A steel cable with a cross-sectional area $3\;cm^2$ has an elastic limit of $2.4\;*\;10^8\;pa$. Find the maximum upward acceleration that can be given a $1200\;kg$ elevator supported by the cable if the stress does not exceed one-third of the elastic limit.
Given,
Area of the cross-section of cable $(A) = 3\;cm^2 = 3\,*\,10^{-4}\;m^2$
Elastic limit $ = 2.4 * 10^8\;pa$
Stress $ = \frac{1}{3} * 2.4 * 10^8 \;pa$
Mass $(M) = 1200\;kg$
Upward Acceleration $(a) = ?$
We have,
$F = Stress * A = \frac{1}{3}* 2.4 * 10^8 * 3 * 10^{-4} = 2.4 * 10^4\;N$
⇒ $a = $ $\frac{F}{M} = \frac{2.4 \; * \; 10^4}{1200}$ $ = 20\;m/s^2$
Area of the cross-section of cable $(A) = 3\;cm^2 = 3\,*\,10^{-4}\;m^2$
Elastic limit $ = 2.4 * 10^8\;pa$
Stress $ = \frac{1}{3} * 2.4 * 10^8 \;pa$
Mass $(M) = 1200\;kg$
Upward Acceleration $(a) = ?$
We have,
$F = Stress * A = \frac{1}{3}* 2.4 * 10^8 * 3 * 10^{-4} = 2.4 * 10^4\;N$
⇒ $a = $ $\frac{F}{M} = \frac{2.4 \; * \; 10^4}{1200}$ $ = 20\;m/s^2$
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