An acrobat of mass 53.2 kg is going to hang by her teeth from a steel wire and she does not want the wire to stretch beyond its elastic limit. The elastic limit for the wire is 2.37 × 108 Pa. What is the minimum diameter the wire should have to support her?

asked by Kim

7 years ago

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1 answer

Force = m g = 53.2*9.81
area = pi r^2
so
2.37*10^8 = 53.2*9.81/(pi R^2)

answered by Damon

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Question

An acrobat of mass 53.2 kg is going to hang by her teeth from a steel wire and she does not want the wire to stretch beyond its elastic limit. The elastic limit for the wire is 2.37 × 108 Pa. What is the minimum diameter the wire should have to support her?

1 answer

Force = m g = 53.2*9.81
area = pi r^2
so
2.37*10^8 = 53.2*9.81/(pi R^2)

Damon · Answer

Force = m g = 53.2*9.81 area = pi r^2 so 2.37*10^8 = 53.2*9.81/(pi R^2)