Question

An elevator mass of 2600 kg falls from a height of 25 m after a sudden failure in the hoisting cable. The mass is stopped by a spring at the bottom of the shaft having a spring constant of 90 kN/m. Determine the spring's compression distance (in m) that brings the falling elevator momentarily to rest.

An elevator mass of 2600 kg falls from a height of 25 m after a sudden failure in the hoisting cable. The mass is stopped by a spring at the bottom of the shaft having a spring constant of 90 kN/m. Determine the peak magnitude of the acceleration in m/s/s that the elevator undergoes when brought to rest by the spring.

10p

An elevator mass of 2600 kg falls from a height of 25 m after a sudden failure in the hoisting cable. The mass is stopped by a spring at the bottom of the shaft. Determine the spring constant (in kN/m) necessary to bring the elevator and occupants to rest without exceeding an acceleration of 2 g's.

Answer 1

Let, & be the spring's Compression distance, and elevator fall from a height h So, equation of motion, ng (h+2) = 1į kar [ k= Spring KY - 2 ugn – 2 rugh = 0 x = mg + J mr gr + 2 mghk Constant 2600x9-8 + 7(2600%(9. B) 42x2 600X7.8x25 600X9.8x25 x90x103) 90 x103 x = 4.056 m Go, Springs Compression distance is 4.056 m

relocity of elevator before hitting the spring is v 1. Ź morzagh v = √2gh of a be the magnitude of acceleration So, r = 2ax or a . 202 294 or a = 2 X 2 x 25 mg a=gh - 9.8425 4.056 => 2 = 60.4 m/s
With regards. Thanks you.