---
title: "Balancing four masses in one plane"
subject: "Theory of Mechanism and Machine II"
chapter: 5
key: "semester_7/theory_of_mechanism_and_machine_ii:balancing-masses-same-plane-numerical"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/balancing-masses-same-plane-numerical
site: BCT Engineering
---

# Balancing four masses in one plane

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

- Asked in: year not given, 6 marks
- Video: [Balancing of Several Masses in the Same Plane (Graphical Method) with Numerical](https://www.youtube.com/watch?v=1HXxF7JG-RU) — 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-same-plane-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-5

## Question

Four masses A, B, C and D of 12 kg, 8 kg, 15 kg and 10 kg rotate in the same plane at radii of 150 mm, 200 mm, 100 mm and 180 mm respectively. The angular positions measured from mass A in the same sense are B $60^\circ$, C $135^\circ$ and D $270^\circ$. Find the magnitude and angular position of the mass to be placed at a radius of 200 mm to balance the system.

## Short answer

Resolving the mr vectors: ΣH=1.539 kg·m, ΣV=0.646 kg·m. Resultant R=1.670 kg·m at 22.8° from mass A. The balancing mass must be opposite: m_b = 1.670/0.2 = 8.35 kg at 202.8° from A.

## Answer

For masses in one plane, balance requires $\sum mr = 0$ (vector sum). Take mass A along the x-axis ($0^\circ$) and find the resultant of the $mr$ vectors by resolving.

| Mass | $m$ (kg) | $r$ (m) | $mr$ (kg·m) | $\theta$ (deg) | $mr\cos\theta$ | $mr\sin\theta$ |
|---|---|---|---|---|---|---|
| A | 12 | 0.15 | 1.800 | 0 | +1.800 | 0.000 |
| B | 8 | 0.20 | 1.600 | 60 | +0.800 | +1.386 |
| C | 15 | 0.10 | 1.500 | 135 | -1.061 | +1.061 |
| D | 10 | 0.18 | 1.800 | 270 | 0.000 | -1.800 |

$$
\sum H = \sum mr\cos\theta = 1.539\ \text{kg·m}, \qquad \sum V = \sum mr\sin\theta = 0.646\ \text{kg·m}
$$

#### Resultant
$$
R = \sqrt{(\sum H)^2 + (\sum V)^2} = \sqrt{(1.539)^2 + (0.646)^2} = 1.670\ \text{kg·m}
$$

$$
\tan\alpha = \frac{\sum V}{\sum H} = \frac{0.646}{1.539} \Rightarrow \alpha = 22.8^\circ \text{ (measured from A)}
$$

#### Balancing mass
The balancing mass $m_b$ must cancel this resultant, so it acts opposite to it ($180^\circ$ away):

$$
m_b r_b = R \Rightarrow m_b = \frac{1.670}{0.2} = 8.35\ \text{kg}
$$

$$
\theta_b = 22.8^\circ + 180^\circ = 202.8^\circ \text{ from mass A, in the same sense as B, C, D}
$$

**Answer:** Balancing mass $= 8.35$ kg at 200 mm radius, at $202.8^\circ$ from mass A.

## Check yourself

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

1. For masses in a single plane, balance requires:
   A. The vector sum of m r to be zero
   B. The sum of masses to be zero
   C. The sum of m r^2 to be zero
   D. Equal radii
   Answer: A. Centrifugal forces m r w^2 are in one plane, so the vector sum of m r must vanish. (Key point: Condition)

2. The method used to find the resultant is:
   A. Resolve each m r into horizontal and vertical components and sum
   B. Add the magnitudes of m r
   C. Multiply masses and radii
   D. Average the angles
   Answer: A. Sum H = 1.539 and sum V = 0.646 kg.m give R = 1.670 kg.m. (Key point: Resolving)

3. With resultant R = 1.670 kg.m, the mass needed at 200 mm radius is:
   A. 8.35 kg
   B. 1.67 kg
   C. 16.7 kg
   D. 0.33 kg
   Answer: A. m_b = R / r_b = 1.670/0.2 = 8.35 kg. (Key point: Balancing mass)

4. The resultant is at 22.8 degrees from A. The balancing mass is placed at:
   A. 22.8 deg
   B. 202.8 deg
   C. 112.8 deg
   D. 157.2 deg
   Answer: B. The balancing mass cancels the resultant, so it acts 180 degrees away: 202.8 deg. (Key point: Position)

## Related in Theory of Mechanism and Machine II

- [Dynamic balancing with two correction planes](https://www.bctengineer.com/question.md?key=semester_7%2Ftheory_of_mechanism_and_machine_ii%3Abalancing-masses-two-planes-numerical) — Theory of Mechanism and Machine II, chapter 5
- [Balancing one mass using two planes](https://www.bctengineer.com/question.md?key=semester_7%2Ftheory_of_mechanism_and_machine_ii%3Asingle-mass-two-plane-balance-derivation) — Theory of Mechanism and Machine II, chapter 5

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