Interpret the spring balance reading in the stationary lift

When the lift is stationary, the spring balance reads the person's apparent mass as $60\,\text{kg}$.

Hence the actual mass of the person is $m = 60\,\text{kg}$

Find the actual weight of the person

Actual weight is $W = mg = 60 \times 10 = 600\,\text{N}$

Lift descending with downward acceleration

If the lift accelerates downward with acceleration $a = 1.8\,\text{m/s}^2$, then the apparent weight (normal reaction by the spring balance) becomes $N = m(g-a)$

Substituting the values: $N = 60(10-1.8)$ $N = 60(8.2)$ $N = 492\,\text{N}$

Final answer

Therefore, the weight measured in the descending lift is $\boxed{492\,\text{N}}$