Solution 14.2
From Mechanics
(Difference between revisions)
(New page: Taking moments about the edge of the quay gives: <math>\begin{align} & 1\times 60\times 9\textrm{.}8=1\textrm{.}5\times m\times 9\textrm{.}8 \\ & m=\frac{60}{1\textrm{.}5}=40\text{ kg} \...) |
(New page: Taking moments about the edge of the quay gives: <math>\begin{align} & 1\times 60\times 9\textrm{.}8=1\textrm{.}5\times m\times 9\textrm{.}8 \\ & m=\frac{60}{1\textrm{.}5}=40\text{ kg} \...) |
Current revision
Taking moments about the edge of the quay gives:
\displaystyle \begin{align} & 1\times 60\times 9\textrm{.}8=1\textrm{.}5\times m\times 9\textrm{.}8 \\ & m=\frac{60}{1\textrm{.}5}=40\text{ kg} \\ \end{align}