---
title: "Inertia force and lift problem"
subject: "Engineering Mechanics"
chapter: 2
key: "semester_3/engineering_mechanics:inertia-force-lift-problem"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/inertia-force-lift-problem
site: BCT Engineering
---

# Inertia force and lift problem

Engineering Mechanics · Chapter 2: Kinetics of Particles: Force, Mass and Acceleration

- Asked in: year not given, 3+5 marks
- Video: [Lift Problem (Apparent Weight) | Engineering Mechanics | GATE & ESE 2023 CE/ME Exam Preparation](https://www.youtube.com/watch?v=zQSIu4OPTnM) — BYJU'S Exam Prep GATE & ESE: CE, ME & XE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/inertia-force-lift-problem
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-2

## Question

(a) Explain dynamic equilibrium and the concept of inertia force (D'Alembert's principle). (b) A man of mass 75 kg stands on a weighing scale fixed in a lift. Find the force indicated by the scale when the lift (i) accelerates upward at 1.5 m/s², (ii) accelerates downward at 2 m/s², (iii) falls freely after the cable breaks. Also find the cable tension in cases (i) and (ii) if the total mass of the lift with the man is 575 kg.

## Short answer

Inertia force is –m a, added to real forces to treat dynamics as statics (D'Alembert's principle). Scale reading R = m(g+a): (i) 848.3 N, (ii) 585.8 N, (iii) 0. Cable tension T = M(g+a): (i) 6503 N, (ii) 4491 N.

## Answer

#### (a) Dynamic equilibrium and inertia force
**Inertia force** is a fictitious force equal to the mass times the acceleration of the body and directed opposite to that acceleration:

$$
\mathbf{F}_i = -m\mathbf{a}
$$

**D'Alembert's principle** states that if the inertia force is added to the real forces acting on a body, the body can be treated as being in (dynamic) equilibrium, so that

$$
\Sigma \mathbf{F} - m\mathbf{a} = 0 \quad (\text{i.e. } \Sigma\mathbf{F} + \mathbf{F}_i = 0)
$$

A dynamics problem is thus turned into a statics problem, and the equations of statics (including moment equations) can be used. For a rotating body the inertia couple $-I\alpha$ is also added.

#### (b) Scale reading in the lift
Forces on the man: weight $mg = 75 \times 9.81 = 735.75$ N downward and scale reaction $R$ upward. Take upward as positive and let $a$ be the lift's upward acceleration; then $R - mg = ma$.

| Case | $a$ (up +) | $R = m(g + a)$ |
|---|---|---|
| (i) Up at 1.5 m/s² | +1.5 | 75(9.81 + 1.5) = 848.25 N |
| (ii) Down at 2 m/s² | -2 | 75(9.81 - 2) = 585.75 N |
| (iii) Free fall | -9.81 | 0 (weightless) |

#### Cable tension (lift + man = 575 kg)
The same equation applies to the whole system: $T - Mg = Ma$, so $T = M(g + a)$.

- (i) $T = 575(9.81 + 1.5) = 6503.3$ N
- (ii) $T = 575(9.81 - 2) = 4490.8$ N

**Answer: Scale reading = 848.3 N (i), 585.8 N (ii), 0 (iii); cable tension = 6503 N (i), 4491 N (ii).**

## Check yourself

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

1. The inertia force of a body of mass m and acceleration a is:
   A. m a in the direction of a
   B. -m a, opposite to the acceleration
   C. m g downward
   D. m v
   Answer: B. Inertia force is the fictitious force -ma, opposite to the acceleration. (Key point: Inertia force)

2. What does D'Alembert's principle allow us to do?
   A. Ignore friction in dynamics
   B. Treat a moving body as in dynamic equilibrium by adding the inertia force, turning dynamics into statics
   C. Replace weight with mass
   D. Conserve energy in all motions
   Answer: B. Adding the inertia force makes sum F - ma = 0, so the equations of statics apply. (Key point: D'Alembert's principle)

3. A lift accelerates upward at 1.5 m/s^2 with a 75 kg man. The scale reads:
   A. 735.75 N
   B. 848.25 N
   C. 623.25 N
   D. 112.5 N
   Answer: B. R = m(g + a) = 75(9.81 + 1.5) = 848.25 N. (Key point: Scale reading)

4. Which pair is right for free fall and for the cable tension when the lift (575 kg) accelerates downward at 2 m/s^2?
   A. Scale 735.75 N; tension 4490.8 N
   B. Scale 0; tension 4490.8 N
   C. Scale 0; tension 6503 N
   D. Scale 585.75 N; tension 0
   Answer: B. In free fall a = -g so R = 0. For downward 2 m/s^2, T = M(g + a) = 575(9.81 - 2) = 4490.8 N. (Key point: Free fall and cable tension)

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