Question Number 43945 by ajfour last updated on 18/Sep/18
Commented by ajfour last updated on 18/Sep/18
$${The}\:{rectangular}\:{plate}\:{can}\:{swing} \\ $$$${freely}\:{about}\:{the}\:{diagonal}\:{but}\:{is} \\ $$$${at}\:{balance},\:{not}\:{turning}\:{any}\:{more}. \\ $$$$\boldsymbol{{Find}}\:\boldsymbol{{p}},\:\boldsymbol{{h}}\:\boldsymbol{{in}}\:\boldsymbol{{terms}}\:\boldsymbol{{of}}\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{d}}. \\ $$
Answered by MrW3 last updated on 19/Sep/18
$${G}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }={length}\:{of}\:{diagonal} \\ $$$${H}=\sqrt{{G}^{\mathrm{2}} −{d}^{\mathrm{2}} }=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{d}^{\mathrm{2}} }={height}\:{of}\:{topmost}\:{corner} \\ $$$$\frac{{e}}{{b}}=\frac{{b}}{{G}}\Rightarrow{e}=\frac{{b}^{\mathrm{2}} }{{G}} \\ $$$$\frac{{f}}{{a}}=\frac{{a}}{{G}}\Rightarrow{f}=\frac{{a}^{\mathrm{2}} }{{G}} \\ $$$${p}=\frac{{e}}{{G}}×{H}=\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{d}^{\mathrm{2}} } \\ $$$${h}=\frac{{f}}{{G}}×{H}=\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{d}^{\mathrm{2}} } \\ $$
Commented by MrW3 last updated on 19/Sep/18
Commented by ajfour last updated on 19/Sep/18
$${Great}\:{simplification}\:{Sir}. \\ $$$${Awesome}\:{solution}.\: \\ $$