Ncert Exercises & Solutions

Q. 35 Six point masses of mass m each are at the vertices of a regular hexagon of side l. Calculate the force on any of the masses.

Hint: Apply Newton's law of gravitation.

Consider the diagram below in which six point masses are Placed at six vertices A, B, C, D, E and F.

$AC = AG + GC = 2 AG = 2 lcos 30 ° = 2 l \times \sqrt{3} / 2 = \sqrt{3} l = AE AD = AH + HJ + JD = lsin 30 ° + l + lsin 30 ° = 2 l$

Step 1: Find the forces on any mass due to other masses.

Force on mass m at A due to mass m at B is,

$F_{1} = \frac{Gmm}{l^{2}}$ along with AB.

Force on mass m at A due to mass m at C is,

$F_{2} = \frac{Gm \times m}{{(\sqrt{3} l)}^{2}} = \frac{{Gm}^{2}}{2 l^{2}}$ along with AC.

$[∵ AC = \sqrt{3} l]$

Force on mass m at A due to mass mat D is,

$F_{3} = \frac{Gm \times m}{(\sqrt{2 l})} = \frac{{Gm}^{2}}{4 l^{2}}$ along with AD.

$[∵ AD = 2 l]$

Force on mass m at A due to mass mat E is,

$F_{4} = \frac{Gm \times m}{{(\sqrt{3 l})}^{2}} = \frac{{Gm}^{2}}{3 l^{2}}$ along with AE.

Force on mass m at A due to mass m at F is,

$F_{5} = \frac{Gm \times m}{l^{2}} = \frac{{Gm}^{2}}{l^{2}}$ along with AF.

Step 2: Find the resultant force on mass m at A.

Resultant forces on mass m due to

$F_{1}$ and

$F_{5}, F_{1}' = \sqrt{F_{1}^{2} + F_{5}^{2} + 2 F_{1} F_{5} \cos 120 °} = \frac{{Gm}^{2}}{l^{2}}$ along with AD.

$[∵$ Angle between

$F_{1} and F_{5} = 120 °$ ]

$Resultant force due to F_{2} and F_{4}, F_{2}' = \sqrt{F_{2}^{2} + F_{4}^{2} + 2 F_{2} F_{4} \cos 60 °} = \frac{\sqrt{3} {Gm}^{2}}{3 l^{2}} = \frac{{Gm}^{2}}{\sqrt{3} l^{2}}$ along with AD.

So the net force along AD

$= F_{1}' + F_{2}' + F_{3} = \frac{{Gm}^{2}}{l^{2}} + \frac{{Gm}^{2}}{\sqrt{3 l^{2}}} + \frac{{Gm}^{2}}{4 l^{2}} = \frac{{Gm}^{2}}{l^{2}} (1 + \frac{1}{\sqrt{3}} + \frac{1}{4})$

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions