Question Number 32444 by Tinkutara last updated on 25/Mar/18
Commented by ajfour last updated on 25/Mar/18
Commented by Tinkutara last updated on 25/Mar/18
But if we simply apply the force parallel to (3) up the incline then isn't it will be minimum and also stops the block?
Commented by ajfour last updated on 25/Mar/18
$${N}=\frac{{mg}}{\mathrm{2}}−{F}\mathrm{sin}\:\theta \\ $$$${F}\mathrm{cos}\:\theta+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\left(\frac{{mg}}{\mathrm{2}}−{F}\mathrm{sin}\:\theta\right)=\frac{{mg}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{F}=\frac{{mg}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)}{\mathrm{cos}\:\theta−\frac{\mathrm{sin}\:\theta}{\sqrt{\mathrm{3}}}} \\ $$$${F}\:=\:\frac{{mg}}{\sqrt{\mathrm{3}}\mathrm{cos}\:\theta−\mathrm{sin}\:\theta}=\frac{{mg}}{\mathrm{2sin}\:\left(\mathrm{60}°−\theta\right)} \\ $$$${F}\:{is}\:{minimum}\:{for}\:\:\theta=−\mathrm{30}° \\ $$$$\Rightarrow\:\:{along}\:{the}\:{vector}\:\:{inclined}\:{at} \\ $$$$\mathrm{30}°\:{to}\:{horizontal}.\:\:{That}\:{is}\:{along} \\ $$$$\hat {{i}}+\frac{\hat {{j}}}{\sqrt{\mathrm{3}}}\:\:\:\:{or}\:{along}\:\:\:\sqrt{\mathrm{3}}\hat {{i}}+\hat {{j}}\:. \\ $$$${F}_{{min}} =\frac{{mg}}{\mathrm{2}} \\ $$$$..................... \\ $$$${if}\:{force}\:{is}\:{applied}\:{parallel}\:{to} \\ $$$${incline}\:{then} \\ $$$${F}+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\left(\frac{{mg}}{\mathrm{2}}\right)=\frac{{mg}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${F}={mg}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)\:=\:\frac{{mg}}{\sqrt{\mathrm{3}}}\:\:>\:\frac{{mg}}{\mathrm{2}}\:. \\ $$
Commented by Tinkutara last updated on 25/Mar/18
Answer given is 1⃣
Commented by ajfour last updated on 25/Mar/18
yes sorry for the error.
Commented by Tinkutara last updated on 25/Mar/18
Thank you very much Sir! I got the answer. ��������