Masses {rm{8, 2, 4, 2 }}kg{rm{ }} are p | vedclass

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Question

Let corner $A$ of square $ABCD$ is at the origin and the mass $8\ kg $ is placed at this corner (given in problem) Diagonal of square $d = a\sqrt 2&nb

Vedclass · Accepted Answer

$30$

Vedclass · Answer

Let corner $A$ of square $ABCD$ is at the origin and the mass $8\ kg $ is placed at this corner (given in problem) Diagonal of square $d = a\sqrt 2  = 80\,cm$

$&rArr;$ $a = 40\sqrt 2\ cm$, ${m_1} = 8\ kg,\,$${m_2} = 2\ kg,$ ${m_3} = 4\ kg,$ ${m_4} = 2\ kg$

Let ${\overrightarrow {\,r} _1},\,{\overrightarrow {\,r} _2},\,{\overrightarrow {\,r} _3},\,{\overrightarrow {\,r} _4}$ are the position vectors of respective masses    ${\overrightarrow {\,r} _1} = 0\hat i + 0\hat j$,  $\overrightarrow {{r_2}}  = a\hat i + 0\hat j$, $\overrightarrow {{r_3}}  = a\hat i + a\hat j$, $\overrightarrow {{r_4}}  = 0\hat i + a\hat j$

From the formula of centre of mass $\overrightarrow {r\,}  = \frac{{{m_1}\overrightarrow {{r_1}}  + {m_2}\overrightarrow {{r_2}}  + {m_3}\overrightarrow {{r_3}}  + {m_4}\overrightarrow {{r_4}} }}{{{m_1} + {m_2} + {m_3} + {m_4}}}$$ = 15\sqrt 2 i + 15\sqrt 2 \hat j$

$	herefore $ co-ordinates of centre of mass $ = (15\sqrt 2 ,\,15\sqrt 2 )$ and co-ordination of the corner $ = [0,0]$ From the formula of distance between two points $({x_1},\,\,{y_1})$ and $({x_2},\,{y_2})$

distance $ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $= $\sqrt {{{(15\sqrt 2  - 0)}^2} + {{(15\sqrt 2  - 0)}^2}} $ =$\sqrt {900} $ = $30\ cm$

Masses ${\rm{8, 2, 4, 2 }}kg{\rm{ }}$ 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 ........ $cm$

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