Newton's Laws of Motion: Real-World Scenarios (12)

hard 3 min read

Question

A 60 kg passenger stands on a weighing scale inside an elevator. The elevator accelerates upward at 2m/s22\,\text{m/s}^2 for 3 seconds, moves with constant velocity for 4 seconds, and then decelerates at 3m/s23\,\text{m/s}^2 before stopping. Take g=10m/s2g = 10\,\text{m/s}^2.

What does the scale read in each phase, and during which phase will the passenger feel “heaviest”?

Solution — Step by Step

The scale reads the normal force NN it exerts on the passenger, not their true weight. By Newton’s third law, that equals the passenger’s push on the scale.

So we apply Newton’s second law to the passenger and solve for NN in each phase.

Taking upward as positive, the net force equation for the passenger is:

Nmg=maN - mg = ma

N=m(g+a)=60×(10+2)=720NN = m(g + a) = 60 \times (10 + 2) = 720\,\text{N}

Apparent weight is greater than the actual 600 N. The passenger feels heavier — this matches the “lift starting up” sensation.

When acceleration is zero, Newton’s second law gives:

N=mg=60×10=600NN = mg = 60 \times 10 = 600\,\text{N}

The scale shows the true weight. The lift feels like ordinary ground.

The lift is still moving up but slowing down, so acceleration points downward. Substitute a=3m/s2a = -3\,\text{m/s}^2:

N=m(g+a)=60×(103)=420NN = m(g + a) = 60 \times (10 - 3) = 420\,\text{N}

The scale reads less than the true weight — the passenger feels lighter, as if briefly floating.

Comparing 720 N, 600 N, and 420 N, the passenger feels heaviest in Phase 1 (accelerating up) at 720 N.

Final readings: 720 N, 600 N, 420 N. Heaviest in Phase 1.

Why This Works

Newton’s second law is written in the inertial ground frame, but the passenger’s lived experience matches the apparent weight NN. The brain interprets normal force as “how heavy I feel” — that is why a free-falling passenger (where N=0N = 0) feels weightless.

The key insight: deceleration while moving up is mathematically identical to acceleration while moving down. Direction of motion does not matter — only the direction of acceleration matters for NN.

Alternative Method

In the non-inertial elevator frame, we add a pseudo-force ma-ma acting on the passenger. Then equilibrium gives:

N=mg+maliftN = mg + ma_{\text{lift}}

Same answers, but you must be careful with signs of alifta_{\text{lift}}.

Common Mistake

Students assume that during deceleration the passenger feels heavier because the lift is “pushing harder against the floor.” Wrong — the direction of acceleration decides apparent weight, not the direction of velocity. When the lift moves up but slows down, aa points down, so N<mgN < mg.

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