Question

Three equal masses of 1 kg each are placed at the vertices of an equilateral triangle of side 1 m then the gravitational force on the mass is due to the other masses

Answer 1

Answer:

11.45$\times10^{-11}\ \rm N$

Explanation:

Given :

Mass of each particle at the vertex of the triangle= 1 kg

Length of side of equilateral triangle= 1 m

We know that from Newton's Law of gravitation the  gravitational force between two particles of masses $m_1\ and\ m_2$ and having distance r between their centres is given by

$F=\dfrac{Gm_1m_2}{r^2}$

The force exerted by first on second and second on first is equal but opposite in direction.

$\vec F_{12}=-\vec F_{21}$

Let consider the force exerted by particle and particle 2 on third particle  then according to the question we have

$F_{13}=\dfrac{G\times 1\times 1}{1^2}=G\\F_{23}=\dfrac{G\times 1\times 1}{1^2}=G$

Since it is from the geometry it can be seen that that angle between $\vec F_{12}$ and $F_{13}$ is $60^\circ$

Let $F_{net}$ be the resultant force magnitude given by

$F_{net}=\sqrt{G^2+G^2+2G^2\times \cos60^\circ}\\F=G\sqrt{3}\\F=6.67\times10^{-11}\times 1.73\\F=11.45\times10^{-11}\ \rm N$

Since the magnitude of each mass is same and they are placed a the same distance from each other so magnitude of net force on each particle by other two is same

Answer 2

Explanation:

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