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A steel cable with Cross Sectional Area 3.00cm² has an elastic limit of 2.40 x 10^8pascals. Find the maximum upward acceleration that can be given a 1200kg elevated supported by the cable if the stress is not to exceed one-third of the elastic limit.​

Sagot :

Answer:

Stress = F / A       force per unit area

A = 3.00 cm^2 = 3 E-4  m^2

F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N    max force applied

F/3 = 2.4E4 N  if force not to exceed limit   (= f)

f = M a

a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2      about 2 g

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