A steel cable with cross-sectional area of 3.04cm^2 has an elastic limit of 2.42×10^8 Pa . Find the maximum upward acceleration that can be given to a 1230kg elevator supported by the cable if the stress is not to exceed one-third of the elastic limit.
Any help would be appreciated! Thanks!
Any help would be appreciated! Thanks!
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The cable limit is 242,000,000N/m^2.
1cm^2 = 1/10,000th. of 1m^2.
This cable's limit is (242,000/10) x 3.04 = 73,568N.
Divide by 3, = max. tension of 24,522.67N.
Acceleration = (f/m) = 24,522.67/1230, = 19.937m/sec^2.
Max. up acceleration = (19.937 - 9.8) = 10.137m/sec^2.
1cm^2 = 1/10,000th. of 1m^2.
This cable's limit is (242,000/10) x 3.04 = 73,568N.
Divide by 3, = max. tension of 24,522.67N.
Acceleration = (f/m) = 24,522.67/1230, = 19.937m/sec^2.
Max. up acceleration = (19.937 - 9.8) = 10.137m/sec^2.