A steel wire with cross-sectional area $15\,\text{mm}^2$ is given an ultimate tensile stress of $4.5\times10^8\,\text{N m}^{-2}$.
(a)[2]
Calculate the greatest tension that the wire can withstand.
(b)[3]
The density of the wire’s steel is $7800\,\text{kg m}^{-3}$. The wire is suspended vertically. Calculate the greatest length of steel wire that could hang vertically before its own weight causes it to break.
Worked solution & mark scheme
This 5-mark question has a full step-by-step worked solution and mark scheme. One marking point: “stress, or $\sigma = F/A$; maximum tension $= \mathrm{UTS}\times A = 4.5\times10^8 \times 15\times10^{-6} = 6800\ (6750)\ \mathrm{N}$” …