Question

Ex-4. Three masses, each equal to M are palced at the three comers of a square
of side a. Calculate the force of attraction on unit mass at the fourth corner.​

Answer 1

Solution:

$\sf{Newton's\;universal\;law\;of\;gravitation:}$

$\sf{F=\dfrac{Gm_{1}\;m_{2}}{r^{2}}}$

$\sf{Force\;on\;mass\;at\;D\;due\;to\;mass\;at\;A:}$

$\sf{F_{1}=\dfrac{Gm^{2}}{L^{2}}}$

$\sf{Force\;on\;mass\;at\;D\;due\;to\;mass\;at\;C:}$

$\sf{F_{2}=\dfrac{Gm^{2}}{L^{2}}}$

$\sf{Force\;on\;mass\;at\;D\;due\;to\;mass\;at\;B:}$

$\sf{F_{3}= \dfrac{Gm^{2}}{\big[\sqrt{2}L\big]^{2}} = \dfrac{Gm^{2}}{2L^{2}}}$

$\sf{Net\;force\;at\;D\;due\;to\;A\;and\;C:}$

$\sf{F_{4}=\sqrt{F_{1}^{2}+F_{2}^{2}}}$

$\sf{=\sqrt{2}\dfrac{Gm^{2}}{L^{2}}}$

$\sf{Net\;force\;at\;D:}$

$\sf{F_{net} = F_{4}+F_{3}}$

$\sf{=\sqrt{2}\dfrac{Gm^{2}}{L^{2}}+\dfrac{Gm^{2}}{2L^{2}}}$

$\bf{=\dfrac{Gm^{2}}{L^{2}}\;\Bigg[\sqrt{2}+\dfrac{1}{2}\Bigg]}$