Question

If three equal masses m are placed at the three vertices of an equilateral triangle of side 1/m then what force acts on a particle of mass 2m placed at the centroid?

Options

• Gm²

• 2Gm²

• Zero

• -Gm²

Answer 1

Zero

Explanation:

F_GA = `("Gm"(2"m"))/1hat"j"`

F_GB = `("Gm"(2"m"))/1(-hat"i"cos30°-  hat"j"sin30°)`

F_GC = `("Gm"(2"m"))/1(hat"i"cos30°-  hat"j"sin30°)`

∴ Resultant force on (2m) is F_R

= F_GA + F_GB + F_GC 

= `2"Gm"^2  hat"j"+2"Gm"^2  hat"i"(-cos30°+cos30°)+2"Gm"^2  hat"j"(-sin30°sin30°)`

= `2"Gm"^2  hat"j".2"Gm"^2  hat"j"(-2xx1/2)`

= `2"Gm"^2  hat"j"-2"Gm"^2  hat"j"`

= 0.