A steel cable with cross-sectional area 3.00 cm^2 has an elastic limit of 2.40×10^8 Pa.
Find the maximum upward acceleration that can be given an elevator of mass 1300 supported by the cable if the stress is not to exceed one-third of the elastic limit.
2006-11-24 19:17:48 · 3 answers · asked by googoosh g 1 in Science & Mathematics ➔ Physics
one third of the elastic limit is 8x10^7 Pa and the total area is 3x10^(-4) m² so the total force is 2.4x10^4 N. It takes 12,740 N to hold the elevator stationary agains the acceleration of gravity which leaves 11,260 N available to accelerate the elevator upwards, so the maximum acceleration would be about 8.66 m/s².
Doug
2006-11-24 19:28:39 · answer #1 · answered by doug_donaghue 7 · 4⤊ 0⤋
It's an F = ma problem, and you are given F and m and asked to solve for a. F = 2.4E8 x 3E-4 x 0.333 newtons, and you can take it from there.
2006-11-24 19:31:20 · answer #2 · answered by Anonymous · 0⤊ 1⤋
18Pa
2006-11-24 19:22:23 · answer #3 · answered by Anonymous · 0⤊ 1⤋