---
title: "Balancing one mass using two planes"
subject: "Theory of Mechanism and Machine II"
chapter: 5
key: "semester_7/theory_of_mechanism_and_machine_ii:single-mass-two-plane-balance-derivation"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/single-mass-two-plane-balance-derivation
site: BCT Engineering
---

# Balancing one mass using two planes

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

- Asked in: year not given, 6 marks
- Video: [Balancing Of Single Rotating Mass By Two Balancing Mass](https://www.youtube.com/watch?v=eVAI3nPGY3E) — TutorialsPoint
- 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/single-mass-two-plane-balance-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-5

## Question

A mass $m$ rotates at radius $r$ in plane A of a shaft. Show how it can be balanced by two masses rotating in two other planes L and M on the same shaft. Obtain the magnitudes and positions of the balancing masses when plane A lies (i) between L and M and (ii) outside L and M.

## Short answer

A single rotating mass in plane A is balanced by two masses in planes L and M. (i) A between L and M: m_M r_M = m r a/b, m_L r_L = m r c/b, both opposite m. (ii) A outside (beyond M): m_M r_M = m r a/b opposite m, m_L r_L = m r (a-b)/b in same direction as m.

## Answer

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
```text
     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 $m$.

## Check yourself

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

1. Why can a mass in plane A not be balanced by a single mass in another plane?
   A. A couple would remain
   B. The force would not balance
   C. Two masses are always lighter
   D. Centrifugal force is zero
   Answer: A. The balancing mass in a different plane balances the force but creates an unbalanced couple, so two planes are needed. (Key point: Why two planes)

2. With A between L and M (L-A = a, L-M = b), taking moments about L gives:
   A. m_M r_M = m r (a/b)
   B. m_M r_M = m r (b/a)
   C. m_M r_M = m r (c/b)
   D. m_M r_M = m r
   Answer: A. m_M r_M b = m r a so m_M r_M = m r a/b. (Key point: Case (i) couple)

3. With A between L and M, the balancing mass in L is:
   A. m_L r_L = m r (c/b)
   B. m_L r_L = m r (a/b)
   C. m_L r_L = m r (b/c)
   D. m_L r_L = 0
   Answer: A. m_L r_L = m r - m r a/b = m r c/b, with both balancing masses opposite to m. (Key point: Case (i) force)

4. When A lies outside L and M, the balancing masses are:
   A. The nearer plane's mass opposite to m, the other in the same direction as m
   B. Both opposite to m
   C. Both in the same direction as m
   D. Both zero
   Answer: A. m_M is opposite to m and larger than m r, while m_L = m r (a-b)/b acts in the same direction as m. (Key point: Case (ii) directions)

## 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 four masses in one plane](https://www.bctengineer.com/question.md?key=semester_7%2Ftheory_of_mechanism_and_machine_ii%3Abalancing-masses-same-plane-numerical) — Theory of Mechanism and Machine II, chapter 5
- [Static versus dynamic balancing](https://www.bctengineer.com/question.md?key=semester_7%2Ftheory_of_mechanism_and_machine_ii%3Astatic-dynamic-balancing-differentiate) — Theory of Mechanism and Machine II, chapter 5

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