Question Number 213043 by ajfour last updated on 29/Oct/24
Commented by ajfour last updated on 29/Oct/24
$${Find}\:\beta\:{such}\:{that}\:{it}\:{remains}\:{constant} \\ $$$${while}\:{the}\:{motion}\:{of}\:{frame}\:{down}\:{the} \\ $$$${slope}\:{of}\:{inclination}\:\alpha. \\ $$
Answered by mr W last updated on 29/Oct/24
Commented by mr W last updated on 29/Oct/24
$${since}\:\beta\:{remains}\:{constant},\:{the}\:{frame} \\ $$$${has}\:{no}\:{rotation}\:{motion},\:{that}\:{means} \\ $$$${all}\:{three}\:{forces}\:{acting}\:{on}\:{the}\:{frame} \\ $$$${must}\:{intersect}\:{at}\:{the}\:{same}\:{point}, \\ $$$${i}.{e}.\:{on}\:{its}\:{center}\:{of}\:{mass}.\:{that}\:{means} \\ $$$$\varphi=\mathrm{0},\:{or}\:\beta=\alpha. \\ $$
Commented by ajfour last updated on 29/Oct/24
$${mg}\left(\frac{{b}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\mathrm{90}°−\alpha−\varphi\right) \\ $$$$\:\:\:\:=\left({mg}\mathrm{sin}\:\alpha\right)\left(\frac{{b}}{\mathrm{2}}\right)\mathrm{cos}\:\varphi \\ $$$$\Rightarrow\:\mathrm{sin}\:\left(\alpha+\varphi\right)=\mathrm{sin}\:\alpha\mathrm{cos}\:\varphi \\ $$$$\Rightarrow\:\mathrm{cos}\:\alpha\mathrm{sin}\:\varphi=\mathrm{0} \\ $$$$\Rightarrow\:\varphi=\mathrm{0} \\ $$
Commented by ajfour last updated on 29/Oct/24
$${must}\:{be}\:{so},\:{thank}\:{you}\:{sir}. \\ $$