sphere B is $2.13\times {10}^{-8}N$

b) The net gravitational force on sphere B is $60.0°$atwith positive x-axis.

Step-by-step solution

The given data

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

Understanding the concept of Newton’s law of gravitation

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)

a) Calculating the magnitude of the force on the sphere B

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$2.13\times {10}^{-8}N$

b) Calculating the direction of the net force on sphere B

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.