Question

Three equal masses of $mkg$ each are fixed at the vertices of an equilateral triangle $AB{C}^{\prime }$ and a mass $2m$ is placed at centroid $G$ of the triangle as shown below.

The force acting on a mass $2m$ placed at the centroid $G$ of the triangle is

• A. $\frac{2G{m}^2}{\sqrt{2}}$
• B. $2G{m}^2$
• C. $6G{m}^2$
• D. zero

Answer 1

Answer: (D)

The angle between $GC$ and positive $X$-axis is $30^{\circ }$ and so is the angle between $GB$ and negative $X$-axis.

From figure, resolving the forces on $2m$ due to masses at $B$ and $C$ along $X$-axis.
$\Rightarrow \left|F_{x\text{ (net) }}\right|=\left|F\cos 30^{\circ }\hat{i}-F\cos 30^{\circ }\hat{i}\right|=0$
Similarly along $Y$-axis,
$\left|F_{y\text{ (net) })}\right|=\left|F\hat{j}-\left(F\sin 30^{\circ }+F\sin 30^{\circ }\right)\hat{j}\right|$
$=\left|F\hat{j}-\left(\frac{F}{2}+\frac{F}{2}\right)\hat{j}\right|=|F\hat{j}-F\hat{j}|=0$
$F_{\text{net }}=$ Resultant force on mass at $G$ due to masses at $A,B$ and $C=\sqrt{F_{x\text{ (net) }}^2+F_{y\text{ (net) }}^2}=0$