Find the minimum diameter of an alloy cable, tensile stress75 MPa (75,000,000 Pa), needed to support a Force of 15 kN (15,000 N)?

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Answer 1 · 2018-05-22T12:26:25

Given tensile strength #=75 xx10^6\ MPa=>75 xx10^6\ Nm^-2#
Force to be supported #F=15xx10^3\ N#

Corresponding area #a=F/"tensile strength"#

#=>a=(15xx10^3)/(75xx10^6) = 2xx10^-4\ m^2#

We know that for a circular cable of diameter #d#

#a=(pid^2)/4#
#=>d=sqrt((4a)/pi)#

Inserting calculated value we get

#d=sqrt((4xx2xx10^-4)/pi)#
#d=0.016\ m#