Question Number 102753 by Dwaipayan Shikari last updated on 11/Jul/20
$${A}\:{wedge}\:{has}\:{a}\:{weight}\:{of}\:\mathrm{9}{kg}\:.\:{And}\:{a}\:{block}\:{has}\:{a}\:{weight}\:{of}\:\mathrm{2}{kg} \\ $$$${If}\:{the}\:{block}\:{starts}\:{sliding}\:{with}\:{an}\:{angle}\:{of}\:\mathrm{45}°\:{with}\:{the} \\ $$$${horizontal}\:{then}\:{what}\:{is}\:{accelaration}\:{of}\:{the}\:{wedge}? \\ $$$$ \\ $$$$ \\ $$
Answered by mr W last updated on 11/Jul/20
$${N}={mg}\:\mathrm{cos}\:\theta−{ma}\:\mathrm{sin}\:\theta \\ $$$${Ma}={N}\:\mathrm{sin}\:\theta \\ $$$${Ma}=\left({mg}\:\mathrm{cos}\:\theta−{ma}\:\mathrm{sin}\:\theta\right)\:\mathrm{sin}\:\theta \\ $$$$\left({M}+{m}\:\mathrm{sin}^{\mathrm{2}} \:\theta\right){a}={mg}\:\mathrm{sin}\:\theta\mathrm{cos}\:\theta \\ $$$$\Rightarrow{a}=\frac{{mg}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{{M}+{m}\:\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$=\frac{\mathrm{2}\:{g}×\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{9}+\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}}=\frac{{g}}{\mathrm{10}} \\ $$