Balancing one mass using two planes

A single rotating mass in one plane cannot be balanced by one mass in a different plane because a couple remains. It needs two masses in two planes, so that both force and couple are balanced.

Set-up

     L        A         M
     |        |         |
     |  a     |   c     |
   --+--------+---------+-----> shaft axis
     m_L      m         m_M
     r_L      r         r_M
        |---------b----------|

Let the distances be $a$ (L to A), $c$ (A to M) and $b = a + c$ (L to M). The balancing masses are $m_L$ at radius $r_L$ and $m_M$ at radius $r_M$. All three masses lie in one axial plane through the shaft (same line).

Case (i) A between L and M

Force balance:

$$m_L r_L + m_M r_M = m r \quad(\text{balancing masses on the side opposite to } m)$$

Couple balance about plane L:

$$m_M r_M\, b = m r\, a \;\Rightarrow\; m_M r_M = m r\,\frac{a}{b}$$

Substituting in the force equation,

$$m_L r_L = m r - m r\frac{a}{b} = m r\,\frac{c}{b}$$

Both balancing masses act in the same direction, opposite to $m$ (at $180^\circ$ to it).

Case (ii) A outside L and M

Say A lies beyond M (order L, M, A), with A at distance $a$ from L and M at $b$ ($a > b$). Couple balance about plane L:

$$m_M r_M\, b = m r\, a \;\Rightarrow\; m_M r_M = m r\,\frac{a}{b}\ (\text{larger than } mr)$$

Let the direction of $m$ be positive. The couple equation fixes $m_M$ opposite to $m$. For the forces, with signs ($+$ same direction as $m$):

$$m_L r_L - m_M r_M + m r = 0 \;\Rightarrow\; m_L r_L = m_M r_M - m r = m r\,\frac{a-b}{b}$$

The result is positive, so $m_L$ acts in the same direction as $m$, and $m_M$ acts opposite to $m$.

Summary: The balancing masses follow from $\sum F = 0$ and $\sum M = 0$. When A is between the planes, both balancing masses are opposite to $m$. When A is outside, the plane nearer to A carries a larger mass opposite to $m$, and the farther plane carries a mass in the same direction as $m$.