Question Number 54360 by ajfour last updated on 02/Feb/19
Commented by ajfour last updated on 02/Feb/19
$${Find}\:{minimum}\:{mass}\:{of}\:{block} \\ $$$${that}\:{should}\:{be}\:{glued}\:{at}\:{the}\:{top}\:{of} \\ $$$${wedge}\:{to}\:{cause}\:{it}\:{to}\:{topple}\:{down}. \\ $$
Answered by ajfour last updated on 02/Feb/19
$${should}\:{it}\:{be}\:\:\:\:\:{m}\geqslant\:\frac{{M}}{\mathrm{3}}\left(\frac{{a}}{{b}\mathrm{cos}\:\alpha}−\mathrm{1}\right)\:? \\ $$
Answered by mr W last updated on 02/Feb/19
![mgb cos α≥Mg[(2/3)((a/2)+b cos θ)−b cos α] mb cos α≥M[((a−b cos α)/3)] ⇒m≥(M/3)[(a/(b cos α))−1]](Q54365.png)
$${mgb}\:\mathrm{cos}\:\alpha\geqslant{Mg}\left[\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{{a}}{\mathrm{2}}+{b}\:\mathrm{cos}\:\theta\right)−{b}\:\mathrm{cos}\:\alpha\right] \\ $$$${mb}\:\mathrm{cos}\:\alpha\geqslant{M}\left[\frac{{a}−{b}\:\mathrm{cos}\:\alpha}{\mathrm{3}}\right] \\ $$$$\Rightarrow{m}\geqslant\frac{{M}}{\mathrm{3}}\left[\frac{{a}}{{b}\:\mathrm{cos}\:\alpha}−\mathrm{1}\right] \\ $$