Answer:
for mass M, x_cm = 1.33 L
for mass 2M, x_cm = 0.33 L
for mass 3M, x_cm = 0.67 L
Explanation:
The system of mass image in the attachment
The x-coordinate of the mass center from the mass M
x_{cm} = [ M(0) + (2M)L + (3M)(2L) ]÷(M + 2M + 3M)
= [ 2LM + 6LM ] / (6M)
= 1.33 L
The x-coordinate of mass center from the mass 2M is
x_{cm} = [ M(-L) + (2M)(0) + (3M)(L) ] ÷ (M + 2M + 3M)
= [ - LM + 3LM ] / (6M)
= 0.33 L
The x-coordinate of mass center from the mass 3M is
xcm = [ M(-2L) + (2M)(-L) + (3M)(0) ]÷(M + 2M + 3M)
= [ - 2LM - 2LM ] / (6M)
= 0.67 L