Question

A steel cable with cross-sectional area 3.00 cm² has an elastic limit of 2.40 x 10⁸ Pa. Find the maximum upward acceleration that can be given a 1200-kg elevator supported by the cable if the stress is not to exceed one-third of the elastic limit.

Short answer

The maximum upward acceleration that can be given in a 1200 kg elevator supported by the cable is$10.2\mathrm{m}/{\mathrm{s}}^2$

Step-by-step solution

The given data

The elastic limit of the cable $=2.40\times {10}^8$Pa

The cross-sectional area of steel cable, $\mathrm{A}=3{\mathrm{m}}^2$

Mass of elevator, m = 1200 kg

Formula used

Stress of wire $\mathbf{\phi }\mathbf{=}\frac{{\mathbf{F}}_{\mathbf{\perp }}}{\mathbf{A}}$

Where ${\mathbf{F}}_{\mathbf{\perp }}$is the tensile force applied to an object (tension in the cable), A is the cross-sectional area

Net force, $\mathbf{\sum }_{\mathbf{}}\mathbf{F}\mathbf{=}\mathbf{ma}$

Where m is the mass of the object and a is the acceleration

Find the tension in the cable

Since stress cannot exceed one-third of the elastic limit, so stress

$\begin{array}{r}\phi =\frac{1}{3}\left(2.4\times {10}^8\mathrm{Pa}\right)\\ =0.8\times {10}^8\mathrm{Pa}\end{array}$

Now, tension ${\mathrm{F}}_{\perp }=\mathrm{A\phi }$

$$\begin{gathered} {\mathrm{F}}_{\perp }=\left(3\times {10}^{-4}{\mathrm{m}}^2\right)\left(0.8\times {10}^8\mathrm{Pa}\right) \\ =2.4\times {10}^4\mathrm{N} \end{gathered}$$

Hence,${\mathrm{F}}_{\perp }=2.4\times {10}^4\mathrm{N}$

Find the acceleration

Since net force is $\sum _{}\mathrm{F}=\mathrm{ma}$

So

${\mathrm{F}}_{\perp }-\mathrm{mg}=\mathrm{ma}\mathrm{a}=\frac{{\mathrm{F}}_{\perp }}{\mathrm{m}}-\mathrm{g}$

Hence acceleration is

$\begin{array}{r}a=\frac{2.4\times {10}^4\mathrm{N}}{1200\mathrm{kg}}-9.8\mathrm{m}/{\mathrm{s}}^2\\ =10.2\mathrm{m}/{\mathrm{s}}^2\end{array}$

Therefore, the maximum upward acceleration that can be given in a 1200-kg elevator supported by the cable is $10.2\mathrm{m}/{\mathrm{s}}^2$