Dynamic balancing with two correction planes

Method: Take plane L as the reference plane. For complete balance, $\sum mr = 0$ (forces) and $\sum mr\,x = 0$ (couples). Use the angle of A as $0^\circ$. Unknowns: $m_L$, $m_M$ and their angles.

Couple about plane L

$$\sum mrx\cos\theta = -2.478, \qquad \sum mrx\sin\theta = -0.413\ \ (\text{kg·m}^2)$$

The couple of the balancing mass in M about L must cancel this. Plane M is at $x_M = 1.70$ m, so

$$m_M r_M x_M = \sqrt{(-2.478)^2 + (-0.413)^2} = 2.512\ \text{kg·m}^2$$ $$m_M r_M = \frac{2.512}{1.70} = 1.478\ \text{kg·m} \Rightarrow m_M = \frac{1.478}{0.15} = 9.85\ \text{kg}$$

The couple of $m_M$ must point opposite to the resultant couple, i.e. along $(+2.478,\ +0.413)$:

$$\theta_M = \tan^{-1}\frac{0.413}{2.478} = 9.5^\circ$$

Force balance

The four given masses give $\sum H = -0.381$ and $\sum V = 1.782$ (kg·m). Adding plane M gives $1.077$ and $2.025$, so plane L must supply

$$m_L r_L\cos\theta_L = -1.077, \qquad m_L r_L\sin\theta_L = -2.025$$ $$m_L r_L = \sqrt{(-1.077)^2 + (-2.025)^2} = 2.294\ \text{kg·m} \Rightarrow m_L = \frac{2.294}{0.15} = 15.29\ \text{kg}$$

Both components are negative, so the angle is in the third quadrant:

$$\theta_L = 180^\circ + \tan^{-1}\frac{2.025}{1.077} = 242.0^\circ$$

Answer: Plane L: $m_L = 15.29$ kg at $242.0^\circ$ from A. Plane M: $m_M = 9.85$ kg at $9.5^\circ$ from A (both at 150 mm radius, same sense as the given angles).