Question

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
(a) 20 cm (b) 30 cm
(c) 40 cm (d) 60 cm

Answer 1

$\Large\boxed{\sf{(b)\quad\!\!30\ cm}}$

Let 'd' be the distance of the center of mass from A.

Let 'a' be the side of the square so that its diagonal is,

$\longrightarrow\sf{a\sqrt2=80\ cm}$

$\longrightarrow\sf{a=40\sqrt2\ cm}$

Let an XY coordinate plane is formed by taking AB and AD as X and Y axes respectively, so that,

• 'A' will be at (0, 0).

• 'B' will be at (40√2, 0).

• 'C' will be at (40√2, 40√2).

• 'D' will be at (0, 40√2).

So the X coordinate of the center of mass will be,

$\longrightarrow\sf{\bar x=\dfrac{8(0)+2(40\sqrt2)+4(40\sqrt2)+2(0)}{8+2+4+2}\ cm}$

$\longrightarrow\sf{\bar x=\dfrac{0+80\sqrt2+160\sqrt2+0}{16}\ cm}$

$\longrightarrow\sf{\bar x=\dfrac{240\sqrt2}{16}\ cm}$

$\longrightarrow\sf{\bar x=15\sqrt2\ cm}$

And the Y coordinate of the center of mass will be,

$\longrightarrow\sf{\bar y=\dfrac{8(0)+2(0)+4(40\sqrt2)+2(40\sqrt2)}{8+2+4+2}\ cm}$

$\longrightarrow\sf{\bar y=\dfrac{0+0+160\sqrt2+80\sqrt2}{16}\ cm}$

$\longrightarrow\sf{\bar y=\dfrac{240\sqrt2}{16}\ cm}$

$\longrightarrow\sf{\bar y=15\sqrt2\ cm}$

Hence the distance of the center of mass from A is,

$\longrightarrow\sf{d=\sqrt{(\bar x)^2+(\bar y)^2}\ cm}$

$\longrightarrow\sf{d=\sqrt{(15\sqrt2)^2+(15\sqrt2)^2}}\ cm}$

$\longrightarrow\sf{d=\sqrt{450+450}\ cm}$

$\longrightarrow\sf{d=\sqrt{900}\ cm}$

$\longrightarrow\sf{\underline{\underline{d=30\ cm}}}$