Question

four particles of mass 1kg, 2kg, 3kg, 4kg are placed at the four vertices a,b,c,d of a square of side 1m. find the position of the centre of mass.

Answer 1

your answer is (7/10,1/2)
hope it helps

Answer 2

Answer: The position of center of mass is $(X_{cm},Y_{cm}) = (\dfrac{1}{2},\dfrac{7}{10})$

Explanation:

Given that,

Mass of first particle $m_{1} =1 kg$

Mass of second particle $m_{2} =2 kg$

Mass of third particle $m_{3} =3 kg$

Mass of fourth particle $m_{4} =4 kg$

Each side of square = 1m

Four particles are placed at the four vertices a,b,c,d of a square.

The center of mass of X-axis is

$X_{cm} = \dfrac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3}+m_{4}x_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}$

$X_{cm} = \dfrac{1\times0+2\times1+3\times1+4\times0}{1+2+3+4}$

$X_{cm} = \dfrac{1}{2}$

The center of mass of Y-axis is

$Y_{cm} = \dfrac{m_{1}y_{1}+m_{2}y_{2}+m_{3}y_{3}+m_{4}y_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}$

$Y_{cm} = \dfrac{1\times0+2\times0+3\times1+4\times1}{1+2+3+4}$

$Y_{cm} = \dfrac{7}{10}$

Hence, the position of center of mass is $(X_{cm},Y_{cm}) = (\dfrac{1}{2},\dfrac{7}{10})$