---
title: "Dynamic balancing with two correction planes"
subject: "Theory of Mechanism and Machine II"
chapter: 5
key: "semester_7/theory_of_mechanism_and_machine_ii:balancing-masses-two-planes-numerical"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/balancing-masses-two-planes-numerical
site: BCT Engineering
---

# Dynamic balancing with two correction planes

Theory of Mechanism and Machine II · Chapter 5: Balance of Machinery

- Asked in: year not given, 10 marks
- Video: [HOW TO BALANCE SEVERAL MASSES IN DIFFERENT PLANES](https://www.youtube.com/watch?v=sg-_PUMg4GM) — TECH CAD
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/balancing-masses-two-planes-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-5

## Question

A shaft carries four masses A, B, C and D of 18 kg, 24 kg, 30 kg and 22 kg at radii 100 mm, 120 mm, 90 mm and 110 mm respectively, in parallel planes. Measured from plane A, the angular positions of B, C and D are $70^\circ$, $150^\circ$ and $250^\circ$ in the same sense. The planes of A, B, C and D are 0.20 m, 0.55 m, 0.95 m and 1.40 m from a balancing plane L, and a second balancing plane M is 1.70 m from L (on the same side as the other planes, beyond D). The balancing masses in L and M are to rotate at a radius of 150 mm. Find the magnitudes and angular positions of the balancing masses.

## Short answer

Take plane L as reference; couple balance gives plane M mass 9.85 kg at 9.5° from A. Force balance then gives plane L mass 15.29 kg at 242.0° from A, both at 150 mm radius.

## Answer

**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
| Plane | $m$ (kg) | $r$ (m) | $mr$ | $x$ (m) | $mrx$ | $\theta$ | $mrx\cos\theta$ | $mrx\sin\theta$ |
|---|---|---|---|---|---|---|---|---|
| A | 18 | 0.10 | 1.80 | 0.20 | 0.360 | 0 | +0.360 | 0.000 |
| B | 24 | 0.12 | 2.88 | 0.55 | 1.584 | 70 | +0.542 | +1.488 |
| C | 30 | 0.09 | 2.70 | 0.95 | 2.565 | 150 | -2.221 | +1.282 |
| D | 22 | 0.11 | 2.42 | 1.40 | 3.388 | 250 | -1.159 | -3.184 |

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

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
| Plane | $mr$ (kg·m) | $\theta$ | $mr\cos\theta$ | $mr\sin\theta$ |
|---|---|---|---|---|
| A | 1.80 | 0 | +1.800 | 0.000 |
| B | 2.88 | 70 | +0.985 | +2.706 |
| C | 2.70 | 150 | -2.338 | +1.350 |
| D | 2.42 | 250 | -0.828 | -2.274 |
| M | 1.478 | 9.5 | +1.458 | +0.243 |

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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which plane is taken as the reference plane to find m_M first?
   A. Plane L, so m_L has zero couple
   B. Plane A
   C. Plane D
   D. Plane M
   Answer: A. Taking moments about L removes the unknown m_L from the couple equation, so only m_M appears. (Key point: Reference plane choice)

2. From the couple polygon the quantity m_M r_M x_M equals:
   A. The magnitude of the resultant of m r x of A, B, C, D (2.512 kg.m^2)
   B. The sum of all m
   C. m r of D only
   D. Zero
   Answer: A. The balancing couple in M cancels the resultant of the given masses' couples: 2.512 kg.m^2. (Key point: Couple equation)

3. With x_M = 1.70 m and r_M = 0.15 m, m_M is:
   A. 9.85 kg
   B. 15.29 kg
   C. 1.48 kg
   D. 98.5 kg
   Answer: A. m_M r_M = 2.512/1.70 = 1.478, so m_M = 1.478/0.15 = 9.85 kg. (Key point: Mass in M)

4. After m_M is known, m_L is found from:
   A. The force balance: remaining sum of m r vector cancelled by L
   B. A second couple equation about D
   C. Setting m_L = m_M
   D. Averaging masses
   Answer: A. Sum of forces including M is resolved, and the remainder to be cancelled gives m_L r_L = 2.294, so m_L = 15.29 kg. (Key point: Mass in L)

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