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Inertia force and lift problem

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.

(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.

(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:

Fi=−ma\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

ΣF−ma=0(i.e. ΣF+Fi=0)\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 −Iα-I\alpha is also added.

(b) Scale reading in the lift

Forces on the man: weight mg=75×9.81=735.75mg = 75 \times 9.81 = 735.75 N downward and scale reaction RR upward. Take upward as positive and let aa be the lift's upward acceleration; then R−mg=maR - mg = ma.

Caseaa (up +)R=m(g+a)R = m(g + a)
(i) Up at 1.5 m/s²+1.575(9.81 + 1.5) = 848.25 N
(ii) Down at 2 m/s²-275(9.81 - 2) = 585.75 N
(iii) Free fall-9.810 (weightless)

Cable tension (lift + man = 575 kg)

The same equation applies to the whole system: T−Mg=MaT - Mg = Ma, so T=M(g+a)T = M(g + a).

  • (i) T=575(9.81+1.5)=6503.3T = 575(9.81 + 1.5) = 6503.3 N
  • (ii) T=575(9.81−2)=4490.8T = 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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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