Q11-41E - University Physics with Modern Physics

Question

A steel cable with cross-sectional area 3.00 cm² has an elastic limit of $\mathbf{2}\mathbf{.}\mathbf{40}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}$ 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.

Step-by-Step Solution

Answer

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

Step 1: The given data

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

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

Mass of elevator, $m=1200 kg$

Step 2: Formula used

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

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

Net force, $\mathbf{\sum }\mathit{F}\mathbf{=}\mathit{m}\mathit{a}$

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

Step 3: Find the tension in the cable

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

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

Now, tension $F_{\perp }=A\phi$

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

Hence, $F_{\perp }=2.4\times {10}^4 N$

Step 4: Find the acceleration

Since net force is $\sum F=ma$

So

$$\begin{gathered} F_{\perp }-mg=ma \\ a=\frac{F_{\perp }}{m}-g \end{gathered}$$

Hence acceleration is 

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

Therefore, the maximum upward acceleration that can be given in a 1200-kg elevator supported by the cable is  10.2m/s²