Commit d1b7cceb by Olivier

### Exercises 1: fix answer to E1.5

The answer to E1.5 introduced in commit ae3510d5
carried text over from the previous versions of the exercise; this is now fixed.
No numerical result is changed – only the detail to the solution.
parent 7777b353
 \renewcommand{\lastedityear}{2018} \renewcommand{\lasteditmonth}{04} \renewcommand{\lasteditmonth}{06} \renewcommand{\lasteditday}{18} \renewcommand{\numberofthischapter}{1} \renewcommand{\titleofthischapter}{Effects of pressure} ... ... @@ -257,8 +257,8 @@ A system of lock doors is set up to allow boats to travel up the side of a hill \tab 2) $M_\net = \SI{7,79}{\kilo\newton\meter}$; \tab 3) observe the equation used to calculate $M_\net$ to answer this question. If needed, ask for help in class! \item [\ref{exo_lock}]% \tab 2) $F_\text{net, door A} = \rho g L \left[\frac{1}{2} z^2\right]_{\SI{5}{\metre}}^{\SI{9}{\metre}} = \SI{+686,7}{\kilo\newton}$ (positive in right direction) \tab 3) $M_\text{net, door A, about bottom axis} = F_\net \frac{L}{2} = \SI{2,4689}{\mega\newton\metre}$ (positive clockwise): $\vec F_\text{net, door A}$ exerts at $R_2 = \SI{3,595}{\metre}$ from the bottom. \tab 2) $F_\text{net, door A} = \rho g W \left[\frac{1}{2} l^2\right]_{\SI{5}{\metre}}^{\SI{9}{\metre}} = \SI{+686,7}{\kilo\newton}$ (positive in right direction) \tab 3) $M_\text{net, door A, about bottom axis} = \rho g W \left[\frac{1}{6} l^3\right]_{\SI{5}{\metre}}^{\SI{9}{\metre}} = \SI{2,4689}{\mega\newton\metre}$ (positive clockwise): $\vec F_\text{net, door A}$ exerts at $R_2 = \SI{3,595}{\metre}$ from the bottom. \item [\ref{exo_buoyancy_tin_can}]% \tab 1) $F_\text{vertical} = \SI{4,625}{\milli\newton}$ upwards; \tab 2) $F_\text{vertical} = \SI{3,775}{\newton}$ in both cases; ... ...
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