Question Number 224293 by fantastic last updated on 31/Aug/25
Commented by fantastic last updated on 31/Aug/25
Answered by fantastic last updated on 31/Aug/25
Commented by fantastic last updated on 31/Aug/25
$${mass}\:{of}\:{the}\:{bug}={m} \\ $$$${as}\:{per}\:{the}\:{Q}. \\ $$$${mg}\mathrm{sin}\:\alpha=\mu{n} \\ $$$${or}\:{mg}\mathrm{sin}\:\alpha=\mathrm{0}.\mathrm{75}{mg}\mathrm{cos}\:\alpha \\ $$$${or}\:{tan}\alpha=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\therefore\:\mathrm{cos}\:\alpha=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\therefore\frac{\mathrm{1}−{h}}{\mathrm{1}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\Rightarrow{h}=\frac{\mathrm{1}}{\mathrm{5}}{m}=\mathrm{0}.\mathrm{2}\:{m}\checkmark \\ $$
Answered by mr W last updated on 31/Aug/25
Commented by mr W last updated on 31/Aug/25
$$\mathrm{tan}\:\phi=\mu\:\Rightarrow\mathrm{cos}\:\phi=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }} \\ $$$${h}={r}\left(\mathrm{1}−\mathrm{cos}\:\phi\right)={r}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}\right) \\ $$$$\:\:\:=\mathrm{1}×\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{0}.\mathrm{75}^{\mathrm{2}} }}\right)=\mathrm{0}.\mathrm{2}\:{m}\:\checkmark \\ $$
Commented by fantastic last updated on 31/Aug/25
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