Answer
(a) The stress is $2.8\times 10^7~N/m^2$
(b) The strain is $4.67\times 10^{-4}$
(c) The column is compressed by $9.34\times 10^{-4}~m$
(d) The maximum load is $5.0\times 10^5~N$
Work Step by Step
(a) We can find the stress:
$\frac{F}{A} = \frac{7.0\times 10^4~N}{25\times 10^{-4}~m^2}$
$\frac{F}{A} = 2.8\times 10^7~N/m^2$
The stress is $2.8\times 10^7~N/m^2$
(b) We can find the strain:
$\frac{\Delta L}{L} =\frac{F/A}{Y}$
$\frac{\Delta L}{L} = \frac{2.8\times 10^7~N/m^2}{6.0\times 10^{10}~Pa}$
$\frac{\Delta L}{L} = 4.67\times 10^{-4}$
The strain is $4.67\times 10^{-4}$
(c) We can find the change in length:
$\frac{\Delta L}{L} = 4.67\times 10^{-4}$
$\Delta L = (4.67\times 10^{-4})(L)$
$\Delta L = (4.67\times 10^{-4})(2.0~m)$
$\Delta L = 9.34\times 10^{-4}~m$
The column is compressed by $9.34\times 10^{-4}~m$
(d) We can find the maximum load:
$\frac{F}{A} = 2.0\times 10^8~Pa$
$F = (2.0\times 10^8~Pa)~(A)$
$F = (2.0\times 10^8~Pa)~(25\times 10^{-4}~m^2)$
$F = 5.0\times 10^5~N$
The maximum load is $5.0\times 10^5~N$
