Question

33. What is the minimum diameter of the wire which is connected with lift having
mass of 1000kg and accelerating uniformly at 1.2 ms 2 rates? The wire can bear
maximum stress of 1.4 x 108 Nm. Take g = 9.8 ms.​

Answer 1

Explanation:

Tension in wire = T= 1000 (9.8 + 1.2) = 11000N

maximum stress = T/ λr min² = 110000 x 7 /22 x r min²

r min² = 50²/ (10⁴)²

minimum radius = 50 x 10⁻⁴

minimum diameter= 2 x rmin

= 2 x 50 x 10⁻⁴

= 100 x 10⁻⁴m

= 0.01m 

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