The position of center of mass of three masses $2$ kg, $3$ kg and $15$ kg placed with respect to mid point ($p$) of normal bisector, as shown in the figure is ______.
Answer: (A) $\left(\dfrac{\sqrt{3}}{4}, 1.25\right)$
Concept. Write each mass's coordinates with the origin at $P$, the midpoint of the altitude, then use $\vec r_{cm}=\sum m\vec r/\sum m$.
Step 1. Geometry. Sides $10$ m with an apex angle of $120^\circ$: altitude $10\cos60^\circ=5$ m, half-base $10\sin60^\circ=5\sqrt3$ m. With $P$ at the middle of the altitude: apex $(0,2.5)$, base corners $(-5\sqrt3,-2.5)$ (2 kg) and $(5\sqrt3,-2.5)$ (3 kg), as in the figure.
Step 2. $x$-coordinate. $\dfrac{-2(5\sqrt3)+3(5\sqrt3)+0}{20}=\dfrac{5\sqrt3}{20}=\dfrac{\sqrt3}{4}$ m.
Step 3. $y$-coordinate. $\dfrac{15(2.5)+5(-2.5)}{20}=\dfrac{25}{20}=1.25$ m.
Answer. $\left(\dfrac{\sqrt3}{4},1.25\right)$ (Option A).
Solution by Sreeraj P, M.Sc Physics
