Three equal masses of m kg each are fixed at the vetrices of an equilateral triangle ABC (a) What is the force acting on a mass 2m placed at the centriod G of the triangle. (b) What is the force if the mass at the vertex A is doubled ? Take AG = BG = 1m

Three equal masses of m kg each are fixed at the vetrices of an equila

1.	Three equal masses of m kg each are fixed at the vetrices of an equilateral triangle ABC (a) What is the force acting on a mass 2m placed at the centriod G of the triangle. (b) What is the force if the mass at the vertex A is doubled ? Take AG = BG = 1m
Answer» <html><body><p></p>Solution :The individual forces in vector notation are <br/> `F_(GA)=(<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>(m)(<a href="https://interviewquestions.tuteehub.com/tag/2m-300757" style="font-weight:bold;" target="_blank" title="Click to know more about 2M">2M</a>))/1hatj` <br/> `F_(GB)=(G(m)(2m))/1 (-hat1 cos 30^@- hatj <a href="https://interviewquestions.tuteehub.com/tag/sin-1208945" style="font-weight:bold;" target="_blank" title="Click to know more about SIN">SIN</a> 30^@)` <br/> `F_(GC)=(G(m)(2m))/1 (hat1 cos 30^@ -hatj sin 30^@)` <br/> From the principle of superposition and the law of vector addition the resultant gravitational force `F_R` on(2m) is <br/> `F_R=F_(GA)+F_(GB)+F_(GC)` <br/> `F_R=2Gm^2hatj + 2Gm^2 (-hati cos 30^@ - hatj sin 30^@) + 2Gm^2 (hati cos 30^@ - hatj sin 30^@) =0` <br/> (b) By symmetry, the X-component of the force cancels out. The Y-component survives. <br/> `therefore F_R=4Gm^2 hatj - 2Gm^2 hatj =2Gm^2 hatj`</body></html>