Question

The net gravitational force on any of the masses at the vertices of the hexagon is

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

Option 0

Answer 2

To find the net gravitational force on any of the masses at the vertices of a regular hexagon, we can choose one vertex, say A, and calculate the vector sum of the gravitational forces exerted by the other five masses on the mass at A.

Let 'm' be the mass of each point mass and 'l' be the side length of the regular hexagon.

The distances from vertex A to the other vertices are:

1. Adjacent vertices (B and F): The distance is 'l'. The forces are $F_{AB} = \frac{Gm^2}{l^2}$ and $F_{AF} = \frac{Gm^2}{l^2}$.
2. Alternate vertices (C and E): The distance is $\sqrt{3}l$ (length of the shorter diagonal). The forces are $F_{AC} = \frac{Gm^2}{(\sqrt{3}l)^2} = \frac{Gm^2}{3l^2}$ and $F_{AE} = \frac{Gm^2}{(\sqrt{3}l)^2} = \frac{Gm^2}{3l^2}$.
3. Opposite vertex (D): The distance is $2l$ (length of the main diagonal). The force is $F_{AD} = \frac{Gm^2}{(2l)^2} = \frac{Gm^2}{4l^2}$.

Due to the symmetry of the hexagon, the net force will be directed along the line connecting the chosen vertex (A) to the opposite vertex (D). Let's call this direction the AD-axis.

We can find the resultant force by summing the components along the AD-axis.

1. Resultant of forces from adjacent masses ($F_{AB}$ and $F_{AF}$): The angle between $F_{AB}$ and $F_{AF}$ is $120^\circ$. Their magnitudes are equal: $F_1 = \frac{Gm^2}{l^2}$. The resultant force $F_{R1}$ is given by the parallelogram law of vector addition: $F_{R1} = \sqrt{F_1^2 + F_1^2 + 2F_1 F_1 \cos(120^\circ)}$ $F_{R1} = \sqrt{2F_1^2 + 2F_1^2 (-\frac{1}{2})} = \sqrt{2F_1^2 - F_1^2} = \sqrt{F_1^2} = F_1 = \frac{Gm^2}{l^2}$. This resultant force $F_{R1}$ acts along the AD-axis.

2. Resultant of forces from alternate masses ($F_{AC}$ and $F_{AE}$): The angle between $F_{AC}$ and $F_{AE}$ is $60^\circ$. Their magnitudes are equal: $F_2 = \frac{Gm^2}{3l^2}$. The resultant force $F_{R2}$ is given by: $F_{R2} = \sqrt{F_2^2 + F_2^2 + 2F_2 F_2 \cos(60^\circ)}$ $F_{R2} = \sqrt{2F_2^2 + 2F_2^2 (\frac{1}{2})} = \sqrt{3F_2^2} = \sqrt{3}F_2 = \sqrt{3} \frac{Gm^2}{3l^2} = \frac{Gm^2}{\sqrt{3}l^2}$. This resultant force $F_{R2}$ also acts along the AD-axis.

3. Force from the opposite mass ($F_{AD}$): This force $F_{AD} = \frac{Gm^2}{4l^2}$ already acts along the AD-axis.

Net gravitational force: Since all the resultant forces and the direct force from D are along the same direction (AD-axis), the net force is the scalar sum of their magnitudes: $F_{net} = F_{R1} + F_{R2} + F_{AD}$ $F_{net} = \frac{Gm^2}{l^2} + \frac{Gm^2}{\sqrt{3}l^2} + \frac{Gm^2}{4l^2}$ Factor out $\frac{Gm^2}{l^2}$: $F_{net} = \frac{Gm^2}{l^2} \left( 1 + \frac{1}{\sqrt{3}} + \frac{1}{4} \right)$