Question Number 228213 by ajfour last updated on 02/Apr/26
Commented by ajfour last updated on 02/Apr/26
$${A}\:{cylindee}\:{against}\:{a}\:{frictionless} \\ $$$$\:{bump}\:{and}\:{a}\:{wall}. \\ $$$${If}\:\:{mg}\:{is}\:{given}\:{and}\:{N}/{R}={b}/{a} \\ $$$${Find}\:{a}/{b}\:{and}\:{N}\:{and}\:{R}. \\ $$
Answered by mr W last updated on 03/Apr/26
Commented by mr W last updated on 03/Apr/26
$$\frac{{N}}{{R}}=\frac{{b}+{a}}{{b}−{a}}=\frac{{b}}{{a}} \\ $$$${let}\:\frac{{a}}{{b}}=\lambda \\ $$$$\frac{\mathrm{1}}{\lambda}=\frac{\lambda+\mathrm{1}}{\mathrm{1}−\lambda} \\ $$$$\lambda^{\mathrm{2}} +\mathrm{2}\lambda−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\lambda=\sqrt{\mathrm{2}}−\mathrm{1}=\frac{{a}}{{b}} \\ $$$$\frac{{N}}{{mg}}=\frac{{b}+{a}}{\:\mathrm{2}\sqrt{{ba}}}=\frac{\mathrm{1}+\lambda}{\mathrm{2}\sqrt{\lambda}}=\frac{\mathrm{1}+\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}=\frac{\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}}{\:\mathrm{2}} \\ $$$$\Rightarrow{N}=\frac{{mg}\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}}\approx\mathrm{1}.\mathrm{099}{mg} \\ $$$$\frac{{R}}{{mg}}=\frac{{b}−{a}}{\mathrm{2}\sqrt{{ba}}}=\frac{\mathrm{1}−\lambda}{\mathrm{2}\sqrt{\lambda}}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}=\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\sqrt{\sqrt{\mathrm{2}}+\mathrm{1}}}{\mathrm{2}} \\ $$$$\Rightarrow{R}=\frac{{mg}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\sqrt{\sqrt{\mathrm{2}}+\mathrm{1}}}{\mathrm{2}}\approx\mathrm{0}.\mathrm{455}{mg} \\ $$
Commented by ajfour last updated on 03/Apr/26
$${Yes}\:{sir},\:{this}\:{is}\:{it}.\:{Thanks} \\ $$