Commit ae3510d5 authored by Olivier's avatar Olivier

Exercises 1: added solution to 1.5 (door lock)

Since the exercise was changed in commit 5237e16c , the solution for
the height at which the force is exerting was missing; this is added
parent 8f69f6b4
\renewcommand{\titleofthischapter}{Effects of pressure}
......@@ -257,9 +257,8 @@ A system of lock doors is set up to allow boats to travel up the side of a hill
\tab 2) $M_\net = \SI{7,79}{\mega\newton\meter}$;
\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) $F_\net = \rho g L \left[\frac{1}{2} z^2\right]_{\SI{5}{\metre}}^{\SI{9}{\metre}} = \SI{686,7}{\kilo\newton}$ per door;
%\tab 3) $M = F_\net \frac{L}{2} = \SI{858,3}{\kilo\newton\metre}$ per hinge.
%\tab \tab 4) yes — look at the analytical expression for the moment to find out why.
\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.
\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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