Question

Three masses 3 kg, 4 kg and 5 kg are located at the corners of an equilateral triangle
of side 1m. Locate the centre of mass of the system.
please explain step by step

Answer 1

$\purple{\underline \bold{Given :}}$

$\begin{gathered} \\ \tt: \implies mass \: on \: vertex \: a \: = 3 \: kg \end{gathered}$

$\begin{gathered} \\ \tt: \implies mass \: on \: vertex \: B \: = 4 \: kg \end{gathered}$

$\begin{gathered} \\ \tt: \implies mass \: on \: vertex \: C \: = 5 \: kg \end{gathered}$

$\blue{\underline \bold{★ Finding \: \: B \: co - ordinate \: of \: triangle :}}$

$\begin{gathered} \\ \tt: \implies {x}^{2} = \: {1}^{2} - {(0.5)}^{2} \end{gathered}$

$\begin{gathered} \\ \tt: \implies {x}^{2} = \: 1 - 0.25 \end{gathered}$

$\begin{gathered} \\ \tt: \implies {x}^{2} = \: 0.75 \end{gathered}$

$\begin{gathered} \\ \tt: \implies x \: = \sqrt{0.75} \end{gathered}$

$\begin{gathered} \\ \tt: \implies x \: = 0.86\end{gathered}$

$\pink{\underline \bold{★ Now \: \:Using :}}$

$\begin{gathered} \\ \tt: \implies x_{m} = \frac{m_{1}r_{1} + m_{2} r_{2}+ m_{3}r_{3}}{m_{1} + m_{2} + m_{3}} \end{gathered}$

$\begin{gathered} \\ \tt: \implies x_{m} = \frac{3(0) + 4(1) + 5(0.5)}{3 + 4 + 5} \end{gathered}$

$\begin{gathered} \\ \tt: \implies x_{m} = \frac{6.5}{12} \end{gathered}$

$\green{\underline \bold{★ Now \: \:Using :}}$

$\begin{gathered} \\ \tt: \implies y_{m} = \frac{m_{1}y_{1} + m_{2} y_{2}+ m_{3}y_{3}}{m_{1} + m_{2} + m_{3}} \end{gathered}$

$\begin{gathered} \\ \tt: \implies y_{m} = \frac{4(0) + 3(0) + 5(0.86)}{4 + 3 + 5} \end{gathered}$

$\begin{gathered} \\ \tt: \implies y_{m} = \frac{4.3}{12} \end{gathered}$

$\orange{\underline \bold{★ \: COM \: \: : \: \: ( \frac{6.5}{12} ,\frac{4.3}{12} )}}$

Answer 2

$\large{\red{\frak{ Given}}}\begin{cases}\textsf{ Three masses of 3kg , 4 kg and 5kg are on the corners of the triangle. }\\\textsf{ The side of the equilateral triangle is 1m. }\end{cases}$

$\large{\red{\frak{ To \ Find }}}\begin{cases}\textsf{The coordinates of the centre of mass .}\end{cases}$

Here we need to find the coordinates of the centre of mass of the system . There are masses of 3kg , 4kg and 5kg at the corners of the triangle and the side of the triangle is 1m .

So lets make a graph of the system, where one of the vertex of the triangle is at the origin that is (0,0) . And hence the second coordinate is (1,0 ). The coordinates of the midpoint of this line will be (0.5,0) . Let's find out the coordinates of the third coordinate of triangle using the Pythagoras Theorem. For its steps kindly refer to the attachment. Hence we see that the ,

$\sf:\implies\boxed{ \green{\sf Third \ Coordinate \ = \ \bigg(\dfrac{1}{2},\dfrac{\sqrt3}{2}\bigg) }}$

$\purple{\bigstar}\underline{\boldsymbol{ According\ to\ the \ Question :-}}$

$\sf:\implies \pink{ ( x , y )=\bigg( \dfrac{mx_1 + mx_2+mx_3}{m_1+m_2+m_3} , \dfrac{my_1 + my_2+my_3}{m_1+m_2+m_3} \bigg) }\\\\\sf:\implies (x,y)= \bigg( \dfrac{ (3)(0)+(4)(1)+(5)(0.5)}{3+4+5},\dfrac{(3)(0)+(4)(0)+(5)\bigg(\dfrac{\sqrt3}{2}\bigg) }{3+4+5}\bigg)\\\\\sf:\implies (x,y)=\bigg( \dfrac{6.5}{12},\dfrac{5\sqrt3}{24}\bigg)\\\\\sf:\implies\underset{\blue{\sf Required\ coordinate }}{\underbrace{ \boxed{\pink{\frak {(x,y)= (0.54,0.36)}}}}}$