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
Return to Main Menu
Comments
Post a Comment