Let the +y-axis be upward. The free-body diagrams for the elevator plus student and the student alone are shown in the figure below. The total mass of student plus elevator is \(\displaystyle{M}={850}{k}{g}\). The tension in the cable is denoted by T.

The normal force exerted by the scale on the student is denoted by n. By Newton's third law, this force is equal in magnitude to the downward pressure exerted by the student's feet on the scale (the scale reading), that is n represents the scale’s reading.

The mass of the student alone, whose actual weight is \(\displaystyle{W}_{{{s}}}={550}{N},{i}{s}{m}_{{{s}}}=\frac{{W}_{{{s}}}}{{g}}={550}\frac{{N}}{{9.80}}\frac{{m}}{{s}^{{{2}}}}={56.1}{k}{g}\)

(a)Application of the y-equation of Newton's second law, \(\displaystyle{E}{F}_{{{y}}}={m}{a}_{{{y}}}\) gives

\(\displaystyle{n}-{m}_{{{s}}}{g}={m}_{{{s}}}{a}_{{{y}}}.\)

The scale’s reading corresponds to \(\displaystyle{n}={450}{N}\). Solving for dy, we find

\(\displaystyle{a}_{{{y}}}={\frac{{{n}-{m}_{{{s}}}{g}}}{{{m}_{{{s}}}}}}={\frac{{{450}{N}-{550}{N}}}{{{56.1}{k}{g}}}}=-{1},{78}\frac{{m}}{{s}^{{{2}}}}\).

So the student, and thus the elevator, has an acceleration of magnitude 1.78 m/ s, which is directed downward.

(b) Repeating part (a) using the new scale reading, \(\displaystyle{n}={670}{N}\), we obtain

\(\displaystyle{a}_{{{y}}}={\frac{{{670}{N}-{550}{N}}}{{{56.1}{k}{g}}}}=-{2.14}\frac{{m}}{{s}^{{{2}}}}\),

which represents an acceleration of magnitude 2.14 m/s” and upward direction

(c) In case the scale's reading was zero, that isn = 0, we have

\(\displaystyle{a}_{{{y}}}={\frac{{{0}-{m}_{{{s}}}{g}}}{{{m}_{{{s}}}}}}=-{g}\).

The elevator is in free fall...Trouble!

(d) To find the tension T in the cable, we must apply Newton's second law to the student plus elevator this time.

Application of \(\displaystyle{E}{F},={M}{a}_{{{y}}}\) gives

\(\displaystyle{T}—{M}{g}={M}{a}_{{{y}}}\), NSH which leads to \(\displaystyle{T}={M}{\left({a}_{{{y}}}+{9}\right)}\).

In part (a), \(\displaystyle{a}_{{{y}}}=—{1.78}\frac{{m}}{{s}^{{{2}}}}\). Hence,

\(T = (850 kg)(—1.78 m/s^{2} + 9.80 m/s^{2}) = 6817 N.\)

In part (c), \(\displaystyle{a}_{{{y}}}=—{g}\). Hence,

\(\displaystyle{T}={M}{\left(-{g}+{9}\right)}={0}\).

Yes, exactly! That is the trouble!