Total moment `= 10 × 2 + 5 × 4 + 7 × 5 = 75\ "kg.m"`

If we put the masses together, we have: `10 + 5 + 7 = 22\ "kg"`

For an equivalent moment, we need:

`22 times bar(d)=75`

where `bar(d)` is the distance from the center of mass to the point of rotation.

i.e. `bar(d)=75/22 approx 3.4\ text[m]`

So our equivalent system (with one mass of `22\ "kg"`) would have: