Question

three pieces of iron of uniform thickness and mass m,m and 2m respectively are placed at the three corners of a triangle having co-ordinate (2.5, 1.5)( 3.5, 1.5) and (3,3) respectively find the centre of the mass of the system​

Answer 1

Hope this help u....

Answer 2

The centre of the mass of the sysytem is (3,2.25)

To find the centre of the mass of the system we will use the following formula:

$mr_{com} = \frac{mr_{1} + mr_{2} + mr_{3} }{m_{1} + m_{2} + m_{3} }$

here,

$m_{1$= m , $m_{2}$ = m, $m_{3}$= 2m

$x_{1}$=2.5, $x_{2}$=3.5, $x_{3}$=3,

$y_{1}$=1.5, $y_{2}$=1.5, $y_{3}$=3.

So for the x coordinates we have,

$mx_{com} = \frac{mx_{1} + mx_{2} + mx_{3} }{m_{1} + m_{2} + m_{3} }$

$mx_{com} = \frac{m2.5 + m3.5 + 2m3 }{m + m + 2m }$

           =  12÷4

           = 3

For the y coordnates we have,

$my_{com} = \frac{my_{1} + my_{2} + my_{3} }{m_{1} + m_{2} + m_{3} }$

$my_{com} = \frac{m1.5 + m1.5 + 2m3 }{m + m + 2m }$

          = 9÷4

          = 2.25