Question Number 225726 by mr W last updated on 08/Nov/25
Commented by mr W last updated on 08/Nov/25
Answered by mr W last updated on 15/Nov/25
Commented by mr W last updated on 15/Nov/25
$${T}_{{A}} =\left(\mu\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)\rho{gb} \\ $$$${s}\rho{g}={T}_{{A}} \:\mathrm{sin}\:\theta=\left(\mu\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)\rho{gb}\:\mathrm{sin}\:\theta \\ $$$${s}=\left(\mu\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right){b}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\frac{{b}}{{s}}=\frac{\mathrm{1}}{\left(\mu\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)\:\mathrm{sin}\:\theta} \\ $$$${fraction}\:{of}\:{free}\:{hanging}\:{part}: \\ $$$$\lambda=\frac{\mathrm{2}{s}}{{l}}=\frac{\mathrm{2}{s}}{\mathrm{2}{s}+\mathrm{2}{b}}=\frac{\mathrm{1}}{\mathrm{1}+\frac{{b}}{{s}}} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\left(\mu\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)\:\mathrm{sin}\:\theta}}\:\checkmark \\ $$