Question

Three particles each of mass m are placed at the three corners of an equilateral triangle. The centre of the triangle is at a distance x from either corner. If a mass M be placed at the centre, what will be the net gravitational force on it (a) Zero (b) 3GMm /x2 (c) 2GMa /x2 (d) GM x2​

Answer 1

★ The magnitude of gravitational force between two point masses is directly proportional to the product of masses and inversely proportional to the square of the distance between them. This is known as newton's law of gravitation.

Mathematically, $\underline{\boxed{\bf{\orange{F=\dfrac{Gm_1m_2}{r^2}}}}}$

A] Force on mass M at O due to mass m at A is

$\sf:\implies\:F_{OA}=\dfrac{GMm}{x^2}\longrightarrow Along\:OA$

B] Force on mass M at O due to mass m at B is

$\sf:\implies\:F_{OB}=\dfrac{GMm}{x^2}\longrightarrow Along\:OB$

C] Force on mass M at O due to mass m at C is

$\sf:\implies\:F_{OC}=\dfrac{GMm}{x^2}\longrightarrow Along\:OC$

Resolving $\sf\overrightarrow{F}_{OB}$ and $\sf\overrightarrow{F}_{OC}$ into two components.

Components acting along OP and OQ are equal in magnitude and opposite in direction. So they will cancel out while the components acting along OD will add up.

∴ The resultant force on the mass M at O is

$\sf:\implies\:F_R=F_{OA}-(F_{OB}sin30^{\circ}+F_{OC}sin30^{\circ})$

$\sf:\implies\:F_R=\dfrac{GMm}{x^2}-\left[\dfrac{GMm}{x^2}\times\dfrac{1}{2}+\dfrac{GMm}{x^2}\times\dfrac{1}{2}\right]$

$\sf:\implies\:F_R=\dfrac{GMm}{x^2}-\dfrac{GMm}{x^2}$

$:\implies\:\underline{\boxed{\bf{\gray{F_R=0}}}}$