Question

Four bodies have been arranged at the corner of a rectangle find the coordinates of the centre of the mass of system cirners are 2m 3m 2m m

Answer 1

Explanation:

Here is your answer.....

Answer 2

The coordinates of the center of mass in the given system of rectangle with length "a" and breadth "b" is (5a/8, 3b/8).

Given :

Mass placed at 4 corners of rectangle: 2m, 3m, 2m, m

To Find :

Coordinates of the center of mass of the system

Solution :

Let we have a rectangle ABCD with the length "a" and breadth "b". So, the coordinates and mass placed on each corner of the rectangle is :

Corner A (0,0) with mass "2m"

Corner B (a,0) with mass "3m"

Corner C (a,b) with mass "2m"

Corner D (0,b) with mass "m"

The total mass placed on all corners is :

                      $M = m_{1} +m_{2} +m_{3} +m_{4}$

                           $= 2m + 3m + 2m + m$

                           $= 8m$

Calculating the X-Coordinate for the center of mass :

               $X_{cm} = (x_{1}m_{1} + x_{2}m_{2} +x_{3}m_{3} +x_{4}m_{4} ) / (m_{1} +m_{2} +m_{3} +m_{4} )$

                        $= ( 0*2m + a*3m + a*2m + 0*m) / M$

                         $= (0 + 3ma + 2ma+0 )/8m$

                         $= 5ma/8m$

                         $= 5a/8$

Calculating the Y-Coordinate for the center of mass :

               $Y_{cm} = (y_{1}m_{1} + y_{2}m_{2} +y_{3}m_{3} +y_{4}m_{4} ) / (m_{1} +m_{2} +m_{3} +m_{4} )$

                        $= ( 0*2m + 0*3m + b*2m + b*m) / M$

                         $= (0 + 0 + 2mb + mb )/8m$

                         $= 3mb/8m$

                         $= 3b/8$

Hence, the center of mass of the given system of particles is at (5a/8, 3b/8) where "a" and "b" are the length and breadth of the rectangle.

To learn more about the Centre of Mass, visit

https://brainly.in/question/19616311

#SPJ2