Question

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

Answer 1

From Figure, the gravitational force on mass $2m$ at G due to mass at A is
$F_1=G\frac{m\times 2m}{1^2}=2G{m}^2$ along GA
Gravitational force on mass $2$m at G due to mass at B is
$F_2=G\frac{m\times 2m}{1^2}=2G{m}^2$ along BG
Gravitational force on mass $2m$ at G due to mass at C is $F_3=G\frac{m\times 2m}{1^2}=2G{m}^2$ along GC
Draw DE parallel to BC passing through point G. Then $\angle$EGC$={30}^o=\angle$DGB.
Resolving $\overrightarrow{F_2}$ and $\overrightarrow{F_3}$ into two rectangular components, we have
$F_2\cos {30}^o$ along GD and $F_2\sin {30}^o$ along GH
$F_3\cos {30}^o$ along GE and $F_2\sin {30}^o$ along GH
Here, $F_2\cos {30}^o$ and $F_3\sin {30}^o$ are equal in magnitude and acting in opposite directions, and cancel out each other. The resultant force on mass $2m$ at G is
$F_1-(F_2\sin {30}^o+F_3\sin {30}^o)$
$=2G{m}^2-(2G{m}^2\times \frac{1}{2}+2G{m}^2\times \frac{1}{2})$
$=0$
b. When mass at A is $2$m, then gravitational force on mass $2m$ at G due to mass $2m$ at A is
$F_1^{\prime }=G\frac{2m\times 2m}{1^2}=4G{m}^2$ along GA
The resultant force on mass $2m$ at G due to masses at A, B and C is
$F_1=(F_2\sin {30}^o+F_3\sin {30}^o)$
$=4G{m}^2-(2G{m}^2\times \frac{1}{2}+2G{m}^2\times \frac{1}{2})$
$=2G{m}^2$ along GA.