A wire of radius 0.3cm is used to lift a block of 1.5kg. Calculate the stress introduced into the wire [ take g = 10m/s\(^2\)]

Answer Details

Stress is the force acting per unit cross-sectional area of the wire:

\[\sigma = \frac{F}{A},\]

measured in \(\text{N m}^{-2}\) (pascals). Two quantities must be prepared before substituting: the stretching force and the area of the circular cross-section.

The force is the weight of the block:

\[F = mg = 1.5\times 10 = 15\ \text{N}.\]

The radius must be converted from centimetres to metres, since the answer is required in \(\text{N m}^{-2}\):

\[r = 0.3\ \text{cm} = 0.3\times 10^{-2}\ \text{m} = 3.0\times 10^{-3}\ \text{m},\]\[A = \pi r^2 = \pi (3.0\times 10^{-3})^2 = 2.83\times 10^{-5}\ \text{m}^2.\]

Therefore

\[\sigma = \frac{15}{2.83\times 10^{-5}} = 5.3\times 10^{5}\ \text{N m}^{-2} = 53\times 10^{4}\ \text{N m}^{-2}.\]

Note that \(53\times 10^{4}\) and \(5.3\times 10^{5}\) are the same number written differently, so compare powers of ten carefully rather than glancing only at the digits.

Three traps are set here. Using the diameter in place of the radius quarters the stress. Forgetting to square the \(10^{-2}\) when converting the radius, so that the area comes out a hundred times too large, produces a figure a hundred times too small. And a negative power of ten in the answer should be rejected on sight: a force of \(15\ \text{N}\) spread over an area far smaller than \(1\ \text{m}^2\) must give a stress much larger than \(15\ \text{N m}^{-2}\), not a tiny fraction of it. Always convert lengths to metres before squaring.