Masses 8, 2, 4, 2 kg are placed at the corners A, B, C, D respectively of a square ABCD of diagonal 80 cm . The distance of centre of mass from A will be

Correct Answer: 30 cm

Step-by-Step Explanation:

Let us set up a coordinate system to solve the problem easily. Let the corner A of the square ABCD be at the origin of our coordinate system.

Let $a$ represent the side length of the square. The diagonal of the square is given as $d=80\text{cm}$.
Since the diagonal of a square with side length $a$ is given by $d=a\sqrt{2}$, we can write:
$a=\frac{80}{\sqrt{2}}\text{cm}=40\sqrt{2}\text{cm}$

Now, let us assign coordinates to each corner of the square ABCD with respect to corner A situated at $(0,0)$:

• Corner A: Coordinates are $(0,0)$ with mass $m_A=8\text{kg}$
• Corner B: Coordinates are $(a,0)$ with mass $m_B=2\text{kg}$
• Corner C: Coordinates are $(a,a)$ with mass $m_C=4\text{kg}$
• Corner D: Coordinates are $(0,a)$ with mass $m_D=2\text{kg}$

The total mass $M$ of the system is the sum of the individual corner masses:
$M=8+2+4+2=16\text{kg}$

The x-coordinate of the center of mass ($X_{\mathrm{cm}}$) is computed as:
$X_{\mathrm{cm}}=\frac{m_A{x}_A+m_B{x}_B+m_C{x}_C+m_D{x}_D}{M}$
$X_{\mathrm{cm}}=\frac{8\left(0\right)+2\left(a\right)+4\left(a\right)+2\left(0\right)}{16}=\frac{6a}{16}=\frac{3a}{8}$

Similarly, the y-coordinate of the center of mass ($Y_{\mathrm{cm}}$) is computed as:
$Y_{\mathrm{cm}}=\frac{m_A{y}_A+m_B{y}_B+m_C{y}_C+m_D{y}_D}{M}$
$Y_{\mathrm{cm}}=\frac{8\left(0\right)+2\left(0\right)+4\left(a\right)+2\left(a\right)}{16}=\frac{6a}{16}=\frac{3a}{8}$

The distance $r$ of the center of mass $(X_{\mathrm{cm}},Y_{\mathrm{cm}})$ from A $(0,0)$ is:
$r=\sqrt{X_{\mathrm{cm}}^2+Y_{\mathrm{cm}}^2}=\sqrt{{\left(\frac{3a}{8}\right)}^2+{\left(\frac{3a}{8}\right)}^2}=\frac{3a}{8}\sqrt{2}$

Now, substitute the value of $a=40\sqrt{2}\text{cm}$ into the distance equation:
$r=\frac{3\left(40\sqrt{2}\right)}{8}\sqrt{2}=15\cdot 2=30\text{cm}$

Thus, the distance of the center of mass from corner A is 30 cm.