Center of MassMedium
Question
The position of center of mass of three masses $2\text{ }kg,3\text{ }kg$ and 15 kg placed with respect to mid point (p) of normal bisector, as shown in the figure is $\_\_\_\_$。
Options
A.$\left( \frac{\sqrt{3}}{4},1.25 \right)$
B.$\left( \frac{\sqrt{3}}{4},1.0 \right)$
C.$(0,0)$
D.$(1.25,0)$
Solution
Sol. [Key Concept: Place the three masses by geometry (2 kg and 3 kg on the base at $30^{\circ}$, 15 kg at the apex) and compute $X_{cm},Y_{cm}$ about $P$.]
$$\begin{matrix} & \\ & \left. \ X_{cm} = \frac{2 \times ( - 5\sqrt{3}) + 3(5\sqrt{3}) + 15 \times 0}{20} = \frac{- 5}{2} \right) \\ & Y_{cm} = \frac{2 \times ( - 5/2) + 3(5/2) + 15 \times (5/2)}{20} \\ & Y_{cm} = 1.25\text{ }cm \end{matrix}$$
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