Please solve $16\mathrm{.}16$ & $16\mathrm{.}17.$ Refer to Figure $16\mathrm{.}17.$ Given $H=6m,=18\mathrm{.}5k\frac{N}{m^3},{}^{\text{'}}=33,{}^{\text{'}}=22,c^{\text{'}}=0,=10,$ and $=80,$ use Eqs. $(16\mathrm{.}50)$ and $(16\mathrm{.}51)$ to determine the magnitude of the active thrust on the wall, assuming earthquake loading with the following parameters: $k_v=0$ and $k_h=0\mathrm{.}15.$

$P_{ae}=\frac{1}{2}{H}^2(1-k_v)K_{ae}$

$(16\mathrm{.}50)$

$K_{ae}=$ active earth pressure coefficient

$=\frac{si{n}^2({}^{\text{'}}+-{}^{\text{'}})}{cos{}^{\text{'}}si{n}^{2}sin(-{}^{\text{'}}-{}^{\text{'}})[1+\sqrt[\phantom{2}]{\frac{sin({}^{\text{'}}+{}^{\text{'}})sin({}^{\text{'}}-{}^{\text{'}}-)}{sin(-{}^{\text{'}}-{}^{\text{'}})sin(+)}}{]}^2}$

$16\mathrm{.}17.$ Redo Problem $16\mathrm{.}16$ using Eqs. $(16\mathrm{.}52)$ and $(16\mathrm{.}56)$ to determine the magnitude and location of the active thrust on the wall.

$P_{ae}=\frac{1}{2}{H}^2(1-k_v)[K_a({}^{\text{'}},{}^{\text{'}})](\frac{si{n}^2{}^{\text{'}}}{cos{}^{\text{'}}si{n}^2})$

Figure $16\mathrm{.}17$

${}^{\text{'}}=-{}^{\text{'}}$

$P_{ae}=P_{ae}-,P_a\frac{?}{b}$

$ar(z)=\frac{(0\mathrm{.}6H)(P_{ae})+(\frac{H}{3})(P_a)}{P_{ae}}$