Question

Four bodies of masses 2,3,5 and 8 kg are placed at the four corners of a square of side 2m. The position of CM will be

Answer 1

hope this will help you.

Answer 2

Answer:

Position of CM = ($\frac{8}{9}$, $\frac{13}{9}$)

Explanation:

Given:-

     Masses are; 2kg, 3kg, 5kg, & 8 kg.

Length of Side of Square, a = 2m

Refer to the co-ordinates of the given system from the attached image.

Now, we know that;-

         Xcm = $\frac{m1x2 + m2x2 + m3x3 + m4x4}{m1 + m2 + m3+ m4}$

         Xcm = $\frac{2(0) + 6 + 10 + 0}{2 + 3 + 5 + 8}$

         Xcm = $\frac{16}{18}$

        Xcm = $\frac{8}{9}$  _______(1)

Also, we know that;-

         Ycm = $\frac{m1y1 + m2y2 + m3y3 + m4y4}{m1 + m2 + m3 + m4}$

         Ycm = $\frac{2(0) + 3(0) + 5(2) + 8(2)}{2+3+5 + 8}$

         Ycm = $\frac{26}{18}$

         Ycm = $\frac{13}{9}$ _______(2)

From (1) and (2), we conclude;-

         Rcm = ($\frac{8}{9}$, $\frac{13}{9}$)

Hence, the position of CM will be ($\frac{8}{9}$, $\frac{13}{9}$).

Hope it helps! ;-))